# Bond order of carbon monoxide and nitrosyl cation

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:

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.

• 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? Sep 10, 2023 at 17:52
• 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. Sep 10, 2023 at 18:07

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 $$x=0$$.