Question

What is the formula of a compound in which the element Y forms hcp lattice and atoms of X occupy $(\frac{2}{3}{)}^{rd}$ of tetrahedral voids?

Answer 1

Suppose the number of atoms Y in hcp lattice = n. As the number of tetrahedral voids is double the number of atoms in close packing, the number of tetrahedral voids = 2n. As atoms X occupy $(\frac{2}{3}{)}^{rd}$ of the tetrahedral voids, the number of atoms X in the lattice :

$\frac{2}{3}$$\times 2n=$$\frac{4n}{3}$

$\therefore \text{Ratio of}X:Y=\frac{4n}{3}:n=\frac{4}{3}:1=4:3$

Hence, the formula of the compound is $X_4{Y}_3$.