Counting d-electrons in M(I) transition metals

[Mayhem]

Formally, d-electron count is calculated by subtracting the group number of the metal by its oxidation state.

If we limit ourselves to the first row transition metals, the group number - 2 = #d-electrons

Co(I) (group 9) only loses one electron. Its electron configuration becomes [Ar]3d⁷4s¹.
Thus our d-electron "count" involves an electron in the s-orbital. This cannot be insignificant. In practice, is the 4s electron counted as a d-electron formally or is it "demoted" into a d-orbital when it is unpaired via oxidation?

 

[Borek]

Sorry for being late, I hoped someone will have better answer than my somewhat handwavy treatment.

My understanding (which can be wrong, QM was never my forte) is that energies of these orbitals are so close there is no way to say where the electron "resides" and detailed discussion is a moot.

In a way assigning electrons to orbitals is just a simplification stemming from the single electron model. Schroedinger equation gives a single wavefunction describing all electrons at once, and it doesn't put them in separate boxes. Yes, thinking in terms of electron configuration is a valuable tool, but as every simplified model it sometimes fails.

 

[Mayhem]

Thank you Borek.

Indeed, the reality of electrons is much stranger than how we model it in chemistry.

But does this have any implications for spectroscopy, i.e. would Co(I) complexes have different allowed electronic transitions than other d⁸ equivalent complexes.

 

[Borek]

In complexes it is much better to speak about MO, not AO.

Generally speaking, if you were trying to predict spectra that could be the case, but my bet is that this is a bit different situation - in complexes ligand field "distorts the orbitals"* and changes their relative energies, so electron assignment to these "orbitals" is much easier and much better defined, that's why these complexes exist.

*Quotes to avoid being inconsistent and write "orbitals exist" and "orbitals don't exist" in neighboring posts

 

[DrDu]

Somewhat oversimplified, I think you can assume that all valence electrons in d-block elements are d-electrons if the atom has a positive oxidation number. The reason is that with increasing atomic charge, the atom gets more hydrogen like and the aufbau principle will be followed (i.e. d will be filled before s of the next shell. A semiquantitative understanding is possible applying the so-called "Slater-rules".