Provide the electron configuration, bond order, magnetic property, and energy level diagram for the Boron $(B_{2})$ molecule, and discuss its existence.

(N/A) $B_{2}$ $(Z=5)$ has the atomic configuration $1s^{2} 2s^{2} 2p^{1}$. Thus, the total number of electrons in $B_{2}$ is $10$.
The molecular orbital $(MO)$ electron configuration for $B_{2}$ is: $KK(\sigma_{2s})^{2}(\sigma^{*}_{2s})^{2}(\pi 2p_{x})^{1}(\pi 2p_{y})^{1}$. Note that for $B_{2}$, the energy of $\pi 2p$ orbitals is lower than that of the $\sigma 2p_{z}$ orbital due to $s-p$ mixing.
Bond order $= \frac{1}{2} (N_{b} - N_{a}) = \frac{1}{2} (6 - 4) = 1$.
Since there are two unpaired electrons in the $\pi 2p$ orbitals, the $B_{2}$ molecule is paramagnetic.
It possesses a single bond, which makes it stable in the gas phase.
The bond length is $159 \ pm$ and the bond dissociation energy is $290 \ kJ \ mol^{-1}$.
The energy level diagram is provided below.