1. Use MO theory and bond order formula

For diatomic molecules/ions,

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

Also:

• Higher bond order $\Rightarrow$ shorter bond length
• Unpaired electrons $\Rightarrow$ paramagnetic
• All electrons paired $\Rightarrow$ diamagnetic

2. Option A: $C_2^{2-}$ is expected to be diamagnetic

First count electrons:

• Carbon has 6 electrons, so $C_2$ has $12$
• $C_2^{2-}$ has $14$ electrons

For molecules up to $N_2$, the MO order is:

Filling 14 electrons:

All electrons are paired, so $C_2^{2-}$ is diamagnetic.

So, A is correct.

3. Option B: $O_2^{2+}$ expected to have a longer bond length than $O_2$

First compare bond orders.

For $O_2$, the MO configuration in valence shell is:

So,

Now $O_2^{2+}$ has 2 fewer electrons, removed from highest occupied antibonding $\pi^*$ orbitals:

Thus,

Since bond order increases from 2 to 3, bond length decreases, not increases.

So, B is incorrect.

4. Option C: $N_2^+$ and $N_2^-$ have the same bond order

$N_2$ has 14 electrons. For molecules up to $N_2$, the relevant valence MO order is:

For $N_2$:

Hence bond order is

For $N_2^+$

Remove one electron from highest occupied MO, i.e. $\sigma(2p_z)$:

For $N_2^-$

Add one electron to next MO, which is antibonding $\pi^*$:

Thus both have the same bond order.

So, C is correct.

5. Option D: $He_2^+$ has the same energy as two isolated He atoms

Count electrons:

• $He_2^+$ has $4 - 1 = 3$ electrons

MO configuration:

Bond order:

Since bond order is positive, $He_2^+$ is more stable than two isolated He atoms and can exist as a weakly bound species. Therefore it does not have the same energy as two isolated He atoms.

So, D is incorrect.

6. Final selection

Correct statements are:

7. Comparison with stored correct answer

Stored correct answer: $A, C$

Our derived answer matches the stored answer exactly.