Question

In a binary compound, atoms of element A form a hcp structure and those of element M occupy 2/3 of the tetrahedral voids of the hcp structure. The formula of the binary compound is:

• A. $M_2{A}_3$
• B. $M{A}_4$
• C. $M_{4}A$
• D. $M_4{A}_3$

Answer 1

The correct option is D $M_4{A}_3$

Answer : D
In hcp, there are 6 atoms per unit cell.
$\therefore N=6$
Since, (A) forms hexagonal lattice
$\text{No. of A atoms}=6$

$\text{Number of octahedral voids}=N=6$
$\text{Numer of tetrahedral voids}=2N=2\times 6=12$

$M$ occupies $\frac{2}{3}$ of the tetrahedral voids in the unit cell.
$\therefore$ $\text{No. of M atoms}=12\times \frac{2}{3}=8$
$\therefore$

$\text{Emperical formula of the compound}=M_8{A}_{6}or{M}_4{A}_3$