Question

In a metallic oxide, oxide ions are arranged in cubic close packing. One sixth of the tetrahedral voids are occupied by cations $P$ and one third of octahedral voids are occupied by the cation $Q$. Deduce the formula of the compound.

Answer 1

In CCP, there would be $2$ tetrahedral voids and $1$ octahedral void.

No. of $O^{-2}$ ions $=4$

One-sixth tetrahedral voids are occupied by $P$, i.e.,

No. of $P$ ions $=\frac{1}{6}\times 8=\frac{4}{3}$

One-third octahedral voids are occupied by $Q$ while one

No. of $Q$ ions $=\frac{1}{3}\times 4=\frac{4}{3}$

Therefore,

Formula of compound $=X_{\frac{4}{3}}{Y}_{\frac{4}{3}}{O}_4=XY{O}_3$

Hence the formula of compound is $XY{O}_3$.