Consider the species N2^+, O2^+, NO, O2^-, C2, CN^-, NO^+, B2 and He2^+. The number of species that are paramagnetic and have a bond order strictly…

Fill molecular orbitals using the correct ordering: for species with 14 or fewer total electrons ($\mathrm{B_2}$, $\mathrm{C_2}$, $\mathrm{N_2^+}$, $\mathrm{CN^-}$, $\mathrm{NO^+}$) the $\pi_{2p}$ orbitals lie below $\sigma_{2p}$; for the rest ($\mathrm{O_2^\pm}$, $\mathrm{NO}$) $\sigma_{2p}$ is lower. Results (bond order, magnetism): $\mathrm{N_2^+}$ 2.5 paramagnetic; $\mathrm{O_2^+}$ 2.5 paramagnetic; $\mathrm{NO}$ 2.5 paramagnetic (one electron in $\pi^*$); $\mathrm{O_2^-}$ 1.5 paramagnetic but not above 1.5; $\mathrm{C_2}$ 2 diamagnetic; $\mathrm{CN^-}$ 3 diamagnetic; $\mathrm{NO^+}$ 3 diamagnetic; $\mathrm{B_2}$ 1 paramagnetic (two unpaired $\pi$ electrons); $\mathrm{He_2^+}$ 0.5 paramagnetic. Only $\mathrm{N_2^+}$, $\mathrm{O_2^+}$ and $\mathrm{NO}$ meet both conditions (counted by a small program), so the answer is 3.