I’m having a hard time trying to memorize this, so writing a small article here in the hopes it will help.

When filling electrons into atomic orbitals, three rules govern how nature arranges them to minimise energy. These are Hund’s rules, and understanding them requires first revisiting quantum numbers and the Pauli exclusion principle.

Quantum numbers

Each electron in an atom is described by four quantum numbers:

The Pauli exclusion principle states that no two electrons in the same atom can share all four quantum numbers. This is ultimately a consequence of electrons being fermions (particles with half-integer spin), whose many-body wavefunction must be antisymmetric under exchange.

A subshell with quantum number $\ell$ has $2\ell + 1$ orbitals, and each orbital can hold at most 2 electrons (spin up and spin down), so the maximum occupancy of a subshell is $2(2\ell+1)$.

Subshell $\ell$ Orbitals Max electrons
$s$ 0 1 2
$p$ 1 3 6
$d$ 2 5 10
$f$ 3 7 14

The Aufbau principle

Before stating Hund’s rules, recall the Aufbau principle: subshells are filled in order of increasing energy. The approximate ordering is given by the $(n + \ell)$ rule (Madelung’s rule): lower $n + \ell$ first; when equal, lower $n$ first:

$$ 1s \to 2s \to 2p \to 3s \to 3p \to 4s \to 3d \to 4p \to 5s \to 4d \to \ldots $$

Hund’s rules apply within a partially filled subshell once we start placing electrons into its degenerate orbitals.

Hund’s rules

Rule 1: Maximise total spin $S$

For a given electron configuration, the term with the maximum total spin $S$ (equivalently, the highest spin multiplicity $2S+1$) has the lowest energy.

Electrons with the same spin cannot occupy the same orbital (Pauli), so they end up in different orbitals and avoid each other spatially. This reduces electron-electron repulsion, lowering the energy. In practice this means: fill each orbital with one electron (all spin-up) before pairing any spins.

Rule 2: Maximise total orbital angular momentum $L$

Among terms with the same $S$, the one with the largest total orbital angular momentum $L$ has the lowest energy.

Classically, electrons orbiting in the same direction avoid each other more effectively than electrons orbiting in opposite directions, so larger $L$ means less repulsion and lower energy.

$L$ is obtained by combining the individual $m_\ell$ values:

$$ L = \left| \sum_i m_{\ell_i} \right|_{\text{max}} $$

(More precisely, $L$ is the maximum value of the total $M_L = \sum_i m_{\ell_i}$ over the valid assignments consistent with Pauli and with the $S$ from Rule 1.)

Rule 3: $J$ depends on subshell filling

Given $S$ and $L$, the total angular momentum $J = |L - S|$ for a less than half-filled subshell, and $J = L + S$ for a more than half-filled (or exactly half-filled if $L \neq 0$) subshell.

The total angular momentum $\mathbf{J} = \mathbf{L} + \mathbf{S}$ can range from $|L - S|$ to $L + S$ in integer steps. Rule 3 picks the ground-state $J$ based on the sign of the spin-orbit coupling constant:

  • Less than half-filled: spin-orbit coupling is positive, so the state with antiparallel $\mathbf{L}$ and $\mathbf{S}$ (i.e., $J = |L-S|$) has the lowest energy.
  • More than half-filled: spin-orbit coupling is negative (the subshell can be thought of in terms of positive “holes”), so parallel alignment ($J = L+S$) has the lowest energy.
  • Exactly half-filled with $L = 0$ (e.g., the $p^3$ or $d^5$ configurations): $J = S$.

Spectroscopic symbol

The spectroscopic term symbol summarising all three rules is written:

$$ {}^{2S+1}L_J $$

where $L$ is denoted by a capital letter ($S, P, D, F, \ldots$ for $L = 0,1,2,3,\ldots$).

Examples

Carbon (C), $1s^2, 2s^2, 2p^2$

The two $2p$ electrons go into the three $p$ orbitals ($m_\ell = -1, 0, +1$).

Rule 1: maximise $S$. Place one electron per orbital, both spin-up:

$$ m_\ell: \quad \boxed{\uparrow} \quad \boxed{\uparrow} \quad \boxed{\phantom{\uparrow}} $$

Total spin: $S = \tfrac{1}{2} + \tfrac{1}{2} = 1$, multiplicity $2S+1 = 3$ (triplet).

Rule 2: maximise $L$. The maximum $M_L$ consistent with the above is $m_\ell = +1$ and $m_\ell = 0$, giving $M_L = 1$, so $L = 1$ ($P$ state).

Rule 3: subshell is less than half-filled ($2 < 3$), so $J = |L - S| = |1-1| = 0$.

Ground state of Carbon: ${}^3P_0$.

Nitrogen (N), $1s^2, 2s^2, 2p^3$

Three $2p$ electrons, one per orbital.

Rule 1: all spin-up gives $S = \tfrac{3}{2}$, multiplicity $4$ (quartet).

$$ m_\ell: \quad \boxed{\uparrow} \quad \boxed{\uparrow} \quad \boxed{\uparrow} $$

Rule 2: $M_L = (+1) + 0 + (-1) = 0$, so $L = 0$ ($S$ state).

Rule 3: half-filled, $L = 0$, so $J = S = \tfrac{3}{2}$.

Ground state of Nitrogen: ${}^4S_{3/2}$.

Oxygen (O), $1s^2, 2s^2, 2p^4$

Four $2p$ electrons, one subshell past half-filling.

Rule 1: three go in spin-up (one per orbital), the fourth must pair with one of them (spin-down):

$$ m_\ell: \quad \boxed{\uparrow\downarrow} \quad \boxed{\uparrow} \quad \boxed{\uparrow} $$

Net $S = 1$, multiplicity $3$ (triplet).

Rule 2: the paired electron (at $m_\ell = +1$, say) contributes $+1 + (-1) = 0$ to $M_L$ with its partner; the remaining two contribute $0 + (-1) = -1$… but we want the maximum $M_L$. The best assignment puts the doubly occupied orbital at $m_\ell = -1$: $M_L = (-1-1) + 0 + 1 = -1$, i.e., $|M_L|_\text{max} = 1$, giving $L = 1$ ($P$ state).

Rule 3: more than half-filled ($4 > 3$), so $J = L + S = 1 + 1 = 2$.

Ground state of Oxygen: ${}^3P_2$.

Iron (Fe), $[\text{Ar}], 3d^6, 4s^2$

Note that $2s^2$ is full, so the outer orbital is actually $3d^6$. Six $3d$ electrons, five $d$ orbitals ($m_\ell = -2,-1,0,+1,+2$).

Rule 1: five spin-up, one pairing (spin-down) in the $m_\ell = -2$ orbital:

$$ m_\ell{:} \quad \boxed{\uparrow\downarrow} \quad \boxed{\uparrow} \quad \boxed{\uparrow} \quad \boxed{\uparrow} \quad \boxed{\uparrow} $$

$S = 2$, multiplicity $5$ (quintet).

Rule 2: $M_L = (-2-2) + (-1) + 0 + 1 + 2 = -2$… again choosing to minimise cancellation: putting the extra electron at $m_\ell = +2$ gives $M_L = (2+2) + 1 + 0 + (-1) + (-2) = 2$, so $L = 2$ ($D$ state).

Rule 3: more than half-filled ($6 > 5$), so $J = L + S = 2 + 2 = 4$.

Ground state of Iron: ${}^5D_4$.

LS coupling vs jj coupling

There’s two main ways to couple angular momentum for multielectronic atoms, LS coupling(also called Russell-Saunders coupling) and jj coupling. The two schemes differ in which interactions are treated as dominant.

LS coupling (Russell-Saunders)

In LS coupling, electron-electron repulsion is much stronger than spin-orbit interaction. The hierarchy of interactions is:

  1. Electron-electron repulsion couples the individual orbital angular momenta $\ell_i$ into a total $L$, and the individual spins $s_i$ into a total $S$.
  2. Spin-orbit interaction (weaker) then couples $L$ and $S$ into the total $J$.

This is the regime where Hund’s rules apply. A good quantum number set is $(L, S, J, M_J)$, and term symbols ${}^{2S+1}L_J$ are meaningful.

LS coupling works well for light atoms (roughly $Z \lesssim 30$). In these atoms, spin-orbit coupling is a small perturbation on top of the electron-electron structure.

jj coupling

In jj coupling, spin-orbit interaction is much stronger than electron-electron repulsion. The hierarchy is reversed:

  1. Spin-orbit interaction couples each electron’s own $\ell_i$ and $s_i$ into an individual $j_i = \ell_i + s_i$.
  2. Residual electron-electron repulsion (weaker) then couples the individual $j_i$ into the total $J$.

$L$ and $S$ are no longer good quantum numbers separately. The relevant set is the collection of individual $j_i$ values and the total $J$. States are labelled by $(j_1, j_2, \ldots)_J$ rather than by a term symbol.

jj coupling describes heavy atoms (roughly $Z \gtrsim 60$), such as lead (Pb, $Z=82$) or bismuth (Bi, $Z=83$), where relativistic effects amplify spin-orbit coupling enormously.

Why spin-orbit coupling grows with $Z$

For a hydrogen-like ion the spin-orbit coupling constant scales as:

$$ \xi \propto \frac{Z^4}{n^3} $$

where $n$ is the principal quantum number. Electron-electron repulsion scales only as $Z$ (shielded nuclear charge). So for large $Z$ the ratio $\xi / E_{\text{ee}}$ grows rapidly and spin-orbit coupling wins.