# 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?

Mar 15, 2017

Because of how there IS only one electron in $\text{H}$ 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 $3 s$ and $3 p$...

$\text{H}$ atom:

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" "" "" "underbrace(" "" "" "" "" "" "" "" "" ")
$3 s \text{ "" "" "" "" "" } 3 p$

For multi-electron atoms:

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" "" "" "underbrace(" "" "" "" "" "" "" "" "" ")
$\underline{\textcolor{w h i t e}{\uparrow \downarrow}} \text{ "" "" "" "" } 3 p$
$3 s$

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

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