6.1: Lattice Energies

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Skills to Develop

• Explain the formation of cations, anions, and ionic compounds
• Describe the energetics of ionic bond formation and breakage
• Use the Born-Haber cycle to compute lattice energies for ionic compounds

A bond’s strength describes how strongly each atom is joined to another atom, and therefore how much energy is required to break the bond between the two atoms. In a previous section, you learned about the bond strength of covalent bonds. Now we will compare that to the strength of ionic bonds, which is related to the lattice energy of a compound. But first, let's explore how ionic bonds form.

The Formation of Ionic Compounds

Video $$\PageIndex{1}$$: Why do atoms bond?

Video $$\PageIndex{2}$$: And how do atoms bond?

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

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

Neutral atoms and their associated ions have very different physical and chemical properties. Sodium atoms form sodium metal, a soft, silvery-white metal that burns vigorously in air and reacts explosively with water. Chlorine atoms form chlorine gas, Cl2, a yellow-green gas that is extremely corrosive to most metals and very poisonous to animals and plants. The vigorous reaction between the elements sodium and chlorine forms the white, crystalline compound sodium chloride, common table salt, which contains sodium cations and chloride anions (Figure Figure $$\PageIndex{1}$$). The compound composed of these ions exhibits properties entirely different from the properties of the elements sodium and chlorine. Chlorine is poisonous, but sodium chloride is essential to life; sodium atoms react vigorously with water, but sodium chloride simply dissolves in water.

Figure $$\PageIndex{1}$$: (a) Sodium is a soft metal that must be stored in mineral oil to prevent reaction with air or water. (b) Chlorine is a pale yellow-green gas. (c) When combined, they form white crystals of sodium chloride (table salt). (credit a: modification of work by “Jurii”/Wikimedia Commons)

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

As all substances must be electrically neutral, the total number of positive charges on the cations of an ionic compound must equal the total number of negative charges on its anions. The formula of an ionic compound represents the simplest ratio of the numbers of ions necessary to give identical numbers of positive and negative charges. For example, the formula for aluminum oxide, Al2O3, indicates that this ionic compound contains two aluminum cations, Al3+, for every three oxide anions, O2− [thus, (2 × +3) + (3 × –2) = 0].

It is important to note, however, that the formula for an ionic compound does not represent the physical arrangement of its ions. It is incorrect to refer to a sodium chloride (NaCl) “molecule” because there is not a single ionic bond, per se, between any specific pair of sodium and chloride ions. The attractive forces between ions are isotropic—the same in all directions—meaning that any particular ion is equally attracted to all of the nearby ions of opposite charge. This results in the ions arranging themselves into a tightly bound, three-dimensional lattice structure. Sodium chloride, for example, consists of a regular arrangement of equal numbers of Na+ cations and Cl anions (Figure Figure $$\PageIndex{2}$$).

Figure $$\PageIndex{2}$$: The atoms in sodium chloride (common table salt) are arranged to (a) maximize opposite charges interacting. The smaller spheres represent sodium ions, the larger ones represent chloride ions. In the expanded view (b), the geometry can be seen more clearly. Note that each ion is “bonded” to all of the surrounding ions—six in this case.

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

$\ce{NaCl}(s)⟶\ce{Na+}(g)+\ce{Cl-}(g)\hspace{20px}ΔH=\mathrm{769\:kJ}$

Video $$\PageIndex{3}$$: A review of ionic bonds from last term.

Ionic Bond Strength and Lattice Energy

An ionic compound is stable because of the electrostatic attraction between its positive and negative ions. The lattice energy of a compound is a measure of the strength of this attraction. The lattice energy ($$ΔH_{lattice}$$) of an ionic compound is defined as the energy required to separate one mole of the solid into its component gaseous ions. For the ionic solid MX, the lattice energy is the enthalpy change of the process:

$MX_{(s)}⟶Mn^+_{(g)}+X^{n−}_{(g)} \;\;\;\;\; ΔH_{lattice} \label{EQ6}$

Note that we are using the convention where the ionic solid is separated into ions, so our lattice energies will be endothermic (positive values). Some texts use the equivalent but opposite convention, defining lattice energy as the energy released when separate ions combine to form a lattice and giving negative (exothermic) values. Thus, if you are looking up lattice energies in another reference, be certain to check which definition is being used. In both cases, a larger magnitude for lattice energy indicates a more stable ionic compound. For sodium chloride, ΔHlattice = 769 kJ. Thus, it requires 769 kJ to separate one mole of solid NaCl into gaseous Na+ and Cl ions. When one mole each of gaseous Na+ and Cl ions form solid NaCl, 769 kJ of heat is released.

The lattice energy $$ΔH_{lattice}$$ of an ionic crystal can be expressed by the following equation (derived from Coulomb’s law, governing the forces between electric charges):

$ΔH_{lattice}=\dfrac{C(Z^+)(Z^−)}{R_o} \label{EQ7}$

in which

• $$\ce{C}$$ is a constant that depends on the type of crystal structure;
• $$Z^+$$ and $$Z^–$$ are the charges on the ions; and
• $$R_o$$ is the interionic distance (the sum of the radii of the positive and negative ions).

Thus, the lattice energy of an ionic crystal increases rapidly as the charges of the ions increase and the sizes of the ions decrease. When all other parameters are kept constant, doubling the charge of both the cation and anion quadruples the lattice energy. For example, the lattice energy of LiF (Z+ and Z = 1) is 1023 kJ/mol, whereas that of MgO (Z+ and Z = 2) is 3900 kJ/mol (Ro is nearly the same—about 200 pm for both compounds).

Different interatomic distances produce different lattice energies. For example, we can compare the lattice energy of MgF2 (2957 kJ/mol) to that of MgI2 (2327 kJ/mol) to observe the effect on lattice energy of the smaller ionic size of F as compared to I.

Example $$\PageIndex{2}$$: Lattice Energy Comparisons

The precious gem ruby is aluminum oxide, Al2O3, containing traces of Cr3+. The compound Al2Se3 is used in the fabrication of some semiconductor devices. Which has the larger lattice energy, Al2O3 or Al2Se3?

Solution

In these two ionic compounds, the charges Z+ and Z are the same, so the difference in lattice energy will mainly depend upon Ro. The O2– ion is smaller than the Se2– ion. Thus, Al2O3 would have a shorter interionic distance than Al2Se3, and Al2O3 would have the larger lattice energy.

Exercise $$\PageIndex{2}$$

Zinc oxide, ZnO, is a very effective sunscreen. How would the lattice energy of ZnO compare to that of NaCl?

ZnO would have the larger lattice energy because the Z values of both the cation and the anion in ZnO are greater, and the interionic distance of ZnO is smaller than that of NaCl.

The Born-Haber Cycle

It is not possible to measure lattice energies directly. However, the lattice energy can be calculated using the equation given in the previous section or by using a thermochemical cycle. The Born-Haber cycle is an application of Hess’s law that breaks down the formation of an ionic solid into a series of individual steps:

• $$ΔH^\circ_\ce f$$, the standard enthalpy of formation of the compound
• IE, the ionization energy of the metal
• EA, the electron affinity of the nonmetal
• $$ΔH^\circ_s$$, the enthalpy of sublimation of the metal
• D, the bond dissociation energy of the nonmetal
• ΔHlattice, the lattice energy of the compound

Figure $$\PageIndex{3}$$ diagrams the Born-Haber cycle for the formation of solid cesium fluoride.

Figure $$\PageIndex{3}$$: The Born-Haber cycle shows the relative energies of each step involved in the formation of an ionic solid from the necessary elements in their reference states.

We begin with the elements in their most common states, Cs(s) and F2(g). The $$ΔH^\circ_\ce s$$ represents the conversion of solid cesium into a gas, and then the ionization energy converts the gaseous cesium atoms into cations. In the next step, we account for the energy required to break the F–F bond to produce fluorine atoms. Converting one mole of fluorine atoms into fluoride ions is an exothermic process, so this step gives off energy (the electron affinity) and is shown as decreasing along the y-axis. We now have one mole of Cs cations and one mole of F anions. These ions combine to produce solid cesium fluoride. The enthalpy change in this step is the negative of the lattice energy, so it is also an exothermic quantity. The total energy involved in this conversion is equal to the experimentally determined enthalpy of formation, $$ΔH^\circ_\ce f$$, of the compound from its elements. In this case, the overall change is exothermic.

Hess’s law can also be used to show the relationship between the enthalpies of the individual steps and the enthalpy of formation. Table $$\PageIndex{1}$$ shows this for cesium fluoride, CsF.

Enthalpy of sublimation of Cs(s) $$\ce{Cs}(s)⟶\ce{Cs}(g)\hspace{20px}ΔH=ΔH^\circ_s=\mathrm{77\:kJ/mol}$$ $$\dfrac{1}{2}\ce{F2}(g)⟶\ce{F}(g)\hspace{20px}ΔH=\dfrac{1}{2}D=\mathrm{79\:kJ/mol}$$ $$\ce{Cs}(g)⟶\ce{Cs+}(g)+\ce{e-}\hspace{20px}ΔH=IE=\ce{376\:kJ/mol}$$ $$\ce{F}(g)+\ce{e-}⟶\ce{F-}(g)\hspace{20px}ΔH=−EA=\ce{-328\:kJ/mol}$$ $$\ce{Cs+}(g)+\ce{F-}(g)⟶\ce{CsF}(s)\hspace{20px}ΔH=−ΔH_\ce{lattice}=\:?$$ $$ΔH=ΔH^\circ_f=ΔH^\circ_s+\dfrac{1}{2}D+IE+(−EA)+(−ΔH_\ce{lattice})$$ $$\ce{Cs}(s)+\dfrac{1}{2}\ce{F2}(g)⟶\ce{CsF}(s)=\ce{-554\:kJ/mol}$$

Thus, the lattice energy can be calculated from other values. For cesium chloride, using this data, the lattice energy is:

$ΔH_\ce{lattice}=\mathrm{(411+109+122+496+368)\:kJ=770\:kJ}$

The Born-Haber cycle may also be used to calculate any one of the other quantities in the equation for lattice energy, provided that the remainder is known. For example, if the relevant enthalpy of sublimation $$ΔH^\circ_s$$, ionization energy (IE), bond dissociation enthalpy (D), lattice energy ΔHlattice, and standard enthalpy of formation $$ΔH^\circ_\ce f$$ are known, the Born-Haber cycle can be used to determine the electron affinity of an atom.

Lattice energies calculated for ionic compounds are typically much larger than bond dissociation energies measured for covalent bonds. Whereas lattice energies typically fall in the range of 600–4000 kJ/mol (some even higher), covalent bond dissociation energies are typically between 150–400 kJ/mol for single bonds. Keep in mind, however, that these are not directly comparable values. For ionic compounds, lattice energies are associated with many interactions, as cations and anions pack together in an extended lattice. For covalent bonds, the bond dissociation energy is associated with the interaction of just two atoms.

Summary

The strength of a covalent bond is measured by its bond dissociation energy, that is, the amount of energy required to break that particular bond in a mole of molecules. Multiple bonds are stronger than single bonds between the same atoms. The enthalpy of a reaction can be estimated based on the energy input required to break bonds and the energy released when new bonds are formed. For ionic bonds, the lattice energy is the energy required to separate one mole of a compound into its gas phase ions. Lattice energy increases for ions with higher charges and shorter distances between ions. Lattice energies are often calculated using the Born-Haber cycle, a thermochemical cycle including all of the energetic steps involved in converting elements into an ionic compound.

Key Equations

• Lattice energy for a solid MX: $$\ce{MX}(s)⟶\ce M^{n+}(g)+\ce X^{n−}(g)\hspace{20px}ΔH_\ce{lattice}$$
• Lattice energy for an ionic crystal: $$ΔH_\ce{lattice}=\mathrm{\dfrac{C(Z^+)(Z^-)}{R_o}}$$

Footnotes

1. This question is taken from the Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service.

Glossary

Born-Haber cycle
thermochemical cycle relating the various energetic steps involved in the formation of an ionic solid from the relevant elements
lattice energy (ΔHlattice)
energy required to separate one mole of an ionic solid into its component gaseous ions

Contributors

• Adelaide Clark, Oregon Institute of Technology
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