According to my books the bond order of $\ce{CO+}$ is $3.5$. But shouldn't it be $2.5$? On googling this, I found the following answer that is on Stack Exchange but its only talks about the bond length.

I am unable to understand why it is $3.5$ as I am in class 11.


1 Answer 1


For a long time it was taught in school and universities that the HOMO of carbon monoxide is anti-bonding. Without more context it was also often taught that the bond order in CO is three, since there are eight electrons in bonding orbitals and two in anti-bonding orbitals. $$\text{Bond order} = \frac12(\text{bonding} - \text{anti-bonding})$$ By assuming that the HOMO is anti-bonding (it is not!) and taking away one electron, the bond order has to increase to 3.5. This is wrong.

When we have a look at the MO diagram, a calculated version can be found here, we know that the HOMO, i.e. 3σ, is a bonding orbital, while the anti-bonding orbital is the 2σ. Upon ionisation, we would indeed remove one bonding electron and therefore the bond order has to decrease to 2.5 as you suggested.
However, it is not that easy. Strictly speaking the below MO scheme is, as well as MO theory itself, an approximation, and only one possible configuration. While we do not have to use resonance structures with MO theory, we have to consider other configurations (analogous to excited states). So naturally the bond order of CO is not strictly 3. And removing an electron does not mean we are removing it from only one orbital, rather than decreasing the electron density. Therefore we cannot accurately predict the bond order with these simple considerations.
Experimental observations and theoretical calculations suggest that the bond indeed becomes stronger when removing an electron. See the linked question and Philipp's answer within for more detail. (Don't look at the other answers, they are as wrong as they could be.)

In short: The bond order of $\ce{CO}$ is not exactly 3 and removing an electron will not increase the bond order to 3.5. In both cases, the observed bond order is probably closer to 2.5, while experiments suggest that the bond is stronger in $\ce{CO+}$.

MO of CO

An orbital with bonding character has no node perpendicular to the bond axis; an orbital with anti-bonding character has at least one node perpendicular to the bond axis (electron density is zero). Strictly speaking there are no non-bonding orbitals.

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    $\begingroup$ Shouldn’t the last sentence be ‘… in carbon monoxide’? Iirc, there are a few nonbonding orbitals (due to symmetry) e.g. in $\ce{HCl}$. $\endgroup$
    – Jan
    Commented Jan 24, 2017 at 21:39
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    $\begingroup$ @Jan That's why I said strictly, one of the two categories will fit any orbital. What we usually classify as non-bonding orbitals are linear combinations that "don't change in energy". That's simply not possible due to an external field. In HCl the non-bonding orbitals don't have a node perpendicular to the bond axis (you caught that I forgot that), so they are classifiable as bonding. $\endgroup$ Commented Jan 25, 2017 at 1:30
  • $\begingroup$ @Martin-マーチン as shown, doesn't the $\mathrm{3\sigma}$ orbital have 2 nodes perpendicular to bond axis? (or maybe they are not nodes; just wanted to clarify) $\endgroup$ Commented Aug 27, 2020 at 10:57
  • $\begingroup$ @Aniruddha you are right, and my wording is maybe a bit off. The perpendicular nodes pass through the nuclei, therefore there is no node on the bonding axis between the respective nuclei that span it. If you look at the infinity of that axis, you are absolutely correct. I should probably retired that. Thanks for catching it. $\endgroup$ Commented Aug 27, 2020 at 15:47

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