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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
$C_2A_3$
B
$C_3A_2$
C
$C_3A_4$
D
$C_4A_3$
Detailed Solution
Anions (A) form hcp, so number of anions per unit cell $=6$, and number of octahedral voids $=6$.
Cations (C) occupy 75% of octahedral voids: number of cations $=6\times\frac{3}{4}=\frac{18}{4}=\frac{9}{2}$
Formula: $C_{9/2}A_6\Rightarrow C_9A_{12}\Rightarrow C_3A_4$