Question

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

• A. $A\to \text{hcp lattice}$;
  $B\to \frac{2}{3}\text{tetrahedral voids}$
• B. $A\to \text{hcp lattice}$;
  $B\to \frac{1}{3}\text{tetrahedral voids}$
• C. $B\to \text{hcp lattice}$;
  $A\to \frac{1}{3}\text{tetrahedral voids}$
• D. $B\to \text{hcp lattice}$;
  $A\to \frac{2}{3}\text{tetrahedral voids}$

Answer 1

The correct option is C $B\to \text{hcp lattice}$;

$A\to \frac{1}{3}\text{tetrahedral voids}$
$\text{Total effective number of atoms in hcp unit lattice}=6$
$\text{Number of octahedral voids in hcp}=6$
$\therefore$
$\text{Number of tetrahedral voids in hcp}=2\times 6=12$

We can solve this by trial and error method.

Options	hcp lattice	Tetrahedral voids	Ratio A : B	Formula
(a)	$A=6$	$B=\frac{2}{3}\times 12=8$	3 : 4	$A_3{B}_4$
(b)	$A=6$	$B=\frac{1}{3}\times 12=4$	3 : 2	$A_3{B}_2$
(c)	$B=6$	$A=\frac{1}{3}\times 12=4$	2 : 3	$A_2{B}_3$
(d)	$B=6$	$A=\frac{2}{3}\times 12=8$	4 : 3	$A_4{B}_3$

Thus, option (c) is correct.