# Why is ZnFe2O4 a normal spinel?

Considering that $\ce{Zn^2+}$ is $\mathrm{d}^{10}$, $\ce{Fe^3+}$ is $\mathrm{d^5}$, and $\ce{O^2-}$ is a weak field ligand, the CFSE values of both the normal and inverse spinel structures should be $0$.

Why is it then that the normal spinel, with $\ce{Zn^2+}$ in tetrahedral holes and $\ce{Fe^3+}$ in octahedral holes, is formed?

Mixed Zn ferrites of the type Zn$_{x}$M$_{3-x}$O$_{4}$ where M is a magnetic ion have been the subject of numerous investigations in the past, especially with regard to their magnetic properties.$^{1}$ The metallic ions occupy the tetrahedral A or octahedral B sites in the spinel crystal structure. The Zn ions preferentially occupy the A sites because of their tendency to form covalent bonds involving sp' orbitals.
Reference number 1, on the first sentence, is a book [J. Smit and H.P.J. Wihn, Ferrites (Wiley, New York, 1959)]. The statement on $sp3$ bonding is not backed up by a reference. I can only assume that it was considered common knowledge in the spinel community in 1980.