Use Molecular Orbital Theory

A molecule is diamagnetic if all electrons are paired in its molecular orbitals. A molecule is paramagnetic if it has one or more unpaired electrons.

Check each molecule

A. $N_2$

Total electrons in $N_2$: $7+7=14$

For $N_2$ (up to nitrogen), the MO order is: $\sigma_{1s},\ \sigma^*_{1s},\ \sigma_{2s},\ \sigma^*_{2s},\ \pi_{2p_x}=\pi_{2p_y},\ \sigma_{2p_z}$

Filling 14 electrons gives: $\sigma_{1s}^2\,\sigma_{1s}^{*2}\,\sigma_{2s}^2\,\sigma_{2s}^{*2}\,(\pi_{2p_x})^2(\pi_{2p_y})^2\,(\sigma_{2p_z})^2$

All electrons are paired.

Therefore, $N_2$ is diamagnetic.

B. $O_2$

Total electrons in $O_2$: $8+8=16$

For $O_2$, the MO order is: $\sigma_{1s},\ \sigma^*_{1s},\ \sigma_{2s},\ \sigma^*_{2s},\ \sigma_{2p_z},\ \pi_{2p_x}=\pi_{2p_y},\ \pi^*_{2p_x}=\pi^*_{2p_y}$

Filling 16 electrons gives: $\sigma_{1s}^2\,\sigma_{1s}^{*2}\,\sigma_{2s}^2\,\sigma_{2s}^{*2}\,\sigma_{2p_z}^2\,(\pi_{2p_x})^2(\pi_{2p_y})^2\,(\pi^*_{2p_x})^1(\pi^*_{2p_y})^1$

There are two unpaired electrons.

Therefore, $O_2$ is paramagnetic.

C. $S_2$

Sulfur is in the same group as oxygen, so $S_2$ has a similar valence MO pattern.

Total valence electrons in $S_2$: $6+6=12$

MO filling in valence shell is analogous to $O_2$: $\sigma_{3s}^2\,\sigma_{3s}^{*2}\,\sigma_{3p_z}^2\,(\pi_{3p_x})^2(\pi_{3p_y})^2\,(\pi^*_{3p_x})^1(\pi^*_{3p_y})^1$

There are two unpaired electrons.

Therefore, $S_2$ is paramagnetic.

D. $C_2$

Total electrons in $C_2$: $6+6=12$

For $C_2$ (up to nitrogen), the MO order is: $\sigma_{1s},\ \sigma^*_{1s},\ \sigma_{2s},\ \sigma^*_{2s},\ \pi_{2p_x}=\pi_{2p_y},\ \sigma_{2p_z}$

Filling 12 electrons gives: $\sigma_{1s}^2\,\sigma_{1s}^{*2}\,\sigma_{2s}^2\,\sigma_{2s}^{*2}\,(\pi_{2p_x})^2(\pi_{2p_y})^2$

All electrons are paired.

Therefore, $C_2$ is diamagnetic.

Final selection

Diamagnetic molecules are: $N_2 \text{ and } C_2$

So the correct options are A and D.