The following question was asked in JEE Mains exam in 2022, a competitive exam for engineering in India.

The difference between bond order of CO and NO$^+$ is $\frac{x}{2}$ where $x$ = ____. (Round off to the nearest integer).

Here, I have taken the bond order of CO as 2.6 (which is cited as the actual bond order of CO) as per:

  1. This Stack Exchange answer
  2. Vocation India
  3. Kayson Education
  4. Research Gate

and many other sources, the reason being the anti-bonding character in the non - bonding orbital in the Molecular Orbital diagram.

Upon solving we get:

$$3 - 2.6 = \frac{x}{2}\\ 0.4 = \frac{x}{2}\\ x = 0.8\\ x = 1$$

But the answer given for said question is 0. They have taken the bond order of CO and NO$^+$, both as 3. So their solution is:

$$ 3 - 3 = \frac{x}{2}\\ 0 = \frac{x}{2}\\ x = 0$$

On a technical level, the bond order of CO should be ~2.6, so my answer should be correct. Is the solution given wrong? The views of others on this topic would be welcome.

  • 5
    $\begingroup$ At this level they both have 3. If you want to nitpick it, then what makes you think NO cation would have effective bond order of 3 while CO doesn't? $\endgroup$
    – Mithoron
    Sep 10, 2023 at 17:52
  • $\begingroup$ Well, the difference between $\ce{CO}$ and $\ce{NO+}$ is about a proton in the nucleus of nitrogen. They are isoelectric and at that level of approximation, that means they are basically the same. $\endgroup$ Sep 10, 2023 at 18:07

1 Answer 1


I think at this level you are supposed to make the simplifying assumption that bond order is $(1/2)×$(the number of electrons in bonding orbitals minus the number of electrons in antibonding orbitals); the $1/2$ coming from needing two electrons to make a bond. Ignore the "effective" bond order formalism on which the $2.6$ value for carbon monoxide is based (if we were to assess the "effective bond order" in the nitrosyl cation the same way that probably would also be less than $3.0$).

Here, you should verify that the molecular orbital structures for both species give eight bonding electrons and two antibonding ones, so a bond order of $(8-2)/2=3$ for both and that's where they get $x=0$.


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