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

Answer: (C)

$C_3{A}_4$

Anions (A) are in hcp, so the number of anions $\left(A\right)=6$

Cations (C) are in $75\%O.V.,$ so number of cations (C)
$=6\times \frac{75}{100}$

$=\frac{18}{4}$

$=\frac{9}{2}$

$\Rightarrow$ So, the formula of compound will be

$C_{\frac{9}{2}}{A}_6\Rightarrow C_9{A}_{12}$

$C_9{A}_{12}\Rightarrow C_3{A}_4$