Question

If the anion (A) forms hexagonal closest packing and cation (C) occupies only $\frac{2}{3}rd$ of octahedral voids in it, what will be the general formula of the compound?

• A. $CA$
• B. $C{A}_2$
• C. $C_2{A}_3$
• D. $C_3{A}_2$

Answer 1

The correct option is C $C_2{A}_3$

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 $\frac{2}{3}$ of the octahedral voids in the unit cell.
$\therefore$ $\text{No. of C atoms}=6\times \frac{2}{3}=4$
$\therefore$
$\text{Emperical formula of the compound}=C_4{A}_{6}or{C}_2{A}_3$