319. Exercises - 319.3. Development of Quantum Theory
31. How are the Bohr model and the quantum mechanical model of the hydrogen atom similar? How are they different?
32. What are the allowed values for each of the four quantum numbers: $n, l, m_{l}$, and $m_{s}$ ?
33. Describe the properties of an electron associated with each of the following four quantum numbers: $n, l$, $m_{l}$, and $m_{s}$.
34. Answer the following questions:
(a) Without using quantum numbers, describe the differences between the shells, subshells, and orbitals of an atom.
(b) How do the quantum numbers of the shells, subshells, and orbitals of an atom differ?
35. Identify the subshell in which electrons with the following quantum numbers are found:
(a) $n=2, l=1$
(b) $n=4, l=2$
(c) $n=6, l=0$
36. Which of the subshells described in the previous question contain degenerate orbitals? How many degenerate orbitals are in each?
37. Identify the subshell in which electrons with the following quantum numbers are found:
(a) $n=3, l=2$
(b) $n=1, l=0$
(c) $n=4, l=3$
38. Which of the subshells described in the previous question contain degenerate orbitals? How many degenerate orbitals are in each?
39. Sketch the boundary surface of a $d_{x^{2}-y^{2}}$ and a $p_{y}$ orbital. Be sure to show and label the axes.
40. Sketch the $p_{X}$ and $d_{X Z}$ orbitals. Be sure to show and label the coordinates.
41. Consider the orbitals shown here in outline.

(x)

(y)


319. Exercises - 319.3. Development of Quantum Theory
(x)

(y)

(z)
(a) What is the maximum number of electrons contained in an orbital of type (x)? Of type (y)? Of type (z)?
(b) How many orbitals of type (x) are found in a shell with $n=2$ ? How many of type (y)? How many of type (z)?
(c) Write a set of quantum numbers for an electron in an orbital of type (x) in a shell with $n=4$. Of an orbital of type (y) in a shell with $n=2$. Of an orbital of type (z) in a shell with $n=3$.
(d) What is the smallest possible $n$ value for an orbital of type (x)? Of type (y)? Of type (z)?
(e) What are the possible 1 and $m_{l}$ values for an orbital of type (x)? Of type (y)? Of type (z)?
42. State the Heisenberg uncertainty principle. Describe briefly what the principle implies.
43. How many electrons could be held in the second shell of an atom if the spin quantum number $m_{S}$ could have three values instead of just two? (Hint: Consider the Pauli exclusion principle.)
44. Which of the following equations describe particle-like behavior? Which describe wavelike behavior? Do any involve both types of behavior? Describe the reasons for your choices.
(a) $c=\lambda \nu$
(b) $E=\frac{m v^{2}}{2}$
(c) $r=\frac{n^{2} a_{0}}{Z}$
(d) $E=h \nu$
(e) $\lambda=\frac{h}{m v}$
45. Write a set of quantum numbers for each of the electrons with an $n$ of 4 in a Se atom.


319. Exercises - 319.4. Electronic Structure of Atoms (Electron Configurations)
46. Read the labels of several commercial products and identify monatomic ions of at least four transition elements contained in the products. Write the complete electron configurations of these cations.
47. Read the labels of several commercial products and identify monatomic ions of at least six main group elements contained in the products. Write the complete electron configurations of these cations and anions.
48. Using complete subshell notation (not abbreviations, $1 s^{2} 2 s^{2} 2 p^{6}$, and so forth), predict the electron configuration of each of the following atoms:
(a) C
(b) P
(c) V
(d) Sb
(e) Sm
49. Using complete subshell notation $\left(1 s^{2} 2 s^{2} 2 p^{6}\right.$, and so forth), predict the electron configuration of each of the following atoms:
(a) N
(b) Si
(c) Fe
(d) Te
(e) Tb
50. Is $1 s^{2} 2 s^{2} 2 p^{6}$ the symbol for a macroscopic property or a microscopic property of an element? Explain your answer.
51. What additional information do we need to answer the question "Which ion has the electron configuration $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6 "}$ ?
52. Draw the orbital diagram for the valence shell of each of the following atoms:
(a) C
(b) P
(c) V
(d) Sb
(e) Ru
53. Use an orbital diagram to describe the electron configuration of the valence shell of each of the following atoms:
(a) N
(b) Si
(c) Fe
(d) Te
(e) Mo
54. Using complete subshell notation $\left(1 s^{2} 2 s^{2} 2 p^{6}\right.$, and so forth), predict the electron configurations of the following ions.
(a) $\mathrm{N}^{3-}$
(b) $\mathrm{Ca}^{2+}$
(c) $\mathrm{S}^{-}$
(d) $\mathrm{Cs}^{2+}$
(e) $\mathrm{Cr}^{2+}$


319. Exercises - 319.4. Electronic Structure of Atoms (Electron Configurations)
(b) $\mathrm{Ca}^{2+}$
(c) $\mathrm{S}^{-}$
(d) $\mathrm{Cs}^{2+}$
(e) $\mathrm{Cr}^{2+}$
(f) $\mathrm{Gd}^{3+}$
55. Which atom has the electron configuration $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2} 4 d^{2}$ ?
56. Which atom has the electron configuration $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}$ ?
57. Which ion with a +1 charge has the electron configuration $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6}$ ? Which ion with a -2 charge has this configuration?
58. Which of the following atoms contains only three valence electrons: $\mathrm{Li}, \mathrm{B}, \mathrm{N}, \mathrm{F}, \mathrm{Ne}$ ?
59. Which of the following has two unpaired electrons?
(a) Mg
(b) Si
(c) S
(d) Both Mg and S
(e) Both Si and S.
60. Which atom would be expected to have a half-filled $6 p$ subshell?
61. Which atom would be expected to have a half-filled $4 s$ subshell?
62. In one area of Australia, the cattle did not thrive despite the presence of suitable forage. An investigation showed the cause to be the absence of sufficient cobalt in the soil. Cobalt forms cations in two oxidation states, $\mathrm{Co}^{2+}$ and $\mathrm{Co}^{3+}$. Write the electron structure of the two cations.


319. Exercises - 319.4. Electronic Structure of Atoms (Electron Configurations)
63. Thallium was used as a poison in the Agatha Christie mystery story "The Pale Horse." Thallium has two possible cationic forms, +1 and +3 . The +1 compounds are the more stable. Write the electron structure of the +1 cation of thallium.
64. Write the electron configurations for the following atoms or ions:
(a) $\mathrm{B}^{3+}$
(b) $\mathrm{O}^{-}$
(c) $\mathrm{Cl}^{3+}$
(d) $\mathrm{Ca}^{2+}$
(e) Ti
65. Cobalt-60 and iodine-131 are radioactive isotopes commonly used in nuclear medicine. How many protons, neutrons, and electrons are in atoms of these isotopes? Write the complete electron configuration for each isotope.
66. Write a set of quantum numbers for each of the electrons with an $n$ of 3 in a Sc atom.


319. Exercises - 319.5. Periodic Variations in Element Properties
67. Based on their positions in the periodic table, predict which has the smallest atomic radius: $\mathrm{Mg}, \mathrm{Sr}, \mathrm{Si}, \mathrm{Cl}$, I.
68. Based on their positions in the periodic table, predict which has the largest atomic radius: Li, Rb, N, F, I.
69. Based on their positions in the periodic table, predict which has the largest first ionization energy: $\mathrm{Mg}, \mathrm{Ba}$, $\mathrm{B}, \mathrm{O}, \mathrm{Te}$.
70. Based on their positions in the periodic table, predict which has the smallest first ionization energy: Li, Cs, N, F, I.
71. Based on their positions in the periodic table, rank the following atoms in order of increasing first ionization energy: $\mathrm{F}, \mathrm{Li}, \mathrm{N}, \mathrm{Rb}$
72. Based on their positions in the periodic table, rank the following atoms in order of increasing first ionization energy: $\mathrm{Mg}, \mathrm{O}, \mathrm{S}, \mathrm{Si}$
73. Atoms of which group in the periodic table have a valence shell electron configuration of $n s^{2} n p^{3}$ ?
74. Atoms of which group in the periodic table have a valence shell electron configuration of $n s^{2}$ ?
75. Based on their positions in the periodic table, list the following atoms in order of increasing radius: Mg , $\mathrm{Ca}, \mathrm{Rb}, \mathrm{Cs}$.
76. Based on their positions in the periodic table, list the following atoms in order of increasing radius: $\mathrm{Sr}, \mathrm{Ca}$, Si, Cl.
77. Based on their positions in the periodic table, list the following ions in order of increasing radius: $\mathrm{K}^{+}, \mathrm{Ca}^{2+}$, $\mathrm{Al}^{3+}, \mathrm{Si}^{4+}$.
78. List the following ions in order of increasing radius: $\mathrm{Li}^{+}, \mathrm{Mg}^{2+}, \mathrm{Br}^{-}, \mathrm{Te}^{2-}$.


319. Exercises - 319.5. Periodic Variations in Element Properties
78. List the following ions in order of increasing radius: $\mathrm{Li}^{+}, \mathrm{Mg}^{2+}, \mathrm{Br}^{-}, \mathrm{Te}^{2-}$.
79. Which atom and/or ion is (are) isoelectronic with $\mathrm{Br}^{+}: \mathrm{Se}^{2+}, \mathrm{Se}, \mathrm{As}^{-}, \mathrm{Kr}, \mathrm{Ga}^{3+}, \mathrm{Cl}^{-}$?
80. Which of the following atoms and ions is (are) isoelectronic with $\mathrm{S}^{2+}: \mathrm{Si}^{4+}, \mathrm{Cl}^{3+}, \mathrm{Ar}, \mathrm{As}^{3+}, \mathrm{Si}, \mathrm{Al}^{3+}$ ?
81. Compare both the numbers of protons and electrons present in each to rank the following ions in order of increasing radius: $\mathrm{As}^{3-}, \mathrm{Br}^{-}, \mathrm{K}^{+}, \mathrm{Mg}^{2+}$.
82. Of the five elements $\mathrm{Al}, \mathrm{Cl}, \mathrm{I}, \mathrm{Na}, \mathrm{Rb}$, which has the most exothermic reaction? (E represents an atom.) What name is given to the energy for the reaction? Hint: Note the process depicted does not correspond to electron affinity.)
$\mathrm{E}^{+}(g)+\mathrm{e}^{-} \longrightarrow \mathrm{E}(g)$
83. Of the five elements $\mathrm{Sn}, \mathrm{Si}, \mathrm{Sb}, \mathrm{O}, \mathrm{Te}$, which has the most endothermic reaction? (E represents an atom.) What name is given to the energy for the reaction? $\mathrm{E}(g) \longrightarrow \mathrm{E}^{+}(g)+\mathrm{e}^{-}$


319. Exercises - 319.5. Periodic Variations in Element Properties
84. The ionic radii of the ions $\mathrm{S}^{2-}, \mathrm{Cl}^{-}$, and $\mathrm{K}^{+}$are $184,181,138 \mathrm{pm}$ respectively. Explain why these ions have different sizes even though they contain the same number of electrons.
85. Which main group atom would be expected to have the lowest second ionization energy?
86. Explain why Al is a member of group 13 rather than group 3 ?


319. Exercises - 319.6. The Periodic Table
87. Using the periodic table, classify each of the following elements as a metal or a nonmetal, and then further classify each as a main-group (representative) element, transition metal, or inner transition metal:
(a) uranium
(b) bromine
(c) strontium
(d) neon
(e) gold
(f) americium
(g) rhodium
(h) sulfur
(i) carbon
(j) potassium
88. Using the periodic table, classify each of the following elements as a metal or a nonmetal, and then further classify each as a main-group (representative) element, transition metal, or inner transition metal:
(a) cobalt
(b) europium
(c) iodine
(d) indium
(e) lithium
(f) oxygen
(g) cadmium
(h) terbium
(i) rhenium
89. Using the periodic table, identify the lightest member of each of the following groups:
(a) noble gases
(b) alkaline earth metals
(c) alkali metals
(d) chalcogens
90. Using the periodic table, identify the heaviest member of each of the following groups:
(a) alkali metals
(b) chalcogens
(c) noble gases
(d) alkaline earth metals
91. Use the periodic table to give the name and symbol for each of the following elements:
(a) the noble gas in the same period as germanium
(b) the alkaline earth metal in the same period as selenium
(c) the halogen in the same period as lithium
(d) the chalcogen in the same period as cadmium
92. Use the periodic table to give the name and symbol for each of the following elements:
(a) the halogen in the same period as the alkali metal with 11 protons
(b) the alkaline earth metal in the same period with the neutral noble gas with 18 electrons
(c) the noble gas in the same row as an isotope with 30 neutrons and 25 protons
(d) the noble gas in the same period as gold
93. Write a symbol for each of the following neutral isotopes. Include the atomic number and mass number for each.
(a) the alkali metal with 11 protons and a mass number of 23


319. Exercises - 319.6. The Periodic Table
93. Write a symbol for each of the following neutral isotopes. Include the atomic number and mass number for each.
(a) the alkali metal with 11 protons and a mass number of 23
(b) the noble gas element with 75 neutrons in its nucleus and 54 electrons in the neutral atom
(c) the isotope with 33 protons and 40 neutrons in its nucleus
(d) the alkaline earth metal with 88 electrons and 138 neutrons
94. Write a symbol for each of the following neutral isotopes. Include the atomic number and mass number for each.
(a) the chalcogen with a mass number of 125
(b) the halogen whose longest-lived isotope is radioactive
(c) the noble gas, used in lighting, with 10 electrons and 10 neutrons
(d) the lightest alkali metal with three neutrons


319. Exercises - 319.7. Ionic and Molecular Compounds
95. Using the periodic table, predict whether the following chlorides are ionic or covalent: $\mathrm{KCl}, \mathrm{NCl}_{3}, \mathrm{ICl}$, $\mathrm{MgCl}_{2}, \mathrm{PCl}_{5}$, and $\mathrm{CCl}_{4}$.
96. Using the periodic table, predict whether the following chlorides are ionic or covalent: $\mathrm{SiCl}_{4}, \mathrm{PCl}_{3}, \mathrm{CaCl}_{2}$, $\mathrm{CsCl}, \mathrm{CuCl}_{2}$, and $\mathrm{CrCl}_{3}$.
97. For each of the following compounds, state whether it is ionic or covalent. If it is ionic, write the symbols for the ions involved:
(a) $\mathrm{NF}_{3}$
(b) BaO ,
(c) $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}$
(d) $\mathrm{Sr}\left(\mathrm{H}_{2} \mathrm{PO}_{4}\right)_{2}$
(e) IBr
(f) $\mathrm{Na}_{2} \mathrm{O}$
98. For each of the following compounds, state whether it is ionic or covalent, and if it is ionic, write the symbols for the ions involved:
(a) $\mathrm{KClO}_{4}$
(b) $\mathrm{Mg}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}$
(c) $\mathrm{H}_{2} \mathrm{~S}$
(d) $\mathrm{Ag}_{2} \mathrm{~S}$
(e) $\mathrm{N}_{2} \mathrm{Cl}_{4}$
(f) $\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}$
99. For each of the following pairs of ions, write the symbol for the formula of the compound they will form.
(a) $\mathrm{Ca}^{2+}, \mathrm{S}^{2-}$
(b) $\mathrm{NH}_{4}{ }^{+}, \mathrm{SO}_{4}{ }^{2-}$


319. Exercises - 319.7. Ionic and Molecular Compounds
(a) $\mathrm{Ca}^{2+}, \mathrm{S}^{2-}$
(b) $\mathrm{NH}_{4}{ }^{+}, \mathrm{SO}_{4}{ }^{2-}$
(c) $\mathrm{Al}^{3+}, \mathrm{Br}^{-}$
(d) $\mathrm{Na}^{+}, \mathrm{HPO}_{4}{ }^{2-}$
(e) $\mathrm{Mg}^{2+}, \mathrm{PO}_{4}{ }^{3-}$
100. For each of the following pairs of ions, write the symbol for the formula of the compound they will form.
(a) $\mathrm{K}^{+}, \mathrm{O}^{2-}$
(b) $\mathrm{NH}_{4}{ }^{+}, \mathrm{PO}_{4}{ }^{3-}$
(c) $\mathrm{Al}^{3+}, \mathrm{O}^{2-}$
(d) $\mathrm{Na}^{+}, \mathrm{CO}_{3}{ }^{2-}$
(e) $\mathrm{Ba}^{2+}, \mathrm{PO}_{4}{ }^{3-}$


319. Exercises - 319.7. Ionic and Molecular Compounds
174 3•Exercises


320. CHAPTER 4 <br> Chemical Bonding and Molecular Geometry -
Figure 4.1 Nicknamed "buckyballs," buckminsterfullerene molecules ( $\mathrm{C}_{60}$ ) contain only carbon atoms (left) arranged to form a geometric framework of hexagons and pentagons, similar to the pattern on a soccer ball (center). This molecular structure is named after architect R. Buckminster Fuller, whose innovative designs combined simple geometric shapes to create large, strong structures such as this weather radar dome near Tucson, Arizona (right). (credit middle: modification of work by "Petey21"/Wikimedia Commons; credit right: modification of work by Bill Morrow)


321. CHAPTER OUTLINE - 321.1. Ionic Bonding
4.2 Covalent Bonding
4.3 Chemical Nomenclature
4.4 Lewis Symbols and Structures
4.5 Formal Charges and Resonance
4.6 Molecular Structure and Polarity

INTRODUCTION It has long been known that pure carbon occurs in different forms (allotropes) including graphite and diamonds. But it was not until 1985 that a new form of carbon was recognized:
buckminsterfullerene. This molecule was named after the architect and inventor R. Buckminster Fuller (1895-1983), whose signature architectural design was the geodesic dome, characterized by a lattice shell structure supporting a spherical surface. Experimental evidence revealed the formula, $\mathrm{C}_{60}$, and then scientists determined how 60 carbon atoms could form one symmetric, stable molecule. They were guided by bonding theory-the topic of this chapter-which explains how individual atoms connect to form more complex structures.


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

- Explain the formation of cations, anions, and ionic compounds
- Predict the charge of common metallic and nonmetallic elements, and write their electron configurations

As you have learned, ions are atoms or molecules bearing an electrical charge. A cation (a positive ion) forms when a neutral atom loses one or more electrons from its valence shell, and an anion (a negative ion) forms when a neutral atom gains one or more electrons in its valence shell.

Compounds composed of ions are called ionic compounds (or salts), and their constituent ions are held together by ionic bonds: electrostatic forces of attraction between oppositely charged cations and anions. The properties of ionic compounds shed some light on the nature of ionic bonds. Ionic solids exhibit a crystalline structure and tend to be rigid and brittle; they also tend to have high melting and boiling points, which suggests that ionic bonds are very strong. Ionic solids are also poor conductors of electricity for the same reason-the strength of ionic bonds prevents ions from moving freely in the solid state. Most ionic solids, however, dissolve readily in water. Once dissolved or melted, ionic compounds are excellent conductors of electricity and heat because the ions can move about freely.

Neutral atoms and their associated ions have very different physical and chemical properties. Sodium atoms form sodium metal, a soft, silvery-white metal that burns vigorously in air and reacts explosively with water. Chlorine atoms form chlorine gas, $\mathrm{Cl}_{2}$, a yellow-green gas that is extremely corrosive to most metals and very poisonous to animals and plants. The vigorous reaction between the elements sodium and chlorine forms the white, crystalline compound sodium chloride, common table salt, which contains sodium cations and chloride anions (Figure 4.2). The compound composed of these ions exhibits properties entirely different from the properties of the elements sodium and chlorine. Chlorine is poisonous, but sodium chloride is essential to life; sodium atoms react vigorously with water, but sodium chloride simply dissolves in water.


322. LEARNING OBJECTIVES -
FIGURE 4.2 (a) Sodium is a soft metal that must be stored in mineral oil to prevent reaction with air or water. (b) Chlorine is a pale yellow-green gas. (c) When combined, they form white crystals of sodium chloride (table salt). (credit a: modification of work by "Jurii"/Wikimedia Commons)


323. The Formation of Ionic Compounds -
Binary ionic compounds are composed of just two elements: a metal (which forms the cations) and a nonmetal (which forms the anions). For example, NaCl is a binary ionic compound. We can think about the formation of such compounds in terms of the periodic properties of the elements. Many metallic elements have relatively low ionization potentials and lose electrons easily. These elements lie to the left in a period or near the bottom of a group on the periodic table. Nonmetal atoms have relatively high electron affinities and thus readily gain electrons lost by metal atoms, thereby filling their valence shells. Nonmetallic elements are found in the upper-right corner of the periodic table.

As all substances must be electrically neutral, the total number of positive charges on the cations of an ionic compound must equal the total number of negative charges on its anions. The formula of an ionic compound represents the simplest ratio of the numbers of ions necessary to give identical numbers of positive and negative charges. For example, the formula for aluminum oxide, $\mathrm{Al}_{2} \mathrm{O}_{3}$, indicates that this ionic compound contains two aluminum cations, $\mathrm{Al}^{3+}$, for every three oxide anions, $\mathrm{O}^{2-}$ [thus, $(2 \times+3)+(3 \times-2)=0$ ].

It is important to note, however, that the formula for an ionic compound does not represent the physical arrangement of its ions. It is incorrect to refer to a sodium chloride ( NaCl ) "molecule" because there is not a
single ionic bond, per se, between any specific pair of sodium and chloride ions. The attractive forces between ions are isotropic-the same in all directions-meaning that any particular ion is equally attracted to all of the nearby ions of opposite charge. This results in the ions arranging themselves into a tightly bound, threedimensional lattice structure. Sodium chloride, for example, consists of a regular arrangement of equal numbers of $\mathrm{Na}^{+}$cations and $\mathrm{Cl}^{-}$anions (Figure 4.3).


323. The Formation of Ionic Compounds -
FIGURE 4.3 The atoms in sodium chloride (common table salt) are arranged to (a) maximize opposite charges interacting. The smaller spheres represent sodium ions, the larger ones represent chloride ions. In the expanded view (b), the geometry can be seen more clearly. Note that each ion is "bonded" to all of the surrounding ions-six in this case.

The strong electrostatic attraction between $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}$ions holds them tightly together in solid NaCl . It requires 769 kJ of energy to dissociate one mole of solid NaCl into separate gaseous $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}$ions:

$$
\mathrm{NaCl}(s) \longrightarrow \mathrm{Na}^{+}(g)+\mathrm{Cl}^{-}(g) \quad \Delta H=769 \mathrm{~kJ}
$$


324. Electronic Structures of Cations -
When forming a cation, an atom of a main group element tends to lose all of its valence electrons, thus assuming the electronic structure of the noble gas that precedes it in the periodic table. For groups 1 (the alkali metals) and 2 (the alkaline earth metals), the group numbers are equal to the numbers of valence shell electrons and, consequently, to the charges of the cations formed from atoms of these elements when all valence shell electrons are removed. For example, calcium is a group 2 element whose neutral atoms have 20 electrons and a ground state electron configuration of $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2}$. When a Ca atom loses both of its valence electrons, the result is a cation with 18 electrons, a $2+$ charge, and an electron configuration of $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$. The $\mathrm{Ca}^{2+}$ ion is therefore isoelectronic with the noble gas Ar.

For groups 13-17, the group numbers exceed the number of valence electrons by 10 (accounting for the possibility of full $d$ subshells in atoms of elements in the fourth and greater periods). Thus, the charge of a cation formed by the loss of all valence electrons is equal to the group number minus 10. For example, aluminum (in group 13) forms $3+$ ions $\left(\mathrm{Al}^{3+}\right)$.


324. Electronic Structures of Cations -
Exceptions to the expected behavior involve elements toward the bottom of the groups. In addition to the expected ions $\mathrm{Tl}^{3+}, \mathrm{Sn}^{4+}, \mathrm{Pb}^{4+}$, and $\mathrm{Bi}^{5+}$, a partial loss of these atoms' valence shell electrons can also lead to the formation of $\mathrm{Tl}^{+}, \mathrm{Sn}^{2+}, \mathrm{Pb}^{2+}$, and $\mathrm{Bi}^{3+}$ ions. The formation of these $1+, 2+$, and $3+$ cations is ascribed to the inert pair effect, which reflects the relatively low energy of the valence $s$-electron pair for atoms of the heavy elements of groups 13,14 , and 15 . Mercury (group 12) also exhibits an unexpected behavior: it forms a diatomic ion, $\mathrm{Hg}_{2}{ }^{2+}$ (an ion formed from two mercury atoms, with an $\mathrm{Hg}-\mathrm{Hg}$ bond), in addition to the expected monatomic ion $\mathrm{Hg}^{2+}$ (formed from only one mercury atom).


324. Electronic Structures of Cations -
Transition and inner transition metal elements behave differently than main group elements. Most transition metal cations have $2+$ or $3+$ charges that result from the loss of their outermost $s$ electron(s) first, sometimes followed by the loss of one or two $d$ electrons from the next-to-outermost shell. For example, iron $\left(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6} 4 s^{2}\right)$ forms the ion $\mathrm{Fe}^{2+}\left(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6}\right)$ by the loss of the $4 s$ electrons and the ion $\mathrm{Fe}^{3+}\left(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{5}\right)$ by the loss of the $4 s$ electron and one of the $3 d$ electrons. Although the $d$ orbitals of
the transition elements are-according to the Aufbau principle-the last to fill when building up electron configurations, the outermost $s$ electrons are the first to be lost when these atoms ionize. When the inner transition metals form ions, they usually have a $3+$ charge, resulting from the loss of their outermost $s$ electrons and a $d$ or $f$ electron.


326. Determining the Electronic Structures of Cations -
There are at least 14 elements categorized as "essential trace elements" for the human body. They are called "essential" because they are required for healthy bodily functions, "trace" because they are required only in small amounts, and "elements" in spite of the fact that they are really ions. Two of these essential trace elements, chromium and zinc, are required as $\mathrm{Cr}^{3+}$ and $\mathrm{Zn}^{2+}$. Write the electron configurations of these cations.


327. Solution -
First, write the electron configuration for the neutral atoms:
$\mathrm{Zn}:[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 s^{2}$
Cr: $[\mathrm{Ar}] 3 d^{5} 4 s^{1}$
Next, remove electrons from the highest energy orbital. For the transition metals, electrons are removed from the $s$ orbital first and then from the $d$ orbital. For the $p$-block elements, electrons are removed from the $p$ orbitals and then from the $s$ orbital. Zinc is a member of group 12, so it should have a charge of $2+$, and thus loses only the two electrons in its $S$ orbital. Chromium is a transition element and should lose its $s$ electrons and then its $d$ electrons when forming a cation. Thus, we find the following electron configurations of the ions:
$\mathrm{Zn}^{2+}:[\mathrm{Ar}] 3 d^{10}$
$\mathrm{Cr}^{3+}:[\mathrm{Ar}] 3 d^{3}$


328. Check Your Learning -
Potassium and magnesium are required in our diet. Write the electron configurations of the ions expected from these elements.


329. Answer: -
$\mathrm{K}^{+}$: $[\mathrm{Ar}], \mathrm{Mg}^{2+}$ : $[\mathrm{Ne}]$


330. Electronic Structures of Anions -
Most monatomic anions form when a neutral nonmetal atom gains enough electrons to completely fill its outer $s$ and $p$ orbitals, thereby reaching the electron configuration of the next noble gas. Thus, it is simple to determine the charge on such a negative ion: The charge is equal to the number of electrons that must be gained to fill the $s$ and $p$ orbitals of the parent atom. Oxygen, for example, has the electron configuration $1 s^{2} 2 s^{2} 2 p^{4}$, whereas the oxygen anion has the electron configuration of the noble gas neon ( Ne ), $1 s^{2} 2 s^{2} 2 p^{6}$. The two additional electrons required to fill the valence orbitals give the oxide ion the charge of $2-\left(\mathrm{O}^{2-}\right)$.


332. Determining the Electronic Structure of Anions -
Selenium and iodine are two essential trace elements that form anions. Write the electron configurations of the anions.


333. Solution -
$\mathrm{Se}^{2-}:[\mathrm{Ar}] 3 d^{10} 4 s^{2} 4 p^{6}$
$I^{-}:[\mathrm{Kr}] 4 d^{10} 5 s^{2} 5 p^{6}$


334. Check Your Learning -
Write the electron configurations of a phosphorus atom and its negative ion. Give the charge on the anion.


335. Answer: -
P: [Ne] $3 s^{2} 3 p^{3} ; \mathrm{P}^{3-}:[\mathrm{Ne}] 3 s^{2} 3 p^{6}$


335. Answer: - 335.1. Covalent Bonding
LEARNING OBJECTIVES
By the end of this section, you will be able to:

- Describe the formation of covalent bonds
- Define electronegativity and assess the polarity of covalent bonds

Ionic bonding results from the electrostatic attraction of oppositely charged ions that are typically produced by the transfer of electrons between metallic and nonmetallic atoms. A different type of bonding results from the mutual attraction of atoms for a "shared" pair of electrons. Such bonds are called covalent bonds. Covalent bonds are formed between two atoms when both have similar tendencies to attract electrons to themselves (i.e., when both atoms have identical or fairly similar ionization energies and electron affinities). For example, two hydrogen atoms bond covalently to form an $\mathrm{H}_{2}$ molecule; each hydrogen atom in the $\mathrm{H}_{2}$ molecule has two electrons stabilizing it, giving each atom the same number of valence electrons as the noble gas He .

Compounds that contain covalent bonds exhibit different physical properties than ionic compounds. Because the attraction between molecules, which are electrically neutral, is weaker than that between electrically charged ions, covalent compounds generally have much lower melting and boiling points than ionic compounds. In fact, many covalent compounds are liquids or gases at room temperature, and, in their solid states, they are typically much softer than ionic solids. Furthermore, whereas ionic compounds are good conductors of electricity when dissolved in water, most covalent compounds are insoluble in water; since they are electrically neutral, they are poor conductors of electricity in any state.


336. Formation of Covalent Bonds -
Nonmetal atoms frequently form covalent bonds with other nonmetal atoms. For example, the hydrogen molecule, $\mathrm{H}_{2}$, contains a covalent bond between its two hydrogen atoms. Figure 4.4 illustrates why this bond is formed. Starting on the far right, we have two separate hydrogen atoms with a particular potential energy, indicated by the red line. Along the $x$-axis is the distance between the two atoms. As the two atoms approach each other (moving left along the $x$-axis), their valence orbitals (1s) begin to overlap. The single electrons on each hydrogen atom then interact with both atomic nuclei, occupying the space around both atoms. The strong attraction of each shared electron to both nuclei stabilizes the system, and the potential energy decreases as the bond distance decreases. If the atoms continue to approach each other, the positive charges in the two nuclei begin to repel each other, and the potential energy increases. The bond length is determined by the distance at which the lowest potential energy is achieved.

FIGURE 4.4 The potential energy of two separate hydrogen atoms (right) decreases as they approach each other, and the single electrons on each atom are shared to form a covalent bond. The bond length is the internuclear distance at which the lowest potential energy is achieved.

It is essential to remember that energy must be added to break chemical bonds (an endothermic process), whereas forming chemical bonds releases energy (an exothermic process). In the case of $\mathrm{H}_{2}$, the covalent bond is very strong; a large amount of energy, 436 kJ , must be added to break the bonds in one mole of hydrogen molecules and cause the atoms to separate:

$$
\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}(\mathrm{~g}) \quad \Delta H=436 \mathrm{~kJ}
$$

Conversely, the same amount of energy is released when one mole of $\mathrm{H}_{2}$ molecules forms from two moles of H atoms:

$$
2 \mathrm{H}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2}(\mathrm{~g}) \quad \Delta H=-436 \mathrm{~kJ}
$$


337. Pure vs. Polar Covalent Bonds -
If the atoms that form a covalent bond are identical, as in $\mathrm{H}_{2}, \mathrm{Cl}_{2}$, and other diatomic molecules, then the electrons in the bond must be shared equally. We refer to this as a pure covalent bond. Electrons shared in
pure covalent bonds have an equal probability of being near each nucleus.
In the case of $\mathrm{Cl}_{2}$, each atom starts off with seven valence electrons, and each Cl shares one electron with the other, forming one covalent bond:

$$
\mathrm{Cl}+\mathrm{Cl} \longrightarrow \mathrm{Cl}_{2}
$$

The total number of electrons around each individual atom consists of six nonbonding electrons and two shared (i.e., bonding) electrons for eight total electrons, matching the number of valence electrons in the noble gas argon. Since the bonding atoms are identical, $\mathrm{Cl}_{2}$ also features a pure covalent bond.

When the atoms linked by a covalent bond are different, the bonding electrons are shared, but no longer equally. Instead, the bonding electrons are more attracted to one atom than the other, giving rise to a shift of electron density toward that atom. This unequal distribution of electrons is known as a polar covalent bond, characterized by a partial positive charge on one atom and a partial negative charge on the other. The atom that attracts the electrons more strongly acquires the partial negative charge and vice versa. For example, the electrons in the $\mathrm{H}-\mathrm{Cl}$ bond of a hydrogen chloride molecule spend more time near the chlorine atom than near the hydrogen atom. Thus, in an HCl molecule, the chlorine atom carries a partial negative charge and the hydrogen atom has a partial positive charge. Figure 4.5 shows the distribution of electrons in the $\mathrm{H}-\mathrm{Cl}$ bond. Note that the shaded area around Cl is much larger than it is around H . Compare this to Figure 4.4, which shows the even distribution of electrons in the $\mathrm{H}_{2}$ nonpolar bond.


337. Pure vs. Polar Covalent Bonds -
We sometimes designate the positive and negative atoms in a polar covalent bond using a lowercase Greek letter "delta," $\delta$, with a plus sign or minus sign to indicate whether the atom has a partial positive charge ( $\delta+$ ) or a partial negative charge ( $\delta-$ ). This symbolism is shown for the $\mathrm{H}-\mathrm{Cl}$ molecule in Figure 4.5.

(a)
$\delta+\delta-$
$\mathrm{H}-\mathrm{Cl}$
(b)

FIGURE 4.5 (a) The distribution of electron density in the HCl molecule is uneven. The electron density is greater around the chlorine nucleus. The small, black dots indicate the location of the hydrogen and chlorine nuclei in the molecule. (b) Symbols $\delta+$ and $\delta$ - indicate the polarity of the $\mathrm{H}-\mathrm{Cl}$ bond.


338. Electronegativity -
Whether a bond is nonpolar or polar covalent is determined by a property of the bonding atoms called electronegativity. Electronegativity is a measure of the tendency of an atom to attract electrons (or electron density) towards itself. It determines how the shared electrons are distributed between the two atoms in a bond. The more strongly an atom attracts the electrons in its bonds, the larger its electronegativity. Electrons in a polar covalent bond are shifted toward the more electronegative atom; thus, the more electronegative atom is the one with the partial negative charge. The greater the difference in electronegativity, the more polarized the electron distribution and the larger the partial charges of the atoms.

Figure 4.6 shows the electronegativity values of the elements as proposed by one of the most famous chemists of the twentieth century: Linus Pauling (Figure 4.7). In general, electronegativity increases from left to right across a period in the periodic table and decreases down a group. Thus, the nonmetals, which lie in the upper right, tend to have the highest electronegativities, with fluorine the most electronegative element of all (EN = 4.0). Metals tend to be less electronegative elements, and the group 1 metals have the lowest electronegativities. Note that noble gases are excluded from this figure because these atoms usually do not share electrons with others atoms since they have a full valence shell. (While noble gas compounds such as $\mathrm{XeO}_{2}$ do exist, they can only be formed under extreme conditions, and thus they do not fit neatly into the general model of electronegativity.)

FIGURE 4.6 The electronegativity values derived by Pauling follow predictable periodic trends, with the higher electronegativities toward the upper right of the periodic table.


339. Electronegativity versus Electron Affinity -
We must be careful not to confuse electronegativity and electron affinity. The electron affinity of an element is a measurable physical quantity, namely, the energy released or absorbed when an isolated gas-phase atom acquires an electron, measured in $\mathrm{kJ} / \mathrm{mol}$. Electronegativity, on the other hand, describes how tightly an atom attracts electrons in a bond. It is a dimensionless quantity that is calculated, not measured. Pauling derived the first electronegativity values by comparing the amounts of energy required to break different types of bonds. He chose an arbitrary relative scale ranging from 0 to 4 .


341. Linus Pauling -
Linus Pauling, shown in Figure 4.7, is the only person to have received two unshared (individual) Nobel Prizes: one for chemistry in 1954 for his work on the nature of chemical bonds and one for peace in 1962 for his opposition to weapons of mass destruction. He developed many of the theories and concepts that are foundational to our current understanding of chemistry, including electronegativity and resonance structures.

FIGURE 4.7 Linus Pauling (1901-1994) made many important contributions to the field of chemistry. He was also a prominent activist, publicizing issues related to health and nuclear weapons.

Pauling also contributed to many other fields besides chemistry. His research on sickle cell anemia revealed the cause of the disease-the presence of a genetically inherited abnormal protein in the blood-and paved the way for the field of molecular genetics. His work was also pivotal in curbing the
testing of nuclear weapons; he proved that radioactive fallout from nuclear testing posed a public health risk.


342. Electronegativity and Bond Type -
The absolute value of the difference in electronegativity ( $\Delta \mathrm{EN}$ ) of two bonded atoms provides a rough measure of the polarity to be expected in the bond and, thus, the bond type. When the difference is very small or zero, the bond is covalent and nonpolar. When it is large, the bond is polar covalent or ionic. The absolute values of the electronegativity differences between the atoms in the bonds $\mathrm{H}-\mathrm{H}, \mathrm{H}-\mathrm{Cl}$, and $\mathrm{Na}-\mathrm{Cl}$ are 0 (nonpolar), 0.9 (polar covalent), and 2.1 (ionic), respectively. The degree to which electrons are shared between atoms varies from completely equal (pure covalent bonding) to not at all (ionic bonding). Figure 4.8 shows the relationship between electronegativity difference and bond type.

FIGURE 4.8 As the electronegativity difference increases between two atoms, the bond becomes more ionic.
A rough approximation of the electronegativity differences associated with covalent, polar covalent, and ionic bonds is shown in Figure 4.8. This table is just a general guide, however, with many exceptions. For example, the H and F atoms in HF have an electronegativity difference of 1.9, and the N and H atoms in $\mathrm{NH}_{3}$ a difference of 0.9 , yet both of these compounds form bonds that are considered polar covalent. Likewise, the Na and Cl atoms in NaCl have an electronegativity difference of 2.1, and the Mn and I atoms in $\mathrm{MnI}_{2}$ have a difference of 1.0, yet both of these substances form ionic compounds.

The best guide to the covalent or ionic character of a bond is to consider the types of atoms involved and their relative positions in the periodic table. Bonds between two nonmetals are generally covalent; bonding between a metal and a nonmetal is often ionic.


342. Electronegativity and Bond Type -
Some compounds contain both covalent and ionic bonds. The atoms in polyatomic ions, such as $\mathrm{OH}^{-}, \mathrm{NO}_{3}{ }^{-}$, and $\mathrm{NH}_{4}{ }^{+}$, are held together by polar covalent bonds. However, these polyatomic ions form ionic compounds by combining with ions of opposite charge. For example, potassium nitrate, $\mathrm{KNO}_{3}$, contains the $\mathrm{K}^{+}$cation and the polyatomic $\mathrm{NO}_{3}{ }^{-}$anion. Thus, bonding in potassium nitrate is ionic, resulting from the electrostatic attraction between the ions $\mathrm{K}^{+}$and $\mathrm{NO}_{3}{ }^{-}$, as well as covalent between the nitrogen and oxygen atoms in $\mathrm{NO}_{3}{ }^{-}$.


344. Electronegativity and Bond Polarity -
Bond polarities play an important role in determining the structure of proteins. Using the electronegativity values in Figure 4.6, arrange the following covalent bonds-all commonly found in amino acids-in order of increasing polarity. Then designate the positive and negative atoms using the symbols $\delta+$ and $\delta-$ :

C-H, C-N, C-O, N-H, O-H, S-H


345. Solution -
The polarity of these bonds increases as the absolute value of the electronegativity difference increases. The atom with the $\delta$ - designation is the more electronegative of the two. Table 4.1 shows these bonds in order of increasing polarity.

Bond Polarity and Electronegativity Difference

| Bond | $\Delta \mathrm{EN}$ | Polarity |
| :---: | :---: | :---: |
| C-H | 0.4 | $\stackrel{\delta-}{\mathrm{C}}-\stackrel{\delta+}{\mathrm{H}}$ |
| S-H | 0.4 | $\stackrel{\delta-}{S-}-\stackrel{\delta+}{H}$ |
| C-N | 0.5 | $\stackrel{\delta+}{\mathrm{C}}-\stackrel{\delta-}{\mathrm{N}}$ |
| N-H | 0.9 | $\stackrel{\delta-}{\mathrm{N}}-\stackrel{\delta+}{\mathrm{H}}$ |
| C-O | 1.0 | $\stackrel{\delta+}{\mathrm{C}}-\stackrel{\delta-}{\mathrm{O}}$ |
| O-H | 1.4 | $\stackrel{\delta-}{\mathrm{O}}-\stackrel{\delta+}{\mathrm{H}}$ |

TABLE 4.1


346. Check Your Learning -
Silicones are polymeric compounds containing, among others, the following types of covalent bonds: $\mathrm{Si}-\mathrm{O}$, $\mathrm{Si}-\mathrm{C}, \mathrm{C}-\mathrm{H}$, and $\mathrm{C}-\mathrm{C}$. Using the electronegativity values in Figure 4.6, arrange the bonds in order of increasing polarity and designate the positive and negative atoms using the symbols $\delta+$ and $\delta-$.

Answer:

| Bond | Electronegativity Difference | Polarity |
| :--- | :--- | :--- |
| $\mathrm{C}-\mathrm{C}$ | 0.0 | nonpolar |
| $\mathrm{C}-\mathrm{H}$ | 0.4 | $\delta-\stackrel{\delta+}{\mathrm{C}}-\mathrm{H}$ |
| $\mathrm{Si}-\mathrm{C}$ | 0.7 | $\delta+\delta-$
 $\mathrm{Si}-\mathrm{C}$ |
| $\mathrm{Si}-\mathrm{O}$ | 1.7 | $\delta+\delta-$
 $\mathrm{Si}-\mathrm{O}$ |


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

- Derive names for common types of inorganic compounds using a systematic approach

Nomenclature, a collection of rules for naming things, is important in science and in many other situations.

This module describes an approach that is used to name simple ionic and molecular compounds, such as $\mathrm{NaCl}, \mathrm{CaCO}_{3}$, and $\mathrm{N}_{2} \mathrm{O}_{4}$. The simplest of these are binary compounds, those containing only two elements, but we will also consider how to name ionic compounds containing polyatomic ions, and one specific, very important class of compounds known as acids (subsequent chapters in this text will focus on these compounds in great detail). We will limit our attention here to inorganic compounds, compounds that are composed principally of elements other than carbon, and will follow the nomenclature guidelines proposed by IUPAC. The rules for organic compounds, in which carbon is the principle element, will be treated in a later chapter on organic chemistry.


348. Ionic Compounds -
To name an inorganic compound, we need to consider the answers to several questions. First, is the compound ionic or molecular? If the compound is ionic, does the metal form ions of only one type (fixed charge) or more than one type (variable charge)? Are the ions monatomic or polyatomic? If the compound is molecular, does it contain hydrogen? If so, does it also contain oxygen? From the answers we derive, we place the compound in an appropriate category and then name it accordingly.


349. Compounds Containing Only Monatomic Ions -
The name of a binary compound containing monatomic ions consists of the name of the cation (the name of the metal) followed by the name of the anion (the name of the nonmetallic element with its ending replaced by the suffix -ide). Some examples are given in Table 4.2.

| Names of Some Ionic Compounds |  |
| :--- | :--- |
| NaCl, sodium chloride | $\mathrm{Na}_{2} \mathrm{O}$, sodium oxide |
| KBr , potassium bromide | CdS, cadmium sulfide |
| $\mathrm{CaI}_{2}$, calcium iodide | $\mathrm{Mg}_{3} \mathrm{~N}_{2}$, magnesium nitride |
| CsF, cesium fluoride | $\mathrm{Ca}_{3} \mathrm{P}_{2}$, calcium phosphide |
| LiCl, lithium chloride | $\mathrm{Al}_{4} \mathrm{C}_{3}$, aluminum carbide |

TABLE 4.2


350. Compounds Containing Polyatomic lons -
Compounds containing polyatomic ions are named similarly to those containing only monatomic ions, i.e., by naming first the cation and then the anion. Examples are shown in Table 4.3.

| Names of Some Polyatomic lonic Compounds |  |
| :--- | :--- |
| $\mathrm{KC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$, potassium acetate | $\mathrm{NH}_{4} \mathrm{Cl}$, ammonium chloride |
| $\mathrm{NaHCO}_{3}$, sodium bicarbonate | $\mathrm{CaSO}_{4}$, calcium sulfate |
| $\mathrm{Al}_{2}\left(\mathrm{CO}_{3}\right)_{3}$, aluminum carbonate | $\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}$, magnesium phosphate |

TABLE 4.3


352. Ionic Compounds in Your Cabinets -
Every day you encounter and use a large number of ionic compounds. Some of these compounds, where they are found, and what they are used for are listed in Table 4.4. Look at the label or ingredients list on the various products that you use during the next few days, and see if you run into any of those in this table, or find other ionic compounds that you could now name or write as a formula.

Everyday Ionic Compounds

| Ionic Compound | Use |
| :---: | :---: |
| NaCl , sodium chloride | ordinary table salt |
| KI, potassium iodide | added to "iodized" salt for thyroid health |
| NaF, sodium fluoride | ingredient in toothpaste |
| $\mathrm{NaHCO}_{3}$, sodium bicarbonate | baking soda; used in cooking (and as antacid) |
| $\mathrm{Na}_{2} \mathrm{CO}_{3}$, sodium carbonate | washing soda; used in cleaning agents |
| NaOCl , sodium hypochlorite | active ingredient in household bleach |
| $\mathrm{CaCO}_{3}$ calcium carbonate | ingredient in antacids |
| $\mathrm{Mg}(\mathrm{OH})_{2}$, magnesium hydroxide | ingredient in antacids |
| $\mathrm{Al}(\mathrm{OH})_{3}$, aluminum hydroxide | ingredient in antacids |
| NaOH , sodium hydroxide | lye; used as drain cleaner |
| $\mathrm{K}_{3} \mathrm{PO}_{4}$, potassium phosphate | food additive (many purposes) |
| $\mathrm{MgSO}_{4}$, magnesium sulfate | added to purified water |
| $\mathrm{Na}_{2} \mathrm{HPO}_{4}$, sodium hydrogen phosphate | anti-caking agent; used in powdered products |
| $\mathrm{Na}_{2} \mathrm{SO}_{3}$, sodium sulfite | preservative |

TABLE 4.4


353. Compounds Containing a Metal Ion with a Variable Charge -
Most of the transition metals and some main group metals can form two or more cations with different charges. Compounds of these metals with nonmetals are named with the same method as compounds in the first category, except the charge of the metal ion is specified by a Roman numeral in parentheses after the name of the metal. The charge of the metal ion is determined from the formula of the compound and the charge of the anion. For example, consider binary ionic compounds of iron and chlorine. Iron typically exhibits a charge of either 2+ or $3+$ (see Figure 3.40), and the two corresponding compound formulas are $\mathrm{FeCl}_{2}$ and $\mathrm{FeCl}_{3}$. The simplest name, "iron chloride," will, in this case, be ambiguous, as it does not distinguish between
these two compounds. In cases like this, the charge of the metal ion is included as a Roman numeral in parentheses immediately following the metal name. These two compounds are then unambiguously named iron(II) chloride and iron(III) chloride, respectively. Other examples are provided in Table 4.5.

| Some Ionic Compounds with Variably Charged Metal lons |  |
| :--- | :--- |
| Compound | Name |
| $\mathrm{FeCl}_{2}$ | iron(II) chloride |
| $\mathrm{FeCl}_{3}$ | iron(III) chloride |
| $\mathrm{Hg}_{2} \mathrm{O}$ | mercury(I) oxide |
| HgO | mercury(II) oxide |
| $\mathrm{SnF}_{2}$ | tin(II) fluoride |
| $\mathrm{SnF}_{4}$ | tin(IV) fluoride |

TABLE 4.5


353. Compounds Containing a Metal Ion with a Variable Charge -
TABLE 4.5

Out-of-date nomenclature used the suffixes -ic and -ous to designate metals with higher and lower charges, respectively: Iron(III) chloride, $\mathrm{FeCl}_{3}$, was previously called ferric chloride, and iron(II) chloride, $\mathrm{FeCl}_{2}$, was known as ferrous chloride. Though this naming convention has been largely abandoned by the scientific community, it remains in use by some segments of industry. For example, you may see the words stannous fluoride on a tube of toothpaste. This represents the formula $\mathrm{SnF}_{2}$, which is more properly named tin(II) fluoride. The other fluoride of tin is $\mathrm{SnF}_{4}$, which was previously called stannic fluoride but is now named tin(IV) fluoride.


354. Ionic Hydrates -
Ionic compounds that contain water molecules as integral components of their crystals are called hydrates. The name for an ionic hydrate is derived by adding a term to the name for the anhydrous (meaning "not hydrated") compound that indicates the number of water molecules associated with each formula unit of the compound. The added word begins with a Greek prefix denoting the number of water molecules (see Table 4.6) and ends with "hydrate." For example, the anhydrous compound copper(II) sulfate also exists as a hydrate containing five water molecules and named copper(II) sulfate pentahydrate. Washing soda is the common name for a hydrate of sodium carbonate containing 10 water molecules; the systematic name is sodium carbonate decahydrate.

Formulas for ionic hydrates are written by appending a vertically centered dot, a coefficient representing the number of water molecules, and the formula for water. The two examples mentioned in the previous paragraph are represented by the formulas

> copper(II) sulfate pentahydrate $\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}$
> sodium carbonate decahydrate $\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot 10 \mathrm{H}_{2} \mathrm{O}$

Nomenclature Prefixes

| Number |  | Prefix |  | Number |  | Prefix |
| :--- | :--- | :--- | :--- | :--- | :---: | :---: |
| 1 (sometimes omitted) | mono- | 6 | hexa- |  |  |  |
| 2 | di- |  | 7 | hepta- |  |  |
| 2 | tri- |  | 8 | octa- |  |  |
| 3 | tetra- |  | 9 | nona- |  |  |
| 4 | penta- |  | 10 | deca- |  |  |
| 5 |  |  |  |  |  |  |

TABLE 4.6


356. Naming Ionic Compounds -
Name the following ionic compounds:
(a) $\mathrm{Fe}_{2} \mathrm{~S}_{3}$
(b) CuSe
(c) GaN
(d) $\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$
(e) $\mathrm{Ti}_{2}\left(\mathrm{SO}_{4}\right)_{3}$


357. Solution -
The anions in these compounds have a fixed negative charge ( $\mathrm{S}^{2-}, \mathrm{Se}^{2-}, \mathrm{N}^{3-}$, and $\mathrm{SO}_{4}{ }^{2-}$ ), and the compounds must be neutral. Because the total number of positive charges in each compound must equal the total number of negative charges, the positive ions must be $\mathrm{Fe}^{3+}, \mathrm{Cu}^{2+}, \mathrm{Ga}^{3+}, \mathrm{Mg}^{2+}$, and $\mathrm{Ti}^{3+}$. These charges are used in the names of the metal ions:
(a) iron(III) sulfide
(b) copper(II) selenide
(c) gallium(III) nitride
(d) magnesium sulfate heptahydrate
(e) titanium(III) sulfate


358. Check Your Learning -
Write the formulas of the following ionic compounds:
(a) chromium(III) phosphide
(b) mercury(II) sulfide
(c) manganese(II) phosphate
(d) copper(I) oxide
(e) iron(III) chloride dihydrate


359. Answer: -
(a) CrP ; (b) HgS ; (c) $\mathrm{Mn}_{3}\left(\mathrm{PO}_{4}\right)_{2}$; (d) $\mathrm{Cu}_{2} \mathrm{O}$; (e) $\mathrm{FeCl}_{3} \cdot 2 \mathrm{H}_{2} \mathrm{O}$


361. Erin Brokovich and Chromium Contamination -
In the early 1990s, legal file clerk Erin Brockovich (Figure 4.9) discovered a high rate of serious illnesses in the small town of Hinckley, California. Her investigation eventually linked the illnesses to groundwater contaminated by $\mathrm{Cr}(\mathrm{VI})$ used by Pacific Gas \& Electric (PG\&E) to fight corrosion in a nearby natural gas pipeline. As dramatized in the film Erin Brokovich (for which Julia Roberts won an Oscar), Erin and lawyer Edward Masry sued PG\&E for contaminating the water near Hinckley in 1993. The settlement they won in 1996-\$333 million-was the largest amount ever awarded for a direct-action lawsuit in the US at that time.

FIGURE 4.9 (a) Erin Brockovich found that $\mathrm{Cr}(\mathrm{VI})$, used by PG\&E, had contaminated the Hinckley, California, water supply. (b) The $\mathrm{Cr}(\mathrm{VI})$ ion is often present in water as the polyatomic ions chromate, $\mathrm{CrO}_{4}{ }^{2-}$ (left), and dichromate, $\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}$ (right).

Chromium compounds are widely used in industry, such as for chrome plating, in dye-making, as preservatives, and to prevent corrosion in cooling tower water, as occurred near Hinckley. In the environment, chromium exists primarily in either the Cr (III) or Cr (VI) forms. Cr (III), an ingredient of many vitamin and nutritional supplements, forms compounds that are not very soluble in water, and it has low toxicity. But $\mathrm{Cr}(\mathrm{VI})$ is much more toxic and forms compounds that are reasonably soluble in water.
Exposure to small amounts of $\mathrm{Cr}(\mathrm{VI})$ can lead to damage of the respiratory, gastrointestinal, and immune systems, as well as the kidneys, liver, blood, and skin.


361. Erin Brokovich and Chromium Contamination -
Despite cleanup efforts, $\mathrm{Cr}(\mathrm{VI})$ groundwater contamination remains a problem in Hinckley and other locations across the globe. A 2010 study by the Environmental Working Group found that of 35 US cities tested, 31 had higher levels of $\mathrm{Cr}(\mathrm{VI})$ in their tap water than the public health goal of 0.02 parts per billion set by the California Environmental Protection Agency.


362. Molecular (Covalent) Compounds -
The bonding characteristics of inorganic molecular compounds are different from ionic compounds, and they are named using a different system as well. The charges of cations and anions dictate their ratios in ionic compounds, so specifying the names of the ions provides sufficient information to determine chemical formulas. However, because covalent bonding allows for significant variation in the combination ratios of the atoms in a molecule, the names for molecular compounds must explicitly identify these ratios.

Compounds Composed of Two Elements
When two nonmetallic elements form a molecular compound, several combination ratios are often possible.

For example, carbon and oxygen can form the compounds CO and $\mathrm{CO}_{2}$. Since these are different substances with different properties, they cannot both have the same name (they cannot both be called carbon oxide). To deal with this situation, we use a naming method that is somewhat similar to that used for ionic compounds, but with added prefixes to specify the numbers of atoms of each element. The name of the more metallic element (the one farther to the left and/or bottom of the periodic table) is first, followed by the name of the more nonmetallic element (the one farther to the right and/or top) with its ending changed to the suffix -ide. The numbers of atoms of each element are designated by the Greek prefixes shown in Table 4.6.

When only one atom of the first element is present, the prefix mono- is usually deleted from that part. Thus, CO is named carbon monoxide, and $\mathrm{CO}_{2}$ is called carbon dioxide. When two vowels are adjacent, the $a$ in the Greek prefix is usually dropped. Some other examples are shown in Table 4.7.

Names of Some Molecular Compounds Composed of Two Elements


362. Molecular (Covalent) Compounds -
Names of Some Molecular Compounds Composed of Two Elements

| Compound |  |  |  | Name |  | Compound |  | Name |
| :--- | :--- | :--- | :--- | :--- | :---: | :---: | :---: | :---: |
| $\mathrm{SO}_{2}$ | sulfur dioxide |  | $\mathrm{BCl}_{3}$ | boron trichloride |  |  |  |  |
| $\mathrm{SO}_{3}$ | sulfur trioxide |  | $\mathrm{SF}_{6}$ | sulfur hexafluoride |  |  |  |  |
| $\mathrm{NO}_{2}$ | nitrogen dioxide |  | $\mathrm{PF}_{5}$ | phosphorus pentafluoride |  |  |  |  |
| $\mathrm{N}_{2} \mathrm{O}_{4}$ | dinitrogen tetroxide |  | $\mathrm{P}_{4} \mathrm{O}_{10}$ | tetraphosphorus decaoxide |  |  |  |  |
| $\mathrm{N}_{2} \mathrm{O}_{5}$ | dinitrogen pentoxide |  | $\mathrm{IF}_{7}$ | iodine heptafluoride |  |  |  |  |

TABLE 4.7

There are a few common names that you will encounter as you continue your study of chemistry. For example, although NO is often called nitric oxide, its proper name is nitrogen monoxide. Similarly, $\mathrm{N}_{2} \mathrm{O}$ is known as nitrous oxide even though our rules would specify the name dinitrogen monoxide. (And $\mathrm{H}_{2} \mathrm{O}$ is usually called water, not dihydrogen monoxide.) You should commit to memory the common names of compounds as you encounter them.


364. Naming Covalent Compounds -
Name the following covalent compounds:
(a) $\mathrm{SF}_{6}$
(b) $\mathrm{N}_{2} \mathrm{O}_{3}$
(c) $\mathrm{Cl}_{2} \mathrm{O}_{7}$
(d) $\mathrm{P}_{4} \mathrm{O}_{6}$


365. Solution -
Because these compounds consist solely of nonmetals, we use prefixes to designate the number of atoms of each element:
(a) sulfur hexafluoride
(b) dinitrogen trioxide
(c) dichlorine heptoxide
(d) tetraphosphorus hexoxide


366. Check Your Learning -
Write the formulas for the following compounds:
(a) phosphorus pentachloride
(b) dinitrogen monoxide
(c) iodine heptafluoride
(d) carbon tetrachloride


367. Answer: -
(a) $\mathrm{PCl}_{5}$; (b) $\mathrm{N}_{2} \mathrm{O}$; (c) $\mathrm{IF}_{7}$; (d) $\mathrm{CCl}_{4}$


368. LINK TO LEARNING -
The following website (http://openstax.org/l/16chemcompname) provides practice with naming chemical compounds and writing chemical formulas. You can choose binary, polyatomic, and variable charge ionic compounds, as well as molecular compounds.


369. Binary Acids -
Some compounds containing hydrogen are members of an important class of substances known as acids. The chemistry of these compounds is explored in more detail in later chapters of this text, but for now, it will suffice to note that many acids release hydrogen ions, $\mathrm{H}^{+}$, when dissolved in water. To denote this distinct chemical property, a mixture of water with an acid is given a name derived from the compound's name. If the compound is a binary acid (comprised of hydrogen and one other nonmetallic element):

1. The word "hydrogen" is changed to the prefix hydro-
2. The other nonmetallic element name is modified by adding the suffix -ic
3. The word "acid" is added as a second word

For example, when the gas HCl (hydrogen chloride) is dissolved in water, the solution is called hydrochloric acid. Several other examples of this nomenclature are shown in Table 4.8.

Names of Some Simple Acids

| Name of Gas | Name of Acid |
| :--- | :--- |
| $\mathrm{HF}(g)$, hydrogen fluoride | $\mathrm{HF}(a q)$, hydrofluoric acid |
| $\mathrm{HCl}(g)$, hydrogen chloride | $\mathrm{HCl}(a q)$, hydrochloric acid |
| $\mathrm{HBr}(g)$, hydrogen bromide | $\mathrm{HBr}(\mathrm{aq})$, hydrobromic acid |
| $\mathrm{HI}(g)$, hydrogen iodide | $\mathrm{HI}(\mathrm{aq})$, hydroiodic acid |
| $\mathrm{H}_{2} \mathrm{~S}(g)$, hydrogen sulfide | $\mathrm{H}_{2} \mathrm{~S}(\mathrm{aq})$, hydrosulfuric acid |

TABLE 4.8


370. Oxyacids -
Many compounds containing three or more elements (such as organic compounds or coordination compounds) are subject to specialized nomenclature rules that you will learn later. However, we will briefly
discuss the important compounds known as oxyacids, compounds that contain hydrogen, oxygen, and at least one other element, and are bonded in such a way as to impart acidic properties to the compound (you will learn the details of this in a later chapter). Typical oxyacids consist of hydrogen combined with a polyatomic, oxygen-containing ion. To name oxyacids:

1. Omit "hydrogen"
2. Start with the root name of the anion
3. Replace -ate with -ic, or -ite with -ous
4. Add "acid"

For example, consider $\mathrm{H}_{2} \mathrm{CO}_{3}$ (which you might be tempted to call "hydrogen carbonate"). To name this correctly, "hydrogen" is omitted; the -ate of carbonate is replace with -ic; and acid is added-so its name is carbonic acid. Other examples are given in Table 4.9. There are some exceptions to the general naming method (e.g., $\mathrm{H}_{2} \mathrm{SO}_{4}$ is called sulfuric acid, not sulfic acid, and $\mathrm{H}_{2} \mathrm{SO}_{3}$ is sulfurous, not sulfous, acid).

Names of Common Oxyacids

| Formula | Anion Name | Acid Name |
| :--- | :--- | :--- |
| $\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$ | acetate | acetic acid |
| $\mathrm{HNO}_{3}$ | nitrate | nitric acid |
| $\mathrm{HNO}_{2}$ | nitrite | nitrous acid |
| $\mathrm{HClO}_{4}$ | perchlorate | perchloric acid |
| $\mathrm{H}_{2} \mathrm{CO}_{3}$ | carbonate | carbonic acid |
| $\mathrm{H}_{2} \mathrm{SO}_{4}$ | sulfate | sulfuric acid |
| $\mathrm{H}_{2} \mathrm{SO}_{3}$ | sulfite | sulfurous acid |
| $\mathrm{H}_{3} \mathrm{PO}_{4}$ | phosphate | phosphoric acid |


370. Oxyacids -
TABLE 4.9


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

- Write Lewis symbols for neutral atoms and ions
- Draw Lewis structures depicting the bonding in simple molecules

Thus far in this chapter, we have discussed the various types of bonds that form between atoms and/or ions. In all cases, these bonds involve the sharing or transfer of valence shell electrons between atoms. In this section, we will explore the typical method for depicting valence shell electrons and chemical bonds, namely Lewis symbols and Lewis structures.


372. Lewis Symbols -
We use Lewis symbols to describe valence electron configurations of atoms and monatomic ions. A Lewis symbol consists of an elemental symbol surrounded by one dot for each of its valence electrons:

- Ca

Figure 4.10 shows the Lewis symbols for the elements of the third period of the periodic table.

| Atoms | Electronic Configuration | Lewis Symbol |
| :--- | :--- | :--- |
| sodium | $[\mathrm{Ne}] 3 s^{1}$ | $\mathrm{Na} \cdot$ |
| magnesium | $[\mathrm{Ne}] 3 s^{2}$ | $\cdot \mathrm{Mg} \cdot$ |
| aluminum | $[\mathrm{Ne}] 3 s^{2} 3 p^{1}$ | $\cdot \dot{\mathrm{Al} \cdot} \cdot$ |
| silicon | $[\mathrm{Ne}] 3 s^{2} 3 p^{2}$ | $\cdot \dot{\mathrm{Si}} \cdot$ |
| phosphorus | $[\mathrm{Ne}] 3 s^{2} 3 p^{3}$ | $\ddot{\mathrm{P}} \cdot$ |
| sulfur | $[\mathrm{Ne}] 3 s^{2} 3 p^{4}$ | $\ddot{\mathrm{~S}} \cdot$ |
| chlorine | $[\mathrm{Ne}] 3 s^{2} 3 p^{5}$ | $: \ddot{\mathrm{Cl}} \cdot$ |
| argon | $[\mathrm{Ne}] 3 s^{2} 3 p^{6}$ | $: \ddot{\mathrm{Ar}}:$ |

FIGURE 4.10 Lewis symbols illustrating the number of valence electrons for each element in the third period of the periodic table.

Lewis symbols can also be used to illustrate the formation of cations from atoms, as shown here for sodium and calcium:
\$\underset{\substack{sodium

atom}}{\mathrm{Na} \cdot \underset{\substack{sodium

cation}}{\mathrm{Na}^{+}}+\mathrm{e}^{-}}\$| calcium |
| :---: |
| atom |\$>\underset{\substack{calcium

cation}}{\mathrm{Ca}^{2+}}+\mathrm{Ce}^{-}\$


372. Lewis Symbols -
cation}}{\mathrm{Ca}^{2+}}+\mathrm{Ce}^{-}\$

Likewise, they can be used to show the formation of anions from atoms, as shown here for chlorine and sulfur:

Figure 4.11 demonstrates the use of Lewis symbols to show the transfer of electrons during the formation of ionic compounds.

| Metal |  | Nonmetal | Ionic Compound |
| :---: | :---: | :---: | :---: |
| $\mathrm{Na} \cdot$
 sodium atom | $+$ | $\ddot{\mathrm{Cl}}$
 chlorine atom | $\longrightarrow \quad \mathrm{Na}^{+}[: \stackrel{\ddot{\mathrm{C}} \cdot:}{\cdot .}]^{-}$
 sodium chloride (sodium ion and chloride ion) |
| -Mg•
 magnesium atom | + | $\ddot{O}$
 oxygen atom | $\longrightarrow \quad \mathrm{Mg}^{2+}[: \ddot{\mathrm{O}}:]^{2-}$
 magnesium oxide (magnesium ion and oxide ion) |
| - Ca•
 calcium atom | $+$ | $2: \ddot{F}$
 fluorine atoms | $\longrightarrow \quad \mathrm{Ca}^{2+}[: \ddot{\mathrm{F}}:]_{2}^{-}$
 calcium fluoride (calcium ion and two fluoride ions) |

FIGURE 4.11 Cations are formed when atoms lose electrons, represented by fewer Lewis dots, whereas anions are formed by atoms gaining electrons. The total number of electrons does not change.


373. Lewis Structures -
We also use Lewis symbols to indicate the formation of covalent bonds, which are shown in Lewis structures, drawings that describe the bonding in molecules and polyatomic ions. For example, when two chlorine atoms form a chlorine molecule, they share one pair of electrons:

The Lewis structure indicates that each Cl atom has three pairs of electrons that are not used in bonding (called lone pairs) and one shared pair of electrons (written between the atoms). A dash (or line) is sometimes used to indicate a shared pair of electrons:

A single shared pair of electrons is called a single bond. Each Cl atom interacts with eight valence electrons: the six in the lone pairs and the two in the single bond.


374. The Octet Rule -
The other halogen molecules ( $\mathrm{F}_{2}, \mathrm{Br}_{2}, \mathrm{I}_{2}$, and $\mathrm{At}_{2}$ ) form bonds like those in the chlorine molecule: one single bond between atoms and three lone pairs of electrons per atom. This allows each halogen atom to have a noble gas electron configuration. The tendency of main group atoms to form enough bonds to obtain eight valence electrons is known as the octet rule.

The number of bonds that an atom can form can often be predicted from the number of electrons needed to reach an octet (eight valence electrons); this is especially true of the nonmetals of the second period of the periodic table ( $\mathrm{C}, \mathrm{N}, \mathrm{O}$, and F ). For example, each atom of a group 14 element has four electrons in its outermost shell and therefore requires four more electrons to reach an octet. These four electrons can be gained by forming four covalent bonds, as illustrated here for carbon in $\mathrm{CCl}_{4}$ (carbon tetrachloride) and silicon in $\mathrm{SiH}_{4}$ (silane). Because hydrogen only needs two electrons to fill its valence shell, it is an exception to the octet rule. The transition elements and inner transition elements also do not follow the octet rule:

Group 15 elements such as nitrogen have five valence electrons in the atomic Lewis symbol: one lone pair and three unpaired electrons. To obtain an octet, these atoms form three covalent bonds, as in $\mathrm{NH}_{3}$ (ammonia). Oxygen and other atoms in group 16 obtain an octet by forming two covalent bonds:

Double and Triple Bonds
As previously mentioned, when a pair of atoms shares one pair of electrons, we call this a single bond. However, a pair of atoms may need to share more than one pair of electrons in order to achieve the requisite octet. A double bond forms when two pairs of electrons are shared between a pair of atoms, as between the carbon and oxygen atoms in $\mathrm{CH}_{2} \mathrm{O}$ (formaldehyde) and between the two carbon atoms in $\mathrm{C}_{2} \mathrm{H}_{4}$ (ethylene):


374. The Octet Rule -
A triple bond forms when three electron pairs are shared by a pair of atoms, as in carbon monoxide (CO) and the cyanide ion $\left(\mathrm{CN}^{-}\right)$:

```
:C:::O: or :C\equivO: :C:::N\overline{* or :C\equivN:}
```

carbon monoxide cyanide ion


375. Writing Lewis Structures with the Octet Rule -
For very simple molecules and molecular ions, we can write the Lewis structures by merely pairing up the unpaired electrons on the constituent atoms. See these examples:

For more complicated molecules and molecular ions, it is helpful to follow the step-by-step procedure outlined here:

1. Determine the total number of valence (outer shell) electrons. For cations, subtract one electron for each positive charge. For anions, add one electron for each negative charge.
2. Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom. (Generally, the least electronegative element should be placed in the center.) Connect each atom to the central atom with a single bond (one electron pair).
3. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen), completing an
octet around each atom.
4. Place all remaining electrons on the central atom.
5. Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible.

Let us determine the Lewis structures of $\mathrm{SiH}_{4}, \mathrm{CHO}_{2}{ }^{-}, \mathrm{NO}^{+}$, and $\mathrm{OF}_{2}$ as examples in following this procedure:

1. Determine the total number of valence (outer shell) electrons in the molecule or ion.

- For a molecule, we add the number of valence electrons on each atom in the molecule:

> | $\mathrm{SiH}_{4}$ |
| :--- |
| $\mathrm{Si}: 4$ valence electrons/atom $\times 1$ atom $=4$ |
| $+\mathrm{H}: 1$ valence electron $/$ atom $\times 4$ atoms $=4$ |

$$
=8 \text { valence electrons }
$$

- For a negative ion, such as $\mathrm{CHO}_{2}{ }^{-}$, we add the number of valence electrons on the atoms to the number of negative charges on the ion (one electron is gained for each single negative charge):


375. Writing Lewis Structures with the Octet Rule -
$$
\begin{aligned}
& \mathrm{CHO}_{2}{ }^{-} \\
& \mathrm{C}: 4 \text { valence electrons/atom } \times 1 \text { atom }=4 \\
& \mathrm{H}: 1 \text { valence electron/atom } \times 1 \text { atom }=1 \\
& \mathrm{O}: 6 \text { valence electrons/atom } \times 2 \text { atoms }=12 \\
& +\quad 1 \text { additional electron }=1 \\
& \hline
\end{aligned}
$$

- For a positive ion, such as $\mathrm{NO}^{+}$, we add the number of valence electrons on the atoms in the ion and then subtract the number of positive charges on the ion (one electron is lost for each single positive charge) from the total number of valence electrons:

$$
\begin{aligned}
& \mathrm{NO}^{+} \\
& \mathrm{N}: 5 \text { valence electrons/atom } \times 1 \text { atom }=5 \\
& \\
& \begin{aligned}
\mathrm{O}: 6 \text { valence electron/atom } \times 1 \text { atom } & =6 \\
+-1 \text { electron (positive charge) } & =-1 \\
\hline & =10 \text { valence electrons }
\end{aligned}
\end{aligned}
$$

- Since $\mathrm{OF}_{2}$ is a neutral molecule, we simply add the number of valence electrons:

$$
\begin{aligned}
& \mathrm{OF}_{2} \\
& \mathrm{O}: 6 \text { valence electrons/atom } \times 1 \text { atom }=6 \\
&+\mathrm{F}: 7 \text { valence electrons/atom } \times 2 \text { atoms }=14 \\
& \hline=20 \text { valence electrons }
\end{aligned}
$$

2. Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom and connecting each atom to the central atom with a single (one electron pair) bond. (Note that we denote ions with brackets around the structure, indicating the charge outside the brackets:)


375. Writing Lewis Structures with the Octet Rule -
When several arrangements of atoms are possible, as for $\mathrm{CHO}_{2}{ }^{-}$, we must use experimental evidence to choose the correct one. In general, the less electronegative elements are more likely to be central atoms. In $\mathrm{CHO}_{2}{ }^{-}$, the less electronegative carbon atom occupies the central position with the oxygen and hydrogen atoms surrounding it. Other examples include P in $\mathrm{POCl}_{3}, \mathrm{~S}$ in $\mathrm{SO}_{2}$, and Cl in $\mathrm{ClO}_{4}{ }^{-}$. An exception is that hydrogen is almost never a central atom. As the most electronegative element, fluorine also cannot be a central atom.
3. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen) to complete their valence shells with an octet of electrons.

- There are no remaining electrons on $\mathrm{SiH}_{4}$, so it is unchanged:

4. Place all remaining electrons on the central atom.

- For $\mathrm{SiH}_{4}, \mathrm{CHO}_{2}{ }^{-}$, and $\mathrm{NO}^{+}$, there are no remaining electrons; we already placed all of the electrons determined in Step 1.
- For $\mathrm{OF}_{2}$, we had 16 electrons remaining in Step 3, and we placed 12, leaving 4 to be placed on the central atom:

5. Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible.

- $\mathrm{SiH}_{4}$ : Si already has an octet, so nothing needs to be done.
- $\mathrm{CHO}_{2}{ }^{-}$: We have distributed the valence electrons as lone pairs on the oxygen atoms, but the carbon atom lacks an octet:

- $\mathrm{NO}^{+}$: For this ion, we added eight valence electrons, but neither atom has an octet. We cannot add any more electrons since we have already used the total that we found in Step 1, so we must move electrons to form a multiple bond:


375. Writing Lewis Structures with the Octet Rule -
$$
[: \ddot{N}-\ddot{O}:]^{+} \text {gives }[: \ddot{N}=0:]^{+}
$$

This still does not produce an octet, so we must move another pair, forming a triple bond:

$$
[: N \equiv O:]^{+}
$$

- In $\mathrm{OF}_{2}$, each atom has an octet as drawn, so nothing changes.


377. Writing Lewis Structures -
NASA's Cassini-Huygens mission detected a large cloud of toxic hydrogen cyanide (HCN) on Titan, one of Saturn's moons. Titan also contains ethane $\left(\mathrm{H}_{3} \mathrm{CCH}_{3}\right)$, acetylene (HCCH), and ammonia $\left(\mathrm{NH}_{3}\right)$. What are the Lewis structures of these molecules?


378. Solution -
Step 1. Calculate the number of valence electrons.
HCN: $(1 \times 1)+(4 \times 1)+(5 \times 1)=10$
$\mathrm{H}_{3} \mathrm{CCH}_{3}:(1 \times 3)+(2 \times 4)+(1 \times 3)=14$
HCCH: $(1 \times 1)+(2 \times 4)+(1 \times 1)=10$
$\mathrm{NH}_{3}:(5 \times 1)+(3 \times 1)=8$
Step 2. Draw a skeleton and connect the atoms with single bonds. Remember that H is never a central atom:

Step 3. Where needed, distribute electrons to the terminal atoms:

HCN: six electrons placed on N
$\mathrm{H}_{3} \mathrm{CCH}_{3}$ : no electrons remain
HCCH: no terminal atoms capable of accepting electrons
$\mathrm{NH}_{3}$ : no terminal atoms capable of accepting electrons
Step 4. Where needed, place remaining electrons on the central atom:

HCN : no electrons remain
$\mathrm{H}_{3} \mathrm{CCH}_{3}$ : no electrons remain
HCCH: four electrons placed on carbon
$\mathrm{NH}_{3}$ : two electrons placed on nitrogen
Step 5. Where needed, rearrange electrons to form multiple bonds in order to obtain an octet on each atom:
HCN : form two more $\mathrm{C}-\mathrm{N}$ bonds
$\mathrm{H}_{3} \mathrm{CCH}_{3}$ : all atoms have the correct number of electrons
HCCH : form a triple bond between the two carbon atoms
$\mathrm{NH}_{3}$ : all atoms have the correct number of electrons


379. Check Your Learning -
Both carbon monoxide, CO , and carbon dioxide, $\mathrm{CO}_{2}$, are products of the combustion of fossil fuels. Both of these gases also cause problems: CO is toxic and $\mathrm{CO}_{2}$ has been implicated in global climate change. What are the Lewis structures of these two molecules?


380. Answer: -
:c三o: : $\ddot{\mathrm{O}}=\mathrm{c}=\ddot{\mathrm{o}}$ :


382. Fullerene Chemistry -
Carbon, in various forms and compounds, has been known since prehistoric times, . Soot has been used as a pigment (often called carbon black) for thousands of years. Charcoal, high in carbon content, has likewise been critical to human development. Carbon is the key additive to iron in the steelmaking process, and diamonds have a unique place in both culture and industry. With all this usage came significant study, particularly with the emergence of organic chemistry. And even with all the known forms and functions of the element, scientists began to uncover the potential for even more varied and extensive carbon structures.

As early as the 1960s, chemists began to observe complex carbon structures, but they had little evidence to support their concepts, or their work did not make it into the mainstream. Eiji Osawa predicted a spherical form based on observations of a similar structure, but his work was not widely known outside Japan. In a similar manner, the most comprehensive advance was likely computational chemist Elena Galpern's, who in 1973 predicted a highly stable, 60-carbon molecule; her work was also isolated to her native Russia. Still later, Harold Kroto, working with Canadian radio astronomers, sought to uncover the nature of long carbon chains that had been discovered in interstellar space.

Kroto sought to use a machine developed by Richard Smalley's team at Rice University to learn more about these structures. Together with Robert Curl, who had introduced them, and three graduate students-James Heath, Sean O'Brien, and Yuan Liu-they performed an intensive series of experiments that led to a major discovery.

In 1996, the Nobel Prize in Chemistry was awarded to Richard Smalley (Figure 4.12), Robert Curl, and Harold Kroto for their work in discovering a new form of carbon, the $\mathrm{C}_{60}$ buckminsterfullerene molecule (Figure 4.1). An entire class of compounds, including spheres and tubes of various shapes, were discovered based on $\mathrm{C}_{60}$. This type of molecule, called a fullerene, shows promise in a variety of applications. Because of their size and shape, fullerenes can encapsulate other molecules, so they have shown potential in various applications from hydrogen storage to targeted drug delivery systems. They also possess unique electronic and optical properties that have been put to good use in solar powered devices and chemical sensors.


382. Fullerene Chemistry -
FIGURE 4.12 Richard Smalley (1943-2005), a professor of physics, chemistry, and astronomy at Rice University, was one of the leading advocates for fullerene chemistry. Upon his death in 2005, the US Senate honored him as the "Father of Nanotechnology." (credit: United States Department of Energy)


383. Exceptions to the Octet Rule -
Many covalent molecules have central atoms that do not have eight electrons in their Lewis structures. These molecules fall into three categories:

- Odd-electron molecules have an odd number of valence electrons, and therefore have an unpaired electron.
- Electron-deficient molecules have a central atom that has fewer electrons than needed for a noble gas configuration.
- Hypervalent molecules have a central atom that has more electrons than needed for a noble gas configuration.


384. Odd-electron Molecules -
We call molecules that contain an odd number of electrons free radicals. Nitric oxide, NO, is an example of an odd-electron molecule; it is produced in internal combustion engines when oxygen and nitrogen react at high temperatures.

To draw the Lewis structure for an odd-electron molecule like NO, we follow the same five steps we would for other molecules, but with a few minor changes:

1. Determine the total number of valence (outer shell) electrons. The sum of the valence electrons is 5 (from $\mathrm{N})+6($ from O$)=11$. The odd number immediately tells us that we have a free radical, so we know that not every atom can have eight electrons in its valence shell.
2. Draw a skeleton structure of the molecule. We can easily draw a skeleton with an N-O single bond: N-O
3. Distribute the remaining electrons as lone pairs on the terminal atoms. In this case, there is no central atom, so we distribute the electrons around both atoms. We give eight electrons to the more electronegative atom in these situations; thus oxygen has the filled valence shell:

4. Place all remaining electrons on the central atom. Since there are no remaining electrons, this step does not apply.
5. Rearrange the electrons to make multiple bonds with the central atom in order to obtain octets wherever possible. We know that an odd-electron molecule cannot have an octet for every atom, but we want to get each atom as close to an octet as possible. In this case, nitrogen has only five electrons around it. To move closer to an octet for nitrogen, we take one of the lone pairs from oxygen and use it to form a NO double bond. (We cannot take another lone pair of electrons on oxygen and form a triple bond because nitrogen would then have nine electrons:)
$: \dot{N}=\ddot{O}$ :


385. Electron-deficient Molecules -
We will also encounter a few molecules that contain central atoms that do not have a filled valence shell. Generally, these are molecules with central atoms from groups 2 and 13, outer atoms that are hydrogen, or other atoms that do not form multiple bonds. For example, in the Lewis structures of beryllium dihydride, $\mathrm{BeH}_{2}$, and boron trifluoride, $\mathrm{BF}_{3}$, the beryllium and boron atoms each have only four and six electrons, respectively. It is possible to draw a structure with a double bond between a boron atom and a fluorine atom in $\mathrm{BF}_{3}$, satisfying the octet rule, but experimental evidence indicates the bond lengths are closer to that expected for B-F single bonds. This suggests the best Lewis structure has three B-F single bonds and an electron deficient boron. The reactivity of the compound is also consistent with an electron deficient boron. However, the $B-F$ bonds are slightly shorter than what is actually expected for $B-F$ single bonds, indicating that some double bond character is found in the actual molecule.

An atom like the boron atom in $\mathrm{BF}_{3}$, which does not have eight electrons, is very reactive. It readily combines with a molecule containing an atom with a lone pair of electrons. For example, $\mathrm{NH}_{3}$ reacts with $\mathrm{BF}_{3}$ because the lone pair on nitrogen can be shared with the boron atom:


386. Hypervalent Molecules -
Elements in the second period of the periodic table $(n=2)$ can accommodate only eight electrons in their valence shell orbitals because they have only four valence orbitals (one $2 s$ and three $2 p$ orbitals). Elements in the third and higher periods ( $n \geq 3$ ) have more than four valence orbitals and can share more than four pairs of electrons with other atoms because they have empty $d$ orbitals in the same shell. Molecules formed from these elements are sometimes called hypervalent molecules. Figure 4.13 shows the Lewis structures for two hypervalent molecules, $\mathrm{PCl}_{5}$ and $\mathrm{SF}_{6}$.

FIGURE 4.13 In $\mathrm{PCl}_{5}$, the central atom phosphorus shares five pairs of electrons. In $\mathrm{SF}_{6}$, sulfur shares six pairs of electrons.

In some hypervalent molecules, such as $\mathrm{IF}_{5}$ and $\mathrm{XeF}_{4}$, some of the electrons in the outer shell of the central atom are lone pairs:

When we write the Lewis structures for these molecules, we find that we have electrons left over after filling the valence shells of the outer atoms with eight electrons. These additional electrons must be assigned to the central atom.


388. Writing Lewis Structures: Octet Rule Violations -
Xenon is a noble gas, but it forms a number of stable compounds. We examined $\mathrm{XeF}_{4}$ earlier. What are the Lewis structures of $\mathrm{XeF}_{2}$ and $\mathrm{XeF}_{6}$ ?


389. Solution -
We can draw the Lewis structure of any covalent molecule by following the six steps discussed earlier. In this case, we can condense the last few steps, since not all of them apply.

Step 1. Calculate the number of valence electrons:
$\mathrm{XeF}_{2}: 8+(2 \times 7)=22$
$\mathrm{XeF}_{6}: 8+(6 \times 7)=50$
Step 2. Draw a skeleton joining the atoms by single bonds. Xenon will be the central atom because fluorine cannot be a central atom:

Step 3. Distribute the remaining electrons.
$\mathrm{XeF}_{2}$ : We place three lone pairs of electrons around each F atom, accounting for 12 electrons and giving
each F atom 8 electrons. Thus, six electrons (three lone pairs) remain. These lone pairs must be placed on the Xe atom. This is acceptable because Xe atoms have empty valence shell $d$ orbitals and can accommodate more than eight electrons. The Lewis structure of $\mathrm{XeF}_{2}$ shows two bonding pairs and three lone pairs of electrons around the Xe atom:

$\mathrm{XeF}_{6}$ : We place three lone pairs of electrons around each F atom, accounting for 36 electrons. Two electrons remain, and this lone pair is placed on the Xe atom:


390. Check Your Learning -
The halogens form a class of compounds called the interhalogens, in which halogen atoms covalently bond to each other. Write the Lewis structures for the interhalogens $\mathrm{BrCl}_{3}$ and $\mathrm{ICl}_{4}{ }^{-}$.


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

- Compute formal charges for atoms in any Lewis structure
- Use formal charges to identify the most reasonable Lewis structure for a given molecule
- Explain the concept of resonance and draw Lewis structures representing resonance forms for a given molecule

In the previous section, we discussed how to write Lewis structures for molecules and polyatomic ions. As we have seen, however, in some cases, there is seemingly more than one valid structure for a molecule. We can use the concept of formal charges to help us predict the most appropriate Lewis structure when more than one is reasonable.


393. Calculating Formal Charge -
The formal charge of an atom in a molecule is the hypothetical charge the atom would have if we could redistribute the electrons in the bonds evenly between the atoms. Another way of saying this is that formal charge results when we take the number of valence electrons of a neutral atom, subtract the nonbonding electrons, and then subtract the number of bonds connected to that atom in the Lewis structure.

Thus, we calculate formal charge as follows:

$$
\text { formal charge }=\# \text { valence shell electrons (free atom) }-\# \text { lone pair electrons }-\frac{1}{2} \# \text { bonding electrons }
$$

We can double-check formal charge calculations by determining the sum of the formal charges for the whole structure. The sum of the formal charges of all atoms in a molecule must be zero; the sum of the formal charges in an ion should equal the charge of the ion.

We must remember that the formal charge calculated for an atom is not the actual charge of the atom in the molecule. Formal charge is only a useful bookkeeping procedure; it does not indicate the presence of actual charges.


395. Calculating Formal Charge from Lewis Structures -
Assign formal charges to each atom in the interhalogen ion $\mathrm{ICl}_{4}{ }^{-}$.


396. Solution -
Step 1. We divide the bonding electron pairs equally for all I-Cl bonds:

Step 2. We assign lone pairs of electrons to their atoms. Each Cl atom now has seven electrons assigned to it, and the I atom has eight.
Step 3. Subtract this number from the number of valence electrons for the neutral atom:
I: $7-8=-1$
Cl: $7-7=0$
The sum of the formal charges of all the atoms equals -1 , which is identical to the charge of the ion ( -1 ).


397. Check Your Learning -
Calculate the formal charge for each atom in the carbon monoxide molecule:

```
:C\equivO:
```


398. Answer: -
C $-1,0+1$


400. Calculating Formal Charge from Lewis Structures -
Assign formal charges to each atom in the interhalogen molecule $\mathrm{BrCl}_{3}$.


401. Solution -
Step 1. Assign one of the electrons in each $\mathrm{Br}-\mathrm{Cl}$ bond to the Br atom and one to the Cl atom in that bond:

Step 2. Assign the lone pairs to their atom. Now each Cl atom has seven electrons and the Br atom has seven electrons.
Step 3. Subtract this number from the number of valence electrons for the neutral atom. This gives the formal charge:
Br: 7-7 = 0
Cl: 7-7=0
All atoms in $\mathrm{BrCl}_{3}$ have a formal charge of zero, and the sum of the formal charges totals zero, as it must in a neutral molecule.


402. Check Your Learning -
Determine the formal charge for each atom in $\mathrm{NCl}_{3}$.


403. Answer: -
N : 0; all three Cl atoms: 0


404. Using Formal Charge to Predict Molecular Structure -
The arrangement of atoms in a molecule or ion is called its molecular structure. In many cases, following the steps for writing Lewis structures may lead to more than one possible molecular structure-different multiple bond and lone-pair electron placements or different arrangements of atoms, for instance. A few guidelines involving formal charge can be helpful in deciding which of the possible structures is most likely for a particular molecule or ion:

1. A molecular structure in which all formal charges are zero is preferable to one in which some formal charges are not zero.
2. If the Lewis structure must have nonzero formal charges, the arrangement with the smallest nonzero formal charges is preferable.
3. Lewis structures are preferable when adjacent formal charges are zero or of the opposite sign.
4. When we must choose among several Lewis structures with similar distributions of formal charges, the structure with the negative formal charges on the more electronegative atoms is preferable.

To see how these guidelines apply, let us consider some possible structures for carbon dioxide, $\mathrm{CO}_{2}$. We know from our previous discussion that the less electronegative atom typically occupies the central position, but formal charges allow us to understand why this occurs. We can draw three possibilities for the structure: carbon in the center and double bonds, carbon in the center with a single and triple bond, and oxygen in the center with double bonds:

| $\ddot{\mathrm{O}}=\mathrm{C}=\ddot{\mathrm{O}}$ | $: \mathrm{O} \equiv \mathrm{C}-\ddot{\mathrm{O}}:$ | $\ddot{\mathrm{O}}=\mathrm{O}=\ddot{\mathrm{C}}$ | Structure |  |  |  |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 0 | 0 | 0 | +1 | 0 | -1 | 0 |
| $\cdots$ | +2 | -2 | Formal charge |  |  |  |

Comparing the three formal charges, we can definitively identify the structure on the left as preferable because it has only formal charges of zero (Guideline 1).


404. Using Formal Charge to Predict Molecular Structure -
Comparing the three formal charges, we can definitively identify the structure on the left as preferable because it has only formal charges of zero (Guideline 1).

As another example, the thiocyanate ion, an ion formed from a carbon atom, a nitrogen atom, and a sulfur atom, could have three different molecular structures: $\mathrm{NCS}^{-}, \mathrm{CNS}^{-}$, or $\mathrm{CSN}^{-}$. The formal charges present in each of these molecular structures can help us pick the most likely arrangement of atoms. Possible Lewis structures and the formal charges for each of the three possible structures for the thiocyanate ion are shown here:

Note that the sum of the formal charges in each case is equal to the charge of the ion ( -1 ). However, the first arrangement of atoms is preferred because it has the lowest number of atoms with nonzero formal charges (Guideline 2). Also, it places the least electronegative atom in the center, and the negative charge on the more electronegative element (Guideline 4).


406. Using Formal Charge to Determine Molecular Structure -
Nitrous oxide, $\mathrm{N}_{2} \mathrm{O}$, commonly known as laughing gas, is used as an anesthetic in minor surgeries, such as the routine extraction of wisdom teeth. Which is the more likely structure for nitrous oxide?


407. Solution -
Determining formal charge yields the following:

| $: N \equiv N=\ddot{O}:$ | $: \ddot{N}=O=\ddot{N}:$ |  |
| ---: | :--- | :--- |
| 0 | $+1-1$ | $-1+2-1$ |

The structure with a terminal oxygen atom best satisfies the criteria for the most stable distribution of formal charge:

The number of atoms with formal charges are minimized (Guideline 2), there is no formal charge with a magnitude greater than one (Guideline 2), the negative formal charge is on the more electronegative element (Guideline 4), and the less electronegative atom is in the center position.


408. Check Your Learning -
Which is the most likely molecular structure for the nitrite $\left(\mathrm{NO}_{2}{ }^{-}\right)$ion?
$[: \ddot{N}=\ddot{O}-\ddot{O}:]^{-}$or $\quad[: \ddot{O}=\ddot{N}-\ddot{O}:]^{-}$

Answer:
$\mathrm{ONO}^{-}$


409. Resonance -
Notice that the more likely structure for the nitrite anion in Example 4.10 may actually be drawn in two different ways, distinguished by the locations of the $\mathrm{N}-\mathrm{O}$ and $\mathrm{N}=\mathrm{O}$ bonds:

If nitrite ions do indeed contain a single and a double bond, we would expect for the two bond lengths to be different. A double bond between two atoms is shorter (and stronger) than a single bond between the same two atoms. Experiments show, however, that both $\mathrm{N}-\mathrm{O}$ bonds in $\mathrm{NO}_{2}{ }^{-}$have the same strength and length, and are identical in all other properties.

It is not possible to write a single Lewis structure for $\mathrm{NO}_{2}{ }^{-}$in which nitrogen has an octet and both bonds are equivalent. Instead, we use the concept of resonance: if two or more Lewis structures with the same arrangement of atoms can be written for a molecule or ion, the actual distribution of electrons is an average of that shown by the various Lewis structures. The actual distribution of electrons in each of the nitrogen-oxygen bonds in $\mathrm{NO}_{2}{ }^{-}$is the average of a double bond and a single bond. We call the individual Lewis structures resonance forms. The actual electronic structure of the molecule (the average of the resonance forms) is called a resonance hybrid of the individual resonance forms. A double-headed arrow between Lewis structures indicates that they are resonance forms.


409. Resonance -
We should remember that a molecule described as a resonance hybrid never possesses an electronic structure described by either resonance form. It does not fluctuate between resonance forms; rather, the actual electronic structure is always the average of that shown by all resonance forms. George Wheland, one of the pioneers of resonance theory, used a historical analogy to describe the relationship between resonance forms and resonance hybrids. A medieval traveler, having never before seen a rhinoceros, described it as a hybrid of a dragon and a unicorn because it had many properties in common with both. Just as a rhinoceros is neither a dragon sometimes nor a unicorn at other times, a resonance hybrid is neither of its resonance forms at any
given time. Like a rhinoceros, it is a real entity that experimental evidence has shown to exist. It has some characteristics in common with its resonance forms, but the resonance forms themselves are convenient, imaginary images (like the unicorn and the dragon).

The carbonate anion, $\mathrm{CO}_{3}{ }^{2-}$, provides a second example of resonance:

One oxygen atom must have a double bond to carbon to complete the octet on the central atom. All oxygen atoms, however, are equivalent, and the double bond could form from any one of the three atoms. This gives rise to three resonance forms of the carbonate ion. Because we can write three identical resonance structures, we know that the actual arrangement of electrons in the carbonate ion is the average of the three structures. Again, experiments show that all three $\mathrm{C}-\mathrm{O}$ bonds are exactly the same.


410. LINK TO LEARNING -
Use this online quiz (http://openstax.org/l/16LewisMake) to practice your skills in drawing resonance structures and estimating formal charges.


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

- Predict the structures of small molecules using valence shell electron pair repulsion (VSEPR) theory
- Explain the concepts of polar covalent bonds and molecular polarity
- Assess the polarity of a molecule based on its bonding and structure

Thus far, we have used two-dimensional Lewis structures to represent molecules. However, molecular structure is actually three-dimensional, and it is important to be able to describe molecular bonds in terms of their distances, angles, and relative arrangements in space (Figure 4.14). A bond angle is the angle between any two bonds that include a common atom, usually measured in degrees. A bond distance (or bond length) is the distance between the nuclei of two bonded atoms along the straight line joining the nuclei. Bond distances are measured in Ångstroms ( $1 \AA=10^{-10} \mathrm{~m}$ ) or picometers ( $1 \mathrm{pm}=10^{-12} \mathrm{~m}, 100 \mathrm{pm}=1 \AA$ ).

FIGURE 4.14 Bond distances (lengths) and angles are shown for the formaldehyde molecule, $\mathrm{H}_{2} \mathrm{CO}$.


412. VSEPR Theory -
Valence shell electron-pair repulsion theory (VSEPR theory) enables us to predict the molecular structure, including approximate bond angles around a central atom, of a molecule from an examination of the number of bonds and lone electron pairs in its Lewis structure. The VSEPR model assumes that electron pairs in the valence shell of a central atom will adopt an arrangement that minimizes repulsions between these electron pairs by maximizing the distance between them. The electrons in the valence shell of a central atom form
either bonding pairs of electrons, located primarily between bonded atoms, or lone pairs. The electrostatic repulsion of these electrons is reduced when the various regions of high electron density assume positions as far from each other as possible.

VSEPR theory predicts the arrangement of electron pairs around each central atom and, usually, the correct arrangement of atoms in a molecule. We should understand, however, that the theory only considers electronpair repulsions. Other interactions, such as nuclear-nuclear repulsions and nuclear-electron attractions, are also involved in the final arrangement that atoms adopt in a particular molecular structure.

As a simple example of VSEPR theory, let us predict the structure of a gaseous $\mathrm{BeF}_{2}$ molecule. The Lewis structure of $\mathrm{BeF}_{2}$ (Figure 4.15) shows only two electron pairs around the central beryllium atom. With two bonds and no lone pairs of electrons on the central atom, the bonds are as far apart as possible, and the electrostatic repulsion between these regions of high electron density is reduced to a minimum when they are on opposite sides of the central atom. The bond angle is $180^{\circ}$ (Figure 4.15).

$$
\because \ddot{\mathrm{F}} \mathrm{Be}_{\mathrm{Be}}^{180^{\circ}}
$$

FIGURE 4.15 The $\mathrm{BeF}_{2}$ molecule adopts a linear structure in which the two bonds are as far apart as possible, on opposite sides of the Be atom.