In a cubic close packed structure (ccp) of mixed oxides, it is found that lattice has `O^(2-)` ions and one half of the octahedral voids are occupied by trivalent cations `(A^(3+))` and one-eighth of the tetrahedral voids are occupied by divalent cations `(B^(2+))`. Derive the formula of the mixed oxide.

Explanation

Number of octahedral voids ion in lattice = 1
Hence, Number of trivalent cations `(A^(3+)) = 1 xx (1)/(2) = (1)/(2)`
Number of tetrahedral voids per ion in lattice = 2
Hence, Number of divalent cations `(B^(2+)) = 2 xx (1)/(8) = (1)/(4)`
Thus, formula is `A_(1//2)O " or " A_(2)BO_(4)`