Suppose a hydrogen atom is excited up to a $\underset{1s}{[\;]} \underset{2s}{[\uparrow]}$ state which takes about 1000 kJ/mol.

Due to electron shielding f orbitals tend to be higher energy than d orbitals which are higher than p orbitals which are higher than s orbitals. But in a hydrogen atom there is only one electron so the 2s and 2p orbitals are equivalent in energy.

So there's no difference in energy with $\underset{1s}{[\;]} \underset{2s}{[\;]} \underset{2p}{[\uparrow \vert \; \vert \;]}$.

Now there are three different symmetrical p orbitals so I guess the end result would still be spherical. But would the radius of the excited hydrogen atom be an overall higher radius then one would typically expect?

I don't know how to square this with How big is an excited hydrogen atom? which seems to be mostly talking about general approximations for very highly excited states or the fine details of the Schrodinger equation which I don't really understand.

I also don't know if this applies to molecules. Are the 2p orbitals and 2s orbitals equivalent in energy when a hydrogen atom is bounded to another atom? Pi bonds are generally weaker than sigma bonds but I don't know how two equal energy sigma bonds would work.


There is a difference in energy between the 2s and 2p levels in hydrogen, which is significant for a wide variety of applications.. The main contribution is called the fine-structure splitting. If you specifically excite to the 2s state you will indeed still be spherically symmetric. The 2P wavefunctions look a bit like dumbbells.

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