2004. LEARNING OBJECTIVES -
Remember that in most main group element compounds, the valence electrons of the isolated atoms combine to form chemical bonds that satisfy the octet rule. For instance, the four valence electrons of carbon overlap with electrons from four hydrogen atoms to form $\mathrm{CH}_{4}$. The one valence electron leaves sodium and adds to the seven valence electrons of chlorine to form the ionic formula unit NaCl (Figure 19.13). Transition metals do not normally bond in this fashion. They primarily form coordinate covalent bonds, a form of the Lewis acid-base interaction in which both of the electrons in the bond are contributed by a donor (Lewis base) to an electron acceptor (Lewis acid). The Lewis acid in coordination complexes, often called a central metal ion (or atom), is often a transition metal or inner transition metal, although main group elements can also form coordination compounds. The Lewis base donors, called ligands, can be a wide variety of chemicals-atoms, molecules, or ions. The only requirement is that they have one or more electron pairs, which can be donated to the central metal. Most often, this involves a donor atom with a lone pair of electrons that can form a coordinate bond to the metal.

(a)

(b)

FIGURE 19.13 (a) Covalent bonds involve the sharing of electrons, and ionic bonds involve the transferring of electrons associated with each bonding atom, as indicated by the colored electrons. (b) However, coordinate covalent bonds involve electrons from a Lewis base being donated to a metal center. The lone pairs from six water molecules form bonds to the scandium ion to form an octahedral complex. (Only the donated pairs are shown.)


2004. LEARNING OBJECTIVES -
The coordination sphere consists of the central metal ion or atom plus its attached ligands. Brackets in a formula enclose the coordination sphere; species outside the brackets are not part of the coordination sphere. The coordination number of the central metal ion or atom is the number of donor atoms bonded to it. The coordination number for the silver ion in $\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}$is two (Figure 19.14). For the copper(II) ion in $\left[\mathrm{CuCl}_{4}\right]^{2-}$, the coordination number is four, whereas for the cobalt(II) ion in $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ the coordination number is six. Each of these ligands is monodentate, from the Greek for "one toothed," meaning that they connect with the central metal through only one atom. In this case, the number of ligands and the coordination number are equal.

FIGURE 19.14 The complexes (a) $\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}$, (b) $\left[\mathrm{Cu}(\mathrm{Cl})_{4}\right]^{2-}$, and (c) $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ have coordination numbers of two, four, and six, respectively. The geometries of these complexes are the same as we have seen with VSEPR theory for main group elements: linear, tetrahedral, and octahedral.


2004. LEARNING OBJECTIVES -
Many other ligands coordinate to the metal in more complex fashions. Bidentate ligands are those in which two atoms coordinate to the metal center. For example, ethylenediamine (en, $\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}$ ) contains two nitrogen atoms, each of which has a lone pair and can serve as a Lewis base (Figure 19.15). Both of the atoms can coordinate to a single metal center. In the complex $\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}$, there are three bidentate en ligands, and the coordination number of the cobalt(III) ion is six. The most common coordination numbers are two, four, and six, but examples of all coordination numbers from 1 to 15 are known.

(a)

(b)

FIGURE 19.15 (a) The ethylenediamine (en) ligand contains two atoms with lone pairs that can coordinate to the metal center. (b) The cobalt(III) complex $\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}$ contains three of these ligands, each forming two bonds to the cobalt ion.

Any ligand that bonds to a central metal ion by more than one donor atom is a polydentate ligand (or "many teeth") because it can bite into the metal center with more than one bond. The term chelate (pronounced "KEY-late") from the Greek for "claw" is also used to describe this type of interaction. Many polydentate ligands are chelating ligands, and a complex consisting of one or more of these ligands and a central metal is a chelate. A chelating ligand is also known as a chelating agent. A chelating ligand holds the metal ion rather like a crab's claw would hold a marble. Figure 19.15 showed one example of a chelate. The heme complex in hemoglobin is another important example (Figure 19.16). It contains a polydentate ligand with four donor atoms that coordinate to iron.


2004. LEARNING OBJECTIVES -
FIGURE 19.16 The single ligand heme contains four nitrogen atoms that coordinate to iron in hemoglobin to form a
chelate.
Polydentate ligands are sometimes identified with prefixes that indicate the number of donor atoms in the ligand. As we have seen, ligands with one donor atom, such as $\mathrm{NH}_{3}, \mathrm{Cl}^{-}$, and $\mathrm{H}_{2} \mathrm{O}$, are monodentate ligands. Ligands with two donor groups are bidentate ligands. Ethylenediamine, $\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}$, and the anion of the acid glycine, $\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2}{ }^{-}$(Figure 19.17) are examples of bidentate ligands. Tridentate ligands, tetradentate ligands, pentadentate ligands, and hexadentate ligands contain three, four, five, and six donor atoms, respectively. The ligand in heme (Figure 19.16) is a tetradentate ligand.

FIGURE 19.17 Each of the anionic ligands shown attaches in a bidentate fashion to platinum(II), with both a nitrogen and oxygen atom coordinating to the metal.

The Naming of Complexes
The nomenclature of the complexes is patterned after a system suggested by Alfred Werner, a Swiss chemist and Nobel laureate, whose outstanding work more than 100 years ago laid the foundation for a clearer understanding of these compounds. The following five rules are used for naming complexes:


2004. LEARNING OBJECTIVES -
1. If a coordination compound is ionic, name the cation first and the anion second, in accordance with the usual nomenclature.
2. Name the ligands first, followed by the central metal. Name the ligands alphabetically. Negative ligands (anions) have names formed by adding -o to the stem name of the group. For examples, see Table 19.1. For most neutral ligands, the name of the molecule is used. The four common exceptions are aqua $\left(\mathrm{H}_{2} \mathrm{O}\right)$, ammine $\left(\mathrm{NH}_{3}\right)$, carbonyl (CO), and nitrosyl (NO). For example, name $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{4}\right]$ as diamminetetrachloroplatinum(IV).

| Examples of Anionic Ligands |  |
| :--- | :--- |
| Anionic Ligand | Name |
| $\mathrm{F}^{-}$ | fluoro |
| $\mathrm{Cl}^{-}$ | chloro |
| $\mathrm{Br}^{-}$ | bromo |
| $\mathrm{I}^{-}$ | iodo |
| $\mathrm{CN}^{-}$ | cyano |
| $\mathrm{NO}_{3}^{-}$ | hitrato |
| $\mathrm{OH}^{-}$ | oxo |
| $\mathrm{O}^{2-}$ | oxalato |
| $\mathrm{C}_{2} \mathrm{O}_{4}^{2-}$ | carbonato |
| $\mathrm{CO}_{3}^{2-}$ |  |


2004. LEARNING OBJECTIVES -
TABLE 19.1
3. If more than one ligand of a given type is present, the number is indicated by the prefixes di- (for two), tri(for three), tetra- (for four), penta- (for five), and hexa- (for six). Sometimes, the prefixes bis- (for two), tris(for three), and tetrakis- (for four) are used when the name of the ligand already includes di-, tri-, or tetra-, or when the ligand name begins with a vowel. For example, the ion bis(bipyridyl)osmium(II) uses bis- to signify that there are two ligands attached to Os, and each bipyridyl ligand contains two pyridine groups $\left(\mathrm{C}_{5} \mathrm{H}_{4} \mathrm{~N}\right)$.

When the complex is either a cation or a neutral molecule, the name of the central metal atom is spelled exactly like the name of the element and is followed by a Roman numeral in parentheses to indicate its oxidation state (Table 19.2 and Table 19.3). When the complex is an anion, the suffix -ate is added to the stem of the name of the metal, followed by the Roman numeral designation of its oxidation state (Table 19.4).
Sometimes, the Latin name of the metal is used when the English name is clumsy. For example, ferrate is used instead of ironate, plumbate instead leadate, and stannate instead of tinate. The oxidation state of the metal is determined based on the charges of each ligand and the overall charge of the coordination compound. For example, in $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Br}$, the coordination sphere (in brackets) has a charge of $1+$ to balance the bromide ion. The water ligands are neutral, and the chloride ligands are anionic with a charge of 1-each. To determine the oxidation state of the metal, we set the overall charge equal to the sum of the ligands and the metal: $+1=-2$ $+x$, so the oxidation state $(x)$ is equal to $3+$.


2004. LEARNING OBJECTIVES -
| Examples in Which the Complex Is a Cation |  |
| :--- | :--- |
| $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]_{\mathrm{Cl}}^{3}$ | hexaamminecobalt(III) chloride |
| $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]^{2+}$ | tetraamminedichloroplatinum(IV) ion |
| $\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}$ | diamminesilver(I) ion |
| $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}$ | tetraaquadichlorochromium(III) chloride |
| $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{3}\right]_{2}\left(\mathrm{SO}_{4}\right)_{3}$ | tris(ethylenediamine)cobalt(III) sulfate |

TABLE 19.2

Examples in Which the Complex Is Neutral

| $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{4}\right]$ | diamminetetrachloroplatinum(IV) |
| :--- | :--- |
| $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{2} \mathrm{Cl}_{2}\right]$ | dichlorobis(ethylenediamine)nickel(II) |

TABLE 19.3


2004. LEARNING OBJECTIVES -
TABLE 19.3

| Examples in Which the Complex Is an Anion |  |
| :--- | :--- |
| $\left[\mathrm{PtCl}_{6}\right]^{2-}$ | hexachloroplatinate(IV) ion |
| $\mathrm{Na}_{2}\left[\mathrm{SnCl}_{6}\right]$ | sodium hexachlorostannate(IV) |

TABLE 19.4


2005. LINK TO LEARNING -
Do you think you understand naming coordination complexes? You can look over more examples and test
yourself with online quizzes (http://openstax.org/l/16namingcomps) at the University of Sydney's site.


2007. Coordination Numbers and Oxidation States -
Determine the name of the following complexes and give the coordination number of the central metal atom.
(a) $\mathrm{Na}_{2}\left[\mathrm{PtCl}_{6}\right]$
(b) $\mathrm{K}_{3}\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]$
(c) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$


2008. Solution -
(a) There are two $\mathrm{Na}^{+}$ions, so the coordination sphere has a negative two charge: $\left[\mathrm{PtCl}_{6}\right]^{2-}$. There are six anionic chloride ligands, so $-2=-6+x$, and the oxidation state of the platinum is $4+$. The name of the complex is sodium hexachloroplatinate(IV), and the coordination number is six. (b) The coordination sphere has a charge of 3 - (based on the potassium) and the oxalate ligands each have a charge of $2-$, so the metal oxidation state is given by $-3=-6+x$, and this is an iron(III) complex. The name is potassium trisoxalatoferrate(III) (note that tris is used instead of tri because the ligand name starts with a vowel). Because oxalate is a bidentate ligand, this complex has a coordination number of six. (c) In this example, the coordination sphere has a cationic charge of $2+$. The $\mathrm{NH}_{3}$ ligand is neutral, but the chloro ligand has a charge of $1-$. The oxidation state is found by $+2=-1+x$ and is $3+$, so the complex is pentaamminechlorocobalt(III) chloride and the coordination number is six.


2009. Check Your Learning -
The complex potassium dicyanoargenate(I) is used to make antiseptic compounds. Give the formula and coordination number.


2010. Answer: -
$\mathrm{K}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]$; coordination number two


2011. The Structures of Complexes -
The most common structures of the complexes in coordination compounds are octahedral, tetrahedral, and square planar (see Figure 19.18). For transition metal complexes, the coordination number determines the geometry around the central metal ion. Table 19.5 compares coordination numbers to the molecular geometry:

FIGURE 19.18 These are geometries of some complexes with coordination numbers of seven and eight.

Coordination Numbers and Molecular Geometry

| Coordination Number | Molecular Geometry | Example |
| :--- | :--- | :--- |
| 2 | linear | $\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}$ |
| 3 | trigonal planar | $\left[\mathrm{Cu}(\mathrm{CN})_{3}\right]^{2-}$ |
| 4 | tetrahedral $\left(d^{0}\right.$ or $\left.d^{10}\right)$, low oxidation states for M | $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ |
| 4 | square planar $\left(d^{8}\right)$ | $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ |
| 5 | trigonal bipyramidal | $\left[\mathrm{CoCl}_{5}\right]^{2-}$ |
| 5 | square pyramidal | $\left[\mathrm{VO}_{3}(\mathrm{CN})_{4}\right]^{2-}$ |
| 6 | octahedral | $\left[\mathrm{CoCl}_{6}\right]^{3-}$ |
| 7 | pentagonal bipyramid | $\left[\mathrm{ZrF}_{7}\right]^{3-}$ |
| 8 | square antiprism | $\left[\mathrm{ReF}_{8}\right]^{2-}$ |
| 8 | dodecahedron | $\left[\mathrm{Mo}^{2-}(\mathrm{CN})_{8}\right]^{4-}$ |
| 9 and above | more complicated structures |  |

TABLE 19.5


2011. The Structures of Complexes -
TABLE 19.5

Unlike main group atoms in which both the bonding and nonbonding electrons determine the molecular shape, the nonbonding $d$-electrons do not change the arrangement of the ligands. Octahedral complexes have a coordination number of six, and the six donor atoms are arranged at the corners of an octahedron around the central metal ion. Examples are shown in Figure 19.19. The chloride and nitrate anions in $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ and $\left[\mathrm{Cr}(\mathrm{en})_{3}\right]\left(\mathrm{NO}_{3}\right)_{3}$, and the potassium cations in $\mathrm{K}_{2}\left[\mathrm{PtCl}_{6}\right]$, are outside the brackets and are not bonded to the metal ion.

$2 \mathrm{Cl}^{-}$

$3 \mathrm{NO}_{3}{ }^{-}$

$2 \mathrm{~K}^{+}$

FIGURE 19.19 Many transition metal complexes adopt octahedral geometries, with six donor atoms forming bond angles of $90^{\circ}$ about the central atom with adjacent ligands. Note that only ligands within the coordination sphere affect the geometry around the metal center.


2011. The Structures of Complexes -
For transition metals with a coordination number of four, two different geometries are possible: tetrahedral or square planar. Unlike main group elements, where these geometries can be predicted from VSEPR theory, a more detailed discussion of transition metal orbitals (discussed in the section on Crystal Field Theory) is required to predict which complexes will be tetrahedral and which will be square planar. In tetrahedral complexes such as $\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{2-}$ (Figure 19.20), each of the ligand pairs forms an angle of $109.5^{\circ}$. In square
planar complexes, such as $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]$, each ligand has two other ligands at $90^{\circ}$ angles (called the cis positions) and one additional ligand at an $180^{\circ}$ angle, in the trans position.

(a)

(b)

FIGURE 19.20 Transition metals with a coordination number of four can adopt a tetrahedral geometry (a) as in $\mathrm{K}_{2}\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]$ or a square planar geometry (b) as shown in $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]$.


2012. Isomerism in Complexes -
Isomers are different chemical species that have the same chemical formula. Transition metal complexes often exist as geometric isomers, in which the same atoms are connected through the same types of bonds but with differences in their orientation in space. Coordination complexes with two different ligands in the cis and trans positions from a ligand of interest form isomers. For example, the octahedral $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]^{+}$ion has two isomers. In the cis configuration, the two chloride ligands are adjacent to each other (Figure 19.21). The other isomer, the trans configuration, has the two chloride ligands directly across from one another.

Violet, cis form

Green, trans form

FIGURE 19.21 The cis and trans isomers of $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right]^{+}$contain the same ligands attached to the same metal ion, but the spatial arrangement causes these two compounds to have very different properties.


2012. Isomerism in Complexes -
Different geometric isomers of a substance are different chemical compounds. They exhibit different properties, even though they have the same formula. For example, the two isomers of $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{NO}_{3}$ differ in color; the cis form is violet, and the trans form is green. Furthermore, these isomers have different dipole moments, solubilities, and reactivities. As an example of how the arrangement in space can influence the molecular properties, consider the polarity of the two $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{NO}_{3}$ isomers. Remember that the polarity of a molecule or ion is determined by the bond dipoles (which are due to the difference in electronegativity of the bonding atoms) and their arrangement in space. In one isomer, cis chloride ligands cause more electron density on one side of the molecule than on the other, making it polar. For the trans isomer, each ligand is directly across from an identical ligand, so the bond dipoles cancel out, and the molecule is nonpolar.


2014. Geometric Isomers -
Identify which geometric isomer of $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]$ is shown in Figure 19.20. Draw the other geometric isomer and give its full name.


2015. Solution -
In the Figure 19.20, the two chlorine ligands occupy cis positions. The other form is shown in Figure 19.22. When naming specific isomers, the descriptor is listed in front of the name. Therefore, this complex is trans-diamminedichloroplatinum(II).

FIGURE 19.22 The trans isomer of $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]$ has each ligand directly across from an adjacent ligand.


2016. Check Your Learning -
Draw the ion trans-diaqua-trans-dibromo-trans-dichlorocobalt(II).


2017. Answer: -
Another important type of isomers are optical isomers, or enantiomers, in which two objects are exact mirror images of each other but cannot be lined up so that all parts match. This means that optical isomers are nonsuperimposable mirror images. A classic example of this is a pair of hands, in which the right and left hand are mirror images of one another but cannot be superimposed. Optical isomers are very important in organic and biochemistry because living systems often incorporate one specific optical isomer and not the other. Unlike geometric isomers, pairs of optical isomers have identical properties (boiling point, polarity, solubility, etc.). Optical isomers differ only in the way they affect polarized light and how they react with other optical isomers. For coordination complexes, many coordination compounds such as $\left[\mathrm{M}(\mathrm{en})_{3}\right]^{\mathrm{n+}}\left[\right.$ in which $\mathrm{M}^{\mathrm{n}+}$ is a central metal ion such as iron(III) or cobalt(II)] form enantiomers, as shown in Figure 19.23. These two isomers will react differently with other optical isomers. For example, DNA helices are optical isomers, and the form that occurs in nature (right-handed DNA) will bind to only one isomer of $\left[\mathrm{M}(\mathrm{en})_{3}\right]^{\mathrm{n}+}$ and not the other.

FIGURE 19.23 The complex $\left[M(e n)_{3}\right]^{n+}\left(M^{n+}=\right.$ a metal ion, en $=$ ethylenediamine) has a nonsuperimposable mirror image.

The $\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}$ion exhibits geometric isomerism (cis/trans), and its cis isomer exists as a pair of optical isomers (Figure 19.24).
cis form (optical isomers)


2017. Answer: -
FIGURE 19.24 Three isomeric forms of $\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}$exist. The trans isomer, formed when the chlorines are positioned at a $180^{\circ}$ angle, has very different properties from the cis isomers. The mirror images of the cis isomer form a pair of optical isomers, which have identical behavior except when reacting with other enantiomers.

Linkage isomers occur when the coordination compound contains a ligand that can bind to the transition metal center through two different atoms. For example, the CN ligand can bind through the carbon atom (cyano) or through the nitrogen atom (isocyano). Similarly, SCN- can be bound through the sulfur or nitrogen
atom, affording two distinct compounds $\left(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SCN}\right]^{2+}\right.$ or $\left.\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NCS}\right]^{2+}\right)$.
Ionization isomers (or coordination isomers) occur when one anionic ligand in the inner coordination sphere is replaced with the counter ion from the outer coordination sphere. A simple example of two ionization isomers are $\left\mathrm{CoCl}_{6}\right$ and $\left\mathrm{CoCl}_{5} \mathrm{Br}\right$.


2018. Coordination Complexes in Nature and Technology -
Chlorophyll, the green pigment in plants, is a complex that contains magnesium (Figure 19.25). This is an example of a main group element in a coordination complex. Plants appear green because chlorophyll absorbs red and purple light; the reflected light consequently appears green. The energy resulting from the absorption of light is used in photosynthesis.

(a)

(b)

FIGURE 19.25 (a) Chlorophyll comes in several different forms, which all have the same basic structure around the magnesium center. (b) Copper phthalocyanine blue, a square planar copper complex, is present in some blue dyes.


2020. Transition Metal Catalysts -
One of the most important applications of transition metals is as industrial catalysts. As you recall from the chapter on kinetics, a catalyst increases the rate of reaction by lowering the activation energy and is regenerated in the catalytic cycle. Over $90 \%$ of all manufactured products are made with the aid of one or more catalysts. The ability to bind ligands and change oxidation states makes transition metal catalysts well suited for catalytic applications. Vanadium oxide is used to produce $230,000,000$ tons of sulfuric acid worldwide each year, which in turn is used to make everything from fertilizers to cans for food. Plastics are made with the aid of transition metal catalysts, along with detergents, fertilizers, paints, and more (see Figure 19.26). Very complicated pharmaceuticals are manufactured with catalysts that are selective, reacting with one specific bond out of a large number of possibilities. Catalysts allow processes to be more economical and more environmentally friendly. Developing new catalysts and better understanding of existing systems are important areas of current research.

FIGURE 19.26 (a) Detergents, (b) paints, and (c) fertilizers are all made using transition metal catalysts. (credit a: modification of work by "Mr. Brian"/Flickr; credit b: modification of work by Ewen Roberts; credit c: modification of work by "osseous"/Flickr)


2022. Deanna D'Alessandro -
Dr. Deanna D'Alessandro develops new metal-containing materials that demonstrate unique electronic, optical, and magnetic properties. Her research combines the fields of fundamental inorganic and physical chemistry with materials engineering. She is working on many different projects that rely on transition metals. For example, one type of compound she is developing captures carbon dioxide waste from power plants and catalytically converts it into useful products (see Figure 19.27).

FIGURE 19.27 Catalytic converters change carbon dioxide emissions from power plants into useful products, and, like the one shown here, are also found in cars.

Another project involves the development of porous, sponge-like materials that are "photoactive." The absorption of light causes the pores of the sponge to change size, allowing gas diffusion to be controlled. This has many potential useful applications, from powering cars with hydrogen fuel cells to making better electronics components. Although not a complex, self-darkening sunglasses are an example of a photoactive substance.

Watch this video (http://openstax.org/l/16DeannaD) to learn more about this research and listen to Dr. D'Alessandro (shown in Figure 19.28) describe what it is like being a research chemist.

FIGURE 19.28 Dr. Deanna D'Alessandro is a functional materials researcher. Her work combines the inorganic and physical chemistry fields with engineering, working with transition metals to create new systems to power cars and convert energy (credit: image courtesy of Deanna D'Alessandro).

Many other coordination complexes are also brightly colored. The square planar copper(II) complex phthalocyanine blue (from Figure 19.25) is one of many complexes used as pigments or dyes. This complex is used in blue ink, blue jeans, and certain blue paints.


2022. Deanna D'Alessandro -
The structure of heme (Figure 19.29), the iron-containing complex in hemoglobin, is very similar to that in chlorophyll. In hemoglobin, the red heme complex is bonded to a large protein molecule (globin) by the attachment of the protein to the heme ligand. Oxygen molecules are transported by hemoglobin in the blood by being bound to the iron center. When the hemoglobin loses its oxygen, the color changes to a bluish red. Hemoglobin will only transport oxygen if the iron is $\mathrm{Fe}^{2+}$; oxidation of the iron to $\mathrm{Fe}^{3+}$ prevents oxygen transport.

FIGURE 19.29 Hemoglobin contains four protein subunits, each of which has an iron center attached to a heme ligand (shown in red), which is coordinated to a globin protein. Each subunit is shown in a different color.

Complexing agents often are used for water softening because they tie up such ions as $\mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}$, and $\mathrm{Fe}^{2+}$, which make water hard. Many metal ions are also undesirable in food products because these ions can catalyze reactions that change the color of food. Coordination complexes are useful as preservatives. For example, the ligand EDTA, $\left(\mathrm{HO}_{2} \mathrm{CCH}_{2}\right)_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{~N}\left(\mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}\right)_{2}$, coordinates to metal ions through six donor atoms and prevents the metals from reacting (Figure 19.30). This ligand also is used to sequester metal ions in paper production, textiles, and detergents, and has pharmaceutical uses.


2022. Deanna D'Alessandro -
FIGURE 19.30 The ligand EDTA binds tightly to a variety of metal ions by forming hexadentate complexes.
Complexing agents that tie up metal ions are also used as drugs. British Anti-Lewisite (BAL), $\mathrm{HSCH}_{2} \mathrm{CH}(\mathrm{SH}) \mathrm{CH}_{2} \mathrm{OH}$, is a drug developed during World War I as an antidote for the arsenic-based war gas Lewisite. BAL is now used to treat poisoning by heavy metals, such as arsenic, mercury, thallium, and chromium. The drug is a ligand and functions by making a water-soluble chelate of the metal; the kidneys eliminate this metal chelate (Figure 19.31). Another polydentate ligand, enterobactin, which is isolated from
certain bacteria, is used to form complexes of iron and thereby to control the severe iron buildup found in patients suffering from blood diseases such as Cooley's anemia, who require frequent transfusions. As the transfused blood breaks down, the usual metabolic processes that remove iron are overloaded, and excess iron can build up to fatal levels. Enterobactin forms a water-soluble complex with excess iron, and the body can safely eliminate this complex.

(a)

(b)

FIGURE 19.31 Coordination complexes are used as drugs. (a) British Anti-Lewisite is used to treat heavy metal poisoning by coordinating metals (M), and enterobactin (b) allows excess iron in the blood to be removed.


2024. Chelation Therapy -
Ligands like BAL and enterobactin are important in medical treatments for heavy metal poisoning. However, chelation therapies can disrupt the normal concentration of ions in the body, leading to serious side effects, so researchers are searching for new chelation drugs. One drug that has been developed is dimercaptosuccinic acid (DMSA), shown in Figure 19.32. Identify which atoms in this molecule could act as donor atoms.

FIGURE 19.32 Dimercaptosuccinic acid is used to treat heavy metal poisoning.


2025. Solution -
All of the oxygen and sulfur atoms have lone pairs of electrons that can be used to coordinate to a metal center, so there are six possible donor atoms. Geometrically, only two of these atoms can be coordinated to a metal at once. The most common binding mode involves the coordination of one sulfur atom and one oxygen atom, forming a five-member ring with the metal.


2026. Check Your Learning -
Some alternative medicine practitioners recommend chelation treatments for ailments that are not clearly related to heavy metals, such as cancer and autism, although the practice is discouraged by many scientific organizations. ${ }^{\frac{1}{~}}$ Identify at least two biologically important metals that could be disrupted by chelation therapy.


2027. Answer: -
$\mathrm{Ca}, \mathrm{Fe}, \mathrm{Zn}$, and Cu

Ligands are also used in the electroplating industry. When metal ions are reduced to produce thin metal coatings, metals can clump together to form clusters and nanoparticles. When metal coordination complexes are used, the ligands keep the metal atoms isolated from each other. It has been found that many metals plate out as a smoother, more uniform, better-looking, and more adherent surface when plated from a bath containing the metal as a complex ion. Thus, complexes such as $\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}$and $\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}$are used extensively in the electroplating industry.

In 1965, scientists at Michigan State University discovered that there was a platinum complex that inhibited cell division in certain microorganisms. Later work showed that the complex was cis-diamminedichloroplatinum(II), $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2}(\mathrm{Cl})_{2}\right]$, and that the trans isomer was not effective. The inhibition of cell division indicated that this square planar compound could be an anticancer agent. In 1978, the US Food and Drug Administration approved this compound, known as cisplatin, for use in the treatment of certain forms of cancer. Since that time, many similar platinum compounds have been developed for the treatment of cancer. In all cases, these are the cis isomers and never the trans isomers. The diammine $\left(\mathrm{NH}_{3}\right)_{2}$ portion is retained with other groups, replacing the dichloro $\left[(\mathrm{Cl})_{2}\right]$ portion. The newer drugs include carboplatin, oxaliplatin, and satraplatin.


2028. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Outline the basic premise of crystal field theory (CFT)
- Identify molecular geometries associated with various d-orbital splitting patterns
- Predict electron configurations of split d orbitals for selected transition metal atoms or ions
- Explain spectral and magnetic properties in terms of CFT concepts

The behavior of coordination compounds cannot be adequately explained by the same theories used for main group element chemistry. The observed geometries of coordination complexes are not consistent with hybridized orbitals on the central metal overlapping with ligand orbitals, as would be predicted by valence bond theory. The observed colors indicate that the $d$ orbitals often occur at different energy levels rather than all being degenerate, that is, of equal energy, as are the three $p$ orbitals. To explain the stabilities, structures, colors, and magnetic properties of transition metal complexes, a different bonding model has been developed. Just as valence bond theory explains many aspects of bonding in main group chemistry, crystal field theory is useful in understanding and predicting the behavior of transition metal complexes.


2029. Crystal Field Theory -
To explain the observed behavior of transition metal complexes (such as how colors arise), a model involving electrostatic interactions between the electrons from the ligands and the electrons in the unhybridized $d$ orbitals of the central metal atom has been developed. This electrostatic model is crystal field theory (CFT). It allows us to understand, interpret, and predict the colors, magnetic behavior, and some structures of coordination compounds of transition metals.

CFT focuses on the nonbonding electrons on the central metal ion in coordination complexes not on the metalligand bonds. Like valence bond theory, CFT tells only part of the story of the behavior of complexes. However, it tells the part that valence bond theory does not. In its pure form, CFT ignores any covalent bonding between ligands and metal ions. Both the ligand and the metal are treated as infinitesimally small point charges.

All electrons are negative, so the electrons donated from the ligands will repel the electrons of the central metal. Let us consider the behavior of the electrons in the unhybridized $d$ orbitals in an octahedral complex. The five $d$ orbitals consist of lobe-shaped regions and are arranged in space, as shown in Figure 19.33. In an octahedral complex, the six ligands coordinate along the axes.

FIGURE 19.33 The directional characteristics of the five $d$ orbitals are shown here. The shaded portions indicate the phase of the orbitals. The ligands (L) coordinate along the axes. For clarity, the ligands have been omitted from the $d_{x^{2}-y^{2}}$ orbital so that the axis labels could be shown.


2029. Crystal Field Theory -
In an uncomplexed metal ion in the gas phase, the electrons are distributed among the five $d$ orbitals in accord with Hund's rule because the orbitals all have the same energy. However, when ligands coordinate to a metal ion, the energies of the $d$ orbitals are no longer the same.
In octahedral complexes, the lobes in two of the five $d$ orbitals, the $d_{z^{2}}$ and $d_{x^{2}-y^{2}}$ orbitals, point toward the ligands (Figure 19.33). These two orbitals are called the $\boldsymbol{e}_{g}$ orbitals (the symbol actually refers to the symmetry of the orbitals, but we will use it as a convenient name for these two orbitals in an octahedral complex). The other three orbitals, the $d_{x y}, d_{x z}$, and $d_{y z}$ orbitals, have lobes that point between the ligands and are called the $\boldsymbol{t}_{\mathbf{2 g}}$ orbitals (again, the symbol really refers to the symmetry of the orbitals). As six ligands approach the metal ion along the axes of the octahedron, their point charges repel the electrons in the $d$ orbitals of the metal ion. However, the repulsions between the electrons in the $e_{g}$ orbitals (the $d_{z^{2}}$ and $d_{x^{2}-y^{2}}$ orbitals) and the ligands are greater than the repulsions between the electrons in the $t_{2 g}$ orbitals (the $d_{z y}, d_{x z}$, and $d_{y z}$ orbitals) and the ligands. This is because the lobes of the $e_{g}$ orbitals point directly at the ligands, whereas the lobes of the $t_{2 g}$ orbitals point between them. Thus, electrons in the $e_{g}$ orbitals of the metal ion in an octahedral complex have higher potential energies than those of electrons in the $t_{2 g}$ orbitals. The difference in energy may be represented as shown in Figure 19.34.


2029. Crystal Field Theory -
FIGURE 19.34 In octahedral complexes, the $e_{g}$ orbitals are destabilized (higher in energy) compared to the $t_{2 g}$ orbitals because the ligands interact more strongly with the $d$ orbitals at which they are pointed directly.

The difference in energy between the $e_{g}$ and the $t_{2 g}$ orbitals is called the crystal field splitting and is symbolized by $\boldsymbol{\Delta}_{\text {oct }}$, where oct stands for octahedral.

The magnitude of $\Delta_{\text {oct }}$ depends on many factors, including the nature of the six ligands located around the central metal ion, the charge on the metal, and whether the metal is using $3 d, 4 d$, or $5 d$ orbitals. Different ligands produce different crystal field splittings. The increasing crystal field splitting produced by ligands is expressed in the spectrochemical series, a short version of which is given here:

$$
\mathrm{I}^{-}<\mathrm{Br}^{-}<\mathrm{Cl}^{-}<\mathrm{F}^{-}<\mathrm{H}_{2} \mathrm{O}<\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}<\mathrm{NH}_{3}<e n<\mathrm{NO}_{2}-<\mathrm{CN}^{-}
$$

a few ligands of the spectrochemical series, in order of increasing field strength of the ligand

In this series, ligands on the left cause small crystal field splittings and are weak-field ligands, whereas those on the right cause larger splittings and are strong-field ligands. Thus, the $\Delta_{\text {oct }}$ value for an octahedral complex with iodide ligands $\left(\mathrm{I}^{-}\right)$is much smaller than the $\Delta_{\text {oct }}$ value for the same metal with cyanide ligands $\left(\mathrm{CN}^{-}\right)$.


2029. Crystal Field Theory -
Electrons in the $d$ orbitals follow the aufbau ("filling up") principle, which says that the orbitals will be filled to give the lowest total energy, just as in main group chemistry. When two electrons occupy the same orbital, the like charges repel each other. The energy needed to pair up two electrons in a single orbital is called the pairing energy ( $\mathbf{P}$ ). Electrons will always singly occupy each orbital in a degenerate set before pairing. $P$ is similar in magnitude to $\Delta_{\text {oct }}$. When electrons fill the $d$ orbitals, the relative magnitudes of $\Delta_{\text {oct }}$ and $P$ determine which orbitals will be occupied.

In $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$, the strong field of six cyanide ligands produces a large $\Delta_{\text {oct. }}$. Under these conditions, the electrons require less energy to pair than they require to be excited to the $e_{g}$ orbitals ( $\Delta_{\text {oct }}>P$ ). The six $3 d$ electrons of the $\mathrm{Fe}^{2+}$ ion pair in the three $t_{2 g}$ orbitals (Figure 19.35). Complexes in which the electrons are paired because of the large crystal field splitting are called low-spin complexes because the number of unpaired electrons (spins) is minimized.

FIGURE 19.35 Iron(II) complexes have six electrons in the $5 d$ orbitals. In the absence of a crystal field, the orbitals are degenerate. For coordination complexes with strong-field ligands such as $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}, \Delta_{\text {oct }}$ is greater than P , and the electrons pair in the lower energy $t_{2 g}$ orbitals before occupying the eg orbitals. With weak-field ligands such as $\mathrm{H}_{2} \mathrm{O}$, the ligand field splitting is less than the pairing energy, $\Delta_{\text {oct }}$ less than P , so the electrons occupy all $d$ orbitals singly before any pairing occurs.


2029. Crystal Field Theory -
In $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$, on the other hand, the weak field of the water molecules produces only a small crystal field splitting ( $\Delta_{\text {oct }}<\mathrm{P}$ ). Because it requires less energy for the electrons to occupy the $e_{g}$ orbitals than to pair together, there will be an electron in each of the five $3 d$ orbitals before pairing occurs. For the six $d$ electrons on the iron(II) center in $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$, there will be one pair of electrons and four unpaired electrons (Figure 19.35). Complexes such as the $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ ion, in which the electrons are unpaired because the crystal field splitting is not large enough to cause them to pair, are called high-spin complexes because the number of unpaired electrons (spins) is maximized.

A similar line of reasoning shows why the $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ ion is a low-spin complex with only one unpaired electron, whereas both the $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ and $\left[\mathrm{FeF}_{6}\right]^{3-}$ ions are high-spin complexes with five unpaired electrons.


2031. High- and Low-Spin Complexes -
Predict the number of unpaired electrons.
(a) $\mathrm{K}_{3}\left[\mathrm{CrI}_{6}\right]$
(b) $\left[\mathrm{Cu}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right] \mathrm{Cl}_{2}$
(c) $\mathrm{Na}_{3}\left[\mathrm{Co}\left(\mathrm{NO}_{2}\right)_{6}\right]$


2032. Solution -
The complexes are octahedral.
(a) $\mathrm{Cr}^{3+}$ has a $d^{3}$ configuration. These electrons will all be unpaired.
(b) $\mathrm{Cu}^{2+}$ is $d^{9}$, so there will be one unpaired electron.
(c) $\mathrm{Co}^{3+}$ has $d^{6}$ valence electrons, so the crystal field splitting will determine how many are paired. Nitrite is a strong-field ligand, so the complex will be low spin. Six electrons will go in the $t_{2 g}$ orbitals, leaving 0 unpaired.


2033. Check Your Learning -
The size of the crystal field splitting only influences the arrangement of electrons when there is a choice between pairing electrons and filling the higher-energy orbitals. For which $d$-electron configurations will there be a difference between high- and low-spin configurations in octahedral complexes?


2034. Answer: -
$d^{4}, d^{5}, d^{6}$, and $d^{7}$


2036. CFT for Other Geometries -
CFT is applicable to molecules in geometries other than octahedral. In octahedral complexes, remember that the lobes of the $e_{g}$ set point directly at the ligands. For tetrahedral complexes, the $d$ orbitals remain in place, but now we have only four ligands located between the axes (Figure 19.36). None of the orbitals points directly at the tetrahedral ligands. However, the $e_{g}$ set (along the Cartesian axes) overlaps with the ligands less than does the $t_{2 g}$ set. By analogy with the octahedral case, predict the energy diagram for the $d$ orbitals in a tetrahedral crystal field. To avoid confusion, the octahedral $e_{g}$ set becomes a tetrahedral e set, and the octahedral $t_{2 g}$ set becomes a $t_{2}$ set.

FIGURE 19.36 This diagram shows the orientation of the tetrahedral ligands with respect to the axis system for the orbitals.


2037. Solution -
Since CFT is based on electrostatic repulsion, the orbitals closer to the ligands will be destabilized and raised in energy relative to the other set of orbitals. The splitting is less than for octahedral complexes because the overlap is less, so $\Delta_{\text {tet }}$ is usually small $\left(\Delta_{\text {tet }}=\frac{4}{9} \Delta_{\text {oct }}\right)$ :

Check Your Learning
Explain how many unpaired electrons a tetrahedral $d^{4}$ ion will have.


2038. Answer: -
4; because $\Delta_{\text {tet }}$ is small, all tetrahedral complexes are high spin and the electrons go into the $t_{2}$ orbitals before pairing

The other common geometry is square planar. It is possible to consider a square planar geometry as an octahedral structure with a pair of trans ligands removed. The removed ligands are assumed to be on the $z$-axis. This changes the distribution of the $d$ orbitals, as orbitals on or near the $z$-axis become more stable, and those on or near the $x$ - or $y$-axes become less stable. This results in the octahedral $t_{2 g}$ and the $e_{g}$ sets splitting and gives a more complicated pattern, as depicted below:

Magnetic Moments of Molecules and Ions
Experimental evidence of magnetic measurements supports the theory of high- and low-spin complexes. Remember that molecules such as $\mathrm{O}_{2}$ that contain unpaired electrons are paramagnetic. Paramagnetic substances are attracted to magnetic fields. Many transition metal complexes have unpaired electrons and hence are paramagnetic. Molecules such as $\mathrm{N}_{2}$ and ions such as $\mathrm{Na}^{+}$and $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$ that contain no unpaired electrons are diamagnetic. Diamagnetic substances have a slight tendency to be repelled by magnetic fields.

When an electron in an atom or ion is unpaired, the magnetic moment due to its spin makes the entire atom or ion paramagnetic. The size of the magnetic moment of a system containing unpaired electrons is related directly to the number of such electrons: the greater the number of unpaired electrons, the larger the magnetic moment. Therefore, the observed magnetic moment is used to determine the number of unpaired electrons present. The measured magnetic moment of low-spin $d^{6}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$ confirms that iron is diamagnetic,
whereas high-spin $d^{6}\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ has four unpaired electrons with a magnetic moment that confirms this arrangement.


2039. Colors of Transition Metal Complexes -
When atoms or molecules absorb light at the proper frequency, their electrons are excited to higher-energy orbitals. For many main group atoms and molecules, the absorbed photons are in the ultraviolet range of the electromagnetic spectrum, which cannot be detected by the human eye. For coordination compounds, the energy difference between the $d$ orbitals often allows photons in the visible range to be absorbed.

The human eye perceives a mixture of all the colors, in the proportions present in sunlight, as white light. Complementary colors, those located across from each other on a color wheel, are also used in color vision. The eye perceives a mixture of two complementary colors, in the proper proportions, as white light. Likewise, when a color is missing from white light, the eye sees its complement. For example, when red photons are absorbed from white light, the eyes see the color green. When violet photons are removed from white light, the eyes see lemon yellow. The blue color of the $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$ ion results because this ion absorbs orange and red light, leaving the complementary colors of blue and green (Figure 19.37).

FIGURE 19.37 (a) An object is black if it absorbs all colors of light. If it reflects all colors of light, it is white. An object has a color if it absorbs all colors except one, such as this yellow strip. The strip also appears yellow if it absorbs the complementary color from white light (in this case, indigo). (b) Complementary colors are located directly across from one another on the color wheel. (c) A solution of $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$ ions absorbs red and orange light, so the transmitted light appears as the complementary color, blue.


2041. Colors of Complexes -
The octahedral complex $\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ has a single $d$ electron. To excite this electron from the ground state $t_{2 g}$ orbital to the $e_{g}$ orbital, this complex absorbs light from 450 to 600 nm . The maximum absorbance corresponds to $\Delta_{\text {oct }}$ and occurs at 499 nm . Calculate the value of $\Delta_{\text {oct }}$ in Joules and predict what color the solution will appear.


2042. Solution -
Using Planck's equation (refer to the section on electromagnetic energy), we calculate:

$$
\begin{gathered}
v=\frac{c}{\lambda} \text { so } \frac{3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}}{\frac{499 \mathrm{~nm} \times 1 \mathrm{~m}}{10^{9} \mathrm{~nm}}}=6.01 \times 10^{14} \mathrm{~Hz} \\
E=h \nu \text { so } 6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{~s} \times 6.01 \times 10^{14} \mathrm{~Hz}=3.99 \times 10^{-19} \text { Joules/ion }
\end{gathered}
$$

Because the complex absorbs 600 nm (orange) through 450 (blue), the indigo, violet, and red wavelengths will be transmitted, and the complex will appear purple.


2043. Check Your Learning -
A complex that appears green, absorbs photons of what wavelengths?


2044. Answer: -
red, 620-800 nm

Small changes in the relative energies of the orbitals that electrons are transitioning between can lead to drastic shifts in the color of light absorbed. Therefore, the colors of coordination compounds depend on many factors. As shown in Figure 19.38, different aqueous metal ions can have different colors. In addition, different oxidation states of one metal can produce different colors, as shown for the vanadium complexes in the link below.

FIGURE 19.38 The partially filled $d$ orbitals of the stable ions $\mathrm{Cr}^{3+}(\mathrm{aq}), \mathrm{Fe}^{3+}(\mathrm{aq})$, and $\mathrm{Co}^{2+}(\mathrm{aq})$ (left, center and right, respectively) give rise to various colors. (credit: Sahar Atwa)

The specific ligands coordinated to the metal center also influence the color of coordination complexes. For example, the iron(II) complex $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{SO}_{4}$ appears blue-green because the high-spin complex absorbs photons in the red wavelengths (Figure 19.39). In contrast, the low-spin iron(II) complex $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ appears pale yellow because it absorbs higher-energy violet photons.

FIGURE 19.39 Both (a) hexaaquairon(II) sulfate and (b) potassium hexacyanoferrate(II) contain $d^{6}$ iron(II) octahedral metal centers, but they absorb photons in different ranges of the visible spectrum.


2045. LINK TO LEARNING -
Watch this video (http://openstax.org/l/16vanadium) of the reduction of vanadium complexes to observe the colorful effect of changing oxidation states.

In general, strong-field ligands cause a large split in the energies of $d$ orbitals of the central metal atom (large $\left.\Delta_{\text {oct }}\right)$. Transition metal coordination compounds with these ligands are yellow, orange, or red because they absorb higher-energy violet or blue light. On the other hand, coordination compounds of transition metals with weak-field ligands are often blue-green, blue, or indigo because they absorb lower-energy yellow, orange, or red light.

A coordination compound of the $\mathrm{Cu}^{+}$ion has a $d^{10}$ configuration, and all the $e_{g}$ orbitals are filled. To excite an electron to a higher level, such as the $4 p$ orbital, photons of very high energy are necessary. This energy corresponds to very short wavelengths in the ultraviolet region of the spectrum. No visible light is absorbed, so the eye sees no change, and the compound appears white or colorless. A solution containing $\left[\mathrm{Cu}(\mathrm{CN})_{2}\right]^{-}$, for example, is colorless. On the other hand, octahedral $\mathrm{Cu}^{2+}$ complexes have a vacancy in the $e_{g}$ orbitals, and electrons can be excited to this level. The wavelength (energy) of the light absorbed corresponds to the visible part of the spectrum, and $\mathrm{Cu}^{2+}$ complexes are almost always colored-blue, blue-green violet, or yellow (Figure 19.40). Although CFT successfully describes many properties of coordination complexes, molecular orbital explanations (beyond the introductory scope provided here) are required to understand fully the behavior of coordination complexes.

(a)

(b)

FIGURE 19.40 (a) Copper(I) complexes with $d^{10}$ configurations such as CuI tend to be colorless, whereas (b) $d^{9}$ copper(II) complexes such as $\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} \cdot 5 \mathrm{H}_{2} \mathrm{O}$ are brightly colored.


2046. Key Terms -
actinide series (also, actinoid series) actinium and the elements in the second row or the $f$-block, atomic numbers 89-103
bidentate ligand ligand that coordinates to one central metal through coordinate bonds from two different atoms
central metal ion or atom to which one or more ligands is attached through coordinate covalent bonds
chelate complex formed from a polydentate ligand attached to a central metal
chelating ligand ligand that attaches to a central metal ion by bonds from two or more donor atoms
cis configuration configuration of a geometrical isomer in which two similar groups are on the same side of an imaginary reference line on the molecule
coordination compound stable compound in which the central metal atom or ion acts as a Lewis acid and accepts one or more pairs of electrons
coordination compound substance consisting of atoms, molecules, or ions attached to a central atom through Lewis acid-base interactions
coordination number number of coordinate covalent bonds to the central metal atom in a complex or the number of closest contacts to an atom in a crystalline form
coordination sphere central metal atom or ion plus the attached ligands of a complex
crystal field splitting ( $\boldsymbol{\Delta}_{\mathbf{o c t}}$ ) difference in energy between the $t_{2 g}$ and $e_{g}$ sets or $t$ and e sets of orbitals
crystal field theory model that explains the energies of the orbitals in transition metals in terms of electrostatic interactions with the ligands but does not include metal ligand bonding
$\boldsymbol{d}$-block element one of the elements in groups $3-11$ with valence electrons in $d$ orbitals
donor atom atom in a ligand with a lone pair of electrons that forms a coordinate covalent bond to a central metal
$\boldsymbol{e}_{\boldsymbol{g}}$ orbitals set of two $d$ orbitals that are oriented on the Cartesian axes for coordination complexes; in octahedral complexes, they are higher in energy than the $t_{2 g}$ orbitals
$\boldsymbol{f}$-block element (also, inner transition element) one of the elements with atomic numbers 58-71 or 90-103 that have valence electrons in $f$ orbitals; they are frequently shown offset below
the periodic table


2046. Key Terms -
the periodic table
first transition series transition elements in the fourth period of the periodic table (first row of the $d$-block), atomic numbers 21-29
fourth transition series transition elements in the seventh period of the periodic table (fourth row of the $d$-block), atomic numbers 89 and 104-111
geometric isomers isomers that differ in the way in which atoms are oriented in space relative to each other, leading to different physical and chemical properties
high-spin complex complex in which the electrons maximize the total electron spin by singly populating all of the orbitals before pairing two electrons into the lower-energy orbitals
hydrometallurgy process in which a metal is separated from a mixture by first converting it into soluble ions, extracting the ions, and then reducing the ions to precipitate the pure metal
ionization isomer (or coordination isomer) isomer in which an anionic ligand is replaced by the counter ion in the inner coordination sphere
lanthanide series (also, lanthanoid series) lanthanum and the elements in the first row or the $f$-block, atomic numbers 57-71
ligand ion or neutral molecule attached to the central metal ion in a coordination compound
linkage isomer coordination compound that possesses a ligand that can bind to the transition metal in two different ways ( $\mathrm{CN}^{-}$vs. $\mathrm{NC}^{-}$)
low-spin complex complex in which the electrons minimize the total electron spin by pairing in the lower-energy orbitals before populating the higher-energy orbitals
monodentate ligand that attaches to a central metal through just one coordinate covalent bond
optical isomer (also, enantiomer) molecule that is a nonsuperimposable mirror image with identical chemical and physical properties, except when it reacts with other optical isomers
pairing energy ( $\mathbf{P}$ ) energy required to place two electrons with opposite spins into a single orbital
platinum metals group of six transition metals consisting of ruthenium, osmium, rhodium, iridium, palladium, and platinum that tend to occur in the same minerals and demonstrate similar chemical properties


2046. Key Terms -
platinum metals group of six transition metals consisting of ruthenium, osmium, rhodium, iridium, palladium, and platinum that tend to occur in the same minerals and demonstrate similar chemical properties
polydentate ligand ligand that is attached to a central metal ion by bonds from two or more donor atoms, named with prefixes specifying how many donors are present (e.g., hexadentate = six coordinate bonds formed)
rare earth element collection of 17 elements including the lanthanides, scandium, and yttrium that often occur together and have similar chemical properties, making separation difficult
second transition series transition elements in the fifth period of the periodic table (second row of the $d$-block), atomic numbers 39-47
smelting process of extracting a pure metal from a molten ore
spectrochemical series ranking of ligands according to the magnitude of the crystal field splitting they induce
steel material made from iron by removing impurities in the iron and adding substances that produce alloys with properties suitable for specific uses
strong-field ligand ligand that causes larger


2047. Summary - 2047.1. Occurrence, Preparation, and Properties of Transition Metals and Their Compounds
The transition metals are elements with partially filled $d$ orbitals, located in the $d$-block of the periodic table. The reactivity of the transition elements varies widely from very active metals such as scandium and iron to almost inert elements, such as the platinum metals. The type of chemistry used in the isolation of the elements from their ores depends upon the concentration of the element in its ore and the difficulty of reducing ions of the elements to the metals. Metals that are more active are more difficult to reduce.

Transition metals exhibit chemical behavior typical of metals. For example, they oxidize in air upon heating and react with elemental halogens to form halides. Those elements that lie above hydrogen in the activity series react with acids, producing salts and hydrogen gas. Oxides, hydroxides, and carbonates of transition metal compounds in low oxidation states are basic. Halides and other salts are generally stable in water, although oxygen must be excluded in some cases. Most transition metals form a variety of stable oxidation states, allowing them to demonstrate a wide range of chemical reactivity.


2047. Summary - 2047.2. Coordination Chemistry of Transition Metals
The transition elements and main group elements can form coordination compounds, or complexes, in
crystal field splittings
superconductor material that conducts electricity with no resistance
$\boldsymbol{t}_{\mathbf{2 g}}$ orbitals set of three $d$ orbitals aligned between the Cartesian axes for coordination complexes; in octahedral complexes, they are lowered in energy compared to the $e_{g}$ orbitals according to CFT
third transition series transition elements in the sixth period of the periodic table (third row of the $d$-block), atomic numbers 57 and 72-79
trans configuration configuration of a geometrical isomer in which two similar groups are on opposite sides of an imaginary reference line on the molecule
weak-field ligand ligand that causes small crystal field splittings
which a central metal atom or ion is bonded to one or more ligands by coordinate covalent bonds. Ligands with more than one donor atom are called polydentate ligands and form chelates. The common geometries found in complexes are tetrahedral and square planar (both with a coordination number of four) and octahedral (with a coordination number of six). Cis and trans configurations are possible in some octahedral and square planar complexes. In addition to these geometrical isomers, optical isomers (molecules or ions that are mirror images but not superimposable) are possible in certain octahedral complexes. Coordination complexes have a wide variety of uses including oxygen transport in blood, water purification, and pharmaceutical use.


2047. Summary - 2047.3. Spectroscopic and Magnetic Properties of Coordination Compounds
Crystal field theory treats interactions between the electrons on the metal and the ligands as a simple electrostatic effect. The presence of the ligands near the metal ion changes the energies of the metal $d$ orbitals relative to their energies in the free ion. Both the color and the magnetic properties of a complex can be attributed to this crystal field splitting. The magnitude of the splitting ( $\Delta_{\text {oct }}$ ) depends on the nature of the ligands bonded to the metal. Strongfield ligands produce large splitting and favor lowspin complexes, in which the $t_{2 g}$ orbitals are completely filled before any electrons occupy the $e_{g}$ orbitals. Weak-field ligands favor formation of highspin complexes. The $t_{2 g}$ and the $e_{g}$ orbitals are singly occupied before any are doubly occupied.


2048. Exercises - 2048.1. Occurrence, Preparation, and Properties of Transition Metals and Their Compounds
1. Write the electron configurations for each of the following elements:
(a) Sc
(b) Ti
(c) Cr
(d) Fe
(e) Ru
2. Write the electron configurations for each of the following elements and its ions:
(a) Ti
(b) $\mathrm{Ti}^{2+}$
(c) $\mathrm{Ti}^{3+}$
(d) $\mathrm{Ti}^{4+}$
3. Write the electron configurations for each of the following elements and its $3+$ ions:
(a) La
(b) Sm
(c) Lu
4. Why are the lanthanoid elements not found in nature in their elemental forms?
5. Which of the following elements is most likely to be used to prepare La by the reduction of $\mathrm{La}_{2} \mathrm{O}_{3}$ : $\mathrm{Al}, \mathrm{C}$, or Fe? Why?
6. Which of the following is the strongest oxidizing agent: $\mathrm{VO}_{4}{ }^{3}, \mathrm{CrO}_{4}{ }^{2-}$, or $\mathrm{MnO}_{4}{ }^{-}$?
7. Which of the following elements is most likely to form an oxide with the formula $\mathrm{MO}_{3}: \mathrm{Zr}, \mathrm{Nb}$, or Mo ?
8. The following reactions all occur in a blast furnace. Which of these are redox reactions?
(a) $3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g)$
(b) $\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \longrightarrow 3 \mathrm{FeO}(s)+\mathrm{CO}_{2}(g)$
(c) $\mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(l)+\mathrm{CO}_{2}(g)$
(d) $\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)$


2048. Exercises - 2048.1. Occurrence, Preparation, and Properties of Transition Metals and Their Compounds
(d) $\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)$
(e) $\mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(g) \longrightarrow 2 \mathrm{CO}(g)$
(f) $\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$
(g) $\mathrm{CaO}(s)+\mathrm{SiO}_{2}(s) \longrightarrow \mathrm{CaSiO}_{3}(l)$
9. Why is the formation of slag useful during the smelting of iron?
10. Would you expect an aqueous manganese(VII) oxide solution to have a pH greater or less than 7.0 ? Justify your answer.
11. Iron(II) can be oxidized to iron(III) by dichromate ion, which is reduced to chromium(III) in acid solution. A $2.5000-\mathrm{g}$ sample of iron ore is dissolved and the iron converted into iron(II). Exactly 19.17 mL of 0.0100 $M \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ is required in the titration. What percentage of the ore sample was iron?
12. How many cubic feet of air at a pressure of 760 torr and $0^{\circ} \mathrm{C}$ is required per ton of $\mathrm{Fe}_{2} \mathrm{O}_{3}$ to convert that $\mathrm{Fe}_{2} \mathrm{O}_{3}$ into iron in a blast furnace? For this exercise, assume air is $19 \%$ oxygen by volume.
13. Find the potentials of the following electrochemical cell:
$\mathrm{Cd}\left|\mathrm{Cd}^{2+}, M=0.10 \| \mathrm{Ni}^{2+}, M=0.50\right| \mathrm{Ni}$


2048. Exercises - 2048.1. Occurrence, Preparation, and Properties of Transition Metals and Their Compounds
$\mathrm{Cd}\left|\mathrm{Cd}^{2+}, M=0.10 \| \mathrm{Ni}^{2+}, M=0.50\right| \mathrm{Ni}$
14. A $2.5624-\mathrm{g}$ sample of a pure solid alkali metal chloride is dissolved in water and treated with excess silver nitrate. The resulting precipitate, filtered and dried, weighs 3.03707 g . What was the percent by mass of chloride ion in the original compound? What is the identity of the salt?
15. The standard reduction potential for the reaction $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}(a q)+\mathrm{e}^{-} \longrightarrow\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}(a q)$ is about 1.8 V . The reduction potential for the reaction $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}(a q)+\mathrm{e}^{-} \longrightarrow\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}(a q)$ is +0.1 V . Calculate the cell potentials to show whether the complex ions, $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ and/or $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$, can be oxidized to the corresponding cobalt(III) complex by oxygen.
16. Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.)
(a) $\mathrm{MnCO}_{3}(s)+\mathrm{HI}(a q) \longrightarrow$
(b) $\mathrm{CoO}(s)+\mathrm{O}_{2}(g)$
(c) $\mathrm{La}(s)+\mathrm{O}_{2}(g) \longrightarrow$


2048. Exercises - 2048.1. Occurrence, Preparation, and Properties of Transition Metals and Their Compounds
(b) $\mathrm{CoO}(s)+\mathrm{O}_{2}(g)$
(c) $\mathrm{La}(s)+\mathrm{O}_{2}(g) \longrightarrow$
(d) $\mathrm{V}(s)+\mathrm{VCl}_{4}(s) \longrightarrow$
(e) $\mathrm{Co}(s)+x s \mathrm{~F}_{2}(g) \longrightarrow$
(f) $\mathrm{CrO}_{3}(s)+\mathrm{CsOH}(a q) \longrightarrow$
17. Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.)
(a) $\mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow$
(b) $\mathrm{FeCl}_{3}(a q)+\mathrm{NaOH}(a q) \longrightarrow$
(c) $\mathrm{Mn}(\mathrm{OH})_{2}(s)+\mathrm{HBr}(a q) \longrightarrow$
(d) $\mathrm{Cr}(s)+\mathrm{O}_{2}(g) \longrightarrow$
(e) $\mathrm{Mn}_{2} \mathrm{O}_{3}(s)+\mathrm{HCl}(a q) \longrightarrow$
(f) $\mathrm{Ti}(s)+x s \mathrm{~F}_{2}(g) \longrightarrow$
18. Describe the electrolytic process for refining copper.
19. Predict the products of the following reactions and balance the equations.
(a) Zn is added to a solution of $\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ in acid.
(b) $\mathrm{FeCl}_{2}$ is added to a solution containing an excess of $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ in hydrochloric acid.
(c) $\mathrm{Cr}^{2+}$ is added to $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ in acid solution.
(d) Mn is heated with $\mathrm{CrO}_{3}$.


2048. Exercises - 2048.1. Occurrence, Preparation, and Properties of Transition Metals and Their Compounds
(d) Mn is heated with $\mathrm{CrO}_{3}$.
(e) CrO is added to $2 \mathrm{HNO}_{3}$ in water.
(f) $\mathrm{FeCl}_{3}$ is added to an aqueous solution of NaOH .
20. What is the gas produced when iron(II) sulfide is treated with a nonoxidizing acid?
21. Predict the products of each of the following reactions and then balance the chemical equations.
(a) Fe is heated in an atmosphere of steam.
(b) NaOH is added to a solution of $\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}$.
(c) $\mathrm{FeSO}_{4}$ is added to an acidic solution of $\mathrm{KMnO}_{4}$.
(d) Fe is added to a dilute solution of $\mathrm{H}_{2} \mathrm{SO}_{4}$.
(e) A solution of $\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}$ and $\mathrm{HNO}_{3}$ is allowed to stand in air.
(f) $\mathrm{FeCO}_{3}$ is added to a solution of $\mathrm{HClO}_{4}$.
(g) Fe is heated in air.
22. Balance the following equations by oxidation-reduction methods; note that three elements change oxidation state.
$\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(s) \longrightarrow \mathrm{Co}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$
23. Dilute sodium cyanide solution is slowly dripped into a slowly stirred silver nitrate solution. A white precipitate forms temporarily but dissolves as the addition of sodium cyanide continues. Use chemical equations to explain this observation. Silver cyanide is similar to silver chloride in its solubility.
24. Predict which will be more stable, $\left[\mathrm{CrO}_{4}\right]^{2-}$ or $\left[\mathrm{WO}_{4}\right]^{2-}$, and explain.


2048. Exercises - 2048.1. Occurrence, Preparation, and Properties of Transition Metals and Their Compounds
24. Predict which will be more stable, $\left[\mathrm{CrO}_{4}\right]^{2-}$ or $\left[\mathrm{WO}_{4}\right]^{2-}$, and explain.
25. Give the oxidation state of the metal for each of the following oxides of the first transition series. (Hint: Oxides of formula $\mathrm{M}_{3} \mathrm{O}_{4}$ are examples of mixed valence compounds in which the metal ion is present in more than one oxidation state. It is possible to write these compound formulas in the equivalent format $\mathrm{MO} \cdot \mathrm{M}_{2} \mathrm{O}_{3}$, to permit estimation of the metal's two oxidation states.)
(a) $\mathrm{Sc}_{2} \mathrm{O}_{3}$
(b) $\mathrm{TiO}_{2}$
(c) $\mathrm{V}_{2} \mathrm{O}_{5}$
(d) $\mathrm{CrO}_{3}$
(e) $\mathrm{MnO}_{2}$
(f) $\mathrm{Fe}_{3} \mathrm{O}_{4}$
(g) $\mathrm{Co}_{3} \mathrm{O}_{4}$
(h) NiO
(i) $\mathrm{Cu}_{2} \mathrm{O}$


2048. Exercises - 2048.2. Coordination Chemistry of Transition Metals
26. Indicate the coordination number for the central metal atom in each of the following coordination compounds:
(a) $\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} \mathrm{Br}_{2}\right]$
(b) $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)(\mathrm{py})(\mathrm{Cl})(\mathrm{Br})\right]\left(\mathrm{py}=\right.$ pyridine, $\left.\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)$
(c) $\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]$
(d) $\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)(\mathrm{py})(\mathrm{Cl})(\mathrm{Br})\right]$
(e) $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right]$
(f) $\left[\mathrm{Fe}(\mathrm{en})_{2}(\mathrm{CN})_{2}\right]^{+}\left(\mathrm{en}=\right.$ ethylenediamine, $\left.\mathrm{C}_{2} \mathrm{H}_{8} \mathrm{~N}_{2}\right)$
27. Give the coordination numbers and write the formulas for each of the following, including all isomers where appropriate:
(a) tetrahydroxozincate(II) ion (tetrahedral)
(b) hexacyanopalladate(IV) ion
(c) dichloroaurate(I) ion (note that aurum is Latin for "gold")
(d) diamminedichloroplatinum(II)
(e) potassium diamminetetrachlorochromate(III)
(f) hexaamminecobalt(III) hexacyanochromate(III)
(g) dibromobis(ethylenediamine) cobalt(III) nitrate
28. Give the coordination number for each metal ion in the following compounds:


2048. Exercises - 2048.2. Coordination Chemistry of Transition Metals
(g) dibromobis(ethylenediamine) cobalt(III) nitrate
28. Give the coordination number for each metal ion in the following compounds:
(a) $\left[\mathrm{Co}\left(\mathrm{CO}_{3}\right)_{3}\right]^{3-}$ (note that $\mathrm{CO}_{3}{ }^{2-}$ is bidentate in this complex)
(b) $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$
(c) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right]_{2}\left(\mathrm{SO}_{4}\right)_{3}$
(d) $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\right]\left[\mathrm{PtCl}_{4}\right]$
(e) $\left[\mathrm{Cr}(\mathrm{en})_{3}\right]\left(\mathrm{NO}_{3}\right)_{3}$
(f) $\left[\mathrm{Pd}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Br}_{2}\right]$ (square planar)
(g) $\mathrm{K}_{3}\left[\mathrm{Cu}(\mathrm{Cl})_{5}\right]$
(h) $\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]$
29. Sketch the structures of the following complexes. Indicate any cis, trans, and optical isomers.
(a) $\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} \mathrm{Br}_{2}\right]$ (square planar)
(b) $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)(\mathrm{py})(\mathrm{Cl})(\mathrm{Br})\right]$ (square planar, py $=$ pyridine, $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}$ )


2048. Exercises - 2048.2. Coordination Chemistry of Transition Metals
(c) $\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}\right]^{+}$(tetrahedral)
(d) $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}\right]^{+}$(square planar)
(e) $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right]$
(f) $\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2} \mathrm{Cl}_{2}\right]^{3-}$ (note that $\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}$ is the bidentate oxalate ion, $-\mathrm{O}_{2} \mathrm{CCO}_{2}{ }^{-}$)
30. Draw diagrams for any cis, trans, and optical isomers that could exist for the following (en is ethylenediamine):
(a) $\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{NO}_{2}\right) \mathrm{Cl}\right]^{+}$
(b) $\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}$
(c) $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{4}\right]$
(d) $\left[\mathrm{Cr}(\mathrm{en})_{3}\right]^{3+}$
(e) $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]$
31. Name each of the compounds or ions given in Exercise 19.28 , including the oxidation state of the metal.
32. Name each of the compounds or ions given in Exercise 19.30.
33. Specify whether the following complexes have isomers.
(a) tetrahedral $\left[\mathrm{Ni}(\mathrm{CO})_{2}(\mathrm{Cl})_{2}\right]$


2048. Exercises - 2048.2. Coordination Chemistry of Transition Metals
33. Specify whether the following complexes have isomers.
(a) tetrahedral $\left[\mathrm{Ni}(\mathrm{CO})_{2}(\mathrm{Cl})_{2}\right]$
(b) trigonal bipyramidal $\left[\mathrm{Mn}(\mathrm{CO})_{4} \mathrm{NO}\right]$
(c) $\left[\mathrm{Pt}(\mathrm{en}){ }_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}_{2}$
34. Predict whether the carbonate ligand $\mathrm{CO}_{3}{ }^{2-}$ will coordinate to a metal center as a monodentate, bidentate, or tridentate ligand.
35. Draw the geometric, linkage, and ionization isomers for $\left\mathrm{CoCl}_{5} \mathrm{CN}\right$.


2048. Exercises - 2048.3. Spectroscopic and Magnetic Properties of Coordination Compounds
36. Determine the number of unpaired electrons expected for $\left[\mathrm{Fe}\left(\mathrm{NO}_{2}\right)_{6}\right]^{3-}$ and for $\left[\mathrm{FeF} \mathrm{F}_{6}\right]^{3-}$ in terms of crystal field theory.
37. Draw the crystal field diagrams for $\left[\mathrm{Fe}\left(\mathrm{NO}_{2}\right)_{6}\right]^{4-}$ and $\left[\mathrm{FeF}_{6}\right]^{3-}$. State whether each complex is high spin or low spin, paramagnetic or diamagnetic, and compare $\Delta_{\text {oct }}$ to $P$ for each complex.
38. Give the oxidation state of the metal, number of $d$ electrons, and the number of unpaired electrons predicted for $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$.
39. The solid anhydrous solid $\mathrm{CoCl}_{2}$ is blue in color. Because it readily absorbs water from the air, it is used as a humidity indicator to monitor if equipment (such as a cell phone) has been exposed to excessive levels of moisture. Predict what product is formed by this reaction, and how many unpaired electrons this complex will have.
40. Is it possible for a complex of a metal in the transition series to have six unpaired electrons? Explain.
41. How many unpaired electrons are present in each of the following?
(a) $\left[\mathrm{CoF}_{6}\right]^{3-}$ (high spin)
(b) $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}$ (low spin)
(c) $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$ (low spin)
(d) $\left[\mathrm{MnCl}_{6}\right]^{4-}$ (high spin)
(e) $\left[\mathrm{RhCl}_{6}\right]^{3-}$ (low spin)


2048. Exercises - 2048.3. Spectroscopic and Magnetic Properties of Coordination Compounds
(d) $\left[\mathrm{MnCl}_{6}\right]^{4-}$ (high spin)
(e) $\left[\mathrm{RhCl}_{6}\right]^{3-}$ (low spin)
42. Explain how the diphosphate ion, $\left[\mathrm{O}_{3} \mathrm{P}-\mathrm{O}-\mathrm{PO}_{3}\right]^{4-}$, can function as a water softener that prevents the precipitation of $\mathrm{Fe}^{2+}$ as an insoluble iron salt.
43. For complexes of the same metal ion with no change in oxidation number, the stability increases as the number of electrons in the $t_{2 g}$ orbitals increases. Which complex in each of the following pairs of complexes is more stable?
(a) $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ or $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$
(b) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ or $\left[\mathrm{CoF}_{6}\right]^{3-}$
(c) $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$ or $\left[\mathrm{MnCl}_{6}\right]^{4-}$
44. Trimethylphosphine, $\mathrm{P}\left(\mathrm{CH}_{3}\right)_{3}$, can act as a ligand by donating the lone pair of electrons on the phosphorus atom. If trimethylphosphine is added to a solution of nickel(II) chloride in acetone, a blue compound that has a molecular mass of approximately 270 g and contains $21.5 \% \mathrm{Ni}, 26.0 \% \mathrm{Cl}$, and $52.5 \% \mathrm{P}\left(\mathrm{CH}_{3}\right)_{3}$ can be isolated. This blue compound does not have any isomeric forms. What are the geometry and molecular formula of the blue compound?


2048. Exercises - 2048.3. Spectroscopic and Magnetic Properties of Coordination Compounds
45. Would you expect the complex $\left[\mathrm{Co}(\mathrm{en})_{3}\right]_{\mathrm{Cl}}^{3}$ to have any unpaired electrons? Any isomers?
46. Would you expect the $\mathrm{Mg}_{3}\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]_{2}$ to be diamagnetic or paramagnetic? Explain your reasoning.
47. Would you expect salts of the gold(I) ion, $\mathrm{Au}^{+}$, to be colored? Explain.
48. $\left[\mathrm{CuCl}_{4}\right]^{2-}$ is green. $\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ is blue. Which absorbs higher-energy photons? Which is predicted to have a larger crystal field splitting?


2049. CHAPTER 20 <br> Nuclear Chemistry -
Figure 20.1 Nuclear chemistry provides the basis for many useful diagnostic and therapeutic methods in medicine, such as these positron emission tomography (PET) scans. The PET/computed tomography scan on the left shows muscle activity. The brain scans in the center show chemical differences in dopamine signaling in the brains of addicts and nonaddicts. The images on the right show an oncological application of PET scans to identify lymph node metastasis.


2050. CHAPTER OUTLINE -
20.1 Nuclear Structure and Stability
20.2 Nuclear Equations
20.3 Radioactive Decay
20.4 Transmutation and Nuclear Energy
20.5 Uses of Radioisotopes
20.6 Biological Effects of Radiation

INTRODUCTION The chemical reactions that we have considered in previous chapters involve changes in the electronic structure of the species involved, that is, the arrangement of the electrons around atoms, ions, or molecules. Nuclear structure, the numbers of protons and neutrons within the nuclei of the atoms involved, remains unchanged during chemical reactions.

This chapter will introduce the topic of nuclear chemistry, which began with the discovery of radioactivity in 1896 by French physicist Antoine Becquerel and has become increasingly important during the twentieth and twenty-first centuries, providing the basis for various technologies related to energy, medicine, geology, and many other areas.


2051. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Describe nuclear structure in terms of protons, neutrons, and electrons
- Calculate mass defect and binding energy for nuclei
- Explain trends in the relative stability of nuclei

Nuclear chemistry is the study of reactions that involve changes in nuclear structure. The chapter on atoms, molecules, and ions introduced the basic idea of nuclear structure, that the nucleus of an atom is composed of protons and, with the exception of ${ }_{1}^{1} \mathrm{H}$, neutrons. Recall that the number of protons in the nucleus is called the atomic number ( Z ) of the element, and the sum of the number of protons and the number of neutrons is the mass number (A). Atoms with the same atomic number but different mass numbers are isotopes of the same element. When referring to a single type of nucleus, we often use the term nuclide and identify it by the notation ${ }_{Z}^{A} \mathrm{X}$, where X is the symbol for the element, A is the mass number, and Z is the atomic number (for example, ${ }_{6}^{14} \mathrm{C}$ ). Often a nuclide is referenced by the name of the element followed by a hyphen and the mass number. For example, ${ }_{6}^{14} \mathrm{C}$ is called "carbon-14."


2051. LEARNING OBJECTIVES -
Protons and neutrons, collectively called nucleons, are packed together tightly in a nucleus. With a radius of about $10^{-15}$ meters, a nucleus is quite small compared to the radius of the entire atom, which is about $10^{-10}$ meters. Nuclei are extremely dense compared to bulk matter, averaging $1.8 \times 10^{14}$ grams per cubic centimeter. For example, water has a density of 1 gram per cubic centimeter, and iridium, one of the densest elements known, has a density of $22.6 \mathrm{~g} / \mathrm{cm}^{3}$. If the earth's density were equal to the average nuclear density, the earth's radius would be only about 200 meters (earth's actual radius is approximately $6.4 \times 10^{6}$ meters, 30,000 times larger). Example 20.1 demonstrates just how great nuclear densities can be in the natural world.


2053. Density of a Neutron Star -
Neutron stars form when the core of a very massive star undergoes gravitational collapse, causing the star's outer layers to explode in a supernova. Composed almost completely of neutrons, they are the densest-known stars in the universe, with densities comparable to the average density of an atomic nucleus. A neutron star in a faraway galaxy has a mass equal to 2.4 solar masses ( 1 solar mass $=M_{\odot}=$ mass of the sun $=1.99 \times 10^{30} \mathrm{~kg}$ ) and a diameter of 26 km .
(a) What is the density of this neutron star?
(b) How does this neutron star's density compare to the density of a uranium nucleus, which has a diameter of about $15 \mathrm{fm}\left(1 \mathrm{fm}=10^{-15} \mathrm{~m}\right)$ ?


2054. Solution -
We can treat both the neutron star and the U-235 nucleus as spheres. Then the density for both is given by:

$$
d=\frac{m}{V} \quad \text { with } \quad V=\frac{4}{3} \pi r^{3}
$$

(a) The radius of the neutron star is $\frac{1}{2} \times 26 \mathrm{~km}=\frac{1}{2} \times 2.6 \times 10^{4} \mathrm{~m}=1.3 \times 10^{4} \mathrm{~m}$, so the density of the neutron star is:

$$
d=\frac{m}{V}=\frac{m}{\frac{4}{3} \pi r^{3}}=\frac{2.4\left(1.99 \times 10^{30} \mathrm{~kg}\right)}{\frac{4}{3} \pi\left(1.3 \times 10^{4} \mathrm{~m}\right)^{3}}=5.2 \times 10^{17} \mathrm{~kg} / \mathrm{m}^{3}
$$

(b) The radius of the $\mathrm{U}-235$ nucleus is $\frac{1}{2} \times 15 \times 10^{-15} \mathrm{~m}=7.5 \times 10^{-15} \mathrm{~m}$, so the density of the U-235 nucleus is:

$$
d=\frac{m}{V}=\frac{m}{\frac{4}{3} \pi r^{3}}=\frac{235 \mathrm{amu}\left(\frac{1.66 \times 10^{-27} \mathrm{~kg}}{1 \mathrm{amu}}\right)}{\frac{4}{3} \pi\left(7.5 \times 10^{-15} \mathrm{~m}\right)^{3}}=2.2 \times 10^{17} \mathrm{~kg} / \mathrm{m}^{3}
$$

These values are fairly similar (same order of magnitude), but the neutron star is more than twice as dense as the U-235 nucleus.


2055. Check Your Learning -
Find the density of a neutron star with a mass of 1.97 solar masses and a diameter of 13 km , and compare it to the density of a hydrogen nucleus, which has a diameter of $1.75 \mathrm{fm}\left(1 \mathrm{fm}=1 \times 10^{-15} \mathrm{~m}\right)$.


2056. Answer: -
The density of the neutron star is $3.4 \times 10^{18} \mathrm{~kg} / \mathrm{m}^{3}$. The density of a hydrogen nucleus is $6.0 \times 10^{17} \mathrm{~kg} / \mathrm{m}^{3}$. The neutron star is 5.7 times denser than the hydrogen nucleus.

To hold positively charged protons together in the very small volume of a nucleus requires very strong attractive forces because the positively charged protons repel one another strongly at such short distances. The force of attraction that holds the nucleus together is the strong nuclear force. (The strong force is one of the four fundamental forces that are known to exist. The others are the electromagnetic force, the gravitational force, and the nuclear weak force.) This force acts between protons, between neutrons, and between protons and neutrons. It is very different from the electrostatic force that holds negatively charged electrons around a positively charged nucleus (the attraction between opposite charges). Over distances less than $10^{-15}$ meters and within the nucleus, the strong nuclear force is much stronger than electrostatic repulsions between protons; over larger distances and outside the nucleus, it is essentially nonexistent.


2057. LINK TO LEARNING -
Visit this website (http://openstax.org/1/16fourfund) for more information about the four fundamental forces.


2058. Nuclear Binding Energy -
As a simple example of the energy associated with the strong nuclear force, consider the helium atom composed of two protons, two neutrons, and two electrons. The total mass of these six subatomic particles may be calculated as:

$$
(2 \times \underset{\text { protons }}{1.0073 \mathrm{amu}})+\underset{\text { neutrons }}{(2 \times 1.0087 \mathrm{amu})}+\underset{\text { electrons }}{1 \times 0.00055 \mathrm{amu})}=4.0 \begin{aligned}
& 4.0331 \mathrm{amu} \\
& (2 \times 0
\end{aligned}
$$

However, mass spectrometric measurements reveal that the mass of an ${ }_{2}^{4} \mathrm{He}$ atom is 4.0026 amu , less than the combined masses of its six constituent subatomic particles. This difference between the calculated and experimentally measured masses is known as the mass defect of the atom. In the case of helium, the mass defect indicates a "loss" in mass of $4.0331 \mathrm{amu}-4.0026 \mathrm{amu}=0.0305 \mathrm{amu}$. The loss in mass accompanying the formation of an atom from protons, neutrons, and electrons is due to the conversion of that mass into energy that is evolved as the atom forms. The nuclear binding energy is the energy produced when the atoms' nucleons are bound together; this is also the energy needed to break a nucleus into its constituent protons and neutrons. In comparison to chemical bond energies, nuclear binding energies are vastly greater, as we will learn in this section. Consequently, the energy changes associated with nuclear reactions are vastly greater than are those for chemical reactions.

The conversion between mass and energy is most identifiably represented by the mass-energy equivalence equation as stated by Albert Einstein:

$$
E=m c^{2}
$$


2058. Nuclear Binding Energy -
The conversion between mass and energy is most identifiably represented by the mass-energy equivalence equation as stated by Albert Einstein:

$$
E=m c^{2}
$$

where $E$ is energy, $m$ is mass of the matter being converted, and $c$ is the speed of light in a vacuum. This equation can be used to find the amount of energy that results when matter is converted into energy. Using this
mass-energy equivalence equation, the nuclear binding energy of a nucleus may be calculated from its mass defect, as demonstrated in Example 20.2. A variety of units are commonly used for nuclear binding energies, including electron volts (eV), with 1 eV equaling the amount of energy necessary to the move the charge of an electron across an electric potential difference of 1 volt, making $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$.


2060. Calculation of Nuclear Binding Energy -
Determine the binding energy for the nuclide ${ }_{2}^{4} \mathrm{He}$ in:
(a) joules per mole of nuclei
(b) joules per nucleus
(c) MeV per nucleus


2061. Solution -
The mass defect for a ${ }_{2}^{4} \mathrm{He}$ nucleus is 0.0305 amu , as shown previously. Determine the binding energy in joules per nuclide using the mass-energy equivalence equation. To accommodate the requested energy units, the mass defect must be expressed in kilograms (recall that $1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{~m}^{2} / \mathrm{s}^{2}$ ).
(a) First, express the mass defect in $\mathrm{g} / \mathrm{mol}$. This is easily done considering the numerical equivalence of atomic mass (amu) and molar mass ( $\mathrm{g} / \mathrm{mol}$ ) that results from the definitions of the amu and mole units (refer to the previous discussion in the chapter on atoms, molecules, and ions if needed). The mass defect is therefore $0.0305 \mathrm{~g} / \mathrm{mol}$. To accommodate the units of the other terms in the mass-energy equation, the mass must be expressed in kg , since $1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{~m}^{2} / \mathrm{s}^{2}$. Converting grams into kilograms yields a mass defect of $3.05 \times 10^{-5} \mathrm{~kg} /$ mol. Substituting this quantity into the mass-energy equivalence equation yields:

$$
\begin{aligned}
& E=m c^{2}=\frac{3.05 \times 10^{-5} \mathrm{~kg}}{\mathrm{~mol}} \times\left(\frac{2.998 \times 10^{8} \mathrm{~m}}{\mathrm{~s}}\right)^{2}=2.74 \times 10^{12} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2} \mathrm{~mol}^{-1} \\
& =2.74 \times 10^{12} \mathrm{~J} \mathrm{~mol}^{-1}=2.74 \mathrm{TJ} \mathrm{~mol}^{-1}
\end{aligned}
$$


2061. Solution -
Note that this tremendous amount of energy is associated with the conversion of a very small amount of matter (about 30 mg , roughly the mass of typical drop of water).
(b) The binding energy for a single nucleus is computed from the molar binding energy using Avogadro's number:

$$
E=2.74 \times 10^{12} \mathrm{~J} \mathrm{~mol}^{-1} \times \frac{1 \mathrm{~mol}}{6.022 \times 10^{23} \text { nuclei }}=4.55 \times 10^{-12} \mathrm{~J}=4.55 \mathrm{pJ}
$$

(c) Recall that $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$. Using the binding energy computed in part (b):

$$
E=4.55 \times 10^{-12} \mathrm{~J} \times \frac{1 \mathrm{eV}}{1.602 \times 10^{-19} \mathrm{~J}}=2.84 \times 10^{7} \mathrm{eV}=28.4 \mathrm{MeV}
$$


2062. Check Your Learning -
What is the binding energy for the nuclide ${ }_{9}^{19} \mathrm{~F}$ (atomic mass: 18.9984 amu ) in MeV per nucleus?


2063. Answer: -
148.4 MeV

Because the energy changes for breaking and forming bonds are so small compared to the energy changes for breaking or forming nuclei, the changes in mass during all ordinary chemical reactions are virtually undetectable. As described in the chapter on thermochemistry, the most energetic chemical reactions exhibit
enthalpies on the order of thousands of $\mathrm{kJ} / \mathrm{mol}$, which is equivalent to mass differences in the nanogram range $\left(10^{-9} \mathrm{~g}\right)$. On the other hand, nuclear binding energies are typically on the order of billions of $\mathrm{kJ} / \mathrm{mol}$, corresponding to mass differences in the milligram range $\left(10^{-3} \mathrm{~g}\right)$.
Nuclear Stability
A nucleus is stable if it cannot be transformed into another configuration without adding energy from the outside. Of the thousands of nuclides that exist, about 250 are stable. A plot of the number of neutrons versus the number of protons for stable nuclei reveals that the stable isotopes fall into a narrow band. This region is known as the band of stability (also called the belt, zone, or valley of stability). The straight line in Figure 20.2 represents nuclei that have a 1:1 ratio of protons to neutrons ( $\mathrm{n}: \mathrm{p}$ ratio). Note that the lighter stable nuclei, in general, have equal numbers of protons and neutrons. For example, nitrogen-14 has seven protons and seven neutrons. Heavier stable nuclei, however, have increasingly more neutrons than protons. For example: iron-56 has 30 neutrons and 26 protons, an n:p ratio of 1.15, whereas the stable nuclide lead- 207 has 125 neutrons and 82 protons, an n:p ratio equal to 1.52. This is because larger nuclei have more proton-proton repulsions, and require larger numbers of neutrons to provide compensating strong forces to overcome these electrostatic repulsions and hold the nucleus together.


2063. Answer: -
FIGURE 20.2 This plot shows the nuclides that are known to exist and those that are stable. The stable nuclides are indicated in blue, and the unstable nuclides are indicated in green. Note that all isotopes of elements with atomic numbers greater than 83 are unstable. The solid line is the line where $\mathrm{n}=\mathrm{Z}$.

The nuclei that are to the left or to the right of the band of stability are unstable and exhibit radioactivity. They change spontaneously (decay) into other nuclei that are either in, or closer to, the band of stability. These nuclear decay reactions convert one unstable isotope (or radioisotope) into another, more stable, isotope. We will discuss the nature and products of this radioactive decay in subsequent sections of this chapter.

Several observations may be made regarding the relationship between the stability of a nucleus and its structure. Nuclei with even numbers of protons, neutrons, or both are more likely to be stable (see Table 20.1). Nuclei with certain numbers of nucleons, known as magic numbers, are stable against nuclear decay. These numbers of protons or neutrons $(2,8,20,28,50,82$, and 126 ) make complete shells in the nucleus. These are similar in concept to the stable electron shells observed for the noble gases. Nuclei that have magic numbers of
both protons and neutrons, such as ${ }_{2}^{4} \mathrm{He},{ }_{8}^{16} \mathrm{O},{ }_{20}^{40} \mathrm{Ca}$, and ${ }_{82}^{208} \mathrm{~Pb}$, are called "double magic" and are particularly stable. These trends in nuclear stability may be rationalized by considering a quantum mechanical model of nuclear energy states analogous to that used to describe electronic states earlier in this textbook. The details of this model are beyond the scope of this chapter.

Stable Nuclear Isotopes

| Number of Stable Isotopes | Proton Number | Neutron Number |
| :--- | :--- | :--- |
| 157 | even | even |
| 53 | even | odd |
| 50 | odd | even |
| 5 | odd | odd |

TABLE 20.1


2063. Answer: -
TABLE 20.1

The relative stability of a nucleus is correlated with its binding energy per nucleon, the total binding energy for the nucleus divided by the number or nucleons in the nucleus. For instance, we saw in Example 20.2 that the binding energy for a ${ }_{2}^{4} \mathrm{He}$ nucleus is 28.4 MeV . The binding energy per nucleon for a ${ }_{2}^{4} \mathrm{He}$ nucleus is therefore:

$$
\frac{28.4 \mathrm{MeV}}{4 \text { nucleons }}=7.10 \mathrm{MeV} / \text { nucleon }
$$

In Example 20.3, we learn how to calculate the binding energy per nucleon of a nuclide on the curve shown in Figure 20.3.

FIGURE 20.3 The binding energy per nucleon is largest for nuclides with mass number of approximately 56.


2065. Calculation of Binding Energy per Nucleon -
The iron nuclide ${ }_{26}^{56} \mathrm{Fe}$ lies near the top of the binding energy curve (Figure 20.3) and is one of the most stable nuclides. What is the binding energy per nucleon (in MeV) for the nuclide ${ }_{26}^{56} \mathrm{Fe}$ (atomic mass of 55.9349 amu )?


2066. Solution -
As in Example 20.2, we first determine the mass defect of the nuclide, which is the difference between the mass of 26 protons, 30 neutrons, and 26 electrons, and the observed mass of an ${ }_{26}^{56} \mathrm{Fe}$ atom:

$$
\begin{aligned}
& \text { Mass defect }=[(26 \times 1.0073 \mathrm{amu})+(30 \times 1.0087 \mathrm{amu})+(26 \times 0.00055 \mathrm{amu})]-55.9349 \mathrm{amu} \\
& =56.4651 \mathrm{amu}-55.9349 \mathrm{amu} \\
& =0.5302 \mathrm{amu}
\end{aligned}
$$

We next calculate the binding energy for one nucleus from the mass defect using the mass-energy equivalence equation:

$$
\begin{aligned}
& E=m c^{2}=0.5302 \mathrm{amu} \times \frac{1.6605 \times 10^{-27} \mathrm{~kg}}{1 \mathrm{amu}} \times\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)^{2} \\
& =7.913 \times 10^{-11} \mathrm{~kg} \cdot \mathrm{~m} / \mathrm{s}^{2} \\
& =7.913 \times 10^{-11} \mathrm{~J}
\end{aligned}
$$

We then convert the binding energy in joules per nucleus into units of MeV per nuclide:

$$
7.913 \times 10^{-11} \mathrm{~J} \times \frac{1 \mathrm{MeV}}{1.602 \times 10^{-13} \mathrm{~J}}=493.9 \mathrm{MeV}
$$

Finally, we determine the binding energy per nucleon by dividing the total nuclear binding energy by the number of nucleons in the atom:

$$
\text { Binding energy per nucleon }=\frac{493.9 \mathrm{MeV}}{56}=8.820 \mathrm{MeV} / \text { nucleon }
$$


2066. Solution -
$$
\text { Binding energy per nucleon }=\frac{493.9 \mathrm{MeV}}{56}=8.820 \mathrm{MeV} / \text { nucleon }
$$

Note that this is almost $25 \%$ larger than the binding energy per nucleon for ${ }_{2}^{4} \mathrm{He}$.
(Note also that this is the same process as in Example 20.1, but with the additional step of dividing the total nuclear binding energy by the number of nucleons.)


2067. Check Your Learning -
What is the binding energy per nucleon in ${ }_{9}^{19} \mathrm{~F}$ (atomic mass, 18.9984 amu )?


2068. Answer: -
7.810 MeV/nucleon


2069. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Identify common particles and energies involved in nuclear reactions
- Write and balance nuclear equations

Changes of nuclei that result in changes in their atomic numbers, mass numbers, or energy states are nuclear reactions. To describe a nuclear reaction, we use an equation that identifies the nuclides involved in the reaction, their mass numbers and atomic numbers, and the other particles involved in the reaction.