Correct statement(s) according to molecular orbital theory (for B2, C2, N2^+, O2^2- use the MO ordering with σ2p above π2p for B2, C2, N2^+) is(are)

For $\mathrm{C_2}$ (12 electrons) the configuration is $\sigma_{1s}^2\sigma^{*2}_{1s}\sigma_{2s}^2\sigma^{*2}_{2s}\pi_{2p}^4$, so bond order is $(8-4)/2=2$ and all electrons are paired: option A is correct. For $\mathrm{B_2}$ (10 electrons) the last two electrons occupy the degenerate $\pi_{2p}$ orbitals singly with parallel spins, giving bond order $1$ but two unpaired electrons, so it is paramagnetic and option B is wrong. $\mathrm{N_2^+}$ has 13 electrons, $\sigma_{2p}$ holds a single electron, giving bond order $(9-4)/2=2.5$ with one unpaired electron: option C is correct. $\mathrm{O_2^{2-}}$ has 18 electrons with $\pi^*_{2p}$ completely filled ($\pi^{*4}$), so bond order is $1$ but it is diamagnetic, making option D wrong.