In a solid, oxide ions form a CCP lattice. One-fourth of the tetrahedral voids are occupied by dival — Solid State Chemistry Question

💡 Solution & Explanation

Let the number of oxide ions in CCP be $N$. Then tetrahedral voids $=2N$ and octahedral voids $=N$. A occupies $\frac14(2N)=\frac N2$ sites, while B occupies all $N$ octahedral sites. The ratio $A:B:O=\frac N2:N:N=1:2:2$, so the formula is $AB_2O_2$.