University Physics IV · Atomic Physics · 11.4
Configurations, Aufbau & Hund's Rules
Two pieces of bookkeeping turn a nuclear charge into a ground state: an order for filling subshells, and a rule for choosing among the many states one open subshell allows. Learn both precisely, and learn where each one fails — the failures are where the physics is.
Build the model
Connect the measurement to the mechanism.
Once the central field lifts the l-degeneracy of the Coulomb problem, an orbital's energy depends on n and l together, and filling becomes a competition: raising n costs energy, but a low-l orbital penetrates the core, sees less screening, and is pulled down. Madelung's rule — fill by increasing n + l, and take the lower n when n + l ties — is a compact summary of where that competition lands across the neutral atoms. It is fitted to spectra, not derived, which is why 4s (n + l = 4) precedes 3d (n + l = 5).
Naming the configuration is only half the job: an open subshell still holds many microstates, degenerate in the central field and split by the residual electron-electron repulsion and by spin-orbit coupling. Hund's three rules pick the lowest of them — maximise S, then L, then take J = |L − S| at or below half filling and L + S above. The cost is that all of this is an empirical ordering of energies that sit within a fraction of an electronvolt of one another.
Chromium and copper already break the filling order; the rules fix only the ground level, never the ordering of excited terms; and the third rule assumes LS coupling, which heavy atoms abandon.
- Simple definition
- The aufbau order is the empirical sequence in which subshells fill — lowest n + l first, and the lower n when n + l ties — and Hund's rules then select, from the states that configuration allows, the term and level that lie lowest.
- Example
- Carbon, Z = 6: n + l gives 1s²2s²2p². The two p electrons take S = 1 and L = 1, and p² is below half filling, so J = |L − S| = 0. The ground level is ³P₀, with ³P₁ and ³P₂ sitting 0.0020 eV and 0.0054 eV above it.
4s has n + l = 4 and 3d has 5, so 4s fills first: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s.
n and l are dimensionless integers with 0 ≤ l ≤ n − 1; fitted to neutral-atom spectra, not derived
s 2, p 6, d 10, f 14 — which is where the period lengths 2, 8, 8, 18, 18, 32, 32 come from.
the 2 counts spin projections and 2l + 1 counts mₗ values; both dimensionless
Fill one electron per mₗ box, spin-up from the top mₗ down, before any pairing begins.
S and L in units of ħ; Pauli caps the parallel count at 2l + 1, one per mₗ box
d⁴ gives ⁵D₀ and d⁶ gives ⁵D₄: the same term, opposite ends of the multiplet.
J in units of ħ; a half-filled subshell has L = 0, so |L − S| and L + S coincide
The sign of A is Hund's third rule: past half filling the whole multiplet inverts.
A is the spin-orbit constant in eV or cm⁻¹; A > 0 below half filling, A < 0 above it
Counts the states a magnetic field separates: ³P₂ splits into 2J + 1 = 5 Zeeman components.
L = 0, 1, 2, 3, 4 are printed S, P, D, F, G; 2S + 1 is the multiplicity
Why the filling order is not just increasing n
In hydrogen, 2s and 2p are degenerate; nothing in a pure 1/r potential distinguishes them. The self-consistent central field of a many-electron atom does. A low-l orbital keeps more amplitude close to the nucleus, so it penetrates inside the screening cloud of the inner electrons, feels a larger effective charge, and is pulled down relative to the high-l orbitals of the same n. Penetration therefore fights the cost of raising n, and it can win: in potassium the 4s orbital lies about 2.7 eV below 3d, so the nineteenth electron goes into 4s and the configuration is [Ar] 4s¹, not [Ar] 3d¹. The aufbau order is nothing more than a record of which side of that competition each subshell falls on, atom by atom — which is exactly why it has exceptions and a derived law would not.
Reading the order off the n + l rule
To reproduce the sequence, group the subshells by the value of n + l and, inside a group, take the smaller n first. n + l = 1 gives 1s; 2 gives 2s; 3 gives 2p then 3s; 4 gives 3p then 4s; 5 gives 3d, 4p, 5s; 6 gives 4d, 5p, 6s; 7 gives 4f, 5d, 6p, 7s. Fill each to its capacity of 2(2l + 1) and keep a running total, and the periodic table appears: 2, 4, 10, 12, 18, 20, 30, 36, and onward. The closures that make noble gases fall at 2, 10, 18, 36, 54 and 86 — after an np subshell, not after a complete shell. Argon is inert at Z = 18 with 3d entirely empty, because the next orbital available, 4s, sits across a large gap.
Hund's first rule is exchange, not magnetism
The rule is often mis-taught as parallel spins repelling magnetically. Estimate that energy: two electron magnetic moments an ångström apart interact with roughly μ₀μB²/4πr³ ≈ 5 × 10⁻⁵ eV. The splitting the rule actually describes is four orders of magnitude larger — carbon's ³P and ¹D terms are 1.26 eV apart. The mechanism is exchange. A two-electron state must be antisymmetric overall, so a symmetric spin function (parallel spins, S = 1) forces an antisymmetric spatial function, and that function vanishes whenever the two electrons sit at the same point. Parallel spins therefore keep the electrons further apart on average and their Coulomb repulsion is smaller. The magnetism of an open-shell atom is a consequence of the alignment, not its cause.
Applying the first two rules in order
The rules are lexicographic: maximise S first, then maximise L among only those states that already have that S. Carbon's 2p² makes the point. Put both electrons spin-up, giving S = 1; being parallel they must take different mₗ values from (+1, 0, −1), and the largest sum available is +1 + 0 = 1, so L = 1 and the term is ³P. You could instead put both electrons in mₗ = +1 with opposite spins and reach L = 2 — but that costs S, giving ¹D, which sits 1.26 eV higher, with ¹S 2.68 eV higher still. The bookkeeping is worth automating: draw the mₗ boxes, fill one electron per box spin-up from the top mₗ downward, then start again pairing from the top.
Hund's third rule is the sign of the spin-orbit constant
Inside a term, spin-orbit coupling splits the levels as EJ = (A/2)[J(J+1) − L(L+1) − S(S+1)], so adjacent levels are separated by A⋅J with J the larger of the two — the Landé interval rule. Everything in Hund's third rule sits in the sign of A. Below half filling A is positive and J = |L − S| lies lowest: carbon's ³P₀ is the ground level, with ³P₁ at 16.4 cm⁻¹ and ³P₂ at 43.4 cm⁻¹ above it, near the 1 : 2 spacing the interval rule predicts. Above half filling the subshell is better counted in holes, A changes sign, and the multiplet inverts: oxygen's 2p⁴ carries the same ³P term, but now ³P₂ is the ground level, with ³P₁ 158 cm⁻¹ and ³P₀ 227 cm⁻¹ above it. Exactly half filled, L = 0 and only one level exists.
Where the order fails, and what fails with it
Chromium is [Ar] 3d⁵ 4s¹ and copper [Ar] 3d¹⁰ 4s¹, not the 3d⁴ 4s² and 3d⁹ 4s² the rule predicts; niobium, ruthenium, rhodium, palladium and platinum break it as well, so the exceptions are not a short list of curiosities. The reason is that the 3d and 4s one-electron energies differ by only a few tenths of an electronvolt in these atoms, so the total energy decides instead: the 3d orbitals are compact, so extra d occupancy costs repulsion, while an extra parallel d electron gains exchange, lifting the count of parallel d pairs from C(4,2) = 6 to C(5,2) = 10 at chromium. Hund's rules are more robust but still bounded: they fix only the lowest term of the ground configuration, say nothing dependable about excited terms, and the third rule assumes LS coupling, which heavy atoms abandon for jj coupling.
Change one variable at a time
Make the relationship visible.
Set q = 4, s = 2 for the n + l prediction at chromium, then q = 5, s = 1 for the real ground state: one electron moves and 2S + 1 jumps from 5 to 7. Then hold s = 2 and push q past 5 — L returns to 0 at half filling, and the ground J switches from |L − S| to L + S.
ELECTRON COUNT24
TOTAL SPIN S3.0 ħ
TOTAL ORBITAL L0 ħ
HUND GROUND J3.0 ħ
Live interpretationELECTRON COUNT: 24. TOTAL SPIN S: 3.0 ħ. TOTAL ORBITAL L: 0 ħ. HUND GROUND J: 3.0 ħ
Catch the common trap
Explain before calculating.
Oxygen has the ground configuration 1s² 2s² 2p⁴. Applying Hund's three rules in order, which term symbol describes its ground level?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyArsenic has Z = 33. Use the n + l filling order to write its ground-state configuration, then apply Hund's rules to give the ground term symbol.
- Order the subshells by increasing n + l, taking the lower n when n + l ties: 1s (1), 2s (2), 2p and 3s (3), 3p and 4s (4), then 3d, 4p and 5s (5).
- Fill each to its capacity 2(2l + 1) and keep a running total: 1s² 2, 2s² 4, 2p⁶ 10, 3s² 12, 3p⁶ 18, 4s² 20, 3d¹⁰ 30. Thirty-three electrons leave three for 4p.
- Configuration: [Ar] 3d¹⁰ 4s² 4p³. Only the open 4p subshell matters — every closed subshell contributes L = 0 and S = 0.
- Hund 1: three electrons in three mₗ boxes go one each with parallel spins, so S = 3/2 and the multiplicity 2S + 1 = 4.
- Hund 2: those parallel spins must take mₗ = +1, 0 and −1, so Σmₗ = 0 and L = 0, an S term. Hund 3: p³ is exactly half filled, so |L − S| = L + S = 3/2 and there is only one level.
Answer[Ar] 3d¹⁰ 4s² 4p³, ground term ⁴S₃/₂ — written ⁴S°₃/₂, the degree marking the odd parity that three p electrons carry.
MediumThe n + l rule predicts [Ar] 3d⁴ 4s² for chromium (Z = 24), but the observed ground configuration is [Ar] 3d⁵ 4s¹. Give the Hund ground term for each configuration, and say what tips the balance between them.
- [Ar] 3d⁴ 4s²: the closed 4s² pair contributes nothing. Hund 1 puts the four d electrons in separate mₗ boxes with parallel spins, so S = 2 and the multiplicity is 5.
- Hund 2: those four take mₗ = +2, +1, 0, −1, so Σmₗ = 2 and L = 2, a D term. Hund 3: d⁴ is below half filling, so J = |L − S| = 0. The predicted term is ⁵D₀.
- [Ar] 3d⁵ 4s¹: five parallel d electrons plus the lone 4s electron give six unpaired spins, so S = 3 and the multiplicity is 7.
- The five d electrons use every mₗ from +2 to −2, so Σmₗ = 0 and L = 0, an S term. With L = 0 there is a single level, J = S = 3, giving ⁷S₃ — which is what chromium's spectrum shows.
- What tips it: promoting one 4s electron into 3d raises the number of parallel-spin d pairs from C(4,2) = 6 to C(5,2) = 10 and breaks up the 4s² pair. Since the 3d and 4s one-electron energies here differ by only a few tenths of an eV, that exchange gain is enough to win.
Answer[Ar] 3d⁴ 4s² would give ⁵D₀; the actual ground state is [Ar] 3d⁵ 4s¹, ⁷S₃. The n + l order loses whenever the exchange gain beats the promotion cost.
HardIron is Z = 26. (a) Write its ground configuration and term symbol. (b) Do the same for Fe²⁺ and Fe³⁺. (c) Give the spin-only magnetic moment of Fe³⁺ in Bohr magnetons.
- (a) The n + l order reaches argon at 18, then 4s² at 20, then fills 3d: iron is [Ar] 3d⁶ 4s². The 4s² pair is closed, so 3d⁶ alone sets the term.
- 3d⁶: the five mₗ boxes each take one parallel electron, and the sixth must pair in mₗ = +2. Four spins are left unpaired, so S = 2 and the multiplicity is 5.
- Σmₗ = (2 + 1 + 0 − 1 − 2) + 2 = 2, so L = 2, a D term. d⁶ is more than half filled, so J = L + S = 4 and the ground level is ⁵D₄.
- (b) Ionise from the outside in, not from the top of the filling order: the diffuse 4s electrons leave first. Fe²⁺ is [Ar] 3d⁶ — the same open subshell, hence the same ⁵D₄.
- Fe³⁺ removes one more, leaving [Ar] 3d⁵: five parallel spins give S = 5/2 and multiplicity 6, every mₗ is used once so Σmₗ = 0 and L = 0, and the half-filled shell has the single level J = 5/2. The term is ⁶S₅/₂.
- (c) With L = 0 there is no orbital contribution to quench or add, so the spin-only form applies: μ = √(n(n + 2)) μB with n = 5 unpaired electrons, giving μ = √35 = 5.92 μB.
AnswerFe [Ar] 3d⁶ 4s², ⁵D₄; Fe²⁺ [Ar] 3d⁶, ⁵D₄; Fe³⁺ [Ar] 3d⁵, ⁶S₅/₂, with a spin-only moment of 5.92 μB.