Arrange the following substances in order of decreasing magnitude of lattice energy.

KCl NaF SrO BaO

The formula is F = (q1 * q2) / r ^ 2

where: q is the individual charges of each ion

r is the distance between the nuclei

The formula is not important but to explain the relationship between the atoms in the compounds and their lattice energy.

From the formula we can first conclude that compounds of ions with greater charges will have a greater lattice energy. This is a direct relationship.

For example, the compounds BaO and SrO, whose ions' charges are ( + 2) and ( - 2) respectively for each, will have greater lattice energies that the compounds NaF and KCl, whose ions' charges are ( + 1) and ( - 1) respectively for each.

So Far: (BaO and SrO) > (NaF and KCl)

The second part required you find the relative distance between the atoms of the compounds. Really, the lattice energy is stronger with smaller atoms, an indirect relationship.

For example, in NaF the ions are smaller than the ions in KCl so it has a greater lattice energy. Because Sr is smaller than Ba, SrO has a greater lattice energy than BaO.

Therefore:

Answer: SrO > BaO > NaF > KCl