In our 3+1 dimensional world covalent bonds occur because electron pair are shared between atoms. How many electrons are needed to form a covalent bond in a n+1 dimensional world?

The reason this is not entirely obvious is that there are more spin quantum numbers in a high-dimensional world. In our world there is just one spin, and hence two electrons with opposite spins are allowed to have otherwise equal quantum numbers, spreading out between the two atoms. Hence we end up with two electrons involved in the bond.

In a $n$ dimensional space there are $\lfloor n/2 \rfloor$ separate angular momenta, and it seems reasonable (but more complex to analyze properly) that there would be more spin quantum numbers. So does that mean that we could have not just 2-electron covalent bonds, but k-electron pair bonds (up to $k=n/2$)? Or do we need $n$ electrons to make a covalent bond? Or even $2^{n/2}$ so each combination of spin quantum numbers is present?

(Ionic bonds appear much easier to model, although the rapid decay of electromagnetic attraction would make them weaker.)


1 Answer 1


What has been bothering me for some time is the binding of spherical harmonics to Cartesian coordinates in QM. This gives an artificially constructed distinction between the first two electrons of the second and third shells and the remaining 6 electrons. Thereby there are solutions of spherical harmonics, which allows an equally distributed arrangement of 8 electrons. And this completely without additional dimensions.

Interestingly, the distribution of electrons in the Cartesian coordinate system leads to a geometric frustration. Although we are always looking for symmetries and equilibrium states.

I run the risk to lose the reader if I use the duplicity of spin and magnetic dipole and ask the reader to consider the distribution of the electrons also as the distribution of their magnetic dipoles. But please, try it only once.

If you distribute the 8 electrons of the second and third shell to two tetrahedra pushed into each other, you get a frustration-free geometry. This is possible if the magnetic dipoles of the electrons are aligned to the nucleus. The 4 electrons of the one tetrahedron are directed with the north poles inwards. The dipoles on the other tetrahedron are then of course directed inwards with the south poles. enter image description here

This brings us back to Gilbert N. Lewis valence theory with his cubical atom.

Here is the corresponding spherical harmonic:

enter image description here

  • $\begingroup$ What does this have to do with my question? $\endgroup$ Commented Feb 5, 2022 at 19:06

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