The bond order in NO is $$2.5$$ while that in $$NO^+$$ is $$3$$. Which of the following statements is true for these two species?

Bond Order vs. Bond Length

In molecular orbital theory, bond order is directly proportional to bond strength and inversely proportional to bond length.

$$\text{Bond Order} \propto \frac{1}{\text{Bond Length}}$$

• A higher bond order indicates a greater number of shared electron pairs between the two nuclei, which pulls the atoms closer together and results in a shorter bond length.
• A lower bond order indicates weaker bonding holding the atoms together, resulting in a longer bond length.

Comparison of the Given Species:

Let's look at the experimentally and theoretically derived values provided in the problem:

1. For $$\text{NO}$$:

   $$\text{Bond Order} = 2.5$$

2. For $$\text{NO}^+$$:

   $$\text{Bond Order} = 3.0$$

Comparing the two values, $$\text{NO}^+$$ has a higher bond order ($$3.0 > 2.5$$) than $$\text{NO}$$. Therefore, the nitrogen-oxygen bond in $$\text{NO}^+$$ must be tighter and shorter, whereas the bond in $$\text{NO}$$ must be longer.

Because its bond order is lower, the bond length in $$\text{NO}$$ is greater than the bond length in $$\text{NO}^+$$.

Correct Option: D — Bond length in NO is greater than in $$\text{NO}^+$$