Question

Zinc sulphide crystallizes with zinc ions occupying half of the tetrahedral holes in a closed packed array of sulphide ions. What is the formula of zinc sulphide?

Answer 1

The zinc sulphide crystals are composed of equal number of the cation $Z{n^2}^ +$ and the anion ${S^{2 - }}$ ions. In a closed packed array two tetrahedral holes are present.

Complete step by step answer: The packing of ccp and hcp lattices indicates that there are two tetrahedral holes present per packing atom. To fill all the all tetrahedral sites, a stoichiometry of either ${M_2}X$ or $M{X_2}$ is required and to fill only half of the sites a stoichiometry of $MX$ are required.
As for example $Ca{F_2}$ belongs to the $M{X_2}$ structure. The $C{a^{2 + }}$ in fluorite is composed of three interpenetrating FCC lattices. On the other hand $L{i_2}O$ belongs to an ${M_2}X$ structure. Both the compounds fill the two tetrahedral voids of the close packed lattices.
Tetrahedrally bonded compounds require a $1:1$ stoichiometry of cations and anions to fill half of the tetrahedral sites. So compounds belonging to the formula $MX$can fill half of the sites in which both the $M$ and the $X$ atoms are tetrahedrally coordinated.
The zincblende containing zinc ions is occupied in these tetrahedral holes. One of the anions and one of cations are present in two interpenetrating FCC lattices. The coordination of $Zn$ and $S$ atoms in close packed lattices is tetrahedral.
Hence the formula of the zinc sulphide is $ZnS$ following the stoichiometry of the compound being $1:1$ and only half of the tetrahedral holes are occupied by $Z{n^2}^ +$ so only alternate tetrahedral holes are occupied by $Z{n^2}^ +$.

Note:
The compounds filling the lattice are considered as hard spheres. If a closed packed array contains N number of spheres then the number of tetrahedral holes present is 2N and the number of octahedral holes is N.