An atom with many electrons is described first by a central field: each electron moves in the spherically symmetric average of the nucleus and all the others, the full charge $Ze$ close to the nucleus and a single charge far out. The levels are still labelled $nl$, but screening removes hydrogen’s degeneracy in $l$: an s electron reaches in close to the nucleus and is bound more tightly than a p electron of the same shell, and so on. Filling the levels in order, two to an orbital, gives the configurations and the periodic table: 1s, 2s, 2p, 3s, 3p, 4s, 3d. What the central field leaves out — the rest of the repulsion — splits a configuration into terms of definite total $L$ and $S$, written ${}^{2S+1}L$, and the spin–orbit coupling splits a term into levels of definite $J$. Hund’s rules, found empirically, say which term lies lowest: the largest $S$, then the largest $L$, then the smallest $J$ if the shell is less than half full and the largest if more.

Seats in a stadium are filled from the front, but which rows count as the front depends on the view: the $s$ seats are closer to the pitch than their row number suggests. Hund’s rules then settle who sits next to whom within a row.

Carbon’s outer configuration $2p^{2}$, split by the repulsion beyond the central field into the terms $^{3}P$, $^{1}D$ and — the triplet lowest, by Hund’s first rule — and by the spin–orbit coupling into levels of definite . The shell is less than half full, so is the ground level, and the gaps above it grow as , by Landé’s interval rule. Not to scale.

In the central-field approximation each electron moves in a spherically symmetric potential, $-Ze^{2}/4\pi\varepsilon_{0}r$ close in and $-e^{2}/4\pi\varepsilon_{0}r$ far out. Its levels fill in the order

• A configuration, such as carbon’s $1s^{2}2s^{2}2p^{2}$, lists how many electrons occupy each subshell; closed subshells have $L=S=0$ and drop out of the rest.

Count the states: two electrons in six spin-orbitals. Then sort them by symmetry. For two orbital angular momenta 1, the total $L=2$ and $L=0$ are symmetric under exchange and $L=1$ antisymmetric; the whole state must be antisymmetric, so the symmetric spin triplet goes with $L=1$ and the antisymmetric singlet with and 2. The states add back up to fifteen.

Two electrons in two different spin-orbitals among the six of $2p$.

The top state of $L=2$, $\ket{1}\ket{1}$, is symmetric and lowering keeps it so; $L=1$ takes the antisymmetric combination at $M=1$; is what is left at , symmetric.

The whole state is antisymmetric: a symmetric spin part needs an antisymmetric orbital part, and the other way round.

$(2S+1)(2L+1)$ states in each term.

Hund’s rules: the largest $S$; the subshell holds two of six, less than half, so $J=|L-S|=0$.

A configuration says how many electrons are in each subshell. Its states split into terms of definite $L$ and $S$, each with $(2S+1)(2L+1)$ states.

Lowest lies the term with the largest $S$, then the largest $L$; the lowest level has $J=|L-S|$ below half filling and $J=L+S$ above.

In a many-electron atom an s electron penetrates the inner shells and feels more of the nucleus than a p electron of the same shell, so the levels depend on $l$ as well as $n$.

Carbon’s $2p^{2}$ gives the terms ${}^{3}P$, ${}^{1}D$ and . The triplet lies lowest, and with the subshell less than half full, .

Great work! You now know how to:

• build configurations in the central-field approximation
• find the terms of two equivalent p electrons
• use Hund’s rules to find a ground level