Why is it wrong to use the concept of hybridization for transition metal complexes?

I have asked a lot of questions on coordination chemistry here before and I have gone through a lot others here as well. Students, including me, attempt to answer those questions using the concept of hybridization because that's what we are taught in class and of course it's easier and more intuitive than crystal field theory/molecular orbital theory. But almost all of the times that I attempted to use the concept of hybridization to explain bonding, somebody comes up and tells that it's wrong.

How do you determine the hybridisation state of a coordinate complex?

This is a link to one such question and the first thing that the person who answered it says: "Again, I feel a bit like a broken record. You should not use hybridization to describe transition metal complexes."

I need to know:

• Why is it wrong? Is it wrong because it's oversimplified?
• Why does it work well while explaining bonding in other compounds?
• What goes wrong in the case of transition metals?

Let's consider, for example, a tetrahedral Ni(II) complex ($$\mathrm{d^8}$$), like $$\ce{[NiCl4]^2-}$$. According to hybridisation theory, the central nickel ion has $$\mathrm{sp^3}$$ hybridisation, the four $$\mathrm{sp^3}$$-type orbitals are filled by electrons from the chloride ligands, and the 3d orbitals are not involved in bonding.

Already there are several problems with this interpretation.

1. The 3d orbitals are very much involved in bonding - a cursory glance at a MO diagram will show that this is the case.

2. If the 3d orbitals are not involved in bonding at all, then they should remain degenerate. OK, maybe they split because of the ligands - but that is crystal field theory, not hybridisation theory. This means that, without bringing in CFT or MOT, we are now unable to rationalise any of the characteristics of such transition metal complexes, e.g. electronic spectra and kinetic stability towards ligand substitution.

3. There is no way to predict whether the geometry will be tetrahedral (e.g. $$\ce{[NiCl4]^2-}$$) or square planar (e.g. $$\ce{[Ni(CN)4]^2-}$$, or $$\ce{[PdCl4]^2-}$$). It is certainly possible to carry out actual experimental work on the compound to determine its geometry, and hence determine its hybridisation. However, there is no way to predict the geometry a priori using hybridisation theory only.

Moving on to Ni(II) octahedral complexes, like $$\ce{[Ni(H2O)6]^2+}$$, the typical explanation is that there is $$\mathrm{sp^3d^2}$$ hybridisation. But all the $$\mathrm{3d}$$ orbitals are already populated, so where do the two d orbitals come from? The $$\mathrm{4d}$$ set, I suppose.

1. Points 1 and 2 from the tetrahedral case above still apply here.

2. The involvement of $$\mathrm{4d}$$ orbitals in bonding is simply not plausible, as these orbitals are energetically inaccessible. On top of that, it is unrealistic to expect that electrons will be donated into the 4d orbitals when there are vacant holes in the 3d orbitals.

For octahedral complexes where there is the possibility for high- and low-spin forms (e.g. $$\mathrm{d^5}$$ $$\ce{Fe^3+}$$ complexes), hybridisation theory becomes even more misleading:

1. Hybridisation theory implies that there is a fundamental difference in the orbitals involved in metal-ligand bonding for the high- and low-spin complexes. However, this is simply not true (an MO diagram will illustrate this point). Again, the notion of 4d orbitals being involved in bonding is simply unrealistic.

2. Again, hybridisation theory provides no way of predicting whether a complex is high- or low-spin. Yes, maybe you could talk about the spectrochemical series, but hybridisation cannot rationalise or predict the spectrochemical series (which requires MOT).

Hybridisation theory is simply inadequate – it does not explain the rich chemistry of the transition metals and their complexes. Even something simple, such as the d-d* transitions responsible for the colours of complexes, cannot be explained by it – let alone complex magnetic behaviour, electronic spectra, reactivity, ... Hybridisation theory is really nothing more than a backwards rationalisation of experimental observations, but it is unable to account for a good deal of said observations.

On top of that, it is actually outright incorrect in certain aspects, as detailed above. In a sense it is similar to the Bohr model. It was the best explanation that we had at some point in time, but it has simply been superseded by better and more complete theories.

All in all, hybridisation theory is a post hoc rationalisation of experimental data. It has little, if any, use in predicting the properties of previously unseen complexes. And when it is correct, it is often correct for the wrong reason, as it makes use of highly unrealistic models of bonding (e.g. the preferential use of 4d orbitals over 3d orbitals in bonding).

Now, a digression. You mentioned that hybridisation works well for "other compounds". That is really not always the case, though. For simple compounds like water, etc. there are already issues associated with the standard VSEPR/hybridisation theory. Superficially, the $$\mathrm{sp^3}$$ hybridisation of oxygen is consistent with the observed bent structure, but that's just about all that can be explained. The photoelectron spectrum of water shows very clearly that the two lone pairs on oxygen are inequivalent, and the MO diagram of water backs this up. Apart from that, hybridisation has absolutely no way of explaining the structures of boranes; Wade's rules do a much better job with the implicitly delocalised bonding.

And these are just Period 2 elements - when you go into the chemistry of the heavier elements, hybridisation generally becomes less and less useful a concept. For example, hypervalency is a huge problem: $$\ce{SF6}$$ is claimed to be $$\mathrm{sp^3d^2}$$ hybridised, but in fact d-orbital involvement in bonding is negligible. On the other hand, non-hypervalent compounds, such as $$\ce{H2S}$$, are probably best described as unhybridised - what happened to the theory that worked so well for $$\ce{H2O}$$? It just isn't applicable here, for reasons beyond the scope of this post.

Hybridisation can sometimes be useful, perhaps most notably in organic chemistry, and especially so when one expands it beyond the simple $$\mathrm{sp}^m\mathrm{d}^n$$ cases where $$m,n$$ are integers. But there are caveats. It is important to recognise that it is not atoms that are hybridised, but rather orbitals: for example, the $$\ce{C-C}$$ bonds in cyclopropane are $$\mathrm{sp^5}$$-hybridised and the $$\ce{C-H}$$ bonds are $$\mathrm{sp^2}$$. Finally, nowadays we have better tools to understand the chemistry of the transition metals, and it would be unwise to continue using hybridisation theory in a setting where it has been shown to be obsolete.

• In case of Ni octahedral complex what do you mean that the 4d orbitals are "energetically inaccessible"? Is the energy gap between the 4s/4p and 4d orbitals that high so that they cannot form hybrid orbitals? – Arishta Jun 25 '17 at 8:24
• No, it is not really to do with hybridisation. Broadly speaking, if you want two orbitals to form a strong bonding interaction, you need two criteria: 1) strong overlap 2) small difference in energy. The 4d orbitals are high in energy and consequently interact very poorly with ligand-based orbitals; that's what is meant by "energetically inaccessible". – orthocresol Jun 25 '17 at 8:29

For the most part, coordination bonds involve d electrons from the transition metal interacting with some other electrons on a separate molecule. Hybrid orbitals are constructed from a single atom's orbitals mixing. If you do hybridization for the metal, then you are mixing its s, p, and d orbitals together in attempt to explain geometry. But ultimately, it is incorrect as really only the d orbitals interact. That's why people get upset when you use hybridization theory for coordination bonds.

Hybridization theory states that methane's four sigma bonds are equal (sp3).

However, if you apply group theory to the four sigma bonds (use the character table for Td point-group), the irreducible representation, if memory serves, = A1 + T2 (A1=single and T2 = a triple degeneracy).

H.T. is a nice tool to help understand introductory molecular geometry and/or o-chem, but, as it is superceded by M.O. theory, it has absolutely no basis in reality.

There is a very good description of a square planar complex above. The simple answer to your question: Almost every d8 transition metal complex is square-planar.

Regarding the rest (this also depends upon the d electron count and is a generalization ), Oh is more favorable than Td, and Td is more favorable than SP. Again, take this with a grain of salt, but it is a decent starting point. Consider the #of ligands and the # of occupied co-ordination sites on the TM center resulting from a L occupying a single co-ordination site--then remember that if a ligand can occupy multiple CN sites, it typically will do so. Regarding the above 'favorability generalisation', remember that lower CN complexes are less stable and, thus, more open to attack because of unoccupied CN sites--think about a medieval town vs a walled medieval town; more L or more occupied co-ordination sites= bigger wall (our TM friend wants to have the largest wall allowed by the system). More often than not, your textbook's and/or prof's examples will rely upon a strict adherence to the 18 electron rule, so keep this in mind when trying to determine whether or not the addition of a MM bond and/or consideration of a dimer is apt (see below).

You should also consider whether or not a M-M bond is favorable (this is typically encountered with odd # d TM complexes--this is not an absolute) because then you would need to consider whether or not dimerization has occured (remember though that poly-dentate ligands may also result in the formation of multiple metal-center complexes that lack MM bonds). H.T. completely fails here--this is why you should learn how to determine a molecule's point-group and ignore H.T.