Using molecular orbital theory for homonuclear diatomic species of second-period elements, choose the correct statement(s).

(A) Correct. $\mathrm{O_2^{2-}}$ has $14$ valence electrons: $\sigma_{2s}^2\sigma_{2s}^{*2}\sigma_{2p}^2\pi_{2p}^4\pi_{2p}^{*4}$, giving bond order $\dfrac{8-6}{2}=1$ with all electrons paired. (B) Incorrect. $\mathrm{C_2}$ has $8$ valence electrons with the $\pi_{2p}$ orbitals lying below $\sigma_{2p}$; the configuration $\sigma_{2s}^2\sigma_{2s}^{*2}\pi_{2p}^4$ has all electrons paired, bond order $2$, so it is diamagnetic. (C) Incorrect. $\mathrm{B_2}$ has $6$ valence electrons, $\sigma_{2s}^2\sigma_{2s}^{*2}\pi_{2p}^{1}\pi_{2p}^{1}$ with parallel spins, so bond order is $1$ and it is paramagnetic. (D) Incorrect. Removing a bonding electron lowers the bond order from $3$ (in $\mathrm{N_2}$) to $2.5$, so $\mathrm{N_2^+}$ has a longer bond than $\mathrm{N_2}$.