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After hearing that hexaamminecobalt(III) and hexafluoridonickelate(IV) are both low-spin due to the central metals' high oxidation states (even though both bear pi-donor ligands), I began to wonder if the percobaltate anion, i.e. tetraoxidocobaltate(V), is also low-spin. It is already known that even tetranorbonyliron(IV) is low-spin (compare: iron(III) is usually high spin even in octahedral environments), so maybe the high oxidation state of cobalt in tetraoxidocobaltate(V) can make an 'exception' to the rule that certain types (i.e. with pi-donors or tetrahedral coordination) of complexes are always high-spin.

Is my reasoning correct, and is there any general piece of data in the literature in this manner?

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The answer, at least for potassium tetraoxidocolbate(V) in dilute tripotassium phosphate solid solution, appears to be high-spin despite the high oxidation state of the cobalt.

Brendel and Klemm[1] oxidized a 3:1 atomic mixture of potassium and cobalt (the metals being in the form of oxides) with $\ce{O2}$ and obtained a product solid-phase soluble in $\ce{K3PO4}$, which is rendered as the isostoichiometric and isostructural $\ce{K3CoO4}$. To assess the electronic spin state, the magnetic moment per cobalt atom was obtained as a function of composition of the solid solution:

$3.2$ a/o* $\ce{Co}\implies 5.1$ Bohr Magnetons

$8.1$ a/o $\ce{Co}\implies 4.3$ Bohr Magnetons

$56$ a/o $\ce{Co}\implies 2.6$ Bohr Magnetons

*a/o = atomic percent, $\ce{Co}/\ce{Co + P}$

The theoretical magnetic moment for four unpaired electrons, meaning high-spin in the tetrahedral $d^4$-core complex, is $\sqrt{24}\approx4.9$ Bohr Magnetons, and the most dilute solution is taken to agree with that figure. More concentrated cobaltate solutions appear sysceptible to antiferromagnetic coupling of cobalt atoms from neighboring $\ce{CoO4^{3-}}$ ions.

The high-spin outcome is unusual for such a highly oxidized metal core, but favored by several factors:

  • The complex is tetrahedral rather than octahedral, which cuts the cryatal field splitting by more than half. It also renders the low-spin configuration for a $d^4$ core with zero rather than two unpaired electrons, which feeds into the next factor.

  • Exchange energy is strongly more favorable for four rather than zero unpaired electrons. This effect is magnified by the quadratic dependence of exchange energy on unpaired electron count.

  • The oxide ion ligands are stronger pi donors than fluoride ligands, due to the greater negative ligand charge and higher orbital energies of oxide versus fluoride. As described here this tends to favor oxo complexes of high-oxidation-state transition elements with few $d$ electrins in their core, and cobalt(V) is a suitable case of this effect.

Reference

  1. Brendel, C. and Klemm, W. (1963), Weitere Versuche zur Darstellung von Kaliumcobaltat (V). Z. anorg. allg. Chem., 320: 59-63. https://doi.org/10.1002/zaac.19633200109
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In tetraoxidocobaltate(V), coordination no. of cobalt is 4, has d4 configuration, strong field ligand as Δt is higher than pairing energy, hence (e)2 (t2)2 splitting configuration so the hybridisation is dsp2 with square planar geometry.

The no. of unpaired electrons is four so the magnetic moment

enter image description here μ = √24B.M.

Therefore it is a high spin complex.

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    $\begingroup$ Unfortunatly this does not seem very coherent. The answer begins with arguments that the complex would be low-spin (strong field ligand, high oxidation stare) but then claims the complex is high spin. $\endgroup$ Oct 19, 2023 at 10:41
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    $\begingroup$ Please see Why it is wrong to use the concept of hybridisation for transition metal complexes $\endgroup$
    – Ian Bush
    Oct 19, 2023 at 11:09
  • $\begingroup$ 🆗 I understood that valence bond theory and crystal field theory cannot always completely explain some complexes but I still have to learn more about molecular orbital theory. $\endgroup$ Oct 20, 2023 at 9:51
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    $\begingroup$ Valence bond theory can definitely explain any complex correctly, but it might not be convenient. Crystal field theory on the other hand is outdated and incomplete. Please don't lump the two together. $\endgroup$ Oct 20, 2023 at 20:14
  • $\begingroup$ But CFT is a damned sight better than hybridisation $\endgroup$
    – Ian Bush
    Oct 21, 2023 at 6:41

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