Question

In a cubic close packed structure of mixed oxides, the lattice is made up of oxide ions, one eighth of tetrahedral voids are occupied by divalent ($X^{2+}$) ions, while one-half of the octahedral voids are occupied by trivalent ions ($Y^{3+}$), then 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 $=8\left(\frac{1}{8}\right)+6\left(\frac{1}{2}\right)=4$ and in ccp the number of tetrahedral voids is $8$

so divalents ions will be $\frac{8}{8}$and number of octahedral voids are $4$ so.one half would be 2 finally,$X{Y}_2{O}_4$.