I'm clear with the concepts of crystal field theory. But I can't figure out the exact reason why the hybridisation of manganese in potassium permanganate ($\ce{KMnO4}$) is $\mathrm{d^3s}$. Can anyone explain please?


This is formally a manganese(VII) compound and hence there are no 3d electrons.

The four $\ce{O^2-}$ ions are considered to be donating two electrons each to the atomic orbitals. Tetrahedral "hybridisation" can be achieved by using the $\mathrm{4s},~\mathrm{3d}_{xy},~\mathrm{3d}_{yz}$, and $\mathrm{3d}_{xz}$ AOs. Since in Mn(VII) the $\mathrm{3d}$ AOs are lower in energy than the $\mathrm{4s}$ and $\mathrm{4p}$ AOs, the "hybridisation" is best considered as $\mathrm{d^3s}$.

I have put "hybridisation" in quotes because it is not used in advanced chemistry (it is an artifact!) but it is correct in saying that the bonding MOs contain mostly $\mathrm{4s},~\mathrm{3d}_{xy},~\mathrm{3d}_{yz}$ character but they'll also probably mix some $\mathrm{4p}$ character in there for good measure!


I feel a bit like a broken record: Do not use hybridisation to describe coordination compounds; it is not helpful.

However, the case of permanganate stands out as something where hybridisation is even more unhelpful than in the traditional ‘do not use it’ cases. Refer to the qualitative molecular orbital scheme of permanganate in figure 1.

orbital scheme tetrahedric complex
Figure 1: Qualitative MO scheme of a tetrahedric complex with σ and π bonding between metal and ligands. Double vertical lines represent electron pairs.

This complex ion cannot be understood well without π contribution of the four oxido ligands. The p-type orbitals of the oxygens transform, as the image shows, as $\mathrm{a_1 + e + t_1 + 2 t_2}$ which means that all of the orbitals of manganese that are not core orbitals ($\mathrm{3d, 4s, 4p}$) will participate in bonding to some extent. Especially the $\mathrm{t_2}$ orbitals are frustrating since oxygen has two energetically degenerate sets of them while manganaese offers another two sets (however non-degenerate; consisting of $\mathrm{3d}$ and $\mathrm{4p}$ orbitals) of them.

On the other hand, however, the analysis of populated orbitals will tell us that all occupied orbitals are ligand-centred orbitals. This again makes it a lot weirder to attempt to define any specific hybridisation for manganese: after all, it is in its $\mathrm{+VII}$, valence-electron less oxidation state.

If you really had to determine a hybridisation, anything short of $\mathrm{d^5sp^3}$ is not in line with the actual picture. That in turn cannot explain the tetrahedral geometry. You lost, go back.

Tl;dr: do not attempt to explain coordination complexes with hybridisation. Ever!

  • 3
    $\begingroup$ Everything here is true, but I disagree somewhat with "it is not helpful." For students who have not yet learned about group theory and delocalized MOs, introducing $d^3s$ hybridization is a way to teach that 1) $d_{xy},d_{yz}$ and $d_{xz}$ can combine with $s$ in a tetrahedral $\sigma$ bond network just as well as $p_x,p_y,p_z$ can, and 2) empty $3d$ orbitals contribute more to $\sigma$ bonding with ligands than $4p$ in tetrahedral metal complexes. The fact that the filled "bonding" orbitals are largely ligand in character and that the $p$ orbitals can mix in is a second-level consideration. $\endgroup$ – Andrew Feb 1 '19 at 11:47
  • 1
    $\begingroup$ @Andrew I want to argue quite the opposite: if the students haven’t even learnt about MO theory, what the heck are they attempting to learn about the bonding situation in these complexes. Don’t teach half-facts which are wrong or lead to false conclusions more often than not. Group theory, on the other hand, isn’t really needed and can be added later as a shortcut technology. $\endgroup$ – Jan Feb 2 '19 at 6:39
  • 3
    $\begingroup$ -It's typical to teach valence bond theory (ie bonds formed by electrons in localized hybrids of AO's) long before any discussion of delocalized MOs and actual symmetry-allowed orbital mixing. Hybridization is deemed useful primarily to explain how structurally equivalent bonds are electronically equivalent and to connect AO shapes to molecular geometries (eg linear = sp, Td = sp3 or sd3, Oh = sp3d2 or sd5, etc). Where you draw the line between pedagogically useful oversimplification and misleadingly false is subjective, but it might be a good discussion for this forum. $\endgroup$ – Andrew Feb 2 '19 at 15:00

yes because there is d3s hybridisation in this complex and in d3s hybridisation only dxy, dyz and dzx atomic orbitals can only participates in this hybridisation so only three AO's out of five are utilised in this and according to CFT energy difference between t2g and next s orbital is less so hybridisation can be easily occurs

  • $\begingroup$ Please see the other answers as to why hybridisation should not be used with d block complexes -and arguably not at all $\endgroup$ – Ian Bush Feb 1 '19 at 8:29

Not the answer you're looking for? Browse other questions tagged or ask your own question.