$Be$ $(\mathrm{Z}=4)$ So, total electron in $\mathrm{Be}_{2}=8$

Electron configuration in $\mathrm{MO}$ for $\mathrm{Be}_{2}$ :

$\mathrm{KK}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}$ OR $(\sigma 1 s)^{2}\left(\sigma^{*} 1 s\right)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}$

Magnetic Property : All electrons are paired. So, diamagnetic

Bond order $\mathrm{BO}=\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)=\frac{1}{2}(2-2)=0 \quad$ OR $\mathrm{BO}=\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)=\frac{1}{2}(4-4)=0$

Bond order is zero, $\mathrm{So}, \mathrm{Be}_{2}$ is unstable and does not exist.

Energy diagram for $\mathrm{Be}_{2}$ molecule is as under.