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

$\mathrm{L}.\mathrm{E}=\frac{{\mathrm{Q}}_1.{\mathrm{Q}}_2}{{\left[{\mathrm{r}}_{\mathrm{c}}+{\mathrm{r}}_{\mathrm{a}}\right]}^2}$

 Q₁ = charge on cation
Q₂ = charge on anion
$\mathrm{NaCl}:\to {\mathrm{Na}}^{+}+{\mathrm{Cl}}^{\ominus }$

$(\mathrm{L}.\mathrm{E}{)}_{\mathrm{NaCl}}=\frac{1\times 1}{{\left[{\mathrm{r}}_{{\mathrm{Na}}^{+}}+{\mathrm{r}}_{{\mathrm{Cl}}^{-}}\right]}^2}="\mathrm{U}"$

⇒ Cao → Ca⁺² + O^-2
$(L.E{)}_{{\mathrm{CaCl}}_2}=\frac{2\times 2}{{\left[r_{C{a}^{+2}}+r_{O^{-2}}\right]}^2}="4\mathrm{U}"$

Given, 

${\left[r_{N{a}^{(+)}}+r_{C{l}^{-}}\right]}^2={\left[r_{C{a}^{(+2)}}+r_{{ }^{O^{-2}}}\right]}^2$

[L.E]_NaCl= U

[L.E]_Cao = 4U