$Be$ $(\mathrm{Z}=4)$, so electron configuration $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2}$.

In $\mathrm{Be}, 4$ electrons and in $\mathrm{Be}_{2}$ total eight electrons.

Electron configuration of $\mathrm{Be}_{2}:\left(\sigma_{1 s}\right)^{2}\left(\sigma_{1 s}^{*}\right)^{2}\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}$

So, $BMO$ $\left(N_{b}\right)=4$ and ABMO $\left(N_{a}\right)=4$

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

Bond order of $\mathrm{Be}_{2}$ is zero, so, $\mathrm{Be}_{2}$ is unstable and not possible.