In a cubic close packed structure of mixed oxides, the lattice is made up of oxide ions, one eight of tetrahedral voids are occupied by divalent `(X^(2+))` ions, while one - half of the octahedral voids are occupied by trivalent ions `(Y^(3+))`, then the formula of the oxide is

Options

A. `XY_(2)O_(4)`
B. `X_(2)YO_(4)`
C. `X_(4)Y_(5)O_(10)`
D. `X_(5)Y_(4)O_(10)`

Answer

C

Explanation

In CCP anions occupy primitives of the cube while cations occupied voids. In CCP there are two tetrahedral voids and one octahedral holes.
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 `(X^(2+))`.
`therefore` Number of divalent cations in tetrahedral voids `=2xx(1)/(5)`
Since half of the octahedral voids are occupied by trivalent cations `(Y^(3+))`
`therefore` Number of trivalent cations `=1xx(1)/(2)`
So the formula of the compound is `X_(2//5)Y_(1//2)OorX_(4)Y_(5)O_(10)`