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Why is a 3s orbital lower in energy than a 3p orbital in all atoms other than a hydrogen atom which only has a single electron?

Md Ariful Islam

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Md Ariful Islam

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Because of how there IS only one electron in $"H"$ atom. That single electron does not introduce orbital anguar momentum, so no matter what value of $l$, the orbital energies for the same $n$ are all the same.

The added electrons in multi-electron atoms are intrinsically correlated in such a way that each electron is influenced by the motions of the others.

This electron correlation introduces the effect of electron-electron repulsion. That in turn generates an energy splitting of the orbitals with the same $n$ but different $l$.

For example, for just the $3s$ and $3p$...

$"H"$ atom:

$ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))"$
$" "" "" "underbrace(" "" "" "" "" "" "" "" "" ")$
$3s" "" "" "" "" "" "3p$

For multi-electron atoms:

$" "" "" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))"$
$" "" "" "underbrace(" "" "" "" "" "" "" "" "" ")$
$ul(color(white)(uarr darr))" "" "" "" "" "3p$
$3s$

We had that the $3s, 3p, 3d$ orbitals are the same energy in the $"H"$ atom, but in higher-electron atoms, we instead have the energy ordering $3s < 3p < 3d$.

In multi-electron atoms, higher $l$ is higher energy for orbitals with the same $n$.

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