Question

For a compound, the anion (X) crystallizes in hexagonal close packing and cation (Y) occupies only $\frac{1}{3}rd$ of octahedral voids in it, what will be the general formula of the compound?

• A. $YX$
• B. $Y_{2}X$
• C. $Y{X}_2$
• D. $Y{X}_3$

Answer 1

The correct option is D $Y{X}_3$

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 (X) form hexagonal lattice
$\text{No. of X atoms}=6$
Cation 'Y' occupies $\frac{1}{3}$ of the octahedral voids in the unit cell.
$\therefore$ $\text{No. of Y atoms}=6\times \frac{1}{3}=2$
$\therefore$
$\text{Emperical formula of the compound}=Y_2{X}_{6}orY{X}_3$