$\mathrm{B}_{2}(\mathrm{Z}=5) 1 s^{2} 2 s^{2} 2 p^{1}$ So, total electron in $\mathrm{B}_{2}=10$

Electron configuration in $\mathrm{MO}$ for $\mathrm{B}_{2}: \mathrm{KK}\left(\sigma_{2 s}\right)^{2}\left(\sigma^{*} 2 s\right)^{2}\left(\pi 2 p_{x}\right)^{1}\left(\pi 2 p_{y}\right)^{1} \quad$ (Because $\mathrm{B}_{2} \quad \pi 2 p<\sigma 2 p_{z}$ ) Bond order $=\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)=\frac{1}{2}(4-2)=1$

Two unpaired electron present in $\mathrm{B}_{2}$, So it is paramagnetic and Single bond so, it is stable. Bond length B – B $(159 \mathrm{pm})$ and

Bond energy $\mathrm{B}_{2} 290 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Energy diagram for $\mathrm{B}_{2}$ Molecule is as under (fig).