Question

In a crystal oxide ions are arranged in fcc and $A^{2+}$ ions are at $1/8^{th}$ of the tetrahedral voids, and ions $B^{3+}$ occupied $1/2$ of the octahedral voids. Calculate the formula of the oxide.

Answer 1

Oxide ions $O^{-2}$ arranged in FCC$⟶O^{-2}$

$B^{3+}$ occupied $\frac{1}{2}$ of the octahedral voids $⟶\frac{1}{2}\times 1=\frac{1}{2}$

$A^{2+}$ occupied ${\frac{1}{8}}^{th}$ of the tetrahedral voids.

$\because$ Number of Tetrahedral voids= $2\times$ Number of octahedral voids

$\Rightarrow 2\times 1=2$

$\therefore {\frac{1}{8}}^{th}$ of $2$ Tetrahedral voids= $\frac{1}{8}\times 2=\frac{1}{4}$

$\therefore \underset{\frac{1}{4}}{A}$ $\underset{\frac{1}{2}}{B}$ $\underset{1}{O}$ Convert into integer $⟶\underset{1}{A}$ $\underset{2}{B}$ $\underset{4}{O}$

Therefore, formula of the oxide is $A{B}_2{O}_4$ .