I know water is in the middle of the spectrochemical series generally being the differentiator between the weak field and strong field ligands.

I have generally seen water being a weak field ligand, but in the case of hexaaquacobalt(II), it pairs up the electrons of $\ce{Co^3+}$ in the complex $\ce{[Co(H2O)6]^3+}.$ Why is water acting as a strong field ligand here?

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    $\begingroup$ Cobalt(II) or (III)? The question seems to mix them up. $\endgroup$ Nov 6, 2022 at 12:35
  • $\begingroup$ If it's Co(III) chemistry.stackexchange.com/questions/112675/… gives the answer, and explain why almost all Co3+, and in fact the vast majority of d6, complexes are low spin $\endgroup$
    – Ian Bush
    Nov 6, 2022 at 18:51

1 Answer 1


Whether a complex is high or low spin does not solely depend on the ligands attached. There are other factors such as the size of the metal, the charge on the metal, the coordination of ligands, and so on, that can determine whether a complex exhibits high or low spin in Crystal Field Theory.

In this case, while $\ce{H2O}$ is lower on the spectrochemical series, that doesn't imply that the complex should have a low splitting and thus be high spin.

For example, $\ce{[Co(H2O)6]^{2+}}$ tends to adopt a high spin configuration. In fact, $\ce{[Co(H2O)_{6-n}(NH3)_n]^2+}$; $n=0-6$ was shown to favour high-spin configurations through DFT calculations even though $\ce{NH3}$ is considered a stronger ligand than water [1].

Therefore, $\ce{[Co(H2O)6]^3+}$ being high spin and diamagnetic doesn't mean that the ligand $\ce{H2O}$ is a strong field ligand. It just implies that all factors considered, the complex is more stable in a low-spin configuration.


[1] https://pubs.acs.org/doi/abs/10.1021/ic0257930


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