Question

In a cubic lattice anion (A) forms hexagonal close packing and cation (C) occupies only $\frac{1}{3}rd$ of octahedral voids in it, what will be the general formula of the compound?

• A. $C{A}_2$
• B. $C{A}_3$
• C. $C_{3}A$
• D. $C_{2}A$

Answer 1

The correct option is B $C{A}_3$

In hexagonal closed packing, there are 6 octahedral voids completely inside the unit cell.
$\text{No. of atoms in hexagonal close packing}=6$
Since, anions (A) forms hexagonal lattice
$\text{No. of A atoms}=6$
Cation 'C' occupies $\frac{1}{3}$ of the octahedral voids in the unit cell.
$\therefore$ $\text{No. of C atoms}=6\times \frac{1}{3}=2$
$\therefore$
$\text{Emperical formula of the compound}=C_2{A}_{6}orC{A}_3$