$N_2$ has the electronic configuration $(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\pi_{2p})^4(\sigma_{2p})^2$ — all electrons are paired → diamagnetic. $O_2$ and $S_2$ have unpaired electrons in $\pi^*$ orbitals → paramagnetic. $C_2$ has all electrons paired in $\pi$ orbitals and is also diamagnetic. Both $C_2$ and $N_2$ are diamagnetic, but $N_2$ is the more standard textbook answer.