1. Idea: Bond length is inversely related to bond order. So we must find:

   • which species is diamagnetic,
   • and among those, which has the highest bond order (hence shortest bond length).

2. Use MO theory for each species.

3. $O_2$

For $O_2$, MO electronic configuration is: $\sigma_{2s}^2\,\sigma_{2s}^{*2}\,\sigma_{2p_z}^2\,(\pi_{2p_x})^2(\pi_{2p_y})^2\,(\pi_{2p_x}^*)^1(\pi_{2p_y}^*)^1$

• Two unpaired electrons are present.
• So $O_2$ is paramagnetic.
• Bond order: $\text{B.O.} = \frac{N_b-N_a}{2} = \frac{8-4}{2}=2$

So $O_2$ is not diamagnetic.

4. $O_2^{2-}$

Adding two electrons to $O_2$ fills the antibonding $\pi^*$ orbitals: $\sigma_{2s}^2\,\sigma_{2s}^{*2}\,\sigma_{2p_z}^2\,(\pi_{2p_x})^2(\pi_{2p_y})^2\,(\pi_{2p_x}^*)^2(\pi_{2p_y}^*)^2$

• All electrons are paired, so it is diamagnetic.
• Bond order: $\text{B.O.} = \frac{8-6}{2}=1$

5. $N_2^{2-}$

Total electrons $= 14+2=16$, same as $O_2$ type count.

For molecules up to nitrogen, MO order is: $\sigma_{2s}<\sigma_{2s}^*<\pi_{2p_x}=\pi_{2p_y}<\sigma_{2p_z}$

Configuration of $N_2^{2-}$: $\sigma_{2s}^2\,\sigma_{2s}^{*2}\,(\pi_{2p_x})^2(\pi_{2p_y})^2\,\sigma_{2p_z}^2\,(\pi_{2p_x}^*)^1(\pi_{2p_y}^*)^1$

• Two unpaired electrons are present.
• Hence paramagnetic.
• Bond order: $\text{B.O.} = \frac{8-4}{2}=2$

So it is not diamagnetic.

6. $C_2^{2-}$

Total electrons $= 12+2=14$, same as $N_2$.

For $C_2^{2-}$, MO configuration: $\sigma_{2s}^2\,\sigma_{2s}^{*2}\,(\pi_{2p_x})^2(\pi_{2p_y})^2\,\sigma_{2p_z}^2$

• All electrons are paired.
• Hence diamagnetic.
• Bond order: $\text{B.O.} = \frac{8-2}{2}=3$

7. Compare only diamagnetic species

• $C_2^{2-}$: diamagnetic, bond order $=3$
• $O_2^{2-}$: diamagnetic, bond order $=1$

Higher bond order means shorter bond length, so $C_2^{2-}$ has the shortest bond length.

8. Final conclusion

The species which is diamagnetic and has the shortest bond length is: $\boxed{C_2^{2-}}$ So the correct option is B.