Hybridisation of cobalt in tris(oxalato)cobaltate(III)

Question

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.

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. — Soumik Das, Apr 04 '18 at 07:25
I would avoid using hybridization to describe transition metal compounds. See https://chemistry.stackexchange.com/questions/76726/why-is-it-wrong-to-use-the-concept-of-hybridization-while-studying-complexes — Ian Bush, Apr 04 '18 at 08:28
@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. — Piyush Maheshwari, Apr 04 '18 at 08:50
@PiyushMaheshwari Then all your textbooks are wrong, as it is completely wrong to use hybridisation for transition metal complexes. — Martin - マーチン, Apr 04 '18 at 12:07
@Martin-マーチン Did you notice the words "at my level"? This was meant to convey that I am not really familiar with the deeper aspects of the concept and its limitations. This is because we are not taught about it in so much detail yet. I don't say that my books are correct either. But to know what is wrong with this theory, we also have to know what is the theory all about. — Piyush Maheshwari, Apr 04 '18 at 18:20
@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 The fact that they still teach this outdated and wrong theory is sad, and if you need it to pass the exam, that does make it worse.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. — Martin - マーチン, Apr 04 '18 at 18:32
@SoumikDas The ion's ability to chelate has nothing to do with its place in the spectrochemical series (i.e. how much splitting it can cause). The chelating ability of ligands has got to do with the stability constants of the complexes they form. — Tan Yong Boon, Apr 08 '19 at 23:06
https://chemistry.stackexchange.com/questions/94582/hybridisation-of-cobalt-in-trisoxalatocobaltateiii — , May 21 '21 at 12:02
Valency on Co is +3 As C2O4^2- is a weak field ligand it cant cause pairing of electons but due to chelating effect So, it is d2sp3 — Dr. Debasish Ghosh, Dec 31 '21 at 05:39
*https://chemistry.stackexchange.com/questions/94582/hybridisation-of-cobalt-in-trisoxalatocobaltateiii* — , Aug 06 '23 at 06:20

score 7 · Accepted Answer · edited May 14 '21 at 11:34

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.

Hybridisation of cobalt in tris(oxalato)cobaltate(III)

1 Answers1