In a crystal lattice, anions $A$ forms hcp array, $2 / 3$ of the tetrahedral voids are occupied by cations…
In a crystal lattice, anions $A$ forms hcp array, $2 / 3$ of the tetrahedral voids are occupied by cations $C$. What is the formula of crystal?
$C_3 A_4$
$\mathrm{C}_4 \mathrm{~A}_3$
$C_2 A_3$
$C_3 A_2$
Solution
For hexagonal close packed ( $h c p)$ crystal, there are 6 atoms per unit cell. That means anion $A=$ number of octahedral voids $=$ number of atoms per unit cell $=6$.
We also know that, tetrahedral voids
$=2 \times$ octahedral voids .
$\therefore \quad C=\frac{2}{3} \times(2 \times 6)=8$
Hence, $C: A=8: 6=4: 3$
Therefore, the formula of the crystal is $C_4 A_3$.