Paramagnetic Diatomic Species Without Two S Two P Mixing

According to Molecular Orbital Theory (MOT), when $2s\text{-}2p$ mixing is not operative (similar to the standard orbital ordering observed in $\mathrm{O}_2$ and $\mathrm{F}_2$), the increasing order of energies of the molecular orbitals is:

$\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < \sigma_{2p_z} < (\pi_{2p_x} = \pi_{2p_y}) < (\pi^*_{2p_x} = \pi^*_{2p_y}) < \sigma^*_{2p_z}$

Now, let us determine the electronic configurations of the given homonuclear diatomic species assuming no $2s\text{-}2p$ mixing:

1. $\mathrm{Be}_2$ (Total electrons $= 8$): \sigma_{1s}^2\ \sigma^*_{1s}^2\ \sigma_{2s}^2\ \sigma^*_{2s}^2

   • Number of unpaired electrons, $n = 0$.
   • Hence, $\mathrm{Be}_2$ is diamagnetic.

2. $\mathrm{B}_2$ (Total electrons $= 10$): \sigma_{1s}^2\ \sigma^*_{1s}^2\ \sigma_{2s}^2\ \sigma^*_{2s}^2\ \sigma_{2p_z}^2

   • Number of unpaired electrons, $n = 0$.
   • Hence, $\mathrm{B}_2$ is diamagnetic.

3. $\mathrm{C}_2$ (Total electrons $= 12$): \sigma_{1s}^2\ \sigma^*_{1s}^2\ \sigma_{2s}^2\ \sigma^*_{2s}^2\ \sigma_{2p_z}^2\ \pi_{2p_x}^1\ \pi_{2p_y}^1

   • According to Hund's rule of maximum multiplicity, the degenerate $\pi_{2p_x}$ and $\pi_{2p_y}$ orbitals each receive one unpaired electron.
   • Number of unpaired electrons, $n = 2$.
   • Hence, $\mathrm{C}_2$ is paramagnetic.

4. $\mathrm{N}_2$ (Total electrons $= 14$): \sigma_{1s}^2\ \sigma^*_{1s}^2\ \sigma_{2s}^2\ \sigma^*_{2s}^2\ \sigma_{2p_z}^2\ \pi_{2p_x}^2\ \pi_{2p_y}^2

   • Number of unpaired electrons, $n = 0$.
   • Hence, $\mathrm{N}_2$ is diamagnetic.

Therefore, the paramagnetic species under the assumption of no $2s\text{-}2p$ mixing is $\mathrm{C}_2$.

Correct Option: (C)