Write the molecular orbital electronic configuration keeping in mind that there is no $2 s-2 p$ mixing, then if highest occupied molecular orbital contain unpaired electron then molecule is paramagnetic otherwise diamagnetic.

Note: Since there is no mixing of $2 s$ and $2 p$ orbitals of each of the atom in a diatomic molecule, the energy of bonding molecular orbital $2 p z$, i.e., $2 p$ is less than that of bonding molecular orbital of $2 p x$ and $2 p y$, i.e., $\left(\pi_{2 p x}^*\right.$ and $\left.\pi_{2 p y}^*\right)$.

$$ \therefore $$ The order of molecular orbital energy levels, i.e., bonding and anti-bonding energy levels is as follows:

$$ \begin{aligned} \left(\sigma_{1 s}\right)\left(\sigma_{1 s}^*\right)\left(\sigma_{2 s}\right)\left(\sigma_{2 s}^*\right)\left(\sigma_{2 p z}\right)\left(\pi_{2 p x}\right. & \left.\equiv \pi_{2 p y}\right) \left(\pi_{2 p x}^*\right. \left.\equiv \pi_{2 p y}^*\right)\left(\sigma_{2 p z}^*\right) \end{aligned} $$

(A) Total number (no.) of electrons in $\mathrm{Be}_2=4+4=$ 8. The electronic configuration of $\mathrm{Be}_2$ is:

$$ \left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2 $$

All the electrons in $\mathrm{Be}_2$ are paired up, $\mathrm{Be}_2$ is diamagnetic.

(B) Total number (no.) of electrons in $\mathrm{B}_2=5+5=$ 10. The electronic configuration of $B_2$ is:

$\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p z}\right)^2$

All the electrons in $B_2$ are paired up, $B_2$ is diamagnetic.

(C) Total no. of electrons in $\mathrm{C}_2=6+6=12$ The electronic configuration of $\mathrm{C}_2$ is:

$$ \left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p z}\right)^2\left(\pi_{2 p x}^{* 1} \equiv \pi_{2 p y}^{* 1}\right) $$

There are two unpaired electrons in doubly degenerate.

$\pi_{2 p x}$ and $\pi_{2 p y} \mathrm{C}_2$ is paramagnetic.

(D) Total no. of electrons in $\mathrm{N}_2=7+7=14$

The electronic configuration of $\mathrm{N}_2$ is:

$$ \left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p z}\right)^2\left(\pi_{2 p x}^{* 2} \equiv \pi_{2 p y}^{* 2}\right) $$

All the electrons in $\mathrm{N}_2$ are paired up; $\mathrm{N}_2$ is diamagnetic.