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

L.E=Q1.Q2rc+ra2 Q1 = charge on cationQ2 = charge on anionNaCl:→Na++Cl⊖(L.E)NaCl=1×1rNa++rCl-2="U"⇒ Cao → Ca+2 + O-2(L.E)CaCl2=2×2rCa+2+rO-22="4U"Given, rNa(+)+rCl-2=rCa(+2)+r O-22[L.E]NaCl= U[L.E]Cao = 4U