According to my notes, octahedral site stabilization energy (OSSE), sterical effects and electronic effects in stability create a confusion in this theory. However, we are able to solve this problem with few rules:
- Tetrahedral splitting energy is nearly equal to 4/9 octahedral splitting energy.
- As the OSSE increases, the $O_\mathrm{h}$ geometries will be more favored. These are useful to decide to whether the structure is spinel or inverse spinel.
I got $\ce{ZnCr2O4}$ ($\ce{Zn^2+}$, $\ce{2Cr^3+}$, $\ce{4O^2-}$):
- $\ce{Zn^2+}$; $\mathrm{d^{10}}$ ion with CFSE = $0$ and thus, there will be no effect on $O_\mathrm{h}$ holes.
- $\ce{Cr^3+}$; $\mathrm{d^3}$ ion with CFSE = $\pu{-8 Dq}$ and OSSE = $\pu{8.45 Dq}$ so, this is thought to occupy $O_\mathrm{h}$ holes.
Since $\ce{Zn^2+}$ can only occupy $T_\mathrm{d}$ holes, then it suits the rules for being normal spinel.
On the other hand, if we got a group II metal with a non-zero CFSE, then what would the result be?