1. We need to identify the species that has no unpaired electrons, i.e. is diamagnetic.

2. Use Molecular Orbital (MO) theory for each species.

Option A: $O_2^{2-}$

For oxygen and beyond, the MO order is:

Total valence electrons in $O_2$:

For $O_2^{2-}$, add 2 electrons:

Filling MOs:

All electrons are paired.

So, $O_2^{2-}$ is diamagnetic.

Option B: $B_2$

For lighter molecules up to nitrogen, MO order is:

Each B has 3 electrons, so total electrons:

Valence electrons = 6. Filling gives:

There are two unpaired electrons.

So, $B_2$ is paramagnetic.

Option C: $N_2^+$

Each N has 7 electrons, so total electrons in $N_2$:

Thus $N_2^+$ has:

Valence electrons in $N_2$ are 10, so in $N_2^+$ there are 9 valence electrons.

For $N_2$, MO filling is:

Removing one electron for $N_2^+$ gives:

There is one unpaired electron.

So, $N_2^+$ is paramagnetic.

Option D: $O_2$

$O_2$ has 12 valence electrons. Filling gives:

There are two unpaired electrons.

So, $O_2$ is paramagnetic.

Conclusion

Only $O_2^{2-}$ has all electrons paired.

Therefore, the molecule/ion that does not contain unpaired electrons is: