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According to my book there are no unpaired electrons in $\ce{NiF6^2-}$. However, that should not be the case since fluoride is a weak field ligand. Why does hexafluoronickelate(IV) behave as a low spin complex?

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This is almost a duplicate of this question. There it is shown that the Crystal Field Stabilization Energy (CFSE) is at a maximum for $\mathrm{d^6}$ ions, and low spin is strongly favoured unless there is a very small splitting energy, as is the case when you have

  • Low charge early first row transition metals - for instance $\ce{Fe^{2+}}$
  • Very weak field ligands

Without experimental evidence or high quality ab initio calculations you can't say a priori whether a given complex is high or low spin, but in practice for $\mathrm{d^6}$ species low spin massively dominates, the only exceptions being some $\ce{Fe^{2+}}$ species and a very small number of fluoro complexes of $\ce{Co^{3+}}$. Thus $\ce{NiF6^2-}$ is low spin for the same reasons that $\ce{Co^{3+}}$ is almost always low spin, namely that the large CFSE for $\mathrm{d^6}$ strongly favours low spin complexes, reinforced by the fact that $\ce{Ni^{4+}}$ is smaller and higher charged than $\ce{Co^{3+}}$, so resulting in a larger splitting energy.

[Yes, I know that $\ce{Ni^{4+}}$ and indeed $\ce{Co^{3+}}$ is unrealistic, but this is all argued within a crystal field framework, an ionic model]

Edit: I've just seen Why is hexafluoridonickelate(IV) diamagnetic? which is a duplicate of this question, and the answer there goes into the theory a bit more deeply than the very simple Crystal Field Theory I have used here

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