Question

$CaO$ 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:

• A. $U/2$
• B. $U$
• C. $2U$
• D. $4U$

Answer 1

Answer: (D)

$4U$

Lattice energy $=-\frac{q_1{q}_2}{r^2}$
Where $q_1$ 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 ${Ca}^{2+}$ and $O^{2-}$ ions is $2$ each, therefore, the lattice energy of $CaO$ is four times the lattice energy of $NaCl$ ie. $4U$.

Hence, option $D$ is correct.