Question

#1) Use CFSE to reason why the inverse and spinel structure is energetically favored.?

Answer 1

Normal vs. inverse spinel structure. For transition metal oxide spinels, the choice of the normal vs. inverse spinel structure is driven primarily by the crystal field stabilization energy (CFSE) of ions in the tetrahedral and octahedral sites. For spinels that contain 3d elements such as Cr, Mn, Fe, Co, and Ni, the electron configuration is typically high spin because O^2- is a weak field ligand.

As an example, we can consider magnetite, Fe₃O₄. This compound contains one Fe²⁺ and two Fe³⁺ ions per formula unit, so we could formulate it as a normal spinel, Fe²⁺(Fe³⁺)₂O₄, or as an inverse spinel, Fe³⁺(Fe²⁺Fe³⁺)O₄. Which one would have the lowest energy?

d-orbital energy diagram for Fe²⁺

First we consider the crystal field energy of the Fe²⁺ ion, which is d⁶. Comparing the tetrahedral and high spin octahedral diagrams, we find that the CFSE in an octahedral field of O^2- ions is [(4)(2/5) - (2)(3/5)]Δ_o - P = 0.4 Δ_o - P. In thetetrahedral field, the CFSE is [(3)(3/5) - (3)(2/5)]Δ_t - P = 0.6 Δ_t - P. Since Δ_o is about 2.25 times larger than Δ_t, the octahedral arrangement has a larger CFSE and is preferred for Fe²⁺.

d-orbital energy diagram for Fe³⁺

In contrast, it is easy to show that Fe³⁺, which is d⁵, would have a CFSE of zero in either the octahedral or tetrahedral geometry. This means that Fe²⁺ has a preference for the octahedral site, but Fe³⁺ has no preference. Consequently, we place Fe²⁺ on octahedral sites and Fe₃O₄ is an inverse spinel, Fe³⁺(Fe²⁺Fe³⁺)O₄.