In a cubic close packed structure of mixed oxide, the oxide ions are in CCP arrangement 1/6 of tetrahedral voids are occupied by cations 'A' and 1/2 of octahedral voids are occupied by cations 'B'. The formula of the oxide is

(O^-2) Oxide ions in ccp = 4

(A)No. of tetrahedral voids = 2n =2(4) =8

1/6 of tetrahedral voids = 1/6 $\times$8 = $\frac{4}{3}$

(B)No of octahedral voids = n =4

1/2 of otahedral voids = 1/2 $\times$4 = 2

A :B :O^-2 =$\frac{4}{3} :2 :4$

               = 4 :6:12

               =2 :3 :6

The formula of the oxide is A₂B₃O₆