In a normal spinel types structure, the oxide ions are arranged in $\mathrm{c}\mathrm{c}\mathrm{p}$ whereas $1/8$ tetrahedral holes are occupied by ${\mathrm{Z}\mathrm{n}}^{2+}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s}$ and $50\mathrm{\%}$ of octahedral holes are occupied by ${\mathrm{F}\mathrm{e}}^{3+}\mathrm{ }\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s}$. The formula of the compound is -