Bond order of carbon monoxide and nitrosyl cation

Question

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.

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 $x=0$.