Ionic Solids
When cations and anions precipitate out of a saturated solution, they crystallize into a lattice arrangement that maximizes the attractive forces between cations and anions while minimizing the repulsive forces between ions of the same charge. For example, shown below is the arrangement of ions found in crystalline sodium chloride.

- Space Filling Model of NaCl Crystal Structure.
The chloride ions are represented as green, and the sodium ions are represented as yellow in this figure. To better understand the energetics of this arrangement, let's examine Coulomb's Law for the potential energy of interaction between two charges:

In NaCl, we have three types of Coulombic interactions, an attractive interaction between sodium cations and chloride anions, and repulsive interactions between sodium cations and between chloride anions.

For a mole of solid NaCl, the sum of all these Coulombic energies for all cations and anions in a crystal is -861 kJ/mole.
Na+(g) + Cl-(g)→NaCl(s) | ΔU = -861 kJ/mole |
Lattice Energy- change in energy when completely gaseous atoms are brought together into 1 mole of a solid ionic substance.
As we see from Coulomb's law, the lattice energy is related to the charge and distance between ions in the solid. Smaller ions will have smaller distances between them in a solid and thus have higher lattice energies.
- Lattice Energies:Quiz
Born-Haber Cycle
Lattice Energies can also be measured experimentally using Hess's law in what is called the Born-Haber Cycle. Recall that Hess's law tells us that the change in energy when going from reactant to products is the same whether the reaction takes place in one step or a series of steps. For example, given the enthalpy changes for the following reactions, we can obtain the enthalpy change for the formation of the crystalline NaCl lattice.
Cl-(g) | → | Cl(g) + e- | ΔH = -E.A.(Cl) = 349 kJ/mole |
Na+(g) + e- | → | Na(g) | ΔH = -E.A.(Na) = -496 kJ/mole |
Na(g) | → | Na(s) | ΔH = -104.8 kJ/mole |
Cl(g) | → | 1/2 Cl2(g) | ΔH = -120.5 kJ/mole |
Na(g) + 1/2 Cl2(g) | → | NaCl(s) | ΔH = -411.2 kJ/mole |
-------------------------------------------------------------------------------------------------------- | |||
Na+(g) + Cl-(g)
→ |
NaCl(s) |
ΔH = -783.5 kJ/mole |
|
That is pretty close to our calculated number of -861 kJ/mole (< 10% error). But, of course, our first the calculation was based on infinitely separated ions at 0 K, and the Born-Haber cycle is based on the reaction enthalpy changes at room temperature and pressure, so we expect some differences.
Homework from Chemisty, The Central Science, 10th Ed.
8.13, 8.15, 8.17, 8.19, 8.21, 8.23, 8.25