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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:

  1. Tetrahedral splitting energy is nearly equal to 4/9 octahedral splitting energy.
  2. 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?

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  • $\begingroup$ Could you state an example of a group II metal ion with a CSFE unequal to 0? $\endgroup$ – Jannis Andreska Apr 10 '14 at 18:01
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Let's take a (2-3) spinel ($\ce{A^{II}B^{III}_{2}O4}$) as an example. If it is normal or inverse can be predicted from the total crystal field stabilization energies. For a normal spinel, the total CFSE wold be: $$\Delta_{T}(A)+2\Delta_{O}(B)=\Delta_{T}(A)+\Delta_{O}(B)+\Delta_{O}(B)$$ and for an inverse spinel $$\Delta_{O}(A)+\Delta_{T}(B)+\Delta_{O}(B)$$ with $\Delta_{T}$ being the tetrahedral and $\Delta_{O}$ being the octahedral splitting energy of $\ce{A^2+}$ and $\ce{B^3+}$, respectively.

If $\Delta_{T}(A)+\Delta_{O}(B)>\Delta_{O}(A)+\Delta_{T}(B)$, the normal spinel structure should be favored, as the resulting CFSE is higher. If it is the other way around, an inverse spinel is to be expected.

To sum it up, if the CFSE gained by placing $\ce{A^2+}$ in the octahedral holes outweighs a potential destabilization caused by putting a part of $\ce{B^3+}$ in the tetrahedral holes, an inverse spinel can be formed. However, a normal spinel is favorable when $\ce{B^3+}$ clearly "prefers" the octahedral and/or $\ce{A^2+}$ the tetrahedral holes.

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