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Will the hybridisation of of $\ce{Co}$ in $\ce{[Co(C2O4)3]}^{3-}$ be $\mathrm{sp^3d^2}$ or $\mathrm{d^2sp^3}$? The question in which I found it has mentioned the answer to be $\mathrm{d^2sp^3}$ while I was thinking of the other one.

According to me, oxalate is a weak field ligand and it should not cause the pairing of electrons in the half-filled $\mathrm d$ orbitals.

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    $\begingroup$ Oxalate is also a chelating ligand, it is not so weak as you might be thinking. It can be thought as very slight strong field ligand. So it is in the competition region of $\Delta _o$ with pairing energy. Unless you have experimental data, you can not tell anything whether it will be inner orbital or Outer orbital complex. $\endgroup$
    – Soumik Das
    Commented Apr 4, 2018 at 7:25
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    $\begingroup$ I would avoid using hybridization to describe transition metal compounds. See chemistry.stackexchange.com/questions/76726/… $\endgroup$
    – Ian Bush
    Commented Apr 4, 2018 at 8:28
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    $\begingroup$ @IanBush Such questions form an important part of undergraduate admission tests here and thus I needed to know what should the answer be. All the textbooks at my level use the concept of hybridisation. $\endgroup$
    – Piyush Maheshwari
    Commented Apr 4, 2018 at 8:50
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    $\begingroup$ @PiyushMaheshwari Then all your textbooks are wrong, as it is completely wrong to use hybridisation for transition metal complexes. $\endgroup$
    – Martin - マーチン ♦
    Commented Apr 4, 2018 at 12:07
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    $\begingroup$ @PiyushMaheshwari of course I did read that, however, I don't need to try to teach you something that is completely wrong, in order to teach you what is correct. (I'm too short on time to do that right now.) The fact that they still teach this outdated and wrong theory is sad, and if you need it to pass the exam, that does make it worse. Please forgive me, but you should immediately forget this approach on hybridisation as soon as you passed this exam. And further on question everything else in your textbooks. $\endgroup$
    – Martin - マーチン ♦
    Commented Apr 4, 2018 at 18:32

1 Answer 1

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As explained in The Chemistry of Iron, Cobalt and Nickel: Comprehensive Inorganic Chemistry, at pages 1104 and 1105:

almost all cobalt(III) complexes are low spin ... only with fluoride ions as ligands are high spin complexes ... found

In table 10, the book specifically lists $\ce{[Co(C2O4)3]}^{3-}$ as low spin and cites to J. Chem. Soc. (A) (1966) 798.

Therefore, according to the historical valence bond theory of transition metal complexes, it would be considered $\mathrm{d^2 sp^3}$ for the following reason:

There are (in the context of the theory) two unoccupied $\ce{3d}$ atomic orbitals in $\ce{Co^{3+}}$.

Six pairs of electrons from the oxalate ions are added to the two $\mathrm{3d}$ orbitals and the $\mathrm{4s}$ and three $\mathrm{4p}$ orbitals.

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