Question

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

Answer 1

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

Anions $\left(A\right)$ are in hcp, so the number of anions $\left(A\right)=6$
Cations $\left(C\right)$ are occuping $75\%$ of the octahedral voids.
So, number of cations $\left(C\right)$
$6\times \frac{75}{100}=\frac{18}{4}=\frac{9}{2}$
So, the formula of compound will be
$C_{9/2}{A}_6\to C_9{A}_{12}\to C_3{A}_4$