Question

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

• A. $C_3A_4$
• B. $C_4A_3$
• C. $C_2A_3$
• D. $C_3A_2$

Answer 1

Answer: (A)

$C_3A_4$

Answer 2

$\bullet$ Anions(A) are in hcp, so number of anions (A) = 6
Cations(C) are in 75% O.V., so number of cations (C)
$= 6 \times \frac{3}{4}$
$= \frac{18}{4}$
$= \frac{9}{2}$
$\bullet$ So formula of compound will be
$C_{\frac{9}{2} } A_6 \Rightarrow \; C_9A_{12}$
$C_9 A_{12} \Rightarrow C_3A_4$