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.

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.

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:

Reaction 1

$$\text{Cs(s)}\to \text{Cs(g)}\Delta H_{\text{1}}=\Delta H_{\text{sub}}=\text{76}\text{.5 kJ/mol}$$

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

$$\text{Cs(g)}\to {\text{Cs}}^{\text{+}}\text{(g)}+{\text{e}}^{\text{–}}\Delta H_{\text{2}}=I_1=\text{375}\text{.7 kJ/mol}$$

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

$$\frac{1}{2}{\text{F}}_2(\text{g})\to \text{F}(\text{g})\Delta H_3=\frac{1}{2}D=79.4\text{ kJ/mol}$$

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

$${\text{F(g) + e}}^{–}\to {\text{F}}^{–}\text{(g)}\Delta H_{\text{4}}\text{= }EA\text{ = –328}\text{.2 kJ/mol}$$

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

$${\text{Cs}}^{\text{+}}{\text{(g) + F}}^{–}\text{(g)}\to \text{CsF(s)}\Delta H_{\text{5}}=\text{–}U$$

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".

 

Reaction 1

$$\text{Ba(s)}\to \text{Ba(g)}\Delta H_{\text{1}}=\Delta H_{\text{sub}}=\text{180}\text{.0 kJ/mol}$$

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

$$\text{Ba(s)}\to {\text{Ba}}^{\text{2+}}\text{(g)}+2{\text{e}}^{\text{–}}\Delta H_{\text{2}}=I_1+I_2=\text{1468}\text{.1 kJ/mol}$$

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

$$\frac{1}{2}{\text{O}}_2(\text{g})\to \text{O}(\text{g})\Delta H_3=\frac{1}{2}D=249.2\text{ kJ/mol}$$

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

$$\text{O(g) }+\mathrm{ 2}{\text{e}}^{\text{–}}\to {\text{O}}^{\text{2–}}\text{(g)}\Delta H_{\text{4}}=E{A}_1+E{A}_2=\text{603 kJ/mol}$$

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

$${\text{Ba}}^{\text{2+}}\text{(g)}+{\text{O}}^{\text{2–}}\text{(g)}\to \text{BaO(s)}\Delta H_{\text{5}}=\text{–}U$$

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

$$\begin{array}{ll}\hfill U& =\Delta H_{\text{sub}}\text{(Ba)}+\left[I_1\text{(Ba)}+I_2\text{(Ba)}\right]+\frac{1}{2}D({\text{O}}_{\text{2}})+\left[E{A}_1\text{(O)}+E{A}_2\text{(O)}\right]-\Delta H_{\text{f}}\text{(BaO)}\hfill \\ \hfill & =180.0\text{ kJ/mol }+\text{ 1468}\text{.1 kJ/mol }+\text{ 249}\text{.2 kJ/mol}+603\text{ kJ/mol}-(\mathrm{-5}48.0\text{ kJ/mol)}\hfill \\ \hfill & =\text{3048 kJ/mol}\hfill \end{array}$$

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.