1. We need to count the species having exactly one unpaired electron among: $\mathrm{O_2},\ \mathrm{O_2^-},\ \mathrm{NO},\ \mathrm{CN^-},\ \mathrm{O_2^{2-}}$

2. Use molecular orbital (MO) theory for each species.

(i) $\mathrm{O_2}$

For $\mathrm{O_2}$, total electrons = $16$.

Its MO configuration in the valence shell is: $\left(\sigma 2s\right)^2\left(\sigma^*2s\right)^2\left(\sigma 2p_z\right)^2\left(\pi 2p_x\right)^2\left(\pi 2p_y\right)^2\left(\pi^* 2p_x\right)^1\left(\pi^* 2p_y\right)^1$

Thus, $\mathrm{O_2}$ has 2 unpaired electrons.

So, it is not counted.

(ii) $\mathrm{O_2^-}$

This has one extra electron compared to $\mathrm{O_2}$, so total electrons = $17$.

The extra electron goes into one of the $\pi^*$ orbitals: $\left(\pi^* 2p_x\right)^2\left(\pi^* 2p_y\right)^1$ (or vice versa)

Hence, $\mathrm{O_2^-}$ has 1 unpaired electron.

So, it is counted.

(iii) $\mathrm{NO}$

Total electrons = $7+8=15$.

Since it has an odd number of electrons, one electron remains unpaired in MO filling.

Hence, $\mathrm{NO}$ has 1 unpaired electron.

So, it is counted.

(iv) $\mathrm{CN^-}$

Total electrons = $6+7+1=14$.

This is isoelectronic with $\mathrm{N_2}$ and CO. All electrons are paired in the molecular orbitals.

Hence, $\mathrm{CN^-}$ has 0 unpaired electrons.

So, it is not counted.

(v) $\mathrm{O_2^{2-}}$

This has two extra electrons compared to $\mathrm{O_2}$, so total electrons = $18$.

Both $\pi^*$ orbitals become fully filled: $\left(\pi^* 2p_x\right)^2\left(\pi^* 2p_y\right)^2$

Hence, $\mathrm{O_2^{2-}}$ has 0 unpaired electrons.

So, it is not counted.

3. Species with exactly one unpaired electron are: $\mathrm{O_2^-},\ \mathrm{NO}$

Therefore, the number of such species is: $\boxed{2}$

4. Comparison with stored correct answer:

• Derived answer = $2$
• Stored correct answer = $2$

They agree.