If we draw the Lewis dot structure of CO like this:

enter image description here

The formal charge on each atom must be 0 But in the actual structure, formal charge on carbon is $-1$ and formal charge on oxygen is $1$.

enter image description here

But I have studied that central atom may have an incomplete octet so that the condition of formal charges being close to $0$ is satisfied.

Then why does carbon form a coordinate bond with oxygen in order to complete it's octet making the formal charges far from 0?

  • 1
    $\begingroup$ Closely related: chemistry.stackexchange.com/questions/51262/… // chemistry.stackexchange.com/questions/30797/… $\endgroup$
    – Poutnik
    May 21, 2023 at 7:22
  • $\begingroup$ @Poutnik , can this be explained without Molecular Orbital Theory as I have not yet studied it? $\endgroup$ May 21, 2023 at 8:36
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    $\begingroup$ @Poutnik This isn't even a question about CO, just about this lowest formal charge misconception. @ OP If there's a need and occasion for a dative bond to complete octet then it happens, end of story. Otherwise a concept of dative bond would be useless. $\endgroup$
    – Mithoron
    May 21, 2023 at 12:02
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    $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    May 22, 2023 at 1:00
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    $\begingroup$ check this paper, it specifically addresses the issue and is open-access: nature.com/articles/srep16307 . Long story short: yes, formal charge is negative for C, charge from population analysis is strongly positive on C and in reality C and O are strongly polarised and concept of atomic charge breaks down. $\endgroup$
    – permeakra
    May 24, 2023 at 21:43

1 Answer 1


I think you are taking a view of the bonding which is very ionic in nature. Consider for a moment the bonding in water. Water has a sp3 oxygen which can be regarded as being a distorted tetrahedral atoms. The water has two sigma bonds between the oxygen atom and the hydrogens and a total of two lone pairs on the oxygen.

What you have in carbon monoxide is a sp oxygen and a sp carbon, you can think of carbon monoxide as being similar to acetylene and dinitrogen.

Between the carbon and the oxygen in CO we have a sigma bond, we also have two pi bonds which are formed from the p orbitals on the carbon and oxygen atoms. These pi bonds are at 90 degrees to each other. It is important to understand that the electrons in a covalent bond are shared between the atoms rather than kept by one atom alone.

It is hard to know what you mean in the biro drawing, do you mean that the oxygen has two lone pairs which are off the axis of the C-O bond or do you mean something else. If this is what you mean then the oxygen would be a sp2 hydridized atom.

If we consider where the lone pair is on a carbon monoxide then it will help to look at S.Niibayashi, K.Mitsui, K.Matsubara, H.Nagashima, Organometallics, 2003, 22, 4885, DOI: 10.1021/om0340701. In this paper they report the strucutre of a metal complex where the lone pair of the oxygen bonds to a titanium.

The Ti-O-C angle is 162 degrees which is only 18 degrees away from a perfect 180 degrees. I think it is likely to be packing effects which have distorted the bond away from 180 degrees. Here is a picture of the complex. Complex showing a carbonyl oxygen bonding to a titanium atom

The differece in electronegativity of the carbon and the oxygen cause the covalent bonds to be polarized such that the oxygen gets more electron density than the carbon. As a result the oxygen end of the carbon monoxide is more negative than the carbon end.

So if anything the oxygen is more negative than the carbon.

  • $\begingroup$ Yes, in my drawing I meant that oxygen has 2 lone pairs(I am unable to understand what was the need of donating one of that lone pair, probably making it an exception to the "atoms should have formal charge close to 0" rule). Btw how can I view that paper? Can you please provide me the link? I copy pasted the name of the paper on my browser but I am not getting the exact paper. $\endgroup$ May 21, 2023 at 9:35
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    $\begingroup$ Thanks for making it clear of what you were thinking, I will edit the answer shortly. $\endgroup$ May 21, 2023 at 9:38

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