CaO and NaCl have the same crystal structure and approximately the same ionic radii. If U is the lattice energy of NaCl, the approximate lattice energy of CaO is

Lattice energy =$-\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}$

Where q₁​ and q_2​ are charges on ions and r is the distance between them.

Since inter-ionic distances in CaO and NaCl are similar, (larger cation has smaller anion and vice-versa) therefore r is almost the same.

Therefore, lattice energy depends only on the charge. Since the magnitude of charge on Na⁺ and Cl⁻ ions is unity and that on Ca²⁺ and O²⁻ ions is 2 each, therefore, the lattice energy of CaO is four times the lattice energy of NaCl ie. 4U.