Metallic bonds are said to be delocalised.
Lithium crystals have each lithium atom in direct contact with eight other lithium atoms. Each atom could therefore be in a hybrid of eight chemical bonds with each other.
But a similar explanation does not work for atoms like beryllium.
Some possible arrangements of electrons that beryllium could be a hybrid of configurations of might be:
$$\ce{Be^{+} + Be^{-}}$$
$$\ce{Be^{2+}_2 + Be^{2-}_2}$$
$$\ce{Be^+ + e^-}$$
$$\ce{Be^{+}_2 + e^-_}$$
$$\ce{Be^{2+}_2 + 2e^{2-}_}$$
$$\ce{Be^{2+}_2 + e^{2-}_2}$$
$$\ce{Be^{2+}_2 + e^{2-}_2}$$
But beryllium is diamagnetic and all these suggest paramagnetic properties or a $e^{2-}_2$ pair which doesn't seem right to me although I'm not fully sure how the free electron hybrids would work.
Another possibility is that the electrons are dragged up into the p orbitals.
For example, the atoms would have valence electron configurations of:
$$\underset{sp}{[\uparrow \vert \uparrow]} \underset{p}{[\; \vert \; ]}$$
which would result in hybrids of $\ce{Be2}$ or $\ce{Be_4}$ I think.
However, the atoms would have to be dragged up really high so that they aren't paramagnetic.
Let me list out the molecular orbital theory bonding and antibonding orbitals for diberyllium.
$$ \underset{\sigma}{[\uparrow \downarrow]} \underset{\sigma^*}{[\uparrow \downarrow]} \underset{\pi}{[\; \vert \;]} \underset{\sigma}{[\; ]} \underset{\pi^*}{[\; \vert \;]} \underset{\sigma^*}{[\;]}$$
If I wanted a bonding material that was not paramagnetic I'd have to lift up the energy to something like:
$$ \underset{\sigma}{[\uparrow \downarrow]} \underset{\sigma^*}{[\;]} \underset{\pi}{[\; \vert \;]} \underset{\sigma}{[ \uparrow \downarrow]} \underset{\pi^*}{[\; \vert \;]} \underset{\sigma^*}{[\;]}$$
which is probably too high energy to matter for any purposes.
