1. We need to count species having exactly one unpaired electron.
   We use Molecular Orbital (MO) theory.

2. Check each species one by one.

(i) $\mathrm{N}_2$

Total electrons $= 7+7=14$.

For $\mathrm{N}_2$ (up to $14$ electrons), MO filling is: $\sigma_{1s}^2\,\sigma_{1s}^{*2}\,\sigma_{2s}^2\,\sigma_{2s}^{*2}\,(\pi_{2p_x}=\pi_{2p_y})^4\,\sigma_{2p_z}^2$ All electrons are paired.

• Unpaired electrons $=0$
• Not counted

(ii) $\mathrm{O}_2$

Total electrons $= 8+8=16$.

For $\mathrm{O}_2$, MO filling is: $\sigma_{1s}^2\,\sigma_{1s}^{*2}\,\sigma_{2s}^2\,\sigma_{2s}^{*2}\,\sigma_{2p_z}^2\,(\pi_{2p_x}=\pi_{2p_y})^4\,(\pi_{2p_x}^*=\pi_{2p_y}^*)^2$ The two electrons in degenerate $\pi^*$ orbitals remain unpaired.

• Unpaired electrons $=2$
• Not counted

(iii) $\mathrm{C}_2^{-}$

Total electrons $= 6+6+1=13$.

For species up to nitrogen type ordering: $\sigma_{1s}^2\,\sigma_{1s}^{*2}\,\sigma_{2s}^2\,\sigma_{2s}^{*2}\,(\pi_{2p_x}=\pi_{2p_y})^4\,\sigma_{2p_z}^1$ So there is exactly one electron in $\sigma_{2p_z}$.

• Unpaired electrons $=1$
• Counted

(iv) $\mathrm{O}_2^{-}$

Total electrons $= 16+1=17$.

Compared to $\mathrm{O}_2$, one extra electron enters a $\pi^*$ orbital: $\ldots (\pi_{2p_x}^*=\pi_{2p_y}^*)^3$ This gives one orbital with a pair and the other with one single electron.

• Unpaired electrons $=1$
• Counted

(v) $\mathrm{O}_2^{2-}$

Total electrons $= 16+2=18$.

Then: $\ldots (\pi_{2p_x}^*=\pi_{2p_y}^*)^4$ All paired.

• Unpaired electrons $=0$
• Not counted

(vi) $\mathrm{H}_2^{+}$

Total electrons = 1+1-1=1?$$ Actually two H atoms contribute 2$electrons, and charge$+1$means one electron removed, so total electrons$=1$.

MO configuration: $\sigma_{1s}^1$ So exactly one unpaired electron is present.

• Unpaired electrons $=1$
• Counted

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

Total electrons $= 6+7+1=14$. This is isoelectronic with $\mathrm{N}_2$.

Hence MO configuration is like $\mathrm{N}_2$: all electrons paired.

• Unpaired electrons $=0$
• Not counted

(viii) $\mathrm{He}_2^{+}$

Total electrons $= 2+2-1=3$.

MO configuration: $\sigma_{1s}^2\,\sigma_{1s}^{*1}$ So exactly one unpaired electron is present.

• Unpaired electrons $=1$
• Counted

3. Count the species with exactly one unpaired electron:

• $\mathrm{C}_2^{-}$
• $\mathrm{O}_2^{-}$
• $\mathrm{H}_2^{+}$
• $\mathrm{He}_2^{+}$

Total number: $4$

4. Comparison with stored correct answer Stored correct answer $=4$. Our derived answer also equals $4$.

So, the stored answer is correct.