Paramagnetic substances are those that contain unpaired electrons. We can determine the presence of unpaired electrons by analyzing the molecular orbital (MO) configuration of each species.

Step 1: Determine the total number of electrons for each species.

• ${\mathrm{O}}_2^{2-}$: $2\times 8+2=18$ electrons
• $\mathrm{NO}$: $7+8=15$ electrons
• $\mathrm{CO}$: $6+8=14$ electrons
• ${\mathrm{CN}}^{-}$: $6+7+1=14$ electrons

Step 2: Write the molecular orbital configuration and identify unpaired electrons.

For species with up to 14 electrons, the MO configuration is $\sigma 1s,{\sigma }^{\ast }1s,\sigma 2s,{\sigma }^{\ast }2s,\pi 2p_x=\pi 2p_y,\sigma 2p_z,{\pi }^{\ast }2p_x={\pi }^{\ast }2p_y,{\sigma }^{\ast }2p_z$.
For species with more than 14 electrons, the order of $\sigma 2p_z$ and $\pi 2p$ orbitals is swapped: $\sigma 1s,{\sigma }^{\ast }1s,\sigma 2s,{\sigma }^{\ast }2s,\sigma 2p_z,\pi 2p_x=\pi 2p_y,{\pi }^{\ast }2p_x={\pi }^{\ast }2p_y,{\sigma }^{\ast }2p_z$.

• A) ${\mathrm{O}}_2^{2-}$ (18 electrons):
  MO configuration: $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2{\pi }^{\ast }2p_x^2{\pi }^{\ast }2p_y^2$. All electrons are paired. Hence, it is diamagnetic.
• B) $\mathrm{NO}$ (15 electrons):
  MO configuration: $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2{\pi }^{\ast }2p_x^1$. It has one unpaired electron in the ${\pi }^{\ast }2p_x$ orbital. Hence, it is paramagnetic.
• C) $\mathrm{CO}$ (14 electrons):
  MO configuration: $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2\pi 2p_y^2\sigma 2p_z^2$. All electrons are paired. Hence, it is diamagnetic.
• D) ${\mathrm{CN}}^{-}$ (14 electrons):
  MO configuration: $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2\pi 2p_y^2\sigma 2p_z^2$. All electrons are paired. Hence, it is diamagnetic.

Therefore, NO is paramagnetic.

Correct Answer: (B)