 Solution

Both Be${}_2$'s occupied bonding and antibonding 2s levels cancel in the simplest diagram. Beryllium's relatively accessible empty 2p orbitals allow mixing and correlated configurations to change that cancellation and produce a weak bond. An s-only single-configuration picture is therefore insufficient; spectroscopy and accurate correlated calculations establish its binding.^[ref]

For He${}_2$, $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2$ similarly gives zero elementary covalent bond order, but the empty-shell excitation gap is much larger and there is no comparable low-energy 2s/2p near-degeneracy. Thus it has no ordinary stable covalent bond.

The literal claim that He${}_2$ does not exist needs qualification: an extraordinarily weak, cold van der Waals helium dimer has been observed. That does not contradict the zero covalent-bond-order diagram.^[ref]

The beryllium dimer and helium dimer illustrate two physically different ways that the elementary bond-order picture can miss weak binding.