Question

A compound of formula $A_2{B}_3$ has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms ?

• A. hcp lattice- A, $\frac{2}{3}$ tetrahedral voids-B
• B. hcp lattice-A, $\frac{1}{3}$ tetrahedral voids-B
• C. hcp lattice-B, $\frac{1}{3}$tetrahedral voids-A
• D. hcp lattice-B, $\frac{2}{3}$tetrahedral voids-A

Answer 1

The correct option is C hcp lattice-B, $\frac{1}{3}$tetrahedral voids-A

Total effective nuinber of atoms in hcp unit lattice
octahedral voids in hcp$=6$
Number of tetrahedral voids (TV) in hcp
$=2\times \text{Number of atoms in hcp lattice}$
As, formula of the lattice is $A_2{B}_3$.
Suppose,

A	B
$\frac{1}{3}\times TV$	hcp
$\frac{1}{3}\times 12$	6
$\frac{2}{3}$	1
2	3

So, A = $\frac{1}{3}$tetrahedral voids, B =hcp lattice