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?


1 Answer 1


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) 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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