I've read that alkali metals form ionic bonds; $\ce{Li}$ is an exception which majorly forms covalent bonds. Wikipedia says dilithium exists. This makes me wonder why $\ce{Li_8}$ doesn't exist. It seems having all 8 electrons shared produces a low energy state (because the valence shell in each of the 8 atoms will be completely filled) ?


3 Answers 3


Yes, they do exist and were characterised spectroscopically, see here (and there is a note on similar clusters for sodium):

Blanc, J.; Bonačić‐Koutecký, V.; Broyer, M.; Chevaleyre, J.; Dugourd, P.; Koutecký, J.; Scheuch, C.; Wolf, J. P.; Wöste, L. Evolution of the electronic structure of lithium clusters between four and eight atoms. J. Chem. Phys. 1992, 96 (3), 1793–1809. DOI: 10.1063/1.462846.

They are not, however, "stable" enough to exist in solid state or even to obtain a pure gas, rather complicated arcane trickery is used to work with (very small amount of) them.

Lithium has rather complicated chemistry, easily forming polyhedral structures when bound to proper partner. For example, methyllithium is a tetramer. The simplest rationalization is that its s- and p-orbitals have very close energies and thus low-energy vacant p-orbitals participate in formation of multicentre electron-deficient bonds.


You may be applying the octet rule to a situation where there is no p orbitals involved, the electronic structure of the (ground state) Li atom. Li has the electronic configuration 1s²2s1. So, the covalent bond is between two 2s orbitals, each with 1 electron. Using the Lewis structure, Li2 is Li:Li and that is a full bonding orbital, there's no p atomic orbitals involved in the bonding which might increase valence to 8 (an octet).


Looking at the electron configuration of lithium ($\ce{(1s)^2(2s)^1}$), it's very difficult for the $\ce{1s}$ electrons to be involved in bond formation/sharing.

The reason is because of the extent of attraction of those electrons in the 1s orbital as they are very close to the nucleus (very stable) and rarely available for any bond formation. Again, say if we were to look at the ionisation energy required to extract those electrons in the $\ce{1s}$ orbital, you would agree with me that it's almost impossible.

Looking at the $\ce{2s}$ orbital, you notice it's not as stable since there is more room for electrons to come and occupy in the $\ce{2s}$ (or $\ce{2p}$) orbital to attain maximum stability.

In short, we can't have a $\ce{Li8}$ molecule.

  • 1
    $\begingroup$ @alphonse . Thanks for the correction, I was basing my explanation on most stable bonds/ compounds formed by Li. Some Li compounds are rare and very difficult to isolate so I had taken that angle. I apologise for this mistake and I do agree with permeakra's explanation $\endgroup$ Mar 18, 2017 at 8:10

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