Question

In a compound, $O^{2-}$ ions are arranged in ccp. One-sixth of tetrahedral voids are occupied by cations A while one-third of octahedral voids are occupied by cation B. The formula of the compound is $AB{O}_3$.
If true enter 1, else enter 0.

Answer 1

Number of ${O_2}^{-}$ ions (ccp or fcc type) $=4$ per unit cell

Number of cation A $=\frac{1}{6}\times$ tetrahedral voids $=\frac{1}{6}\times 8=\frac{4}{3}$ per unit cell

$\frac{1}{6}\times TV=\frac{1}{6}\times 8=\frac{4}{3}$ per unit cell

Number of cation $B=$ $\frac{1}{3}\times OV=\frac{1}{3}\times 4=\frac{4}{3}$ unit cell

$\frac{1}{3}\times OV=\frac{1}{3}\times 4=\frac{4}{3}$ per unit cell

Formula: $A_{\frac{4}{3}}{B}_{\frac{4}{3}}{O}_4\Rightarrow AB{O}_3$