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:
(1) \(C_{2} A_{3}\)
(2) \(\mathrm{C}_{3} \mathrm{~A}_{2}\)
(3) \(C_{3} A_{4}\)
(4) \(\mathrm{C}_{4} \mathrm{~A}_{3}\)

Answer 1

Correct option (3) \(\mathrm{C}_{3} \mathrm{~A}_{4}\)
Explanation:
- Anions \((A)\) are in hcp, so number of anions \((A)=6\)
Cations(C) are in \(75 \%\) O.V., so number of cations (C)
\(=6 \times 3 / 4\)
\(=18 / 4\)
\(=9 / 2\)
- So formula of compound will be
$$
\begin{aligned}
&C_{9 / 2} A_{6}=C_{9} A_{12} \\
&C_{9} A_{12}=C_{3} A_{4}
\end{aligned}
$$