We need to check two things for each ionization process:

1. Whether the bond order increases on removing one electron.
2. Whether the magnetic behaviour changes (paramagnetic $\leftrightarrow$ diamagnetic).

We use molecular orbital (MO) theory.

1. Option A: $C_2 \to C_2^+$

Step 1: MO configuration of $C_2$

For $B_2, C_2, N_2$ (up to nitrogen), the MO order is:

$C_2$ has $12$ electrons total, i.e. $8$ valence electrons. So valence MO filling:

Bond order:

All electrons are paired, so $C_2$ is diamagnetic.

Step 2: MO configuration of $C_2^+$

One electron is removed from the highest occupied MO, i.e. from $\pi(2p)$ bonding orbital. Thus bond order decreases by $\frac12$:

Now one unpaired electron appears, so it becomes paramagnetic.

Conclusion for A

• Bond order decreases, not increases.
• Magnetic behaviour changes.

So A is not correct.

2. Option B: $N_2 \to N_2^+$

Step 1: MO configuration of $N_2$

$N_2$ has $14$ electrons total, i.e. $10$ valence electrons. Configuration:

Bond order:

All electrons are paired, so $N_2$ is diamagnetic.

Step 2: MO configuration of $N_2^+$

One electron is removed from the highest occupied MO, $\sigma(2p_z)$, which is a bonding orbital. Hence bond order decreases by $\frac12$:

There is now one unpaired electron, so it becomes paramagnetic.

Conclusion for B

• Bond order decreases.
• Magnetic behaviour changes.

So B is not correct.

3. Option C: $NO \to NO^+$

Step 1: MO configuration of $NO$

$NO$ has $15$ total electrons, i.e. $11$ valence electrons. It is isoelectronic with $O_2^+$ and has one electron in a $\pi^*$ antibonding orbital. So effectively:

Because of one unpaired electron, $NO$ is paramagnetic.

Step 2: MO configuration of $NO^+$

Removing one electron removes it from the highest occupied orbital, which is the antibonding $\pi^*$ orbital. Removing an antibonding electron increases bond order by $\frac12$:

Now all electrons are paired, so $NO^+$ is diamagnetic.

Conclusion for C

• Bond order increases.
• Magnetic behaviour changes from paramagnetic to diamagnetic.

So C is correct.

4. Option D: $O_2 \to O_2^+$

Step 1: MO configuration of $O_2$

For $O_2$ and $F_2$, the MO order is:

$O_2$ has $12$ valence electrons:

Bond order:

It has two unpaired electrons, so it is paramagnetic.

Step 2: MO configuration of $O_2^+$

One electron is removed from an antibonding $\pi^*$ orbital. Thus bond order increases by $\frac12$:

But $O_2^+$ still has one unpaired electron, so it remains paramagnetic.

Conclusion for D

• Bond order increases.
• Magnetic behaviour does not change (still paramagnetic).

So D is not correct.

Final comparison

Only option C satisfies both conditions:

• bond order increases,
• magnetic behaviour changes.

Therefore, the correct answer is:

The stored correct answer is C, which matches our derived answer.