In a cubic close packed structure of mixed oxides, the lattice is made up of oxide ions, one-fifth of tetrahedral voids are occupied by cations of ‘X’. While one-half of the octahedral voids are occupied by cations of ‘Y’, then the formula of the oxide is

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X₂YO₄

X₄Y₅O₁₀

X₅Y₄O₁₀

XY₂O₄

answer is C.

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Detailed Solution

$O = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4, X = 8 \times \frac{1}{5} = \frac{8}{5} Y = 4 \times \frac{1}{2} = 2 \begin{aligned} X_{\frac{8}{5}} Y_{2} O_{4} \\ X_{8} Y_{10} O_{20} \\ X_{4} Y_{5} O_{10} \end{aligned}$

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