There's a lot of ambiguity in the hybridisation of $\ce{Co^3+}$ in its hexaaqua complex; is it a high spin or low spin complex? As theoretically the CFSE is less than P so it should be low spin, but in a lot of places, it's assumed to be high spin.
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$\begingroup$ Similar question was asked earlier.. $\endgroup$– ZenixCommented May 1, 2020 at 14:33
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1 Answer
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Holleman/Wiberg states (translation by me)
$\ce{Co^{3+}}$ usually forms octahedral low-spin complexes, because it is the only way to achieve a high ligand-field stabilization energy. Octahedral high-spin complexes are typically an exception; they are formed with the "weakest" ligands (fluoride): $\ce{[CoF6]^{3-}}$, $\ce{[CoF3(H2O)3]}$ (the ion $\ce{[Co(H2O)6]^{3+}}$ is in the low-spin state).
and further
That octahedral $\ce{Co^{3+}}$ complexes from a magnetic standpoint are in the low-spin state most of the time, while complexes of the isoelectronic $\ce{Fe^{2+}}$ prefer the high-spin state most of the time, is due to the higher amount of ligand field splitting, following the increased charge on $\ce{Co^{3+}}$.
Source: Holleman, Wiberg: Lehrbuch der Anorganischen Chemie, 101. edition, de Gruyter, 1995.
