Question

Atoms of element B from HCP lattice and those of the element A occupy 2/3 rd of tetrahedral voids. What is the formula of the compound formed by the elements A and B?

Answer 1

Given that, atoms of $B$ forms hcp lattice. So, the atoms at corners are shared by $6$ unit cells. So, its contribution is $\frac{1}{6}$; Face centered atoms contribute $\frac{1}{2}$ and middle layer atoms contribute $1$ each.

So, effective number of atoms in unit cell $=\frac{1}{6}\times 12+\frac{1}{2}\times 2+1\times 3$

$=2+1+3=6$

Number of tetrahedral voids $=2\times$ Number of atoms forming the lattice.

$=2\times 6=12$

So, number of atoms of $A=\frac{2}{3}\times 12=8$

Number of atoms of $B=6$

So, $A:B\Rightarrow A:B$

$8:6$ $4:3$

So, formula is $A_4{B}_3$