1. Idea used: Molecular Orbital (MO) theory
   A molecule will be stabilized by anion formation if adding one electron increases its bond order, i.e. the added electron goes into a bonding MO rather than an antibonding MO.

Bond order is

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

We check each species.

2. Option A: $C_2$
   For $C_2$ (12 total electrons; valence electrons = 8), for B–C–N type ordering:

The next MO is $\sigma(2p)$, which is a bonding MO and is empty in $C_2$.

Bond order of $C_2$:

On forming $C_2^-$, one electron enters the bonding $\sigma(2p)$ MO, so bond order becomes

So anion formation stabilizes $C_2$.

3. Option B: $NO$
   $NO$ has 11 valence electrons. Its highest occupied MO contains one electron in a $\pi^*$ antibonding orbital.

Bond order of $NO$ is:

On forming $NO^-$, the added electron goes into the same $\pi^*$ antibonding orbital, so bond order decreases:

Thus anion formation does not stabilize $NO$.

4. Option C: $O_2$
   For $O_2$, the valence MO configuration ends as

Bond order of $O_2$:

On forming $O_2^-$, the extra electron goes into a $\pi^*$ antibonding MO, so

Hence anion formation reduces stability.

5. Option D: $F_2$
   For $F_2$, bond order is

The added electron in $F_2^-$ goes into the next available $\sigma^*(2p)$ antibonding MO, so bond order becomes

Again, anion formation decreases stability.

6. Conclusion
   Only in $C_2$ does the added electron enter a bonding MO and increase bond order.

Therefore, the molecule expected to be stabilized by anion formation is:

So the correct option is A.