Are the triply-bonded carbons in pyridyne (and benzyne) sp-hybridised?

I was given this molecule (3,4-pyridyne) in a test:


[![3,4-Pyridyne](https://i.stack.imgur.com/BA4yX.png)](https://i.stack.imgur.com/BA4yX.png)


I had to find out if it had any atoms which had a hybridization state other than sp2. The solution said that it doesn't have any. But aren't the triply-bonded carbon atoms sp-hybridized?


Edit:
Benzyne also has a similar structure, with a distorted triple bond -


[![Benzyne](https://i.stack.imgur.com/OLir5.png)](https://i.stack.imgur.com/OLir5.png)




**TL;DR It is sp-hybridised, but maybe not in the way we normally think of alkynes as being.**




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There are a couple of things to say before going into the actual answer.


Firstly, hybridisation is derived from geometry, not from the bonding pattern (in other words, how many σ- and π-bonds there are). In most cases, the bonding pattern dictates the geometry: for example, most saturated carbons are tetrahedral, and a tetrahedral geometry usually signals sp3 hybridisation, so saturated carbons are usually described as being sp3-hybridised. So, *effectively,* the bonding pattern does dictate hybridisation. This is all fine, but only up to a certain point, because it can break down in exceptional cases such as cyclopropane. Because of the geometrical constraints of a three-membered ring, there is not quite enough room for the carbon to be perfectly tetrahedral, and so it is not correct to label the carbons as being sp3-hybridised in the usual sense.


Secondly, hybridisation is usually more nuanced than "sp", "sp2" or "sp3". It is individual orbitals that are hybridised, not atoms; so for example, we can talk about an sp orbital, which is a linear combination (a sum) of one part s-orbital and one part p-orbital. When we say that "a carbon atom is sp-hybridised", what we *really* mean is that the bonding of that carbon atom can be described using four valence orbitals: two sp-hybridised orbitals, and two unhybridised p-orbital. Likewise, an "sp2 carbon atom" uses three sp2-hybridised orbitals and one unhybridised p-orbital; and an "sp3 carbon atom" has four sp3 orbitals.


Again, in the vast majority of cases, these three categories are sufficient to accurately classify all carbon atoms in a molecule. However, as you may guess, this breaks down in exceptional cases: cyclopropane is the most well-known example. Here, each carbon has two sp5(ish) orbitals and two sp2(ish) orbitals. It uses the sp5 orbitals to form σ-bonds to the carbons next to it, and the sp2 orbitals to form σ-bonds to the two hydrogens on it. *"How is it possible to have five p orbitals?"* sp5 does not literally mean one s orbital and five p orbitals; it just means the ratio of s-orbital and p-orbital contribution to this particular hybrid orbital is 1:5.




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With that out of the way, here is how **benzyne** works (PBE0/cc-pVTZ gas phase optimisation + NBO analysis). We're going to look at the orbitals that the triply bonded carbon (C1) uses for bonding to its neighbours (C2 and C6, see structure below):†


* One sp orbital (50.02% s, 49.75% p) for forming a σ-bond to C2;
* One p orbital (0.39% s, 99.38% p) for forming the π-bond to C2 in the plane of the molecule;
* One p orbital (0.00% s, 99.84% p) for forming the π-bond to C2 perpendicular to the plane of the molecule;
* One sp orbital (49.64% s, 50.17% p) for forming a σ-bond to C6.


On their own, the numbers don't tell you which direction the p-orbitals are pointing in, but you can use visualisation software to figure this out. From left to right, these are the (σ- or π-, not the individual hybrid) orbitals, in the same order as listed above.


[![NBO orbitals in benzyne, visualisation with Chemcraft](https://i.stack.imgur.com/C0dH8.png)](https://i.stack.imgur.com/C0dH8.png)


Based on the above, the orbitals can be crudely illustrated as such:


[![NBO orbitals in benzyne, drawn in ChemDraw](https://i.stack.imgur.com/kL9NE.png)](https://i.stack.imgur.com/kL9NE.png)


This is, in fact, pretty much what an organic chemist would describe as an "sp-hybridised carbon atom". **However,** it is not sp-hybridised in the same sense as an ordinary alkyne. In an ordinary alkyne, the σ-bond between the two triply bonded carbons comes from overlap of sp-orbitals. This is also the case here, but there is a distinction. In an ordinary alkyne, the sp-orbitals point directly towards each other. Here, they are not quite aligned in the same way; they instead form what is sometimes called a "bent bond". The same can be said of the C1–C2 π-bond.


This is fundamentally because of the geometry of the ring. Since we have already allocated two p-orbitals to two other bonds, the two sp-orbitals must be derived from the same p-orbital. As a consequence, the two sp-orbitals must point in diametrically opposite directions, 180° away from each other. If C1 were to use an sp-orbital pointing directly towards C2, then the remaining sp-orbital would point rather far away from C6, significantly weakening the corresponding bond with C6. The orbital geometry that you see is a compromise.


The triply bonded carbons in **pyridyne**, for all intents and purposes, can be described in the same way as those in benzyne. The effect of nitrogen on the orbitals above is tiny; qualitatively they look the same (the colours are flipped for the first and last ones relative to the benzyne orbitals earlier, but this is of [no](https://chemistry.stackexchange.com/q/85589/16683) [physical](https://chemistry.stackexchange.com/q/35212/16683) [significance](https://chemistry.stackexchange.com/a/67344/16683)).


[![NBO orbitals of 3,4-pyridyne, visualisation in Chemcraft](https://i.stack.imgur.com/koKET.png)](https://i.stack.imgur.com/koKET.png)




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† The numbers don't add up to exactly 100% because the tiny d- and f-orbital contributions are neglected.


 5

No.


Hybridization derives from geometry, not bond type. For instance, the central carbon in allene ($\ce{CH2=C=CH2}$) has $sp$ hybridization like the carbons in acetylene, despite the absence of triple bonds in the allene, because the linear bonding geometry brings this hybridization about.


In pyridyne the formally triple-bonded carbons do not have the linear bonding geometry corresponding to $sp$ hybridization, rather the geometry corresponds to $sp^2$ hybridization like pyridine or benzene. What seems to be one of the "pi" bonds in the pyridyne structure as drawn on paper, is in fact a "bent bond" formed by the side-by-side overlap of two subsidized orbitals, not a pi bond formed from two $p$ orbitals.


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