Question

In a cubic lattice, the closed packed structure of mixed oxides of the lattice is made up of oxide ions; one eight of the tetrahedral voids are occupied by divalent ions $\left(A^{2+}\right)$ while one half of the octahedral voids are occupied by trivalent ions $\left(B^{3+}\right)$. What is the formula of the oxides?

Answer 1

Let there be $80O^{2-}$ in the crystal.
$\therefore$ Octahedral voids $=80$
Tetrahedral voids $=160$
$A^{2+}$ ions $=\frac{1}{8}\times 160=20$
$B^{3+}ions=\frac{1}{2}\times 80=40$
$A^{2+}:B^{3+}:O^{2-}=20:40:80$
$=1:2:4$.
$\therefore$ Formula is $A{B}_2{O}_4$