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? 




The reason stated in the literature is fairly straightforward. However, actually tracking down the original source is harder than it looks. 


Going back to [Phys. Rev. B22, 115 (1980)](https://journals.aps.org/prb/abstract/10.1103/PhysRevB.22.115) one finds in the opening paragraph the statement 



> 
> 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.


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