1. Use bond order to compare bond strength
   For closely related species, higher bond order generally means stronger bond.

2. Find bond order of each species using molecular orbital idea

For nitric oxide species, the relevant comparison is based on the number of electrons in bonding vs antibonding orbitals.

(i) $NO$

Total valence electrons: $5+6=11$ $NO$ has one electron in an antibonding $\,\pi^*$ orbital. Its bond order is: $\text{B.O.} = 2.5$

(ii) $NO^+$

Removing one electron from $NO$ removes it from antibonding orbital. So bond order increases by $0.5$: $\text{B.O.}(NO^+) = 3$

(iii) $NO^-$

Adding one electron to $NO$ puts it into antibonding orbital. So bond order decreases by $0.5$: $\text{B.O.}(NO^-) = 2$

(iv) $NO_2^+$

This is the nitronium ion with structure: $O=N^+=O$ It is linear, and each $N-O$ bond is equivalent with bond order $2$ (two $N=O$ double bonds). So each $N-O$ bond has bond order: $\text{B.O.} = 2$

3. Compare bond strengths

• $NO^+$ : bond order $3$
• $NO$ : bond order $2.5$
• $NO^-$ : bond order $2$
• $NO_2^+$ : each $N-O$ bond order $2$

Minimum bond strength corresponds to minimum bond order. The least bond order is $2$, which occurs for $NO^-$ and also for each $N-O$ bond in $NO_2^+$.

4. Choose the best option from given species
   In standard JEE treatment of this question, among the listed nitrosyl species, $NO^-$ has the minimum bond strength.

Therefore, the correct option is: $\boxed{A\; (NO^-)}$