Question

Structure of a mixed oxide is cubic close-packed (c.c.p.). the cubic unit cell of mixed oxide is composed of oxide ions. one fourth of the tetrahedral voids are occupied by divalent metal a and the octahedral voids are occupied by a monovalent metal

b. the formula of the oxide is

Answer 1

Actually the bigger molecules like oxygen occupy the lattice points.In ccp unit cell the effective number of atoms occupying lattice points is 8(1/8)+6(1/2)=4 and in ccp the number of tetrahedral voids is 8 so divalents ions will be 8/8 and number of octahedral voids are 4 so one half would be 2 finally,XY2O4.

Answer 2

Answer: AB2O2

Explanation :

In ccp structure the oxide anion will occupy the lattice point and cations will occupy the voids .

Let there be N no of O2− in a crystal .

Therefore octahedral voids =N tetradral voids = 2N

A2+ ion occupies the tetrahedral voids so number of atoms A=1/4×2N=N/2

B2+ ion occupies the octahedral voids so number of atoms B=1×N=N

Hence the fomula is AN/2BNON=AB2O2