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Lets say I have to find hybridisation of $\ce{[Ni(CO)_4]}$ and $\ce{[Ni(CN)_4]^{2-}}$.

The metal atom/ion in these compounds are $\rm Ni$ and $\rm Ni^{2+}$ respectively.

So the outer shell configuration for $\rm Ni$ in $\ce{[Ni(CO)_4]}$ would be $$\rm 4s^2 \ \ 3d_{xy}^2\ 3d_{yz}^1\ 3d_{zx}^1\ 3d_{z^2}^2\ 3d_{x^2 - y^2}^2 \ \ 4p^0 \ \ 4d^0$$

The electrons from s oribital will jump to d orbital and so I expect $\rm CO$ to donate electron pairs in $\rm 4p$ and $\rm 4s$ orbitals and form $\rm sp^3$ hybridisation.

The outer shell configuration for $\rm Ni^{2+}$ in $\ce{[Ni(CN)_4]}^{2-}$ would be $$\rm 4s^2 \ \ 3d_{xy}^0\ 3d_{yz}^1\ 3d_{zx}^1\ 3d_{z^2}^2\ 3d_{x^2 - y^2}^2 \ \ 4p^0 \ \ 4d^0$$

Here also I expect the electrons to jump from s to d and hybridsation to be $\rm sp^3$, but the answer given here is $\rm dsp^2$. Why ?

I am heavily confused with this hybridsation stuff, is there a systematic and logical way of knowing hybridsation of coordination compounds ?


I tried to search this topic on internet and I came across $H = \dfrac12(V + M - I)$ where $V$ is number of valence electron, $M$ is number of single bonds and $I$ is charge on the atom. Here if $H= 2$ then it is $\rm sp$ , $H = 3$ then $\rm sp^2$ and so on.

For $\ce{[Ni(CO)_4]}$, here $V = 10$, $M = 4$ and $I = 0$, so $H = 7$ hence $\rm sp^3d^3$.

Also for $\ce{[Ni(CN)_4]}^{2-}$, here $V = 8$, $M = 4$ and $I = -2$, so $H = 7$ hence $\rm sp^3d^3$.

As it can be seen, this formula is complete failure.

Why this formula does not work for coordination compounds whereas it works for compounds like $\rm PCl_5,PCl_3$ and $\rm SF_6$ ?


In this answer the author makes the claim that $\ce{[NiCl_4]^{2-}}$ has hybridsation $\rm sp^3$ and $\ce{[Ni(H2O)6]^2+}$ has hybridsation $\rm sp^3d^2$. How did he know that ? This is my question not why hybridsation is not a valid theory to study coordination compounds.

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    $\begingroup$ I know there is a debate about whether this is a duplicate of Why is it wrong to use the concept of “hybridization” while studying complexes?. This question has been closed, reopened, and was about to be closed again. Let me state it very clearly: this is not a duplicate. The linked question asks why an incorrect theory is incorrect; this question asks about the incorrect theory itself. While the linked question is obviously relevant, it is not a duplicate. I say this not only as a moderator, but also as the answerer of both questions. $\endgroup$ – orthocresol Aug 26 '17 at 17:29
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Firstly, note that hybridisation theory as applied to transition metal complexes is an incorrect, flawed theory. It is an attempt to rationalise experimental observation (e.g. geometries), but in the process it invokes models of bonding which are highly unrealistic and demonstrably untrue. See: Why is it wrong to use the concept of hybridization for transition metal complexes?

Now, with that disclaimer out of the way, let's examine how that flawed theory works. In traditional hybridisation theory, tetrahedral and square planar 4-coordinate complexes are considered to be $\mathrm{sp^3}$ and $\mathrm{dsp^2}$ respectively. Since $\ce{[Ni(CO)4]}$ is tetrahedral and $\ce{[Ni(CN)4]^2-}$ is square planar, one can assign the hybridisations accordingly. If you have a textbook that still mentions this outdated theory, it should have a table telling you what hybridisation each geometry corresponds to. Here is an example from Housecroft's Inorganic Chemistry 4th ed. (p 667). To their credit, there is a subsequent section debunking this theory.

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These are generally based on post hoc justifications. Essentially, somebody looked at the geometry and then tried to piece together which orbitals could be combined to form such a geometry. So, I do not think there is a good way to predict them a priori. This formula you have dug up does not work very well, even for main group elements. For transition metals the factors deciding geometries are much more complex, and simply cannot be captured in one line of mathematics.


The real problem with the approach above is that you need to know the geometry before you can assign a hybridisation. (Don't even try to do it the other way round - it simply does not work because hybridisation is an incomplete theory. It is like trying to predict wave-particle duality using classical mechanics.)

If you cannot predict the geometry of the complex, then you cannot predict the hybridisation. You could argue that cyanide is a strong field ligand and in conjunction with a $\mathrm{d^8}$ metal, a square planar geometry is preferred. On the other hand, in $\ce{[Ni(CO)4]}$, the central atom is $\mathrm{d^{10}}$ and hence even with a strong-field ligand like carbon monoxide, a tetrahedral geometry is favoured, simply based on steric grounds.

However, this does nothing but introduce more questions. The term "strong-field" has no meaning outside of molecular orbital theory. Hybridisation theory has no inkling of what strong- or weak-field ligands are, and consequently cannot explain why $\mathrm{d^8}$ Ni(II) only prefers to go square planar with strong-field ligands. On top of that, it is not at all obvious from hybridisation theory how the electronic configuration of the central metal should favour a particular geometry. It's simply not possible for your formula to explain any of this.


So, to answer your questions directly:

Is there a systematic and logical way of knowing hybridisation of coordination compounds?

Yes. You use a more advanced theory (MOT) to predict the geometry, and then you use the table above to find the hybridisation. That's a step backwards though, you already used MOT and there's no reason why you should go back to hybridisation.

Why this formula does not work for coordination compounds?

Because hybridisation theory is an incomplete theory and cannot capture the subtle factors influencing geometry of coordination compounds.


Main group compounds

I guess I will elaborate a bit more on main group compounds. The formula you quoted seems to work for main group compounds, but let's get two things out of the way.

  1. It gives the correct hybridisations as espoused by old theory. $\ce{SF6}$ is not $\mathrm{sp^3d^2}$ hybridised because d-orbital involvement in bonding is negligible. So, this formula does not actually work.

  2. It does not predict the hybridisation without predicting the geometry.

Point 1 is important, but let's not go into that now. Point 2 may seem confusing to you. Clearly it is predicting the hybridisation, simply from the number of valence electrons, etc. etc.?

The truth is that this formula only "works" because these quantities (number of valence electrons, number of bond pairs, ...) allow one to predict the geometry, using VSEPR theory. For example, the structure of $\ce{CH4}$ is tetrahedral, and VSEPR theory is sufficient to predict that. Once the geometry has been established as tetrahedral, the hybridisation can then be correctly assigned as $\mathrm{sp^3}$.

So, this formula is skipping a step in between. In fact, the correct sequence is to predict the geometry using VSEPR theory, and then obtain the hybridisation. Now, with transition metal complexes, VSEPR does not work. Thus, it is no surprise that that formula also does not work.

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  • $\begingroup$ If the geometry is trigonal prismatic, then how can we decide between $\mathrm{sd^5}$ and $\mathrm{sp^3d^2}$ ? $\endgroup$ – user8277998 Aug 26 '17 at 15:30
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    $\begingroup$ To be honest I don't actually know. I suspect this is (as I wrote) not entirely possible to deduce, because these are post hoc justifications - meaning that instead of developing a hypothesis and testing it by experiment, people observed the geometry and tried to come up with a hypothesis that explains it - and in this case it just so happens that there is more than one scheme that explains it. $\endgroup$ – orthocresol Aug 26 '17 at 15:57

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