Find bond order and magnetic property of acetylide ion, $\mathrm{C_2^{2-}}$

Each carbon has 6 electrons, so: $\mathrm{C_2}: 6+6=12 \text{ electrons}$ For $\mathrm{C_2^{2-}}$: $12+2=14 \text{ electrons}$

Thus, $\mathrm{C_2^{2-}}$ is isoelectronic with $\mathrm{N_2}$.

MO configuration for 14-electron diatomic species

For molecules up to nitrogen, the MO order is: $\sigma(1s)^2\,\sigma^*(1s)^2\,\sigma(2s)^2\,\sigma^*(2s)^2\,\pi(2p_x)^2=\pi(2p_y)^2\,\sigma(2p_z)^2$

Bonding electrons = 10, antibonding electrons = 4.

So bond order is: $\text{B.O.} = \frac{N_b-N_a}{2} = \frac{10-4}{2}=3$

All electrons are paired, so it is diamagnetic.

Therefore, acetylide ion has:

• Bond order = 3
• Diamagnetic

Check each option

Option A: $\mathrm{O_2^+}$

Neutral $\mathrm{O_2}$ has bond order 2 and is paramagnetic. Removing one electron from antibonding orbital gives: $\text{B.O.} = 2.5$ It still has one unpaired electron, so paramagnetic.

Not same.

Option B: $\mathrm{O_2^-}$

Adding one electron to antibonding orbital gives: $\text{B.O.} = 1.5$ It is paramagnetic.

Not same.

Option C: $\mathrm{N_2^+}$

Neutral $\mathrm{N_2}$ has bond order 3 and is diamagnetic. Removing one electron from bonding $\sigma(2p)$ orbital gives: $\text{B.O.} = 2.5$ It has one unpaired electron, so paramagnetic.

Not same.

Option D: $\mathrm{NO^+}$

Total electrons: $7+8-1=14$ So $\mathrm{NO^+}$ is also isoelectronic with $\mathrm{N_2}$.

Hence:

• Bond order = 3
• Diamagnetic

Same as acetylide ion.

Final answer

The bond order and magnetic property of $\mathrm{C_2^{2-}}$ are same as those of: $\boxed{\mathrm{NO^+}}$ So, the correct option is D.