In a ccp of mixed oxide, it is found that lattice has $O^{2-}$ ions and one half of the octahedral voids are occupied by trivalent cataions ($X^{3+}$) and one eight of the tetrahedral voids by divalent cataions ($Y^{2+}$). What is the formula of the mixed oxide?

$\begin{array}{1 1} X_2YO_4 \\ X_4YO_4 \\ XY_2O_4 \\ X_2Y_2O_2 \end{array}$

Answer: $X_2YO_4$
Number of trivalent catation $= \large\frac{1}{2} $$\times 4 = 2 Number of divalent catation = 8 \times \large\frac{1}{8}$$ = 1$
$\Rightarrow$ the formula is $X_2YO_4$