Question

A compond is formed by cation $C$ and anion $A$. The anions form hexagonal close packed (hcp) lattice and the cations occupy $75\%$ of octahedral voids. The formula of the compound is:

• A. $C_4{A}_3$
• B. $C_2{A}_3$
• C. $C_3{A}_2$
• D. $C_3{A}_4$

Answer 1

The correct option is D $C_3{A}_4$

Number of atoms per unti cell in hcp = $6$
Number of octahedral void in hcp = $6$
Number of anions per unit cell = $6$
Number of octahedral voids occupied by cations $=6\times \frac{75}{100}=\frac{9}{2}$
$\therefore \text{Formula of compound}=C_{9/2}{A}_6=C_3{A}_4$