Using molecular orbital theory (for B2, C2, N2 and the species derived from them the π2p orbitals lie below σ2p; for O2 and its ions σ2p lies below…

Fill the molecular orbitals in the stated order and compute bond order $=\tfrac12(N_b-N_a)$. $\mathrm{C_2}$ (12 electrons) has $\pi_{2p}^4$ as the highest occupied set with no unpaired electrons, so it is diamagnetic with bond order 2: (A) is correct. $\mathrm{B_2}$ (10 electrons) places one electron in each of the two degenerate $\pi_{2p}$ orbitals (Hund's rule), so it is paramagnetic with two unpaired electrons although its bond order is 1: (B) is wrong. Removing an electron from $\mathrm{O_2}$ takes it from the antibonding $\pi^*_{2p}$ orbital, raising the bond order from 2 to 2.5 and shortening the bond; removing one from $\mathrm{N_2}$ takes it from the bonding $\sigma_{2p}$, lowering the order from 3 to 2.5 and lengthening the bond: (C) is correct. $\mathrm{NO}$ has bond order 2.5 with one unpaired electron in $\pi^*$; $\mathrm{NO^{+}}$ (14 electrons, isoelectronic with $\mathrm{N_2}$) has order 3 and no unpaired electrons: (D) is correct. Answer: A, C, D.