Use MO theory without $2s$–$2p$ mixing

Since $2s$–$2p$ mixing is not operative, the molecular orbital order for second-period diatomic molecules is taken as:

$\sigma(2s) < \sigma^*(2s) < \sigma(2p_z) < \pi(2p_x)=\pi(2p_y) < \pi^*(2p_x)=\pi^*(2p_y) < \sigma^*(2p_z)$

A species is paramagnetic if it has one or more unpaired electrons.

Check each option

(A) $Be_2$

Each Be atom has 4 electrons, so $Be_2$ has total $8$ electrons.

MO filling: $\sigma(1s)^2\,\sigma^*(1s)^2\,\sigma(2s)^2\,\sigma^*(2s)^2$

All electrons are paired.

So, $Be_2$ is diamagnetic.

(B) $B_2$

Each B atom has 5 electrons, so $B_2$ has total $10$ electrons.

MO filling: $\sigma(1s)^2\,\sigma^*(1s)^2\,\sigma(2s)^2\,\sigma^*(2s)^2\,\sigma(2p_z)^2$

All electrons are paired.

So, $B_2$ is diamagnetic.

(C) $C_2$

Each C atom has 6 electrons, so $C_2$ has total $12$ electrons.

MO filling: $\sigma(1s)^2\,\sigma^*(1s)^2\,\sigma(2s)^2\,\sigma^*(2s)^2\,\sigma(2p_z)^2\,\pi(2p_x)^1\,\pi(2p_y)^1$

There are two unpaired electrons in the degenerate $\pi$ orbitals.

So, $C_2$ is paramagnetic.

(D) $N_2$

Each N atom has 7 electrons, so $N_2$ has total $14$ electrons.

MO filling: $\sigma(1s)^2\,\sigma^*(1s)^2\,\sigma(2s)^2\,\sigma^*(2s)^2\,\sigma(2p_z)^2\,\pi(2p_x)^2\,\pi(2p_y)^2$

All electrons are paired.

So, $N_2$ is diamagnetic.

Conclusion

The only paramagnetic species is:

$\boxed{C_2}$

Comparison with stored answer

Stored correct answer: $C$

My derived answer is also $C$.

Therefore, the stored answer is correct.