In cubic ZnS, the radii of the Zn and S atoms are 0.83 A⁰ and 1.74 A⁰ respectively. What is the edge length of the unit cell of ZnS in A⁰.

Given, Cubic ZnS
We need to find the edge length of the unit cell of ZnS in A⁰

S2- ions are present in the ZnS lattice as FCC, while Zn2+ ions alternately occupy the tetrahedral gaps.
$\begin{array}{l}\frac{\sqrt{3}\mathrm{a}}{4}=\left({\mathrm{r}}_{{\mathrm{Zn}}^{2+}}+{\mathrm{r}}_{{\mathrm{S}}^{2-}}\right)\\ \mathrm{a}=\frac{4}{\sqrt{3}}(0.83+1.74)=5.94 \mathrm{A}°\end{array}$

Therefore, the correct answer 5.94 A⁰