1. Use molecular orbital idea / total electron count
   For small diatomic species like $CN^-$, $CN^+$, $NO^+$, and $O_2^-$, bond order can be found from

where $N_b$ = bonding electrons and $N_a$ = antibonding electrons.

For these species, comparing isoelectronic species is the quickest method: species with the same total number of electrons generally have the same MO filling pattern and hence same bond order.

2. Find total electrons in each species

• For $CN^-$: $6 + 7 + 1 = 14 \text{ electrons}$

• For $NO^+$: $7 + 8 - 1 = 14 \text{ electrons}$

• For $CN^+$: $6 + 7 - 1 = 12 \text{ electrons}$

• For $O_2^-$: $8 + 8 + 1 = 17 \text{ electrons}$

So,

• $CN^-$ and $NO^+$ are isoelectronic $(14e^-)$
• $CN^+$ has $12e^-$
• $O_2^-$ has $17e^-$

3. Compute bond orders

(i) $CN^-$

It is isoelectronic with $N_2$ (14 electrons), whose bond order is

So,

(ii) $NO^+$

Also has 14 electrons, so it is isoelectronic with $N_2$. Hence,

(iii) $CN^+$

It has 12 electrons, like $C_2$. For $C_2$, bond order is

Thus,

(iv) $O_2^-$

$O_2$ has bond order $2$. Adding one electron to antibonding $\pi^*$ orbital reduces bond order by $\frac{1}{2}$:

4. Check each option

• A: $CN^-$ and $NO^+$ $3,\ 3$ Same bond order ✅

• B: $CN^-$ and $CN^+$ $3,\ 2$ Not same ❌

• C: $O_2^-$ and $CN^-$ $1.5,\ 3$ Not same ❌

• D: $NO^+$ and $CN^+$ $3,\ 2$ Not same ❌

5. Final answer The pair having the same bond order is:

6. Comparison with stored correct answer Stored correct answer: A
   Derived answer: A
   They agree.