With all the knowledge that I have secured reading about resonance, I know that delocalisation of pi electrons is possible only with the p orbitals being in the same plane.

So for two consecutive double bonds, the orbitals wouldn't be in the same plane. So there shouldn't be any hybridization, right? Like in $\ce{CO2}$, the O's are $Sp^2$ hybrids and C is sp, as per the structure, the three unhybridized p orbitals wouldn't be in the same plane, making it not possible to have resonance in it. But that isn't the case with it. It does have resonance structures in it. What am I missing?

  • 2
    $\begingroup$ In a hybridization model of CO2, a better representation is to have the O atoms sp hybridized rather than sp2, and have one lone pair in sp and one in pure p (which is aligned with the C=O bond to the other O). $\endgroup$
    – Andrew
    Sep 7 at 11:44

1 Answer 1


Not only can there be resonance with cumulated double bonds, this is much in evidence with carbon dioxide itself. What is formally the pi bond between the carbon atom and each oxygen atom conjugates with a formal $p$-orbital lone pair on the other oxygen atom, which is in the same plane for each respective bond. Thus the pi electrons become delocalized, producing a series of contributing structures:




The effect of this delocalization is to make the bonds shorter and stronger in carbon dioxide than they would be in an isolated carbon-oxygen double bond: From Wikipedia, we see the difference in bond lengths:

Bond lengths of C=O bonds are around 123 pm in carbonyl compounds. The C=O bond length in carbon dioxide is 116 pm.

  • $\begingroup$ But how can they the orbitals aren't in the same plane ? $\endgroup$
    – Via
    Sep 7 at 11:59
  • $\begingroup$ They are in the same plane. You have an additional $p$ orbital in the same plane as each apparent double bond. $\endgroup$ Sep 7 at 12:11

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