Answer
The four compounds in decreasing lattice energy are as follows: $$LiCl\gt KF\gt NaCl\gt NaBr$$
Work Step by Step
From part a), we find that the cation-anion distance in each compound is as follows:
- $K-F$ distance: $2.71$ angstroms
- $Na-Cl$ distance: $2.83$ angstroms
- $Na-Br$ distance: $2.98$ angstroms
- $Li-Cl$ distance: $2.57$ angstroms
According to Coulomb's law: $$E_{el}=\frac{kQ_1Q_2}{d}$$
the attraction among the oppositely charged ions is inversely proportional with the distance among the ions.
In other words, the smaller the distance among the ions in the compound, the larger the attraction among the ions, and as a result, the higher the lattice energy of the compound.
So, $LiCl$, with the shortest cation-anion distance, has the highest lattice energy, while $NaBr$, with the longest cation-anion distance, has the smallest lattice energy.
In detail, we can arrange the four compounds in decreasing lattice energy as follows: $$LiCl\gt KF\gt NaCl\gt NaBr$$
