# How to use Hund's third rule for excited electronic configurations?

Hund third rule states:

For a given term, in an atom with outermost subshell half-filled or less, the level with the lowest value of the total angular momentum quantum number ${\displaystyle J\,}$ (for the operator ${\displaystyle {\boldsymbol {J}}={\boldsymbol {L}}+{\boldsymbol {S}}}$) lies lowest in energy. If the outermost shell is more than half-filled, the level with the highest value of ${\displaystyle J\,}$m is lowest in energy.

I can't understand how to apply this rule to excited electronic configuration (I'm aware that Hund's do not work perfectly with excited electronic configuration but I wonder if this is the case).

In particular I don't understan what is the "half-filled" outermost "shell" when more than one "subshell" is involved (as in excited configuration). For example for the excited Neon triplet state

$$(1s)^2 (2s)^2 (2p)^5 (3s)^1$$

We have $S=1$ and $L=1$. It turns out that the $J$ to chose between $0$ and $1$ is $J=0$, but I don't understand why, since here there is the subshell $2p$ with one electron missing (and therefore it is "more than half filled"), and the next subshell $3s$ with one electron only (exactly "half filled"). Also if I consider the two subshells (2p+3s) thogheter they are "more than half filled" (there are 6 electron over the 6+2=8 that can be allocated in total)

• I am not too familiar with Hund's rule, but as I interpret it, the outermost subshell in your example, would be the 3s, which means J = L + S. If S = L = 1, then J should be 2. For J to be 0, S or L must be -1. – Guill Aug 15 '18 at 4:51
• One should not use Hund's rules for excited configurations. – user137289 Oct 11 '18 at 21:20