Among the following, the species having the smallest bond is

According to Molecular Orbital Theory, the bond length of a diatomic molecule or ion varies inversely with its bond order:

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

A higher bond order indicates a stronger, tighter covalent interaction, which pulls the nuclei closer together and results in a shorter bond distance.

Step-by-Step Molecular Orbital Analysis:

The formula to calculate the bond order is:

$$\text{Bond Order} = \frac{N_b - N_a}{2}$$

Where $$N_b$$ represents the number of bonding electrons and $$N_a$$ represents the number of antibonding electrons. Let's find the total electron counts and corresponding bond orders for each species:

• Option A: $$\text{NO}^-$$ ($$7 + 8 + 1 = 16\text{ electrons}$$)

  It has an electronic configuration identical to $$\text{O}_2$$.

  $$\text{Bond Order} = \frac{10 - 6}{2} = 2$$



• Option B: $$\text{NO}^+$$ ($$7 + 8 - 1 = 14\text{ electrons}$$)

  It is isoelectronic with $$\text{N}_2$$. All valence electrons are paired up inside the lower-energy bonding molecular orbitals, leaving the anti-bonding $$\pi^*$$ orbitals completely vacant.

  $$\text{Bond Order} = \frac{10 - 4}{2} = 3$$



• Option C: $$\text{O}_2$$ ($$8 + 8 = 16\text{ electrons}$$)

  Two electrons fill the degenerate anti-bonding $$\pi^*_{2p}$$ orbitals singly.

  $$\text{Bond Order} = \frac{10 - 6}{2} = 2$$



• Option D: $$\text{NO}$$ ($$7 + 8 = 15\text{ electrons}$$)

  It contains one unpaired electron residing within an anti-bonding $$\pi^*_{2p}$$ orbital.

  $$\text{Bond Order} = \frac{10 - 5}{2} = 2.5$$

Conclusion:

Comparing the calculated values, the nitrosonium ion ($$\text{NO}^+$$) possesses the highest bond order ($$3$$), which corresponds directly to the strongest chemical linkage and the smallest bond length.

Answer: Option B — $$\text{NO}^+$$