In a cubic close packed structure of mixed oxides, the lattice is made up of oxide ions,1/8th of tetrahedral voids are occupied by divalent ions (A2+) while ½ of octahedral voids are occupied by trivalent ions (B3+). What is the formula of the oxide?

Dear student
In ccp anions occupy primitives of the cube while cations occupy voids.
And there are two tetrahedral voids and one octahedral hole per anion in a cubic close-packed structure.
Now, according to question
No of the divalent cation in tetrahedral voids =$2\times \frac{1}{8}=\frac{1}{4}$
No of trivalent cations= $1\times \frac{1}{2}$
Therefore, a formula of the mixed oxide= ${\mathrm{X}}_{1/4}{\mathrm{Y}}_{1/2}{\mathrm{O}}_1$
which is X₂Y₄O_8
Regards