In a cubic packed structure of mixed oxides, the lattice is made up of oxide ions, one fifth of tetrahedral voids are occupied by divalent (X⁺⁺) ions, while one half of the octahedral voids are occupied by trivalent ions (Y³⁺), then the formula of the oxide is

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X₂YO₄

X₄Y₅O₁₀

X₅Y₄O₁₀

XY₂O₄

answer is C.

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Detailed Solution

In CCP anions oxide ion are forming the lattice reference and cations occupied voids. In ccp there are two tetrahedral voids and one octahedral hole.
For one oxygen atom there are two tetrahedral holes and one octahedral hole.
Since one fifth of the tetrahedral voids are occupied by divalent cations (X²⁺)
$∴$ number of divalent catioins in tetrahedral voids = 2*1/5

Since half of the octahedral voids are occupied by trivalent cations (Y²⁺)
$∴$ number of trivalent cations = $1 x \frac{1}{2}$
So the formula of the compound is X_2/5Y_1/2O or X₄Y₅O₁₀