Chapter 8 Ionic versus Covalent Bonding

Calculating Lattice Energies

The lattice energy of nearly any ionic solid can be calculated rather accurately using a modified form of Equation 8.1:

U, which is always a positive number, represents the amount of energy required to dissociate 1 mol of an ionic solid into the gaseous ions. If we assume that ΔV = 0, then the lattice energy, U, is approximately equal to the change in enthalpy, ΔH (see Chapter 5 "Energy Changes in Chemical Reactions", Section 5.2 "Enthalpy"):

As before, Q₁ and Q₂ are the charges on the ions and r₀ is the internuclear distance. We see from Equation 8.4 that lattice energy is directly related to the product of the ion charges and inversely related to the internuclear distance. The value of the constant k′ depends on the specific arrangement of ions in the solid lattice and their valence electron configurations, topics that will be discussed in more detail in Chapter 12 "Solids". Representative values for calculated lattice energies, which range from about 600 to 10,000 kJ/mol, are listed in Table 8.1 "Representative Calculated Lattice Energies". Energies of this magnitude can be decisive in determining the chemistry of the elements.

Because the lattice energy depends on the product of the charges of the ions, a salt having a metal cation with a +2 charge (M²⁺) and a nonmetal anion with a −2 charge (X²⁻) will have a lattice energy four times greater than one with M⁺ and X⁻, assuming the ions are of comparable size (and have similar internuclear distances). For example, the calculated value of U for NaF is 910 kJ/mol, whereas U for MgO (containing Mg²⁺ and O²⁻ ions) is 3795 kJ/mol.

Because lattice energy is inversely related to the internuclear distance, it is also inversely proportional to the size of the ions. This effect is illustrated in Figure 8.2 "A Plot of Lattice Energy versus the Identity of the Halide for the Lithium, Sodium, and Potassium Halides", which shows that lattice energy decreases for the series LiX, NaX, and KX as the radius of X⁻ increases. Because r₀ in Equation 8.4 is the sum of the ionic radii of the cation and the anion (r₀ = r⁺ + r⁻), r₀ increases as the cation becomes larger in the series, so the magnitude of U decreases. A similar effect is seen when the anion becomes larger in a series of compounds with the same cation.

Figure 8.2 A Plot of Lattice Energy versus the Identity of the Halide for the Lithium, Sodium, and Potassium Halides

Because the ionic radii of the cations decrease in the order K⁺ > Na⁺ > Li⁺ for a given halide ion, the lattice energy decreases smoothly from Li⁺ to K⁺. Conversely, for a given alkali metal ion, the fluoride salt always has the highest lattice energy and the iodide salt the lowest.

The Relationship between Lattice Energies and Physical Properties

The magnitude of the forces that hold an ionic substance together has a dramatic effect on many of its properties. The melting pointThe temperature at which the individual ions in a lattice or the individual molecules in a covalent compound have enough kinetic energy to overcome the attractive forces that hold them together in the solid., for example, is the temperature at which the individual ions have enough kinetic energy to overcome the attractive forces that hold them in place. At the melting point, the ions can move freely, and the substance becomes a liquid. Thus melting points vary with lattice energies for ionic substances that have similar structures. The melting points of the sodium halides (Figure 8.3 "A Plot of Melting Point versus the Identity of the Halide for the Sodium Halides"), for example, decrease smoothly from NaF to NaI, following the same trend as seen for their lattice energies (Figure 8.2 "A Plot of Lattice Energy versus the Identity of the Halide for the Lithium, Sodium, and Potassium Halides"). Similarly, the melting point of MgO is 2825°C, compared with 996°C for NaF, reflecting the higher lattice energies associated with higher charges on the ions. In fact, because of its high melting point, MgO is used as an electrical insulator in heating elements for electric stoves.

Figure 8.3 A Plot of Melting Point versus the Identity of the Halide for the Sodium Halides

The melting points follow the same trend as the magnitude of the lattice energies in Figure 8.2 "A Plot of Lattice Energy versus the Identity of the Halide for the Lithium, Sodium, and Potassium Halides".

The hardnessThe resistance of ionic materials to scratching or abrasion. of ionic materials—that is, their resistance to scratching or abrasion—is also related to their lattice energies. Hardness is directly related to how tightly the ions are held together electrostatically, which, as we saw, is also reflected in the lattice energy. As an example, MgO is harder than NaF, which is consistent with its higher lattice energy.

In addition to determining melting point and hardness, lattice energies affect the solubilities of ionic substances in water. In general, the higher the lattice energy, the less soluble a compound is in water. For example, the solubility of NaF in water at 25°C is 4.13 g/100 mL, but under the same conditions, the solubility of MgO is only 0.65 mg/100 mL, meaning that it is essentially insoluble.

The Born–Haber Cycle

In principle, lattice energies could be measured by combining gaseous cations and anions to form an ionic solid and then measuring the heat evolved. Unfortunately, measurable quantities of gaseous ions have never been obtained under conditions where heat flow can be measured. Instead, lattice energies are found using the experimentally determined enthalpy changes for other chemical processes, Hess’s law, and a thermochemical cycle called the Born–Haber cycleA thermochemical cycle that decribes the process in which an ionic solid is conceptually formed from its component elements in a stepwise manner., similar to those introduced in Chapter 5 "Energy Changes in Chemical Reactions". Developed by Max Born and Fritz Haber in 1919, the Born–Haber cycle describes a process in which an ionic solid is conceptually formed from its component elements in a stepwise manner.

Let’s use the Born–Haber cycle to determine the lattice energy of CsF(s). CsF is a nearly ideal ionic compound because Cs is the least electronegative element that is not radioactive and F is the most electronegative element. To construct a thermochemical cycle for the formation of CsF, we need to know its enthalpy of formation, ΔH_f, which is defined by the following chemical reaction:

Because enthalpy is a state function, the overall ΔH for a series of reactions is the sum of the values of ΔH for the individual reactions. (For more information about state functions and Hess’s law, see Chapter 5 "Energy Changes in Chemical Reactions", Section 5.2 "Enthalpy".) We can therefore use a thermochemical cycle to determine the enthalpy change that accompanies the formation of solid CsF from the parent elements (not ions).

The Born–Haber cycle for calculating the lattice energy of cesium fluoride is shown in Figure 8.4 "The Born–Haber Cycle Illustrating the Enthalpy Changes Involved in the Formation of Solid Cesium Fluoride from Its Elements". This particular cycle consists of six reactions, Equation 8.6 plus the following five reactions:

Figure 8.4 The Born–Haber Cycle Illustrating the Enthalpy Changes Involved in the Formation of Solid Cesium Fluoride from Its Elements

Reaction 1

This equation describes the sublimationThe conversion of a solid directly to a gas (without an intervening liquid phase). of elemental cesium, the conversion of the solid directly to a gas. The accompanying enthalpy change is called the enthalpy of sublimation(ΔH_sub)The enthalpy change that accompanies the conversion of a solid directly to a gas. (Table 8.2 "Selected Enthalpies of Sublimation at 298 K") and is always positive because energy is required to sublime a solid.

Reaction 2

This equation describes the ionization of cesium, so the enthalpy change is the first ionization energy of cesium. Recall from Chapter 7 "The Periodic Table and Periodic Trends" that energy is needed to ionize any neutral atom. Hence, regardless of the compound, the enthalpy change for this portion of the Born–Haber cycle is always positive.

Reaction 3

This equation describes the dissociation of fluorine molecules into fluorine atoms, where D is the energy required for dissociation to occur (Table 8.3 "Selected Bond Dissociation Enthalpies at 298 K"). We need to dissociate only $\frac{1}{2}$ mol of F₂(g) molecules to obtain 1 mol of F(g) atoms. The ΔH for this reaction, too, is always positive because energy is required to dissociate any stable diatomic molecule into the component atoms.

Reaction 4

This equation describes the formation of a gaseous fluoride ion from a fluorine atom; the enthalpy change is the electron affinity of fluorine. Recall from Chapter 7 "The Periodic Table and Periodic Trends" that electron affinities can be positive, negative, or zero. In this case, ΔH is negative because of the highly negative electron affinity of fluorine.

Reaction 5

This equation describes the formation of the ionic solid from the gaseous ions. Because Reaction 5 is the reverse of the equation used to define lattice energy and U is defined to be a positive number, ΔH₅ is always negative, as it should be in a step that forms bonds.

If the enthalpy of formation of CsF from the elements is known (ΔH_f = −553.5 kJ/mol at 298 K), then the thermochemical cycle shown in Figure 8.4 "The Born–Haber Cycle Illustrating the Enthalpy Changes Involved in the Formation of Solid Cesium Fluoride from Its Elements" has only one unknown, the quantity ΔH₅ = −U. From Hess’s law, we can write

We can rearrange Equation 8.7 to give

Substituting for the individual ΔHs, we obtain

Substituting the appropriate values into this equation gives

U is larger in magnitude than any of the other quantities in Equation 8.9. The process we have used to arrive at this value is summarized in Table 8.4 "Summary of Reactions in the Born–Haber Cycle for the Formation of CsF(s)".

Predicting the Stability of Ionic Compounds

Equation 8.7 may be used as a tool for predicting which ionic compounds are likely to form from particular elements. As we have noted, ΔH₁ (ΔH_sub), ΔH₂ (I), and ΔH₃ (D) are always positive numbers, and ΔH₂ can be quite large. In contrast, ΔH₄ (EA) is comparatively small and can be positive, negative, or zero. Thus the first three terms in Equation 8.7 make the formation of an ionic substance energetically unfavorable, and the fourth term contributes little either way. The formation of an ionic compound will be exothermic (ΔH_f < 0) if and only if ΔH₅ (−U) is a large negative number. This means that lattice energy is the most important factor in determining the stability of an ionic compound. Another example is the formation of BaO:

whose Born–Haber cycle is compared with that for the formation of CsF in Figure 8.5 "Comparison of the Enthalpy Changes Involved in the Formation of Solid CsF and BaO from Their Elements".

Figure 8.5 Comparison of the Enthalpy Changes Involved in the Formation of Solid CsF and BaO from Their Elements

The lattice energy of BaO, with a dipositive cation and a dinegative anion, dominates the Born–Haber cycle.

Reaction 1

More than twice as much energy is required to sublime barium metal (180.0 kJ/mol) as is required to sublime cesium (76.5 kJ/mol).

Reaction 2

Nearly four times the energy is needed to form Ba²⁺ ions (I₁ = 502.9 kJ/mol, I₂ = 965.2 kJ/mol, I₁ + I₂ = 1468.1 kJ/mol) as Cs⁺ ions (I₁ = 375.7 kJ/mol).

Reaction 3

Because the bond energy of O₂(g) is 498.4 kJ/mol compared with 158.8 kJ/mol for F₂(g), more than three times the energy is needed to form oxygen atoms from O₂ molecules as is required to form fluorine atoms from F₂.

Reaction 4

Forming gaseous oxide (O²⁻) ions is energetically unfavorable. Even though adding one electron to an oxygen atom is exothermic (EA₁ = −141 kJ/mol), adding a second electron to an O⁻(g) ion is energetically unfavorable (EA₂ = +744 kJ/mol)—so much so that the overall cost of forming O²⁻(g) from O(g) is energetically prohibitive (EA₁ + EA₂ = +603 kJ/mol).

If the first four terms in the Born–Haber cycle are all substantially more positive for BaO than for CsF, why does BaO even form? The answer is the formation of the ionic solid from the gaseous ions (Reaction 5):

Reaction 5

Remember from Equation 8.4 that lattice energies are directly proportional to the product of the charges on the ions and inversely proportional to the internuclear distance. Although the internuclear distances are not significantly different for BaO and CsF (275 and 300 pm, respectively), the larger ionic charges in BaO produce a much higher lattice energy. Substituting values for BaO (ΔH_f = −548.0 kJ/mol) into the equation and solving for U gives

Thus U for BaO is slightly more than four times greater than U for CsF. The extra energy released when BaO forms from its ions more than compensates for the additional energy required to form Ba²⁺(g) and O²⁻(g) ions from Ba(s) and $\frac{1}{2}{\text{O}}_2(\text{g})\text{.}$

If the formation of ionic lattices containing multiply charged ions is so energetically favorable, why does CsF contain Cs⁺ and F⁻ ions rather than Cs²⁺ and F²⁻ ions? If we assume that U for a Cs²⁺F²⁻ salt would be approximately the same as U for BaO, the formation of a lattice containing Cs²⁺ and F²⁻ ions would release 2291 kJ/mol (3048 kJ/mol − 756.9 kJ/mol) more energy than one containing Cs⁺ and F⁻ ions. To form the Cs²⁺ ion from Cs⁺, however, would require removing a 5p electron from a filled inner shell, which calls for a great deal of energy: I₂ = 2234.4 kJ/mol for Cs. Furthermore, forming an F²⁻ ion is expected to be even more energetically unfavorable than forming an O²⁻ ion. Not only is an electron being added to an already negatively charged ion, but because the F⁻ ion has a filled 2p subshell, the added electron would have to occupy an empty high-energy 3s orbital. Cesium fluoride, therefore, is not Cs²⁺F²⁻ because the energy cost of forming the doubly charged ions would be greater than the additional lattice energy that would be gained.

Creating a Lewis Dot Symbol

To write an element’s Lewis dot symbol, we place dots representing its valence electrons, one at a time, around the element’s chemical symbol. Up to four dots are placed above, below, to the left, and to the right of the symbol (in any order, as long as elements with four or fewer valence electrons have no more than one dot in each position). The next dots, for elements with more than four valence electrons, are again distributed one at a time, each paired with one of the first four. Fluorine, for example, with the electron configuration [He]2s²2p⁵, has seven valence electrons, so its Lewis dot symbol is constructed as follows:

The number of dots in the Lewis dot symbol is the same as the number of valence electrons, which is the same as the last digit of the element’s group number in the periodic table. Lewis dot symbols for the elements in period 2 are given in Figure 8.7 "Lewis Dot Symbols for the Elements in Period 2".

Lewis used the unpaired dots to predict the number of bonds that an element will form in a compound. Consider the symbol for nitrogen in Figure 8.7 "Lewis Dot Symbols for the Elements in Period 2". The Lewis dot symbol explains why nitrogen, with three unpaired valence electrons, tends to form compounds in which it shares the unpaired electrons to form three bonds. Boron, which also has three unpaired valence electrons in its Lewis dot symbol, also tends to form compounds with three bonds, whereas carbon, with four unpaired valence electrons in its Lewis dot symbol, tends to share all of its unpaired valence electrons by forming compounds in which it has four bonds.

Figure 8.7 Lewis Dot Symbols for the Elements in Period 2

The Octet Rule

Lewis’s major contribution to bonding theory was to recognize that atoms tend to lose, gain, or share electrons to reach a total of eight valence electrons, called an octet. This so-called octet ruleThe tendency for atoms to lose, gain, or share electrons to reach a total of eight valence electrons. explains the stoichiometry of most compounds in the s and p blocks of the periodic table. We now know from quantum mechanics that the number eight corresponds to one ns and three np valence orbitals, which together can accommodate a total of eight electrons. Remarkably, though, Lewis’s insight was made nearly a decade before Rutherford proposed the nuclear model of the atom. An exception to the octet rule is helium, whose 1s² electron configuration gives it a full n = 1 shell, and hydrogen, which tends to gain or share its one electron to achieve the electron configuration of helium.

Lewis dot symbols can also be used to represent the ions in ionic compounds. The reaction of cesium with fluorine, for example, to produce the ionic compound CsF can be written as follows:

No dots are shown on Cs⁺ in the product because cesium has lost its single valence electron to fluorine. The transfer of this electron produces the Cs⁺ ion, which has the valence electron configuration of Xe, and the F⁻ ion, which has a total of eight valence electrons (an octet) and the Ne electron configuration. This description is consistent with the statement in Chapter 7 "The Periodic Table and Periodic Trends" that among the main group elements, ions in simple binary ionic compounds generally have the electron configurations of the nearest noble gas. The charge of each ion is written in the product, and the anion and its electrons are enclosed in brackets. This notation emphasizes that the ions are associated electrostatically; no electrons are shared between the two elements.

As you might expect for such a qualitative approach to bonding, there are exceptions to the octet rule, which we describe in Section 8.6 "Exceptions to the Octet Rule". These include molecules in which one or more atoms contain fewer or more than eight electrons. In Section 8.5 "Lewis Structures and Covalent Bonding", however, we explain how to form molecular compounds by completing octets.

Using Lewis Dot Symbols to Describe Covalent Bonding

The valence electron configurations of the constituent atoms of a covalent compound are important factors in determining its structure, stoichiometry, and properties. For example, chlorine, with seven valence electrons, is one electron short of an octet. If two chlorine atoms share their unpaired electrons by making a covalent bond and forming Cl₂, they can each complete their valence shell:

Each chlorine atom now has an octet. The electron pair being shared by the atoms is called a bonding pairA pair of electrons in a Lewis structure that is shared by two atoms, thus forming a covalent bond.; the other three pairs of electrons on each chlorine atom are called lone pairsA pair of electrons in a Lewis structure that is not involved in covalent bonding.. Lone pairs are not involved in covalent bonding. If both electrons in a covalent bond come from the same atom, the bond is called a coordinate covalent bondA covalent bond in which both electrons come from the same atom.. Examples of this type of bonding are presented in Section 8.6 "Exceptions to the Octet Rule" when we discuss atoms with less than an octet of electrons.

We can illustrate the formation of a water molecule from two hydrogen atoms and an oxygen atom using Lewis dot symbols:

The structure on the right is the Lewis electron structure, or Lewis structure, for H₂O. With two bonding pairs and two lone pairs, the oxygen atom has now completed its octet. Moreover, by sharing a bonding pair with oxygen, each hydrogen atom now has a full valence shell of two electrons. Chemists usually indicate a bonding pair by a single line, as shown here for our two examples:

The following procedure can be used to construct Lewis electron structures for more complex molecules and ions:

1. Arrange the atoms to show specific connections. When there is a central atom, it is usually the least electronegative element in the compound. Chemists usually list this central atom first in the chemical formula (as in CCl₄ and CO₃²⁻, which both have C as the central atom), which is another clue to the compound’s structure. Hydrogen and the halogens are almost always connected to only one other atom, so they are usually terminal rather than central.

2. Determine the total number of valence electrons in the molecule or ion. Add together the valence electrons from each atom. (Recall from Chapter 6 "The Structure of Atoms" that the number of valence electrons is indicated by the position of the element in the periodic table.) If the species is a polyatomic ion, remember to add or subtract the number of electrons necessary to give the total charge on the ion. For CO₃²⁻, for example, we add two electrons to the total because of the −2 charge.

3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. In H₂O, for example, there is a bonding pair of electrons between oxygen and each hydrogen.

4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). These electrons will usually be lone pairs.

5. If any electrons are left over, place them on the central atom. We explain in Section 8.6 "Exceptions to the Octet Rule" that some atoms are able to accommodate more than eight electrons.

6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. This will not change the number of electrons on the terminal atoms.

Now let’s apply this procedure to some particular compounds, beginning with one we have already discussed.

H₂O

1. Because H atoms are almost always terminal, the arrangement within the molecule must be HOH.

2. Each H atom (group 1) has 1 valence electron, and the O atom (group 16) has 6 valence electrons, for a total of 8 valence electrons.

3. Placing one bonding pair of electrons between the O atom and each H atom gives H:O:H, with 4 electrons left over.

4. Each H atom has a full valence shell of 2 electrons.

5. Adding the remaining 4 electrons to the oxygen (as two lone pairs) gives the following structure:

This is the Lewis structure we drew earlier. Because it gives oxygen an octet and each hydrogen two electrons, we do not need to use step 6.

OCl⁻

1. With only two atoms in the molecule, there is no central atom.

2. Oxygen (group 16) has 6 valence electrons, and chlorine (group 17) has 7 valence electrons; we must add one more for the negative charge on the ion, giving a total of 14 valence electrons.

3. Placing a bonding pair of electrons between O and Cl gives O:Cl, with 12 electrons left over.

4. If we place six electrons (as three lone pairs) on each atom, we obtain the following structure:

Each atom now has an octet of electrons, so steps 5 and 6 are not needed. The Lewis electron structure is drawn within brackets as is customary for an ion, with the overall charge indicated outside the brackets, and the bonding pair of electrons is indicated by a solid line. OCl⁻ is the hypochlorite ion, the active ingredient in chlorine laundry bleach and swimming pool disinfectant.

CH₂O

1. Because carbon is less electronegative than oxygen and hydrogen is normally terminal, C must be the central atom. One possible arrangement is as follows:

2. Each hydrogen atom (group 1) has one valence electron, carbon (group 14) has 4 valence electrons, and oxygen (group 16) has 6 valence electrons, for a total of [(2)(1) + 4 + 6] = 12 valence electrons.

3. Placing a bonding pair of electrons between each pair of bonded atoms gives the following:

Six electrons are used, and 6 are left over.

4. Adding all 6 remaining electrons to oxygen (as three lone pairs) gives the following:

Although oxygen now has an octet and each hydrogen has 2 electrons, carbon has only 6 electrons.

5. There are no electrons left to place on the central atom.

6. To give carbon an octet of electrons, we use one of the lone pairs of electrons on oxygen to form a carbon–oxygen double bond:

Both the oxygen and the carbon now have an octet of electrons, so this is an acceptable Lewis electron structure. The O has two bonding pairs and two lone pairs, and C has four bonding pairs. This is the structure of formaldehyde, which is used in embalming fluid.

An alternative structure can be drawn with one H bonded to O. Formal charges, discussed later in this section, suggest that such a structure is less stable than that shown previously.

Example 4

Write the Lewis electron structure for each species.

1. NCl₃
2. S₂²⁻
3. NOCl

Given: chemical species

Asked for: Lewis electron structures

Strategy:

Use the six-step procedure to write the Lewis electron structure for each species.

Solution:

1. Nitrogen is less electronegative than chlorine, and halogen atoms are usually terminal, so nitrogen is the central atom. The nitrogen atom (group 15) has 5 valence electrons and each chlorine atom (group 17) has 7 valence electrons, for a total of 26 valence electrons. Using 2 electrons for each N–Cl bond and adding three lone pairs to each Cl account for (3 × 2) + (3 × 2 × 3) = 24 electrons. Rule 5 leads us to place the remaining 2 electrons on the central N:

   

   Nitrogen trichloride is an unstable oily liquid once used to bleach flour; this use is now prohibited in the United States.

   

2. In a diatomic molecule or ion, we do not need to worry about a central atom. Each sulfur atom (group 16) contains 6 valence electrons, and we need to add 2 electrons for the −2 charge, giving a total of 14 valence electrons. Using 2 electrons for the S–S bond, we arrange the remaining 12 electrons as three lone pairs on each sulfur, giving each S atom an octet of electrons:

   

3. Because nitrogen is less electronegative than oxygen or chlorine, it is the central atom. The N atom (group 15) has 5 valence electrons, the O atom (group 16) has 6 valence electrons, and the Cl atom (group 17) has 7 valence electrons, giving a total of 18 valence electrons. Placing one bonding pair of electrons between each pair of bonded atoms uses 4 electrons and gives the following:

   

   Adding three lone pairs each to oxygen and to chlorine uses 12 more electrons, leaving 2 electrons to place as a lone pair on nitrogen:

   

   Because this Lewis structure has only 6 electrons around the central nitrogen, a lone pair of electrons on a terminal atom must be used to form a bonding pair. We could use a lone pair on either O or Cl. Because we have seen many structures in which O forms a double bond but none with a double bond to Cl, it is reasonable to select a lone pair from O to give the following:

   

   All atoms now have octet configurations. This is the Lewis electron structure of nitrosyl chloride, a highly corrosive, reddish-orange gas.

   

Exercise

Write Lewis electron structures for CO₂ and SCl₂, a vile-smelling, unstable red liquid that is used in the manufacture of rubber.

Answer:

1. 
   
2. 
   

Using Lewis Electron Structures to Explain Stoichiometry

Lewis dot symbols provide a simple rationalization of why elements form compounds with the observed stoichiometries. In the Lewis model, the number of bonds formed by an element in a neutral compound is the same as the number of unpaired electrons it must share with other atoms to complete its octet of electrons. For the elements of group 17 (the halogens), this number is one; for the elements of group 16 (the chalcogens), it is two; for group 15, three; and for group 14, four. These requirements are illustrated by the following Lewis structures for the hydrides of the lightest members of each group:

Elements may form multiple bonds to complete an octet. In ethylene, for example, each carbon contributes two electrons to the double bond, giving each carbon an octet (two electrons/bond × four bonds = eight electrons). Neutral structures with fewer or more bonds exist, but they are unusual and violate the octet rule.

Allotropes of an element can have very different physical and chemical properties because of different three-dimensional arrangements of the atoms; the number of bonds formed by the component atoms, however, is always the same. As noted at the beginning of the chapter, diamond is a hard, transparent solid; graphite is a soft, black solid; and the fullerenes have open cage structures. Despite these differences, the carbon atoms in all three allotropes form four bonds, in accordance with the octet rule. Elemental phosphorus also exists in three forms: white phosphorus, a toxic, waxy substance that initially glows and then spontaneously ignites on contact with air; red phosphorus, an amorphous substance that is used commercially in safety matches, fireworks, and smoke bombs; and black phosphorus, an unreactive crystalline solid with a texture similar to graphite (Figure 8.10 "The Three Allotropes of Phosphorus: White, Red, and Black"). Nonetheless, the phosphorus atoms in all three forms obey the octet rule and form three bonds per phosphorus atom.

Figure 8.10 The Three Allotropes of Phosphorus: White, Red, and Black

All three forms contain only phosphorus atoms, but they differ in the arrangement and connectivity of their atoms. White phosphorus contains P₄ tetrahedra, red phosphorus is a network of linked P₈ and P₉ units, and black phosphorus forms sheets of six-membered rings. As a result, their physical and chemical properties differ dramatically.

Formal Charges

It is sometimes possible to write more than one Lewis structure for a substance that does not violate the octet rule, as we saw for CH₂O, but not every Lewis structure may be equally reasonable. In these situations, we can choose the most stable Lewis structure by considering the formal chargeThe difference between the number of valence electrons in a free atom and the number of electrons assigned to it in a particular Lewis electron structure. on the atoms, which is the difference between the number of valence electrons in the free atom and the number assigned to it in the Lewis electron structure. The formal charge is a way of computing the charge distribution within a Lewis structure; the sum of the formal charges on the atoms within a molecule or an ion must equal the overall charge on the molecule or ion. A formal charge does not represent a true charge on an atom in a covalent bond but is simply used to predict the most likely structure when a compound has more than one valid Lewis structure.

To calculate formal charges, we assign electrons in the molecule to individual atoms according to these rules:

• Nonbonding electrons are assigned to the atom on which they are located.
• Bonding electrons are divided equally between the bonded atoms.

For each atom, we then compute a formal charge:

To illustrate this method, let’s calculate the formal charge on the atoms in ammonia (NH₃) whose Lewis electron structure is as follows:

A neutral nitrogen atom has five valence electrons (it is in group 15). From its Lewis electron structure, the nitrogen atom in ammonia has one lone pair and shares three bonding pairs with hydrogen atoms, so nitrogen itself is assigned a total of five electrons [2 nonbonding e⁻ + (6 bonding e⁻/2)]. Substituting into Equation 8.11, we obtain

A neutral hydrogen atom has one valence electron. Each hydrogen atom in the molecule shares one pair of bonding electrons and is therefore assigned one electron [0 nonbonding e⁻ + (2 bonding e⁻/2)]. Using Equation 8.11 to calculate the formal charge on hydrogen, we obtain

The hydrogen atoms in ammonia have the same number of electrons as neutral hydrogen atoms, and so their formal charge is also zero. Adding together the formal charges should give us the overall charge on the molecule or ion. In this example, the nitrogen and each hydrogen has a formal charge of zero. When summed the overall charge is zero, which is consistent with the overall charge on the NH₃ molecule.

Typically, the structure with the most charges on the atoms closest to zero is the more stable Lewis structure. In cases where there are positive or negative formal charges on various atoms, stable structures generally have negative formal charges on the more electronegative atoms and positive formal charges on the less electronegative atoms. The next example further demonstrates how to calculate formal charges.

Example 5

Calculate the formal charges on each atom in the NH₄⁺ ion.

Given: chemical species

Asked for: formal charges

Strategy:

Identify the number of valence electrons in each atom in the NH₄⁺ ion. Use the Lewis electron structure of NH₄⁺ to identify the number of bonding and nonbonding electrons associated with each atom and then use Equation 8.11 to calculate the formal charge on each atom.

Solution:

The Lewis electron structure for the NH₄⁺ ion is as follows:

The nitrogen atom shares four bonding pairs of electrons, and a neutral nitrogen atom has five valence electrons. Using Equation 8.11, the formal charge on the nitrogen atom is therefore

$$\text{formal charge (N)}=5-\left(0+\frac{8}{2}\right)=\mathrm{+1}$$

Each hydrogen atom in has one bonding pair. The formal charge on each hydrogen atom is therefore

$$\text{formal charge (H)}=1-\left(0+\frac{2}{2}\right)=0$$

The formal charges on the atoms in the NH₄⁺ ion are thus

Adding together the formal charges on the atoms should give us the total charge on the molecule or ion. In this case, the sum of the formal charges is 0 + 1 + 0 + 0 + 0 = +1.

Exercise

Write the formal charges on all atoms in BH₄⁻.

Answer:

If an atom in a molecule or ion has the number of bonds that is typical for that atom (e.g., four bonds for carbon), its formal charge is zero.

Using Formal Charges to Distinguish between Lewis Structures

As an example of how formal charges can be used to determine the most stable Lewis structure for a substance, we can compare two possible structures for CO₂. Both structures conform to the rules for Lewis electron structures.

CO₂

1. C is less electronegative than O, so it is the central atom.

2. C has 4 valence electrons and each O has 6 valence electrons, for a total of 16 valence electrons.

3. Placing one electron pair between the C and each O gives O–C–O, with 12 electrons left over.

4. Dividing the remaining electrons between the O atoms gives three lone pairs on each atom:

This structure has an octet of electrons around each O atom but only 4 electrons around the C atom.

5. No electrons are left for the central atom.

6. To give the carbon atom an octet of electrons, we can convert two of the lone pairs on the oxygen atoms to bonding electron pairs. There are, however, two ways to do this. We can either take one electron pair from each oxygen to form a symmetrical structure or take both electron pairs from a single oxygen atom to give an asymmetrical structure:

Both Lewis electron structures give all three atoms an octet. How do we decide between these two possibilities? The formal charges for the two Lewis electron structures of CO₂ are as follows:

Both Lewis structures have a net formal charge of zero, but the structure on the right has a +1 charge on the more electronegative atom (O). Thus the symmetrical Lewis structure on the left is predicted to be more stable, and it is, in fact, the structure observed experimentally. Remember, though, that formal charges do not represent the actual charges on atoms in a molecule or ion. They are used simply as a bookkeeping method for predicting the most stable Lewis structure for a compound.

Note the Pattern

The Lewis structure with the set of formal charges closest to zero is usually the most stable.

Example 6

The thiocyanate ion (SCN⁻), which is used in printing and as a corrosion inhibitor against acidic gases, has at least two possible Lewis electron structures. Draw two possible structures, assign formal charges on all atoms in both, and decide which is the preferred arrangement of electrons.

Given: chemical species

Asked for: Lewis electron structures, formal charges, and preferred arrangement

Strategy:

A Use the step-by-step procedure to write two plausible Lewis electron structures for SCN⁻.

B Calculate the formal charge on each atom using Equation 8.11.

C Predict which structure is preferred based on the formal charge on each atom and its electronegativity relative to the other atoms present.

Solution:

A Possible Lewis structures for the SCN⁻ ion are as follows:

B We must calculate the formal charges on each atom to identify the more stable structure. If we begin with carbon, we notice that the carbon atom in each of these structures shares four bonding pairs, the number of bonds typical for carbon, so it has a formal charge of zero. Continuing with sulfur, we observe that in (a) the sulfur atom shares one bonding pair and has three lone pairs and has a total of six valence electrons. The formal charge on the sulfur atom is therefore $6-\left(6+\frac{2}{2}\right)=-1.$ In (b), the sulfur atom has two bonding pairs and two lone pairs, giving it a formal charge of zero. In (c), sulfur has a formal charge of +1. Completing our calculations with nitrogen, in (a) the nitrogen atom has three bonding pairs, giving it a formal charge of zero. In (b), the nitrogen atom has two lone pairs and shares two bonding pairs, giving it a formal charge of $5-\left(4+\frac{4}{2}\right)=-1.$ In (c), nitrogen has a formal charge of −2.

C Which structure is preferred? Structure (b) is preferred because the negative charge is on the more electronegative atom (N), and it has lower formal charges on each atom as compared to structure (c): 0, −1 versus +1, −2.

Exercise

Salts containing the fulminate ion (CNO⁻) are used in explosive detonators. Draw three Lewis electron structures for CNO⁻ and use formal charges to predict which is more stable. (Note: N is the central atom.)

Answer:

The second structure is predicted to be more stable.

Resonance Structures

Sometimes, even when formal charges are considered, the bonding in some molecules or ions cannot be described by a single Lewis structure. Such is the case for ozone (O₃), an allotrope of oxygen with a V-shaped structure and an O–O–O angle of 117.5°.

O₃

1. We know that ozone has a V-shaped structure, so one O atom is central:

2. Each O atom has 6 valence electrons, for a total of 18 valence electrons.

3. Assigning one bonding pair of electrons to each oxygen–oxygen bond gives

with 14 electrons left over.

4. If we place three lone pairs of electrons on each terminal oxygen, we obtain

and have 2 electrons left over.

5. At this point, both terminal oxygen atoms have octets of electrons. We therefore place the last 2 electrons on the central atom:

6. The central oxygen has only 6 electrons. We must convert one lone pair on a terminal oxygen atom to a bonding pair of electrons—but which one? Depending on which one we choose, we obtain either

Which is correct? In fact, neither is correct. Both predict one O–O single bond and one O=O double bond. As you will learn in Section 8.8 "Properties of Covalent Bonds", if the bonds were of different types (one single and one double, for example), they would have different lengths. It turns out, however, that both O–O bond distances are identical, 127.2 pm, which is shorter than a typical O–O single bond (148 pm) and longer than the O=O double bond in O₂ (120.7 pm).

Equivalent Lewis dot structures, such as those of ozone, are called resonance structuresA Lewis electron structure that has different arrangements of electrons around atoms whose positions do not change.. The position of the atoms is the same in the various resonance structures of a compound, but the position of the electrons is different. Double-headed arrows link the different resonance structures of a compound:

The double-headed arrow indicates that the actual electronic structure is an average of those shown, not that the molecule oscillates between the two structures.

Note the Pattern

When it is possible to write more than one equivalent resonance structure for a molecule or ion, the actual structure is the average of the resonance structures.

CO₃²⁻

Like ozone, the electronic structure of the carbonate ion cannot be described by a single Lewis electron structure. Unlike O₃, though, the actual structure of CO₃²⁻ is an average of three resonance structures.

1. Because carbon is the least electronegative element, we place it in the central position:

2. Carbon has 4 valence electrons, each oxygen has 6 valence electrons, and there are 2 more for the −2 charge. This gives 4 + (3 × 6) + 2 = 24 valence electrons.

3. Six electrons are used to form three bonding pairs between the oxygen atoms and the carbon:

4. We divide the remaining 18 electrons equally among the three oxygen atoms by placing three lone pairs on each and indicating the −2 charge:

5. No electrons are left for the central atom.

6. At this point, the carbon atom has only 6 valence electrons, so we must take one lone pair from an oxygen and use it to form a carbon–oxygen double bond. In this case, however, there are three possible choices:

As with ozone, none of these structures describes the bonding exactly. Each predicts one carbon–oxygen double bond and two carbon–oxygen single bonds, but experimentally all C–O bond lengths are identical. We can write resonance structures (in this case, three of them) for the carbonate ion:

The actual structure is an average of these three resonance structures.

Example 7

Benzene is a common organic solvent that was previously used in gasoline; it is no longer used for this purpose, however, because it is now known to be a carcinogen. The benzene molecule (C₆H₆) consists of a regular hexagon of carbon atoms, each of which is also bonded to a hydrogen atom. Use resonance structures to describe the bonding in benzene.

Given: molecular formula and molecular geometry

Asked for: resonance structures

Strategy:

A Draw a structure for benzene illustrating the bonded atoms. Then calculate the number of valence electrons used in this drawing.

B Subtract this number from the total number of valence electrons in benzene and then locate the remaining electrons such that each atom in the structure reaches an octet.

C Draw the resonance structures for benzene.

Solution:

A Each hydrogen atom contributes 1 valence electron, and each carbon atom contributes 4 valence electrons, for a total of (6 × 1) + (6 × 4) = 30 valence electrons. If we place a single bonding electron pair between each pair of carbon atoms and between each carbon and a hydrogen atom, we obtain the following:

Each carbon atom in this structure has only 6 electrons and has a formal charge of +1, but we have used only 24 of the 30 valence electrons.

B If the 6 remaining electrons are uniformly distributed pairwise on alternate carbon atoms, we obtain the following:

Three carbon atoms now have an octet configuration and a formal charge of −1, while three carbon atoms have only 6 electrons and a formal charge of +1. We can convert each lone pair to a bonding electron pair, which gives each atom an octet of electrons and a formal charge of 0, by making three C=C double bonds.

C There are, however, two ways to do this:

Each structure has alternating double and single bonds, but experimentation shows that each carbon–carbon bond in benzene is identical, with bond lengths (139.9 pm) intermediate between those typically found for a C–C single bond (154 pm) and a C=C double bond (134 pm). We can describe the bonding in benzene using the two resonance structures, but the actual electronic structure is an average of the two. The existence of multiple resonance structures for aromatic hydrocarbons like benzene is often indicated by drawing either a circle or dashed lines inside the hexagon:

Exercise

The sodium salt of nitrite is used to relieve muscle spasms. Draw two resonance structures for the nitrite ion (NO₂⁻).

Answer:

Resonance structures are particularly common in oxoanions of the p-block elements, such as sulfate and phosphate, and in aromatic hydrocarbons, such as benzene and naphthalene.

Summary

A plot of the overall energy of a covalent bond as a function of internuclear distance is identical to a plot of an ionic pair because both result from attractive and repulsive forces between charged entities. In Lewis electron structures, we encounter bonding pairs, which are shared by two atoms, and lone pairs, which are not shared between atoms. If both electrons in a covalent bond come from the same atom, the bond is called a coordinate covalent bond. Lewis structures are an attempt to rationalize why certain stoichiometries are commonly observed for the elements of particular families. Neutral compounds of group 14 elements typically contain four bonds around each atom (a double bond counts as two, a triple bond as three), whereas neutral compounds of group 15 elements typically contain three bonds. In cases where it is possible to write more than one Lewis electron structure with octets around all the nonhydrogen atoms of a compound, the formal charge on each atom in alternative structures must be considered to decide which of the valid structures can be excluded and which is the most reasonable. The formal charge is the difference between the number of valence electrons of the free atom and the number of electrons assigned to it in the compound, where bonding electrons are divided equally between the bonded atoms. The Lewis structure with the lowest formal charges on the atoms is almost always the most stable one. Some molecules have two or more chemically equivalent Lewis electron structures, called resonance structures. These structures are written with a double-headed arrow between them, indicating that none of the Lewis structures accurately describes the bonding but that the actual structure is an average of the individual resonance structures.

Key Takeaway

• Lewis dot symbols provide a simple rationalization of why elements form compounds with the observed stoichiometries.

Key Equation

Formal charge on an atom

Equation 8.11: $\text{formal charge}=\text{valence }{\text{e}}^{-}-\left(\text{nonbonding }{\text{e}}^{-}+\frac{\text{bonding }{\text{e}}^{-}}{2}\right)$

Conceptual Problems

1. Compare and contrast covalent and ionic compounds with regard to

   1. volatility.
   2. melting point.
   3. electrical conductivity.
   4. physical appearance.

2. What are the similarities between plots of the overall energy versus internuclear distance for an ionic compound and a covalent compound? Why are the plots so similar?

3. Which atom do you expect to be the central atom in each of the following species?

   1. SO₄²⁻
   2. NH₄⁺
   3. BCl₃
   4. SO₂Cl₂?

4. Which atom is the central atom in each of the following species?

   1. PCl₃
   2. CHCl₃
   3. SO₂
   4. IF₃?

5. What is the relationship between the number of bonds typically formed by the period 2 elements in groups 14, 15, and 16 and their Lewis electron structures?

6. Although formal charges do not represent actual charges on atoms in molecules or ions, they are still useful. Why?

7. Why are resonance structures important?

8. In what types of compounds are resonance structures particularly common?

Numerical Problems

1. Give the electron configuration and the Lewis dot symbol for the following. How many more electrons can each atom accommodate?

   1. Se
   2. Kr
   3. Li
   4. Sr
   5. H

2. Give the electron configuration and the Lewis dot symbol for the following. How many more electrons can each atom accommodate?

   1. Na
   2. Br
   3. Ne
   4. C
   5. Ga

3. Based on Lewis dot symbols, predict the preferred oxidation state of Be, F, B, and Cs.

4. Based on Lewis dot symbols, predict the preferred oxidation state of Br, Rb, O, Si, and Sr.

5. Based on Lewis dot symbols, predict how many bonds gallium, silicon, and selenium will form in their neutral compounds.

6. Determine the total number of valence electrons in the following.

   1. Cr
   2. Cu⁺
   3. NO⁺
   4. XeF₂
   5. Br₂
   6. CH₂Cl₂
   7. NO₃⁻
   8. H₃O⁺

7. Determine the total number of valence electrons in the following.

   1. Ag
   2. Pt²⁺
   3. H₂S
   4. OH⁻
   5. I₂
   6. CH₄
   7. SO₄²⁻
   8. NH₄⁺.

8. Draw Lewis electron structures for the following.

   1. F₂
   2. SO₂
   3. AlCl₄⁻
   4. SO₃²⁻
   5. BrCl
   6. XeF₄
   7. NO⁺
   8. PCl₃

9. Draw Lewis electron structures for the following.

   1. Br₂
   2. CH₃Br
   3. SO₄²⁻
   4. O₂
   5. S₂²⁻
   6. BF₃

10. Draw Lewis electron structures for CO₂, NO₂⁻, SO₂, and NO₂⁺. From your diagram, predict which pair(s) of compounds have similar electronic structures.

11. Write Lewis dot symbols for each pair of elements. For a reaction between each pair of elements, predict which element is the oxidant, which element is the reductant, and the final stoichiometry of the compound formed.

    1. K, S
    2. Sr, Br
    3. Al, O
    4. Mg, Cl

12. Write Lewis dot symbols for each pair of elements. For a reaction between each pair of elements, predict which element is the oxidant, which element is the reductant, and the final stoichiometry of the compound formed.

    1. Li, F
    2. Cs, Br
    3. Ca, Cl
    4. B, F

13. Use Lewis dot symbols to predict whether ICl and NO₄⁻ are chemically reasonable formulas.

14. Draw a plausible Lewis electron structure for a compound with the molecular formula Cl₃PO.

15. Draw a plausible Lewis electron structure for a compound with the molecular formula CH₄O.

16. While reviewing her notes, a student noticed that she had drawn the following structure in her notebook for acetic acid:

    

    Why is this structure not feasible? Draw an acceptable Lewis structure for acetic acid. Show the formal charges of all nonhydrogen atoms in both the correct and incorrect structures.

17. A student proposed the following Lewis structure shown for acetaldehyde.

    

    Why is this structure not feasible? Draw an acceptable Lewis structure for acetaldehyde. Show the formal charges of all nonhydrogen atoms in both the correct and incorrect structures.

18. Draw the most likely structure for HCN based on formal charges, showing the formal charge on each atom in your structure. Does this compound have any plausible resonance structures? If so, draw one.

19. Draw the most plausible Lewis structure for NO₃⁻. Does this ion have any other resonance structures? Draw at least one other Lewis structure for the nitrate ion that is not plausible based on formal charges.

20. At least two Lewis structures can be drawn for BCl₃. Using arguments based on formal charges, explain why the most likely structure is the one with three B–Cl single bonds.

21. Using arguments based on formal charges, explain why the most feasible Lewis structure for SO₄²⁻ has two sulfur–oxygen double bonds.

22. At least two distinct Lewis structures can be drawn for N₃⁻. Use arguments based on formal charges to explain why the most likely structure contains a nitrogen–nitrogen double bond.

23. Is H–O–N=O a reasonable structure for the compound HNO₂? Justify your answer using Lewis electron dot structures.

24. Is H–O=C–H a reasonable structure for a compound with the formula CH₂O? Use Lewis electron dot structures to justify your answer.

25. Explain why the following Lewis structure for SO₃²⁻ is or is not reasonable.

    

26. Draw all the resonance structures for each ion.

    1. HSO₄⁻
    2. HSO₃⁻

Answers

1. 1. [Ar]4s²3d¹⁰4p⁴

      

      Selenium can accommodate two more electrons, giving the Se²⁻ ion.

   2. [Ar]4s²3d¹⁰4p⁶

      

      Krypton has a closed shell electron configuration, so it cannot accommodate any additional electrons.

   3. 1s²2s¹

      

      Lithium can accommodate one additional electron in its 2s orbital, giving the Li⁻ ion.

   4. [Kr]5s²

      

      Strontium has a filled 5s subshell, and additional electrons would have to be placed in an orbital with a higher energy. Thus strontium has no tendency to accept an additional electron.

   5. 1s¹

      

      Hydrogen can accommodate one additional electron in its 1s orbital, giving the H⁻ ion.

3. Be²⁺, F⁻, B³⁺, Cs⁺

7. 1. 11
   2. 8
   3. 8
   4. 8
   5. 14
   6. 8
   7. 32
   8. 8

9.  

   1. 
   2. 
   3. 
   4. 
   5. 
   6. 

11.  

    1. 

       K is the reductant; S is the oxidant. The final stoichiometry is K₂S.

    2. 

       Sr is the reductant; Br is the oxidant. The final stoichiometry is SrBr₂.

    3. 

       Al is the reductant; O is the oxidant. The final stoichiometry is Al₂O₃.

    4. 

       Mg is the reductant; Cl is the oxidant. The final stoichiometry is MgCl₂.

15. The only structure that gives both oxygen and carbon an octet of electrons is the following:

    

17. The student’s proposed structure has two flaws: the hydrogen atom with the double bond has four valence electrons (H can only accommodate two electrons), and the carbon bound to oxygen only has six valence electrons (it should have an octet). An acceptable Lewis structure is

    

    The formal charges on the correct and incorrect structures are as follows:

    

19. The most plausible Lewis structure for NO₃⁻ is:

    

    There are three equivalent resonance structures for nitrate (only one is shown), in which nitrogen is doubly bonded to one of the three oxygens. In each resonance structure, the formal charge of N is +1; for each singly bonded O, it is −1; and for the doubly bonded oxygen, it is 0.

    The following is an example of a Lewis structure that is not plausible:

    

    This structure nitrogen has six bonds (nitrogen can form only four bonds) and a formal charge of –1.

21. With four S–O single bonds, each oxygen in SO₄²⁻ has a formal charge of −1, and the central sulfur has a formal charge of +2. With two S=O double bonds, only two oxygens have a formal charge of –1, and sulfur has a formal charge of zero. Lewis structures that minimize formal charges tend to be lowest in energy, making the Lewis structure with two S=O double bonds the most probable.

23. Yes. This is a reasonable Lewis structure, because the formal charge on all atoms is zero, and each atom (except H) has an octet of electrons.

    

Odd Number of Electrons

Because most molecules or ions that consist of s- and p-block elements contain even numbers of electrons, their bonding can be described using a model that assigns every electron to either a bonding pair or a lone pair.Molecules or ions containing d-block elements frequently contain an odd number of electrons, and their bonding cannot adequately be described using the simple approach we have developed so far. Bonding in these compounds will be discussed in Chapter 23 "The ". There are, however, a few molecules containing only p-block elements that have an odd number of electrons. Some important examples are nitric oxide (NO), whose biochemical importance was described in earlier chapters; nitrogen dioxide (NO₂), an oxidizing agent in rocket propulsion; and chlorine dioxide (ClO₂), which is used in water purification plants. Consider NO, for example. With 5 + 6 = 11 valence electrons, there is no way to draw a Lewis structure that gives each atom an octet of electrons. Molecules such as NO, NO₂, and ClO₂ require a more sophisticated treatment of bonding, which will be developed in Chapter 9 "Molecular Geometry and Covalent Bonding Models".

More Than an Octet of Electrons

The most common exception to the octet rule is a molecule or an ion with at least one atom that possesses more than an octet of electrons. Such compounds are found for elements of period 3 and beyond. Examples from the p-block elements include SF₆, a substance used by the electric power industry to insulate high-voltage lines, and the SO₄²⁻ and PO₄³⁻ ions.

Let’s look at sulfur hexafluoride (SF₆), whose Lewis structure must accommodate a total of 48 valence electrons [6 + (6 × 7) = 48]. If we arrange the atoms and electrons symmetrically, we obtain a structure with six bonds to sulfur; that is, it is six-coordinate. Each fluorine atom has an octet, but the sulfur atom has 12 electrons surrounding it rather than 8.The third step in our procedure for writing Lewis electron structures, in which we place an electron pair between each pair of bonded atoms, requires that an atom have more than 8 electrons whenever it is bonded to more than 4 other atoms.

The octet rule is based on the fact that each valence orbital (typically, one ns and three np orbitals) can accommodate only two electrons. To accommodate more than eight electrons, sulfur must be using not only the ns and np valence orbitals but additional orbitals as well. Sulfur has an [Ne]3s²3p⁴3d⁰ electron configuration, so in principle it could accommodate more than eight valence electrons by using one or more d orbitals. Thus species such as SF₆ are often called expanded-valence moleculesA compound with more than an octet of electrons around an atom.. Whether or not such compounds really do use d orbitals in bonding is controversial, but this model explains why compounds exist with more than an octet of electrons around an atom.

There is no correlation between the stability of a molecule or an ion and whether or not it has an expanded valence shell. Some species with expanded valences, such as PF₅, are highly reactive, whereas others, such as SF₆, are very unreactive. In fact, SF₆ is so inert that it has many commercial applications. In addition to its use as an electrical insulator, it is used as the coolant in some nuclear power plants, and it is the pressurizing gas in “unpressurized” tennis balls.

An expanded valence shell is often written for oxoanions of the heavier p-block elements, such as sulfate (SO₄²⁻) and phosphate (PO₄³⁻). Sulfate, for example, has a total of 32 valence electrons [6 + (4 × 6) + 2]. If we use a single pair of electrons to connect the sulfur and each oxygen, we obtain the four-coordinate Lewis structure (a). We know that sulfur can accommodate more than eight electrons by using its empty valence d orbitals, just as in SF₆. An alternative structure (b) can be written with S=O double bonds, making the sulfur again six-coordinate. We can draw five other resonance structures equivalent to (b) that vary only in the arrangement of the single and double bonds. In fact, experimental data show that the S-to-O bonds in the SO₄²⁻ ion are intermediate in length between single and double bonds, as expected for a system whose resonance structures all contain two S–O single bonds and two S=O double bonds. When calculating the formal charges on structures (a) and (b), we see that the S atom in (a) has a formal charge of +2, whereas the S atom in (b) has a formal charge of 0. Thus by using an expanded octet, a +2 formal charge on S can be eliminated.

Less Than an Octet of Electrons

Molecules with atoms that possess less than an octet of electrons generally contain the lighter s- and p-block elements, especially beryllium, typically with just four electrons around the central atom, and boron, typically with six. One example, boron trichloride (BCl₃) is used to produce fibers for reinforcing high-tech tennis rackets and golf clubs. The compound has 24 valence electrons and the following Lewis structure:

The boron atom has only six valence electrons, while each chlorine atom has eight. A reasonable solution might be to use a lone pair from one of the chlorine atoms to form a B-to-Cl double bond:

This resonance structure, however, results in a formal charge of +1 on the doubly bonded Cl atom and −1 on the B atom. The high electronegativity of Cl makes this separation of charge unlikely and suggests that this is not the most important resonance structure for BCl₃. This conclusion is shown to be valid based on the three equivalent B–Cl bond lengths of 173 pm that have no double bond character. Electron-deficient compounds such as BCl₃ have a strong tendency to gain an additional pair of electrons by reacting with species with a lone pair of electrons.

Example 8

Draw Lewis dot structures for each compound.

1. BeCl₂ gas, a compound used to produce beryllium, which in turn is used to produce structural materials for missiles and communication satellites
2. SF₄, a compound that reacts violently with water

Include resonance structures where appropriate.

Given: two compounds

Asked for: Lewis electron structures

Strategy:

A Use the procedure given earlier to write a Lewis electron structure for each compound. If necessary, place any remaining valence electrons on the element most likely to be able to accommodate more than an octet.

B After all the valence electrons have been placed, decide whether you have drawn an acceptable Lewis structure.

Solution:

1. A Because it is the least electronegative element, Be is the central atom. The molecule has 16 valence electrons (2 from Be and 7 from each Cl). Drawing two Be–Cl bonds and placing three lone pairs on each Cl gives the following structure:

   

   B Although this arrangement gives beryllium only 4 electrons, it is an acceptable Lewis structure for BeCl₂. Beryllium is known to form compounds in which it is surrounded by less than an octet of electrons.

2. A Sulfur is the central atom because it is less electronegative than fluorine. The molecule has 34 valence electrons (6 from S and 7 from each F). The S–F bonds use 8 electrons, and another 24 are placed around the F atoms:

   

   The only place to put the remaining 2 electrons is on the sulfur, giving sulfur 10 valence electrons:

   

   B Sulfur can accommodate more than an octet, so this is an acceptable Lewis structure.

Exercise

Draw Lewis dot structures for XeF₄.

Answer:

Bond Order

When we draw Lewis structures, we place one, two, or three pairs of electrons between adjacent atoms. In the Lewis bonding model, the number of electron pairs that hold two atoms together is called the bond orderThe number of electron pairs that hold two atoms together.. For a single bond, such as the C–C bond in H₃C–CH₃, the bond order is one. For a double bond (such as H₂C=CH₂), the bond order is two. For a triple bond, such as HC≡CH, the bond order is three.

When analogous bonds in similar compounds are compared, bond length decreases as bond order increases. The bond length data in Table 8.5 "Bond Lengths and Bond Dissociation Energies for Bonds with Different Bond Orders in Selected Gas-Phase Molecules at 298 K", for example, show that the C–C distance in H₃C–CH₃ (153.5 pm) is longer than the distance in H₂C=CH₂ (133.9 pm), which in turn is longer than that in HC≡CH (120.3 pm). Additionally, as noted in Section 8.5 "Lewis Structures and Covalent Bonding", molecules or ions whose bonding must be described using resonance structures usually have bond distances that are intermediate between those of single and double bonds, as we demonstrated with the C–C distances in benzene. The relationship between bond length and bond order is not linear, however. A double bond is not half as long as a single bond, and the length of a C=C bond is not the average of the lengths of C≡C and C–C bonds. Nevertheless, as bond orders increase, bond lengths generally decrease.

The Relationship between Bond Order and Bond Energy

As shown in Table 8.5 "Bond Lengths and Bond Dissociation Energies for Bonds with Different Bond Orders in Selected Gas-Phase Molecules at 298 K", triple bonds between like atoms are shorter than double bonds, and because more energy is required to completely break all three bonds than to completely break two, a triple bond is also stronger than a double bond. Similarly, double bonds between like atoms are stronger and shorter than single bonds. Bonds of the same order between different atoms show a wide range of bond energies, however. Table 8.6 "Average Bond Energies (kJ/mol) for Commonly Encountered Bonds at 273 K" lists the average values for some commonly encountered bonds. Although the values shown vary widely, we can observe four trends:

1. Bonds between hydrogen and atoms in the same column of the periodic table decrease in strength as we go down the column. Thus an H–F bond is stronger than an H–I bond, H–C is stronger than H–Si, H–N is stronger than H–P, H–O is stronger than H–S, and so forth. The reason for this is that the region of space in which electrons are shared between two atoms becomes proportionally smaller as one of the atoms becomes larger (part (a) in Figure 8.11 "The Strength of Covalent Bonds Depends on the Overlap between the Valence Orbitals of the Bonded Atoms").

2. Bonds between like atoms usually become weaker as we go down a column (important exceptions are noted later). For example, the C–C single bond is stronger than the Si–Si single bond, which is stronger than the Ge–Ge bond, and so forth. As two bonded atoms become larger, the region between them occupied by bonding electrons becomes proportionally smaller, as illustrated in part (b) in Figure 8.11 "The Strength of Covalent Bonds Depends on the Overlap between the Valence Orbitals of the Bonded Atoms". Noteworthy exceptions are single bonds between the period 2 atoms of groups 15, 16, and 17 (i.e., N, O, F), which are unusually weak compared with single bonds between their larger congeners. It is likely that the N–N, O–O, and F–F single bonds are weaker than might be expected due to strong repulsive interactions between lone pairs of electrons on adjacent atoms. The trend in bond energies for the halogens is therefore

Similar effects are also seen for the O–O versus S–S and for N–N versus P–P single bonds.

3. Because elements in periods 3 and 4 rarely form multiple bonds with themselves, their multiple bond energies are not accurately known. Nonetheless, they are presumed to be significantly weaker than multiple bonds between lighter atoms of the same families. Compounds containing an Si=Si double bond, for example, have only recently been prepared, whereas compounds containing C=C double bonds are one of the best-studied and most important classes of organic compounds.

Figure 8.11 The Strength of Covalent Bonds Depends on the Overlap between the Valence Orbitals of the Bonded Atoms

The relative sizes of the region of space in which electrons are shared between (a) a hydrogen atom and lighter (smaller) vs. heavier (larger) atoms in the same periodic group; and (b) two lighter versus two heavier atoms in the same group. Although the absolute amount of shared space increases in both cases on going from a light to a heavy atom, the amount of space relative to the size of the bonded atom decreases; that is, the percentage of total orbital volume decreases with increasing size. Hence the strength of the bond decreases.

4. Multiple bonds between carbon, oxygen, or nitrogen and a period 3 element such as phosphorus or sulfur tend to be unusually strong. In fact, multiple bonds of this type dominate the chemistry of the period 3 elements of groups 15 and 16. Multiple bonds to phosphorus or sulfur occur as a result of d-orbital interactions, as we discussed for the SO₄²⁻ ion in Section 8.6 "Exceptions to the Octet Rule". In contrast, silicon in group 14 has little tendency to form discrete silicon–oxygen double bonds. Consequently, SiO₂ has a three-dimensional network structure in which each silicon atom forms four Si–O single bonds, which makes the physical and chemical properties of SiO₂ very different from those of CO₂.

The Relationship between Molecular Structure and Bond Energy

Bond energy is defined as the energy required to break a particular bond in a molecule in the gas phase. Its value depends on not only the identity of the bonded atoms but also their environment. Thus the bond energy of a C–H single bond is not the same in all organic compounds. For example, the energy required to break a C–H bond in methane varies by as much as 25% depending on how many other bonds in the molecule have already been broken (Table 8.7 "Energies for the Dissociation of Successive C–H Bonds in Methane"); that is, the C–H bond energy depends on its molecular environment. Except for diatomic molecules, the bond energies listed in Table 8.6 "Average Bond Energies (kJ/mol) for Commonly Encountered Bonds at 273 K" are average values for all bonds of a given type in a range of molecules. Even so, they are not likely to differ from the actual value of a given bond by more than about 10%.

We can estimate the enthalpy change for a chemical reaction by adding together the average energies of the bonds broken in the reactants and the average energies of the bonds formed in the products and then calculating the difference between the two. If the bonds formed in the products are stronger than those broken in the reactants, then energy will be released in the reaction (ΔH_rxn < 0):

The ≈ sign is used because we are adding together average bond energies; hence this approach does not give exact values for ΔH_rxn.

Let’s consider the reaction of 1 mol of n-heptane (C₇H₁₆) with oxygen gas to give carbon dioxide and water. This is one reaction that occurs during the combustion of gasoline:

In this reaction, 6 C–C bonds, 16 C–H bonds, and 11 O=O bonds are broken per mole of n-heptane, while 14 C=O bonds (two for each CO₂) and 16 O–H bonds (two for each H₂O) are formed. The energy changes can be tabulated as follows:

The bonds in the products are stronger than the bonds in the reactants by about 4444 kJ/mol. This means that ΔH_rxn is approximately −4444 kJ/mol, and the reaction is highly exothermic (which is not too surprising for a combustion reaction).

If we compare this approximation with the value obtained from measured ΔHfο values (ΔH_rxn = −4817 kJ/mol), we find a discrepancy of only about 8%, less than the 10% typically encountered. Chemists find this method useful for calculating approximate enthalpies of reaction for molecules whose actual ΔHfο values are unknown. These approximations can be important for predicting whether a reaction is exothermic or endothermic—and to what degree.

Example 10

The compound RDX (Research Development Explosive) is a more powerful explosive than dynamite and is used by the military. When detonated, it produces gaseous products and heat according to the following reaction. Use the approximate bond energies in Table 8.6 "Average Bond Energies (kJ/mol) for Commonly Encountered Bonds at 273 K" to estimate the ΔH_rxn per mole of RDX.

Given: chemical reaction, structure of reactant, and Table 8.6 "Average Bond Energies (kJ/mol) for Commonly Encountered Bonds at 273 K"

Asked for: ΔH_rxn per mole

Strategy:

A List the types of bonds broken in RDX, along with the bond energy required to break each type. Multiply the number of each type by the energy required to break one bond of that type and then add together the energies. Repeat this procedure for the bonds formed in the reaction.

B Use Equation 8.14 to calculate the amount of energy consumed or released in the reaction (ΔH_rxn).

Solution:

We must add together the energies of the bonds in the reactants and compare that quantity with the sum of the energies of the bonds in the products. A nitro group (–NO₂) can be viewed as having one N–O single bond and one N=O double bond, as follows:

In fact, however, both N–O distances are usually the same because of the presence of two equivalent resonance structures.

A We can organize our data by constructing a table:

Bonds Broken (kJ/mol)		Bonds Formed (kJ/mol)
6 C–N	305 × 6 = 1830	3 N≡N	942 × 3 = 2826
6 C–H	411 × 6 = 2466	6 C=O	799 × 6 = 4794
3 N–N	167 × 3 = 501	6 O–H	459 × 6 = 2754
3 N–O	201 × 3 = 603		Total = 10,374
3 N=O	607 × 3 = 1821
1.5 O=O	494 × 1.5 = 741
	Total = 7962

B From Equation 8.14, we have

Δ H rxn ≈ ∑ ( bond energies of bonds broken ) − ∑ ( bond energies of bonds formed ) = 7962 kJ/mol − 1 0,3 7 4 kJ/mol = −2 4 1 2 kJ/mol

Thus this reaction is also highly exothermic.

Exercise

The molecule HCFC-142b, a hydrochlorofluorocarbon used in place of chlorofluorocarbons (CFCs) such as the Freons, can be prepared by adding HCl to 1,1-difluoroethylene:

Use tabulated bond energies to calculate ΔH_rxn.

Answer: −54 kJ/mol

Bond Polarity

The polarity of a bond—the extent to which it is polar—is determined largely by the relative electronegativities of the bonded atoms. In Chapter 7 "The Periodic Table and Periodic Trends", electronegativity (χ) was defined as the ability of an atom in a molecule or an ion to attract electrons to itself. Thus there is a direct correlation between electronegativity and bond polarity. A bond is nonpolar if the bonded atoms have equal electronegativities. If the electronegativities of the bonded atoms are not equal, however, the bond is polarized toward the more electronegative atom. A bond in which the electronegativity of B (χ_B) is greater than the electronegativity of A (χ_A), for example, is indicated with the partial negative charge on the more electronegative atom:

One way of estimating the ionic character of a bond—that is, the magnitude of the charge separation in a polar covalent bond—is to calculate the difference in electronegativity between the two atoms: Δχ = χ_B − χ_A.

To predict the polarity of the bonds in Cl₂, HCl, and NaCl, for example, we look at the electronegativities of the relevant atoms: χ_Cl = 3.16, χ_H = 2.20, and χ_Na = 0.93 (see Figure 7.14 "A Plot of Periodic Variation of Electronegativity with Atomic Number for the First Six Rows of the Periodic Table"). Cl₂ must be nonpolar because the electronegativity difference (Δχ) is zero; hence the two chlorine atoms share the bonding electrons equally. In NaCl, Δχ is 2.23. This high value is typical of an ionic compound (Δχ ≥ ≈1.5) and means that the valence electron of sodium has been completely transferred to chlorine to form Na⁺ and Cl⁻ ions. In HCl, however, Δχ is only 0.96. The bonding electrons are more strongly attracted to the more electronegative chlorine atom, and so the charge distribution is

Remember that electronegativities are difficult to measure precisely and different definitions produce slightly different numbers. In practice, the polarity of a bond is usually estimated rather than calculated.

As with bond energies, the electronegativity of an atom depends to some extent on its chemical environment. It is therefore unlikely that the reported electronegativities of a chlorine atom in NaCl, Cl₂, ClF₅, and HClO₄ would be exactly the same.

Dipole Moments

The asymmetrical charge distribution in a polar substance such as HCl produces a dipole momentThe product of the partial charge $Q$ on the bonded atoms and the distance $r$ between the partial charges: $\mu =Qr$, where $Q$ is measured in coulombs (C) and $r$ in meters (m)., abbreviated by the Greek letter mu (µ). The dipole moment is defined as the product of the partial charge Q on the bonded atoms and the distance r between the partial charges:

where Q is measured in coulombs (C) and r in meters. The unit for dipole moments is the debye (D):

When a molecule with a dipole moment is placed in an electric field, it tends to orient itself with the electric field because of its asymmetrical charge distribution (Figure 8.13 "Molecules That Possess a Dipole Moment Partially Align Themselves with an Applied Electric Field").

Figure 8.13 Molecules That Possess a Dipole Moment Partially Align Themselves with an Applied Electric Field

In the absence of a field (a), the HCl molecules are randomly oriented. When an electric field is applied (b), the molecules tend to align themselves with the field, such that the positive end of the molecular dipole points toward the negative terminal and vice versa.

We can measure the partial charges on the atoms in a molecule such as HCl using Equation 8.16. If the bonding in HCl were purely ionic, an electron would be transferred from H to Cl, so there would be a full +1 charge on the H atom and a full −1 charge on the Cl atom. The dipole moment of HCl is 1.109 D, as determined by measuring the extent of its alignment in an electric field, and the reported gas-phase H–Cl distance is 127.5 pm. Hence the charge on each atom is

By dividing this calculated value by the charge on a single electron (1.6022 × 10⁻¹⁹ C), we find that the charge on the Cl atom of an HCl molecule is about −0.18, corresponding to about 0.18 e⁻:

To form a neutral compound, the charge on the H atom must be equal but opposite. Thus the measured dipole moment of HCl indicates that the H–Cl bond has approximately 18% ionic character (0.1811 × 100), or 82% covalent character. Instead of writing HCl as $\stackrel{{\delta }^{+}}{\text{H}}\text{–}\stackrel{{\delta }^{–}}{\text{Cl}}$ we can therefore indicate the charge separation quantitatively as

Our calculated results are in agreement with the electronegativity difference between hydrogen and chlorine χ_H = 2.20; χ_Cl = 3.16, χ_Cl − χ_H = 0.96), a value well within the range for polar covalent bonds. We indicate the dipole moment by writing an arrow above the molecule.Mathematically, dipole moments are vectors, and they possess both a magnitude and a direction. The dipole moment of a molecule is the vector sum of the dipoles of the individual bonds. In HCl, for example, the dipole moment is indicated as follows:

The arrow shows the direction of electron flow by pointing toward the more electronegative atom.

The charge on the atoms of many substances in the gas phase can be calculated using measured dipole moments and bond distances. Figure 8.14 "A Plot of the Percent Ionic Character of a Bond as Determined from Measured Dipole Moments versus the Difference in Electronegativity of the Bonded Atoms" shows a plot of the percent ionic character versus the difference in electronegativity of the bonded atoms for several substances. According to the graph, the bonding in species such as NaCl(g) and CsF(g) is substantially less than 100% ionic in character. As the gas condenses into a solid, however, dipole–dipole interactions between polarized species increase the charge separations. In the crystal, therefore, an electron is transferred from the metal to the nonmetal, and these substances behave like classic ionic compounds. The data in Figure 8.14 "A Plot of the Percent Ionic Character of a Bond as Determined from Measured Dipole Moments versus the Difference in Electronegativity of the Bonded Atoms" show that diatomic species with an electronegativity difference of less than 1.5 are less than 50% ionic in character, which is consistent with our earlier description of these species as containing polar covalent bonds. The use of dipole moments to determine the ionic character of a polar bond is illustrated in Example 11.

Figure 8.14 A Plot of the Percent Ionic Character of a Bond as Determined from Measured Dipole Moments versus the Difference in Electronegativity of the Bonded Atoms

In the gas phase, even CsF, which has the largest possible difference in electronegativity between atoms, is not 100% ionic. Solid CsF, however, is best viewed as 100% ionic because of the additional electrostatic interactions in the lattice.