Question

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

• A. $\frac{U}{2}$
• B. $U$
• C. $2U$
• D. $4U$

Answer 1

The correct option is D

$4U$

Lattice energy $=\frac{q1q2}{r^2}$

Where q1 and q2 are charges on ions and r is the distance between them. Since interionic distances in CaO and NaCI are similar, (Large cation has smaller anion and vice versa) therefore, r is almost the same. Therefore, lattice energy depends only on charge.

Since the magnitude of charge on $N{a}^{+}$ and $C{l}^{-}$ ions is unity and that on $C{a}^{2+}$ and $O^{2-}$ ions is 2 each, Therefore, the lattice energy of CaO is four times the lattice energy of NaCI, i.e., 4U.