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.

  • $\begingroup$ Where did you get the information that "Lithium crystals have each lithium atom in direct contact with four other lithium atoms."? Everything I find says that its crystal structure at room temperature is body-centered cubic, where each atom has eight neighbors. $\endgroup$
    – f''
    Jul 24, 2016 at 6:43
  • $\begingroup$ That was a typo I meant 8 $\endgroup$ Jul 24, 2016 at 16:08

1 Answer 1


Metallic bonding is essentially delocalisation of covalent bonds to the logical extreme and cannot be considered as "two-centre, two-electron". The metallic lattice is a network of cations immersed in a "sea" of electrons. As the highest energy orbital is the spherical s, there is no directional influence and hence why beryllium has the efficient hexagonal close packed structure.

  • $\begingroup$ But then why is beryllium diamagnetic? If beryllium is like $\ce{Be^+}$ in a sea of $\ce{e^-}$ then it should be paramagnetic. If it is like $\ce{Be^{2+} + 2e^{-}}$ shouldn't it still be paramagnetic? $\endgroup$ Jul 24, 2016 at 19:55
  • $\begingroup$ The two electrons have opposite spins, so there is no overall magnetic moment. $\endgroup$
    – gsurfer04
    Jul 24, 2016 at 21:23
  • $\begingroup$ So it is $\ce{Be^{2+} + e{2-}_2}$! But I thought pairs only happen in superconductors with Cooper pairs? $\endgroup$ Jul 24, 2016 at 21:39
  • 1
    $\begingroup$ Most lanthanoids have $Ln^{3+}$ + 3$e^-$. $\endgroup$
    – gsurfer04
    Jul 25, 2016 at 9:57

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