$\mathrm{CaO}$ and $\mathrm{NaCl}$ have the same crystal structure and approximately the same ionic radii.…

$\mathrm{CaO}$ and $\mathrm{NaCl}$ have the same crystal structure and approximately the same ionic radii. If $\mathrm{U}$ is the lattice energy of $\mathrm{NaCl}$, the approximate lattice energy of $\mathrm{CaO}$ is
  1. $\mathrm{U} / 2$
  2. $\mathrm{U}$
  3. $4 \mathrm{U}$
  4. $2 \mathrm{U}$

Solution

Lattice energy $=\frac{\text { Product of charges }}{\text { interionic distance }}$
In $\mathrm{NaCl}$ the product of charges $=1 \times 1 ;$ In CaO product of charges $=2 \times 2=4$ while the inter ionic distance is almost same in both. Thus lattice energy of $\mathrm{CaO}$ is almost four times the lattice energy of $\mathrm{NaCl}$.

Asked in: JEE-TOPICTESTS-CHEMISTRY

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