Much of chemistry is about bonds, but sometimes it can also be about anti-bonds. It is also true that the simplest of molecules can have quite subtle properties. Thus most undergraduate courses in chemistry deal with how to describe the bonding in the diatomics of the first row of the periodic table. Often, only the series C2 to F2 is covered, so as to take into account the paramagnetism of dioxygen, and the triple bonded nature of dinitrogen (but never mentioning the strongest bond in the universe!). Rarely is diberyllium mentioned, and yet by its strangeness, it can also teach us a lot of chemistry.
The diagram below is what many textbooks show. The diagram can vary (and hence confuse) slightly, in regard to the relative ordering of the σ and π energy levels originating from the overlap of the 2p orbitals. It depends on the atom, and for Be, the σ comes out higher than the π. The other key ordering is that the σ* antibonding orbital resulting from out of phase overlap of the two 2s orbitals is actually lower in energy than the π bonding orbital resulting from in-phase overlap of the 2p orbitals. Yes, an antibonding orbital is more stable than a bonding orbital!
Well, the diagram shows that the pair of occupied molecular orbitals resulting from the two (symmetric and antisymmetric, or g and u) combinations of the 1s orbitals cancel each other, as do the 2s combinations, and we conclude the bond order for this molecule is zero! Actually, if a quantum mechanical calculation is performed (at the ωB97XD/6-311G(d,p) level), the bond length emerges as 2.81Å and a vibrational wavenumber of 167 cm-1 is predicted. Despite the zero bond order, a weak bond IS predicted, and this is the van der Waals or dispersion bond.
Let us now pump this molecule up to a higher energy state by a double excitation of the two electrons in the 2s σ* electrons. We have to split them up, one each, into the next available orbital, which is the π, to form a triplet state (just like di-oxygen).
Well, this (higher energy) state is certainly shorter (a contrast with my item on longer being stronger). The length is now 1.78Å, which is more than 1Å shorter than the original state, despite being ~ 45 kcal/mol higher in energy. The Be-Be stretching wavenumber goes up to 917 cm-1. With four electrons in bonding orbitals, diberyllium has a double bond! One can also pair the π electrons up to form an open shell (excited) singlet, which is ~ 51 kcal/mol higher than the closed shell (unbonded) singlet. This also has a length of 1.78Å and a marginally lower stretch of 909 cm-1. If you want to read more about the doubly excited state of this molecule, see DOI: 10.1139/v96-111.
One might be tempted to make an analogy between physics, and its particles and antiparticles. Yes, electrons can occupy antibonding as well as bonding orbitals. But the overall bond order will be reduced to zero if the total numbers of each are equal. And one can be pretty certain that there is no molecule at all in which the number of antibonding electrons exceeds the bonding ones! Or, if anyone is aware of such an example, do tell!