
Bounded modern-physics survey · Interactive lesson
Many-electron atoms and the exclusion principle
From helium on, every electron repels every other and nothing can be solved exactly. The working model puts each electron alone in one averaged, screened central field and lets Pauli exclusion decide where it may go. Use it to build configurations, test them against measured ionisation energies, and name where the averaging breaks.
UPIII-10.08 · Lesson 8 of 9 in this unit. Position is not a completion score.Bounded modern-physics survey. This modern-physics material is an introductory survey, not a complete course in relativity, quantum, atomic, nuclear or particle physics. Your university determines the required depth and assessment.
Build the model
One averaged field plus Pauli's counting builds the periodic table, almost.
The key idea
In the central-field approximation each electron of a many-electron atom moves independently in one spherically symmetric potential, set by the nucleus and the averaged charge of the other electrons, and the Pauli principle allows at most one electron in each state (n, l, mₗ, mₛ).
Example. Sodium's 3s electron sees Z = 11 screened by ten inner electrons. Its ionisation energy of 5.14 eV matches a hydrogen-like n = 3 level with Zeff = 3√(5.14/13.6) = 1.84, between full screening (1) and none (11).
The complete physical model and assumptions
Hydrogen's solution relied on one electron in a pure 1/r potential. Add a second electron and the Hamiltonian gains e²/(4πε₀r₁₂), coupling the coordinates of every pair and blocking separation of variables; in helium, dropping it misplaces the ground-state energy by 30 eV. The central-field approximation keeps the method by averaging instead: each electron moves in the field of the nucleus plus the spherically averaged charge of all the others, an effective charge running from Z at the nucleus to 1 outside a neutral atom. That potential is still central, so the angular solutions are hydrogen's and l and mₗ survive. It is not 1/r, so the l-degeneracy does not: orbitals that penetrate the inner charge, s before p before d, are bound more tightly. The Pauli principle, which is the antisymmetry of the many-electron wavefunction, admits one electron per set (n, l, mₗ, mₛ), so 2(2l + 1) per subshell. Filling the levels in order gives the shells, the noble-gas closures and the ionisation-energy trends. What the averaging discards is how the repulsion depends on the electrons' relative positions: correlation, and through antisymmetry the exchange energy. That leaves every term of a configuration degenerate, which Hund's rules repair, and where 4s and 3d nearly coincide it reorders configurations: chromium and copper each put one electron in 3d that the n + l rule puts in 4s.
1The repulsion term is too large to drop
For N electrons the Hamiltonian is a sum of hydrogen-like terms, pᵢ²/2mₑ − Ze²/(4πε₀rᵢ) for each electron, plus e²/(4πε₀rᵢⱼ) for every pair. The pair term depends on two electrons' coordinates at once, so the equation no longer separates. Helium shows how big it is. Dropping it puts both electrons in hydrogen-like 1s states with Z = 2, at −2 × 4 × 13.6 eV = −108.8 eV; the measured ground-state energy, minus the sum of both ionisation energies, is −79.0 eV. Averaging the repulsion over that unperturbed state gives (5/4)Z × 13.6 eV = 34.0 eV, and a first-order energy of −74.8 eV. Better, let each electron see a trial charge Zeff and minimise the energy: the optimum is Zeff = Z − 5/16 = 1.69, giving −77.5 eV, within 2% of the measurement. That optimum is the lesson. Each electron behaves as if the other cancels 0.31 of a proton: screening, measured by an energy.
2Antisymmetry is the exclusion principle
Electrons are identical, so swapping two labels cannot change any probability: |ψ(x₂, x₁)|² = |ψ(x₁, x₂)|², where each x includes spin. For electrons, as for every half-integer-spin particle, nature takes the minus sign: ψ(x₂, x₁) = −ψ(x₁, x₂). Build ψ from one-electron states a and b as [a(x₁)b(x₂) − b(x₁)a(x₂)]/√2 and set a = b: ψ vanishes. That is Pauli's rule, no two electrons sharing all of n, l, mₗ and mₛ, and it is a consequence of symmetry, not an extra force. The same sign has a second effect. Two electrons with parallel spins have a symmetric spin part, so their spatial part must be antisymmetric, and it vanishes wherever r₁ = r₂: they keep apart and repel less. Helium's 1s2s configuration measures it. The triplet ³S sits 19.82 eV above the ground state and the singlet ¹S at 20.62 eV, a 0.80 eV splitting produced by the Coulomb force alone. This exchange energy is the origin of Hund's first rule.
3One averaged field keeps mₗ but splits the l levels
The central-field approximation replaces the pair terms by what one electron feels on average from all the others, smeared into a spherically symmetric cloud. Because the cloud is spherical, Gauss's law on a sphere of radius r says only the charge inside r screens the nucleus: the electron feels a force −Zeff(r)e²/(4πε₀r²) with Zeff(r) = Z − N(r), falling from Z at the nucleus to 1 outside a neutral atom, and its potential energy U(r) is the integral of that force from r out to infinity. The cloud depends on the orbitals and the orbitals on the cloud, so the field is found self-consistently, iterating until the two agree: the Hartree method. Because U is central, the angular equation is hydrogen's, and Yₗₘ, l and mₗ carry over unchanged. Because U is not 1/r, energy now depends on l. Near the nucleus Rₙₗ grows as rˡ, so an s orbital has density at the nucleus while a d orbital is held out; the s electron penetrates the core and sees more of Z. Sodium's valence levels show it: 3s at −5.14 eV, 3p at −3.04 eV, 3d at −1.52 eV, against −1.51 eV for hydrogen's n = 3. The 3d electron stays outside the core and sees one net charge, as hydrogen's electron does.
4Filling the levels: capacity, n + l order and closures
Each orbital (n, l, mₗ) takes two electrons of opposite mₛ, so a subshell holds 2(2l + 1). Feeding electrons into the lowest free levels gives an order summarised by the n + l rule: increasing n + l, and for a tie the smaller n first, which reads 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p. It is penetration in bookkeeping form: 4s (n + l = 4) comes before 3d (n + l = 5) because its inner lobes reach through the argon core. The biggest gaps follow each np⁶, so the noble gases close at Z = 2, 10, 18, 36, 54 and 86, not at the 2, 10, 28 and 60 that whole shells of 2n² would give. An open subshell still holds choices. Carbon's two 2p electrons have C(6, 2) = 15 arrangements, all degenerate in the central field. The leftover repulsion splits them into ³P (9 states), ¹D (5) and ¹S (1), and Hund's rules name the lowest: largest S, then largest L, then J = |L − S| for a subshell less than half full, which gives ³P₀.
5Ionisation energy is screening arithmetic
Across period 2 the first ionisation energy climbs from 5.39 eV (Li) to 21.56 eV (Ne). Each added proton is screened only partly by the added electron, which sits in the same shell, about as far out as the one being removed, so Zeff = n√(EI/13.6 eV) rises from 1.26 to 2.52. At sodium the next electron starts n = 3 outside a full n = 2 core, and EI collapses to 5.14 eV. Two dips inside the period are the model's fingerprints. Boron (8.30 eV) falls below beryllium (9.32 eV) because its electron is in 2p, which penetrates less than 2s. Oxygen (13.62 eV) falls below nitrogen (14.53 eV) because nitrogen's three 2p electrons sit in separate orbitals with parallel spins, the ⁴S₃/₂ term, and oxygen's fourth must pair with one of them, adding repulsion and gaining no exchange. The same two dips recur at aluminium and sulfur in period 3. Down a group EI changes slowly, Li 5.39, Na 5.14, K 4.34 eV, because the growth of n is nearly offset by the growth of Zeff: 1.26, 1.84, 2.26.
6Where the averaged field gets the order wrong
The n + l rule predicts chromium [Ar]3d⁴4s² and copper [Ar]3d⁹4s²; spectroscopy finds [Ar]3d⁵4s¹ and [Ar]3d¹⁰4s¹. The cause is what the averaging discarded. The central field ranks configurations by one-electron energies in a single averaged potential, but the ground configuration is the one of lowest total energy, which includes the repulsion and exchange among the valence electrons themselves. Across the 3d series the 4s and 3d orbitals lie close in energy, so that neglected part decides. Moving one 4s electron into 3d gives chromium six parallel spins (S = 3, the ⁷S₃ ground term) instead of four, and gives copper a closed 3d¹⁰ shell. The rejected configuration has not vanished: copper's 3d⁹4s² is a real excited level, ²D₅/₂, 1.39 eV above the ground state. The field also changes with the electron count, so iron, which fills 4s before 3d, loses 4s first: Fe²⁺ is [Ar]3d⁶. About twenty elements break the rule, among them Nb, Mo, Pd, Ag, Pt and Au; palladium, [Kr]4d¹⁰, leaves 5s empty.
Key equations
Read the units, assumptions and limits before substituting.
Exchange antisymmetry (the Pauli principle)
ψ(x₂, x₁) = −ψ(x₁, x₂)
Put both electrons in one state and ψ = −ψ = 0, so no two share (n, l, mₗ, mₛ). Same-spin pairs also get ψ = 0 at r₁ = r₂: they avoid each other, the root of exchange energy.
xᵢ = (rᵢ, mₛ) is electron i's position in m and spin projection ±½; a two-electron ψ has units m⁻³Subshell capacity
Nₘₐₓ = 2(2l + 1)
s, p, d, f hold 2, 6, 10, 14. Summing l = 0 to n − 1 gives 2n² per shell, but real closures follow subshell gaps instead: Z = 2, 10, 18, 36, 54, 86.
l = 0, 1, 2, 3 for s, p, d, f; 2l + 1 values of mₗ times 2 values of mₛ; a dimensionless countScreened central field
F(r) = −Zeff(r) e²/(4πε₀ r²), with Zeff(r) = Z − N(r)
Gauss's law on a sphere of radius r, valid because the averaged cloud is spherically symmetric: charge outside r exerts no net force. So Zeff runs from Z at r → 0 to 1 outside a neutral atom. Central, so l and mₗ survive; not an inverse square, so energy depends on l as well as n.
F radial force on the electron in newtons, negative meaning toward the nucleus; r in m; N(r) the number of other electrons inside radius r; e = 1.602 × 10⁻¹⁹ C; ε₀ = 8.854 × 10⁻¹² F m⁻¹Effective charge from an ionisation energy
Zeff = n √(EI / 13.6 eV)
Turns a measured energy into the charge the electron effectively sees: 1.26 for lithium's 2s, rising to 2.52 for neon's 2p. It is one number for the whole orbital, lumping screening and penetration together, not the field's Zeff(r) at any single radius.
EI measured ionisation energy of the outer electron, in eV; n its principal quantum number; inverts E = −(13.6 eV) Zeff²/n²Quantum defect of a valence level
Eₙₗ = −(13.6 eV)/(n − δₗ)²
Outside the core the electron sees one net charge, so the level looks hydrogenic with n shifted by δₗ. Large δ means deep penetration; d and f orbitals barely enter the core.
δₗ dimensionless quantum defect, nearly independent of n; for sodium δₛ = 1.37, δₚ = 0.88, δd = 0.01; Eₙₗ in eVHund's rules for the lowest term
S maximal, then L maximal, then J = |L − S| (less than half full) or J = L + S (more than half full)
Carbon 2p² gives ³P₀, nitrogen 2p³ ⁴S₃/₂, oxygen 2p⁴ ³P₂. Rule 1 is exchange, not magnetism: same-spin electrons keep apart and repel less.
S, L, J: total spin, orbital and total angular-momentum quantum numbers of one open subshell in LS coupling; dimensionlessInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. At Z = 24 the n + l rule draws 3d⁴4s² with S = 2 and L = 2. Flip the second slider to 1: one 4s electron moves into the empty mₗ = −2 box, S rises to 3 and L falls to 0, chromium's measured ⁷S₃. The flip changes nothing anywhere else except Z = 29, copper.
Live results
- UNPAIRED ELECTRONS
- 4
- TOTAL SPIN S
- 2.0
- TOTAL ORBITAL L
- 2
- ELECTRONS MOVED 4s → 3d
- 0
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Atomic number Z: 24; Filling: 0 = n + l rule, 1 = measured: 0
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The n + l order is not a fixed ladder of orbital energies. Students read “4s fills before 3d” as “4s always lies below 3d”, then predict that iron ionises from 3d, leaving Fe²⁺ as [Ar]3d⁴4s², and file chromium and copper as arbitrary exceptions to memorise. But one-electron energies belong to a self-consistent field, and that field changes with Z and with every electron added or removed. In potassium the 4s level does lie 2.67 eV below 3d. From scandium on, calculations put the 3d orbital energy below 4s, yet the neutral atoms, chromium and copper apart, keep two electrons in 4s, because crowding them into compact 3d orbitals costs extra repulsion. Remove electrons and the balance shifts again: Fe²⁺ is [Ar]3d⁶, the 4s electrons leaving first. The rule predicts the configurations of most neutral atoms; it does not rank orbitals, and chromium and copper are where its averaging shows.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Selenium has Z = 34. Write its ground configuration by the n + l rule, then use Hund's rules to find the number of unpaired electrons and the lowest term, and say how many states that ground level contains.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Fill in n + l order with capacities 2, 2, 6, 2, 6, 2, 10: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ uses 30 electrons, leaving 4 for 4p. The configuration is [Ar]3d¹⁰4s²4p⁴.
- Every subshell except 4p is closed, with S = L = 0, so only the four 4p electrons matter. They have C(6, 4) = 15 possible arrangements; Hund's rules pick the lowest.
- Rule 1: one spin-up electron in each of mₗ = +1, 0, −1, then the fourth must pair, spin down, in mₗ = +1. Two electrons are unpaired, so S = 1.
- Rule 2: with S at its maximum, this arrangement has ML = (+1) + 0 + (−1) + (+1) = 1, the largest available, so L = 1, a P term.
- Rule 3: four electrons out of six is more than half full, so J = L + S = 2. The term is ³P₂; the level holds 2J + 1 = 5 states, and the whole ³P term holds (2S + 1)(2L + 1) = 9 of the 15.
Answer. [Ar]3d¹⁰4s²4p⁴ with 2 unpaired electrons; the ground term is ³P₂, a level of 5 states.
MediumExample 2
Potassium (Z = 19) has ionisation energy 4.34 eV. Its lowest 4p and 3d levels lie 1.61 eV and 2.67 eV above the 4s ground state. Find each level's binding energy, quantum defect and effective charge, taking 13.6 eV for the Rydberg energy, and use them to explain why the 19th electron goes into 4s rather than 3d.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Binding energy is the ionisation energy minus the excitation energy: 4s 4.34 eV, 4p 4.34 − 1.61 = 2.73 eV, 3d 4.34 − 2.67 = 1.67 eV.
- Effective quantum number n* = √(13.6 eV/Eb): 4s gives √(13.6/4.34) = 1.770, so δₛ = 4 − 1.770 = 2.23; 4p gives √(13.6/2.73) = 2.232, so δₚ = 1.77; 3d gives √(13.6/1.67) = 2.854, so δd = 3 − 2.854 = 0.15.
- Effective charge Zeff = n√(Eb/13.6 eV): 4s 4 × 0.565 = 2.26, 4p 4 × 0.448 = 1.79, 3d 3 × 0.350 = 1.05.
- The 3d electron sees almost exactly one net charge and is bound only 0.16 eV more than hydrogen's n = 3 level at 1.51 eV: it stays outside the argon core. The 4s electron, whose inner lobes pass through the core, is bound 3.49 eV more than hydrogen's n = 4 level at 0.85 eV.
- Penetration therefore puts 4s 2.67 eV below 3d, the n + l ordering (4 before 5). As a check, the 4p → 4s photon has λ = 1240 eV nm/1.61 eV = 770 nm, potassium's red resonance line.
Answer. Binding energies 4.34, 2.73 and 1.67 eV; δₛ = 2.23, δₚ = 1.77, δd = 0.15; Zeff = 2.26, 1.79 and 1.05. The penetrating 4s lies 2.67 eV below the non-penetrating 3d, so it fills first.
HardExample 3
Lithium (Z = 3) has one 2s electron outside a 1s² core. Treat each core electron as a hydrogen-like 1s orbital with the two-electron screened charge Zc = Z − 5/16 = 2.6875, so its radial probability density is P(r) = 4(Zc/a₀)³ r² e(−2Zc r/a₀). Find the fraction f(R) of one core electron inside radius R, use Gauss's law to get the effective charge acting on the 2s electron at R = 0.2a₀, a₀ and 4a₀, and compare with the value from lithium's ionisation energy, 5.39 eV.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- The averaged core is spherically symmetric, so Gauss's law on a sphere of radius R counts only the enclosed charge: the 2s electron there feels a force −Zeff(R)e²/(4πε₀R²) with Zeff(R) = 3 − 2f(R), each of the two core electrons contributing the fraction f(R) that lies inside R.
- Substitute x = 2Zc r/a₀ = 5.375r/a₀. Then P dr = ½x²e(−x) dx, and f(R) = ½∫x²e(−x) dx from 0 to X = 5.375R/a₀. Integrating by parts twice gives f(R) = 1 − e(−X)(1 + X + X²/2); as X → ∞, f → 1, so the density is normalised, and as X → 0, f → 0.
- R = 0.2a₀: X = 1.075, f = 1 − 0.3413 × 2.653 = 0.095, so Zeff = 3 − 0.19 = 2.81. Deep inside the core the 2s electron sees almost the bare nucleus.
- R = a₀: X = 5.375, f = 1 − 0.00463 × 20.82 = 0.904, so Zeff = 3 − 1.81 = 1.19. R = 4a₀: X = 21.5, e(−X) = 4.6 × 10⁻¹⁰ and f = 1.000, so Zeff = 1.00, the net charge of Li⁺.
- From the ionisation energy, Zeff = 2√(5.39/13.6) = 1.26: above the 1.00 outside the core, far below the 2.81 deep inside it. Most of the 2s probability lies beyond a₀, which pulls the value toward 1, but the inner lobe reaches where the electron sees nearly 3, and the energy weights those close approaches by 1/r. The energy-derived Zeff describes the whole orbital; it is not the field at any one radius.
Answer. f(R) = 1 − e(−X)(1 + X + X²/2) with X = 5.375R/a₀; Zeff = 2.81 at 0.2a₀, 1.19 at a₀ and 1.00 at 4a₀, bracketing the 1.26 from the ionisation energy, which describes the whole 2s orbital rather than any one radius.
Quick check
Select an answer and explain it before checking.
Review & continue
In the central-field approximation each electron of a many-electron atom moves independently in one spherically symmetric potential, set by the nucleus and the averaged charge of the other electrons, and the Pauli principle allows at most one electron in each state (n, l, mₗ, mₛ).
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.