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List of Hund's rules

In atomic physics, Hund's rules, (occasionally called the "bus seat rule") refer to a simple set of rules used to determine which is the term symbol that corresponds to the ground state of a multi-electron atom. They are named in honour of Friedrich Hund who contributed Hund's Rule, rule two as listed here.

The four rules are:

Electrons can occupy energy levels only above the 2s orbital total orbital.
The term with maximum multiplicity (maximum $S \,$ ) has the lowest energy level.
For a given multiplicity, the term with the largest value of $L \,$ has the lowest energy in an orbital.
For atoms with less than half-filled shells, the level with the lowest value of $J \,$ lies lowest in energy. Otherwise, if the outermost shell is more than half-filled, the term with highest value of $J \,$ is the one with the lowest energy.

The rules deal in a simple way how the usual energy interactions dictate the ground state term. The rules assume that the repulsion between the outer electrons is very much greater than the spin-orbit interaction which is in turn stronger than any other remaining interactions. This is referred to as the LS coupling regime.

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Rule #1

It can be shown that for full orbitals and suborbitals both the residual electrostatic term (repulsion between protons) and the spin-orbit interaction cannot shift all the energy levels together. Thus when determining the ordering of energy levels in general only the inner valence electrons need to be considered.

Rule #2

Due to the Pauli exclusion principle, two electrons cannot share the same set of quantum numbers within the same system. Therefore, there is room for only two electrons in each spatial orbital. One of these electrons must have (for some chosen direction z), $S_Z = 1/2 \,$ , and the other must have $S_Z = -1/2 \,$ . Hund's second rule states that the lowest energy atomic state is the one which maximizes the sum of the $S \,$ values for all of the electrons in the system, maximizing the number of unpaired electrons.

It is often stated that this is the highest energy atomic state because it forces the paired electrons to reside in different spatial orbitals, and this results in a larger average distance between the two electrons, reducing electron-electron repulsion energy. But, in fact, careful calculations have shown that this explanation can be wrong, at least for light systems.

Rule #3

This rule deals again with reducing the repulsion between electrons. It can be understood from the classical picture that if all electrons are orbiting in the same direction (higher orbital angular momentum) they meet less often than if some of them orbit in opposite directions. In that last case the repulsive force increases, which separates electrons. This adds potential energy to them, so their energy level is higher.

Rule #4

This rule considers the energy shifts due to split-orbit coupling. In the case where the spin-orbit coupling is weak compared to the residual electrostatic, where $L \,$ and $S \,$ are still good quantum numbers the splitting is given by:

$\begin{matrix} \Delta E & = & \zeta (L,S) \{ \mathbf{L}\cdot\mathbf{S} \} \\ \ & = & \zeta (L,S) \{ J(J+1)-L(L+1)-S(S+1) \} \end{matrix}$

The value of $\zeta (L,S)\,$ changes from minus to plus for shells less than half full. The first term gives the dependence of the ground state on the magnitude of $J \,$ .

Example

As an example, consider the ground state of silicon. The electronic configuration of Si is $1s^2, 2s^2, 2p^6, 3s^2, 3p^2 \,$ . Applying the first rule, only the outer $3p^2 \,$ electrons need be considered. The possible multiplets are ${}^1\!S ,{}^3\!S ,{}^1\!P ,{}^3\!P , {}^1\!D ,{}^3\!D \,$ ; of those, ${}^3\!S \,$ and ${}^3\!D \,$ are not allowed because of the exclusion principle. The second rule now states that the triplet state ${}^3\!P \,$ with $S = 1 \,$ has the lowest energy. There is no choice of triplets $(S = 1) \,$ states, so the first rule is not required. [If the ${}^3\!D { } (L = 2)\,$ state were allowed, then the third rule would come into force and state that it was more favourable than the ${}^3\!P { } (L = 1) \,$ state.] The triplet ${}^3\!P \,$ consists of three states, $J = 2,1,0 \,$ . With only two of six possible electrons in the shell, it is less than half-full and thus ${}^3\!P_0 \,$ is the ground state.