# What is the hybridization of the carbonyl oxygen in a carboxylic acid?

My professor sent out a review guide. One of the questions on this guide is

The carboxylic acid carbonyl oxygen is:
1. $\ce{sp^2}$
2. $\ce{sp^3}$
3. $\ce{H}$-bond donor
4. rapidly equilibrating between $\ce{sp^3}$ and $\ce{sp^2}$
5. 1 and 3
6. 1 and 4

I chose (4) because of the resonance that can be drawn with a negative oxygen and positive carbon. However apparently the answer is (6). I don't understand how something can be equilibrating between two states and still be one of those states.

Could anyone explain?

• There are almost no occasions where an oxygen atom is sp3 hybridised. In most of the cases, oxygen is sp hybridised (first order approximation) when its coordination number is one, and sp2 hybridised when its coordination number is two. In special occasion, coordination number is higher, it can adopt a different hybridisation. But the flaw of reasoning in the question is the flaw of the hybridisation concept your teacher is trying to teach you, which is at least incomplete and outdated. Sorry. – Martin - マーチン Apr 22 '15 at 17:30
• I don't understand how "6" can be correct. How can the carbonyl oxygen be $\ce{sp^2}$ (1) and also be rapidly equilibrating between $\ce{sp^3}$ and $\ce{sp^2}$ (4)? Sometimes in introductory classes the carbonyl oxygen is presented as being $\ce{sp^2}$ hybridized (even though as @Martin points out this is not really correct either). – ron Apr 22 '15 at 17:50
• I absolutely agree with @ron. I did not even pay much attention to that detail. I guess what is meant is (and in my opinion this is wrong) that within the resonance framework we could describe a doubly bonded oxygen (max sp2) and a singly bonded oxygen (approx sp3), where the former would be predominant. It's 3am here, and I cannot answer this question now, I hope someone else will do that, but if not, pingback to me will work and I'll try to make the time. – Martin - マーチン Apr 22 '15 at 17:57
• 'Equilibrating' is a nonsensical concept regarding wavefunctions. – Lighthart Apr 5 '16 at 4:02

I already pointed it out in the comments, but I believe it is time to give it some more thought and explanation. Let's deal with some conceptual issues first.

Hybridisation is a model, that can be used to describe a bonding situation. It is never cause for a certain geometric arrangement, it is always the result of a bonding arrangement. Strictly speaking it is not even necessary to be invoked, but it can be quite handy. Quoting from Aditya's answer - I could not have said it better:

First of all, 'hybridisation' is a hypothetical concept, i.e. orbitals don't really mix to form new orbitals, but 'hybridisations' are very successful in explaining structures of molecules and their physical properties. So knowing that a compound has $\ce{sp^2}$ hybridisation will just help you arrange the constituent atoms around the central atom. Physically it has no other meaning.
Note: Only orbitals can be hybridised, never compounds.

Having understood that, it is easy to see, that atoms in a molecule could use many different hybridisations schemes. There are some concepts, based on quantum chemical calculations, which would find the most appropriate fit between a Lewis structure and the molecular orbital approach. See for example this fine question: What is natural bond orbital theory used for? There is also a more quantitative approach, which is called Valence Bond theory, but that is beyond the scope of the answer.

Since this is just a mathematical concept, an equilibration between $\ce{sp^2}$ and $\ce{sp^3}$ would not change a thing. We can decide which way we want to look at it. Ron consequently raised the question in the comments:

I don't understand how "6" can be correct. How can the carbonyl oxygen be $\ce{sp^2}$ (1) and also be rapidly equilibrating between $\ce{sp^3}$ and $\ce{sp^2}$ (4)?

As I pointed out, (4) should not even be an option. The premise of the question is clearly flawed. Please do not take hybridisation too seriously.

This is how I come to my next point. Are the oxygen's orbitals in carbon acids really $\ce{sp^2}$ (or even $\ce{sp^3}$) hybridised? Well.

• It is often appropriate to describe oxygen's orbitals as $\ce{sp^2}$, when it is bonded to two other elements. Two $\ce{sp^2}$ lobes are needed for the bonds, the remaining $\ce{sp^2}$ orbital hosts one lone pair, the remaining $\ce{p}$ orbital hosts the other lone pair.
• More often the case is that oxygen has only one bonding partner. In these cases it is more appropriate to describe the oxygen with $\ce{sp}$ orbitals due to symmetry reasons. Consider the local symmetry to have an infinite rotational axis $C_\infty$, hence there have to be two degenerate, perpendicular orbitals. In most of these cases you will also have (partial) π-bonding and/or hyperconjugation.
• There are very few exceptions, like $\ce{H3O+}$, where it is appropriate to describe the oxygen with $\ce{sp^3}$ orbitals.

In all of this it should be considered that hybridisation can be fractional. See the last note here and more consequently "What is Bent's rule?" and "Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?".

An important point to remember is, that when there is a π-bond, the hybridisation can be $\ce{sp^2}$ at maximum.

I believe that answer 6 is really about a wrong interpretation of resonance stabilisation of the molecule. (Please see Why is there a need for resonance? for more information) Carbon acids can be described in the resonance scheme below. Now it appears, that the oxygen employs $\ce{sp^2}$ orbitals in the first structure (A) and $\ce{sp^3}$ in the others (B, C). It is important to recognise, that the bonding situation in this molecule is a superposition of all (possible) resonance structures. Theses structures are not in equilibrium. In conclusion there are several issues with this question and you might want to address this with your teacher.

• What about the notion of this molecule being dissolved in water? If the carbonyl oxygen started out as $\ce{^16O}$, and the hydroxyl oxygen as $\ce{^17O}$, then the hydrogen would flip from on oxygen atom to the other. In that sense there is an equilibrium. – MaxW Nov 13 '15 at 19:48
• @MaxW yes, that is correct and to go even further, this is true for any solution and you don't even have to mark the oxygens. Of course the electronic structure of the molecule will adapt in these situations. But interpreting the question like this is a stretch I think. – Martin - マーチン Nov 14 '15 at 6:31
• A great answer. I'll need to understand why does the oxygen appear to be in the sp3 hybridization state in (B) and (C) though. It has only one linkage to a C atom. It has 7 electrons, due to the -1 charge.. I'll try to wrap my head around it. – CowperKettle Apr 27 '16 at 6:38