1. Use molecular orbital theory and compute bond order using

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

2. Find total electrons in each species

• For ${C_2}^{2-}$: carbon has $6$ electrons, so $2\times 6 + 2 = 14$ total electrons.

• For ${N_2}^{2-}$: nitrogen has $7$ electrons, so $2\times 7 + 2 = 16$ total electrons.

• For ${O_2}^{2-}$: oxygen has $8$ electrons, so $2\times 8 + 2 = 18$ total electrons.

Since core electrons do not affect comparison, we can focus on valence electrons:

• ${C_2}^{2-}$: $10$ valence electrons
• ${N_2}^{2-}$: $12$ valence electrons
• ${O_2}^{2-}$: $14$ valence electrons

3. MO ordering

• For molecules up to nitrogen ($B_2, C_2, N_2$): $\sigma(2s) < \sigma^*(2s) < \pi(2p_x)=\pi(2p_y) < \sigma(2p_z)$

• For oxygen and beyond: $\sigma(2s) < \sigma^*(2s) < \sigma(2p_z) < \pi(2p_x)=\pi(2p_y) < \pi^*(2p_x)=\pi^*(2p_y)$

4. Bond order of ${C_2}^{2-}$

Valence MO filling for $10$ electrons:

Bonding electrons: $2 + 4 + 2 = 8$ Antibonding electrons: $2$ So,

5. Bond order of ${N_2}^{2-}$

Valence MO filling for $12$ electrons:

Bonding electrons: $2 + 4 + 2 = 8$ Antibonding electrons: $2 + 2 = 4$ So,

6. Bond order of ${O_2}^{2-}$

Valence MO filling for $14$ electrons:

Bonding electrons: $2 + 2 + 4 = 8$ Antibonding electrons: $2 + 4 = 6$ So,

7. Compare bond orders

So the correct order is:

8. Match with options

This corresponds to Option B.

9. Verification with stored answer

Stored correct answer = B.

My derived answer also = B, so they agree.