**Correct Option (c) X**_{4} Y_{5} O_{10}

**Explanation: **

In ccp, anions occupy all lattice points of the cube while 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 (X^{2+}).

:. number of divalent cations in tetrahedral voids = 2 x 1/5.

Since half of the octahedral voids are occupied by trivalent cations (Y^{3+})

:. Number of trivalent cations = 1x 1/2.

.So the formula is the compound is (X)_{2 x 1/5 }(Y)_{1/2} (O)_{1}

or X_{2/5} Y_{1/2 }O_{1}

or X_{4}Y_{5}O_{10}