In a spinel-type oxide the oxide ions form a ccp lattice. Cation $A$ occupies one-eighth of the tetrahedral voids and cation $B$ occupies one-half of the octahedral voids. The formula is

Answer: B

Per

O X^{2 -}

ion there are 2 tetrahedral and 1 octahedral void.

A = 2 \times \frac{1}{8} = \frac{1}{4}

;

B = 1 \times \frac{1}{2} = \frac{1}{2}

per oxide. Multiplying by 4:

A : B : O = 1 : 2 : 4

, i.e.

AB X_{2} O X_{4}