Chapters / Chapter 1 · Solid state / Topic 14 of 14

Normal and inverse spinel: CFSE

Chapter 1 · Solid state and crystal structures

Octahedral splitting is larger and inverted relative to tetrahedral splitting.

This is the page in the photograph you sent. It is the last page of the notes, and it states the rule that decides normal against inverse spinel.

From the notes, transcribed: “This generally takes place [in] transition metal oxide. Generally the site selection is governed by the relative change of CFSE (Crystal Field Stabilization Energy) in occupying the Oct and Td holes.”

The crossed-out words in the photograph are “crystal” (before CFSE) and “octa” (before “oct”). The sentence is about spinel oxides, AB₂O₄.

Normal spinel

Named after the mineral spinel, MgAl₂O₄. It consists of a ccp array of oxide ions in which the divalent metal ion (Mg²⁺) occupies 1/8 of the tetrahedral holes and the trivalent metal ion (Al³⁺) occupies 1/2 of the octahedral holes. In a ccp anion sublattice there is one octahedral hole and two tetrahedral holes per anion. There are four oxides per formula unit, and consequently four octahedral holes and eight tetrahedral holes.

Normal: (AII)tet (BIII)2oct O4

Inverse spinel

In an inverse spinel, half of the B ions occupy 1/8 of the tetrahedral holes, and A and the other half of B occupy 1/2 of the octahedral holes in the ccp array of O²⁻.

Inverse: BIII tet [ AII BIII ]oct O4

Normal against inverse: which cation sits in which hole.

Why CFSE decides it

For transition-metal oxide spinels, O²⁻ is a weak-field ligand, so the ions are usually high spin. The structure that forms is the one with the larger total CFSE. The octahedral site preference energy (OSPE) is CFSE(oct) − CFSE(tet), with Δ_t = 4/9 Δ_o. Its size depends on the d-electron count:

d count (high spin)	CFSE oct (Δ_o)	CFSE tet (Δ_o)	OSPE (Δ_o)
d⁰, d⁵, d¹⁰	0	0	0
d¹, d⁶	0.40	0.27	0.13
d², d⁷	0.80	0.53	0.27
d³, d⁸	1.20	0.36	0.84
d⁴, d⁹	0.60	0.18	0.42

So d³ and d⁸ have the strongest octahedral preference, d⁰, d⁵ and d¹⁰ have none, and the rest fall in between.

Worked examples from the notes

Spinel	Distribution	Why
MgAl₂O₄	Normal	Both non-transition metals; CFSE = 0. The more highly charged Al³⁺ prefers the octahedral site.
ZnFe₂O₄	Normal	Zn²⁺ is d¹⁰ and Fe³⁺ is d⁵; both have zero CFSE, and small Zn²⁺ fits the tetrahedral site.
Mn₃O₄	Normal	Mn²⁺ is high-spin d⁵ (CFSE 0); Mn³⁺ is d⁴ and takes the octahedral site.
Fe₃O₄	Inverse: Fe³⁺[Fe²⁺Fe³⁺]O₄	Fe³⁺ is d⁵ (CFSE 0 in either site); Fe²⁺ is d⁶ and prefers octahedral.
CoFe₂O₄, NiFe₂O₄	Inverse	Co²⁺ (d⁷) and Ni²⁺ (d⁸) have a strong octahedral preference over Fe³⁺ (d⁵).
FeCr₂O₄	Normal	Cr³⁺ is d³, with the largest octahedral preference (0.84 Δ_o).
Co₃O₄	Normal	Co³⁺ is low-spin d⁶ with very high CFSE in the octahedral site.

Verified

The rule on this page is correct and is the standard explanation: for transition-metal oxide spinels, normal against inverse is driven primarily by the CFSE of the ions in tetrahedral and octahedral sites. The worked examples above follow the same rule, with the OSPE values taken from the standard high-spin table.