1. Write the total number of electrons in each species

We use MO theory and compare bond orders.

Bond order is

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

2. Count total electrons for each species

• $\mathrm{CN^-}$: $6+7+1=14$ electrons
• $\mathrm{NO^+}$: $7+8-1=14$ electrons
• $\mathrm{O_2}$: $8+8=16$ electrons
• $\mathrm{O_2^+}$: $16-1=15$ electrons
• $\mathrm{O_2^{2+}}$: $16-2=14$ electrons

So, $\mathrm{CN^-}, \mathrm{NO^+}, \mathrm{O_2^{2+}}$ are all 14-electron species.

3. Bond order of 14-electron diatomic species

For second-period 14-electron species like $\mathrm{N_2}$-type systems, the valence MO filling gives bond order $3$.

Thus,

4. Bond order of $\mathrm{O_2}$

For $\mathrm{O_2}$, MO configuration gives two electrons in $\pi^*_{2p}$ antibonding orbitals.

Hence,

5. Bond order of $\mathrm{O_2^+}$

Removing one electron from $\mathrm{O_2}$ removes it from an antibonding $\pi^*_{2p}$ orbital, so bond order increases by $\frac{1}{2}$:

6. List all bond orders

• $\mathrm{CN^-} \rightarrow 3$
• $\mathrm{NO^+} \rightarrow 3$
• $\mathrm{O_2} \rightarrow 2$
• $\mathrm{O_2^+} \rightarrow 2.5$
• $\mathrm{O_2^{2+}} \rightarrow 3$

So, the species having identical bond order are:

Total number of such species $= 3$.

7. Comparison with stored answer

Stored correct answer = $3$

This matches our derived answer.