University Quantum Mechanics II · Identical Particles and Many-Electron Atoms · 7.8
Shell filling, the aufbau order and the periodic table
This is where the periodic table stops being a chart to memorise and becomes a consequence. Pauli's counting fixes how many electrons each subshell seats, penetration fixes the order they are seated in, and measured ionisation energies say whether we got it right. We also mark exactly where the n + l rule is only a rule.
Build the model
Connect the measurement to the mechanism.
Put every electron in the central field of the previous lesson, so each occupies an orbital nlmₗ with a spin mₛ and an energy εₙₗ that depends on n and l only. Antisymmetry then does the counting: a Slater determinant with two equal columns vanishes, so each spin-orbital holds one electron, a subshell holds 2(2l + 1) and a hydrogenic shell 2n². For a sum of one-electron Hamiltonians the ground state is the determinant of the N lowest spin-orbitals, so the whole periodic table hangs on the order of the εₙₗ.
Penetration supplies that order. At fixed n a low-l orbital reaches inside the core and is bound more, so the big gaps fall after each np subshell rather than after each shell: noble gases at Z = 2, 10 and 18, where argon closes 3p with 3d empty, not the n = 3 shell, which would close at nickel, and period lengths 2, 8, 8, 18, 18, 32, 32. First ionisation energies confirm each closure: 24.6 eV at helium against 5.4 eV at lithium, 21.6 eV at neon against 5.1 eV at sodium.
The model stops at the n + l rule. That ordering is empirical, broken at chromium, copper and about eighteen other elements, and it ranks the total energies of neutral configurations, not the orbital energies of a finished atom: ions of the 3d metals lose 4s first, and highly charged ions return to hydrogen's order.
- Simple definition
- The aufbau principle builds a ground-state configuration by seating electrons, one per spin-orbital, in central-field subshells in order of increasing energy, each subshell nl holding 2(2l + 1) of them.
- Example
- Zinc, Z = 30, fills 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰: 2 + 2 + 6 + 2 + 6 + 2 + 10 = 30, every subshell through 3d closed. Its first ionisation energy, 9.39 eV, stands above copper's 7.73 eV and gallium's 6.00 eV on either side.
Derived from antisymmetry: s, p, d and f subshells hold 2, 6, 10 and 14 electrons, the widths of the s, p, d and f blocks of the periodic table.
l is the orbital quantum number; mₗ takes 2l + 1 values and mₛ two, so Nₘₐₓ is a pure count with no unit.
Derived. If whole shells closed, noble gases would sit at Z = 2, 10, 28 and 60. Neutral atoms close at 18 instead, so the l-degeneracy is gone.
n is the principal quantum number and l runs from 0 to n − 1; both sides are dimensionless counts.
Reproduces every period length and block width, and so the shape of the table. It fails in about twenty measured ground states, chromium and copper first.
n and l are the subshell's quantum numbers, both dimensionless; the rule is quoted from measured configurations, not derived.
Derived from the rule: 2, 8, 8, 18, 18, 32, 32, whose running sums put the noble gases at Z = 2, 10, 18, 36, 54, 86 and 118.
p is the period number and Lₚ the number of elements in it; ⌊ ⌋ rounds down. Both are counts.
Quoted from Hartree–Fock: each εᵢ already holds that electron's repulsion from all the others, so Σεᵢ counts every pair twice. Configurations are ranked by E, never by Σεᵢ.
εᵢ Hartree–Fock orbital energies; Jᵢⱼ Coulomb and Kᵢⱼ exchange integrals, K for parallel spins only; all in J, quoted in eV.
Quoted from the previous lesson and extended to ions by ζ²: one measured binding fixes n − δ, so whether 4s or 3d lies lower in an ion becomes a comparison of two numbers.
ζ is the net charge of the core (1 for K, 2 for Ca⁺), δₗ the quantum defect, a pure number; 1 eV = 1.602 × 10⁻¹⁹ J.
Pauli turns orbitals into seats
Quote the central field from the previous lesson: each electron moves in one spherical potential, so its states are Rₙₗ(r)Yₗₘ(θ, φ) times a spin state, with an energy εₙₗ that depends on n and l but not on mₗ or mₛ. Antisymmetry does the counting. A Slater determinant with two equal columns vanishes, so no two electrons share all four of n, l, mₗ and mₛ. At fixed n and l there are 2l + 1 values of mₗ and two of mₛ, so a subshell seats 2(2l + 1): 2, 6, 10 and 14 for s, p, d and f. Summing l from 0 to n − 1 gives 2n² for a shell: 2, 8, 18, 32. As the previous lesson showed, a full subshell is spherical with L = S = 0, so filled subshells keep the field central instead of spoiling it. For H₀ = Σᵢ h(i) the lowest eigenvalue is the sum of the N lowest one-electron energies, each spin-orbital counted once, and its eigenstate is the determinant of those N spin-orbitals. That is the aufbau principle, derived for independent electrons, and all its content sits in the order of the εₙₗ.
Closures sit after np, and ionisation energies show them
A closed subshell is chemically visible only if the next seat lies well above it, so what matters is where the big gaps fall. In hydrogen every l at one n is degenerate and the gaps would follow whole shells, closing at 2, 10, 28 and 60 electrons. Penetration moves them. At fixed n a low-l orbital has inner lobes that see more of the nuclear charge, so ns lies below np, and nd is pushed so high that it no longer travels with its own n. The gaps therefore fall after 1s² at Z = 2, 2p⁶ at Z = 10 and 3p⁶ at Z = 18. Argon's closure is a full 3p subshell, with 8 of the n = 3 shell's 18 seats taken and 3d empty; full shells n = 1 to 3 would need Z = 28, nickel. Measured first ionisation energies decide it: 24.59 eV at helium against 5.39 eV at lithium, 21.56 eV at neon against 5.14 eV at sodium, and 15.76 eV at argon against 4.34 eV at potassium, ratios of 4.56, 4.20 and 3.63. Nickel's 7.64 eV is followed by copper's 7.73 eV, with no sign of a closure at all.
Why potassium's 19th electron takes 4s
After argon the candidates are 3d, with n + l = 5, and 4s, with n + l = 4. Hydrogen puts 3d first, bound by 13.6/9 = 1.51 eV against 0.85 eV for n = 4. Potassium's measured levels, quoted from the NIST tables, reverse that: 4s is bound by 4.34 eV and 3d by 1.67 eV. The 3d orbital carries a centrifugal term l(l + 1)ħ²/(2mₑr²) = 6ħ²/(2mₑr²) that keeps it almost entirely outside the argon core, so it sees a net charge near 1 and is bound only 0.16 eV more than hydrogen's n = 3. The 4s orbital has three radial nodes, and its inner lobes spend time where the charge of 19 protons is barely screened; that lowers it 3.49 eV below hydrogen's n = 4. Potassium's 5s, bound by 1.73 eV, even lies 0.06 eV below its 3d. Calcium repeats the verdict. In Ca⁺ the 3d level sits 1.70 eV above 4s, and neutral calcium's lowest 3d4s level lies 2.52 eV above its 4s² ground state. So electrons 19 and 20 go into 4s, and the 3d row starts at scandium.
The n + l rule summarises the data; it is not a theorem
Madelung's rule orders subshells by n + l and, within equal n + l, by n: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p. It is quoted here, not derived: no derivation from the many-electron Schrödinger equation is known, and semiclassical arguments make it plausible without proving it. Read it as a compact record of measured ground configurations. Grant it and the period lengths follow by counting. Period 1 is 1s alone; period p ≥ 2 runs from ps to pp, taking every subshell with n + l = p + 1 on the way, so it holds 2 + 6 = 8 elements for p = 2 and 3, 2 + 10 + 6 = 18 for p = 4 and 5, and 2 + 14 + 10 + 6 = 32 for p = 6 and 7. The rule fails in about twenty measured ground states. In the 3d row it predicts chromium 4s²3d⁴ and copper 4s²3d⁹, but the measured ground states are 4s¹3d⁵ and 4s¹3d¹⁰, and the lowest levels of the rule's configurations lie 0.96 eV and 1.39 eV higher. Palladium is 4d¹⁰ with 5s empty. Each failure is a small difference between neighbouring configurations, which a one-line rule cannot resolve.
Filling order is not the energy order of the finished atom
The rule says which subshell starts to fill as Z grows along the neutral atoms; it is not a fixed ladder of orbital energies. Ionise a 3d metal and 4s empties first: iron is 4s²3d⁶, Fe⁺ is 4s¹3d⁶ and Fe²⁺ is 3d⁶, and every doubly charged ion from Sc²⁺ to Zn²⁺ is 3dⁿ with 4s empty. The one-electron sequence K, Ca⁺, Sc²⁺ shows why: as the net charge of the argon core goes from 1 to 2 to 3, the 3d level lies 2.67 eV above 4s, then 1.70 eV above it, then 3.15 eV below it. Nuclear charge added without electrons pulls the compact 3d into the core. In the neutral 3d metals 3d is already bound more tightly than 4s, as Hartree–Fock orbital energies (quoted) show, yet 4s stays occupied. The quoted Hartree–Fock total energy resolves this: each εᵢ counts its electron's repulsion from all the others, so Σεᵢ counts every pair twice, and the compact 3d electrons repel strongly. One more 3d electron looks better in Σεᵢ than it is in E. At high ionic charge repulsion is a 1/Z correction and hydrogen's order returns: every 28-electron ion from Cu⁺ upward is 3d¹⁰.
Checks: the energy scale, α, and where the labels blur
Every energy that decides the order is a Coulomb energy of order α²mₑc² = 27.2 eV, the hartree: helium's 24.6 eV is 0.90 of it, and potassium's 2.67 eV gap from 4s to 3d is 0.098 of it. Spin–orbit energies are smaller by a factor between α² = 5.3 × 10⁻⁵ and (Zα)², because a valence electron feels the bare nucleus only on brief visits to the core: potassium's 4p doublet is split by 7.2 meV, 2.6 × 10⁻³ of its 2.73 eV binding, where (Zα)² = 0.019. So relativity cannot reorder 4s and 3d; period 4's filling order is settled at order α²mₑc². In period 6, where (Zα)² reaches 0.33 at gold, quoted relativistic calculations stabilise 6s enough to tip close cases, and are credited with tungsten being 5d⁴6s², as the rule says, where molybdenum above it is 4d⁵5s¹. The configuration label has its own limit: nickel's lowest level, 3d⁸4s² ³F₄, lies only 0.025 eV below the lowest 3d⁹4s level, yet averaged over all their levels 3d⁹4s is the lower configuration, so the answer depends on what is compared.
Change one variable at a time
Make the relationship visible.
Step Z from 18 to 19: the first 4s electron appears and the printed IE(Z − 1) ÷ IE(Z) jumps to 3.63, as it does to 4.20 at Z = 11. At Z = 24 and 29 the shading leaves the dashes. Then set Z = 26 and q = 2: the rule's 4s² 3d⁴ becomes the measured 3d⁶.
ELECTRONS N = Z − q19
MOVED FROM 4s TO 3d VS RULE0
PERIOD OF ELEMENT Z4
n + l OF RULE'S LAST SUBSHELL4
Live interpretationELECTRONS N = Z − q: 19. MOVED FROM 4s TO 3d VS RULE: 0. PERIOD OF ELEMENT Z: 4. n + l OF RULE'S LAST SUBSHELL: 4
Catch the common trap
Explain before calculating.
Neutral titanium (Z = 22) has the ground configuration [Ar]4s²3d², just as the n + l rule predicts. Which statement gives the ground configuration of Ti²⁺ with a reason that survives the measured data?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyMg (Z = 12), Zn (Z = 30) and Kr (Z = 36) each end on a closed subshell. Their measured first ionisation energies are 7.646, 9.394 and 14.000 eV; those of the next elements, Al, Ga and Rb, are 5.986, 5.999 and 4.177 eV. Write each pair's configurations, find IE(Z) ÷ IE(Z + 1) for each pair, and say which closure behaves like a noble gas.
- Fill in the n + l order. Mg is [Ne]3s² and Al is [Ne]3s²3p; Zn is [Ar]4s²3d¹⁰ and Ga adds a 4p electron; Kr is [Ar]4s²3d¹⁰4p⁶ and Rb is [Kr]5s. Seat counts check: 10 + 2 = 12, 18 + 2 + 10 = 30 and 30 + 6 = 36.
- A ratio of two energies is a pure number: Mg to Al gives 7.646/5.986 = 1.277, Zn to Ga gives 9.394/5.999 = 1.566, and Kr to Rb gives 14.000/4.177 = 3.352.
- Only krypton's ratio matches the noble-gas pattern of 3.63 for argon to potassium and 4.20 for neon to sodium. Rubidium's electron starts a new shell: most of its 5s charge lies outside the whole n = 4 shell.
- After Mg and Zn the next electron enters 3p or 4p, at the same n as the closed s subshell and less bound only because it penetrates less. A closed subshell counts as a closure only if the next seat lies far above it.
- Limiting case: in a pure Coulomb field 3s and 3p are degenerate, and magnesium's closure would open no gap at all. The measured ratio of 1.28 comes from the penetration splitting of 3s and 3p, not from a shell closure.
AnswerIE ratios 1.28 for Mg to Al, 1.57 for Zn to Ga and 3.35 for Kr to Rb: only krypton's closed 4p⁶ acts as a noble-gas closure; a filled s or d subshell leaves the next seat close above it.
MediumTake the Madelung n + l rule as given. Check that it reproduces the seven known periods, then use it to find the atomic number at which the first g electron would appear, the length of period 8, and the atomic number of the element that would close it.
- Check the known table first. Running totals of 2, 8, 8, 18, 18, 32 and 32 give 2, 10, 18, 36, 54, 86 and 118, the noble gases helium to oganesson, so the rule reproduces all seven periods.
- List the groups the rule needs next, with l ≤ n − 1: n + l = 8 holds 5f 6d 7p 8s, and n + l = 9 holds 5g 6f 7d 8p 9s. The first g subshell allowed is 5g, with l = 4.
- Period 8 opens with 8s and runs through the n + l = 9 group up to 8p: 8s takes Z = 119 and 120, 5g takes 2(2 × 4 + 1) = 18, Z = 121 to 138, then 6f 14 (139 to 152), 7d 10 (153 to 162) and 8p 6 (163 to 168).
- Length 2 + 18 + 14 + 10 + 6 = 50, which the period formula confirms: 2(⌊8/2⌋ + 1)² = 2 × 25 = 50. A new block opens every second period: p in period 2, d in 4, f in 6 and g in 8.
- What the numbers are worth: the rule is empirical and already fails about twenty times among known elements, and quoted relativistic calculations for Z above 120 predict 8p, 7d, 6f and 5g levels that overlap in energy, so the extrapolated order should not be trusted.
AnswerThe rule puts the first g electron at Z = 121 and a 50-element period 8 closing at Z = 168. That is an extrapolation of an empirical rule, not a derived prediction.
HardK, Ca⁺ and Sc²⁺ each have one electron outside an argon core of net charge ζ = 1, 2 and 3. From the NIST tables, with each 3d energy at its fine-structure centroid: K binds 4s by 4.341 eV with 3d 2.670 eV higher; Ca⁺ binds 4s by 11.872 eV with 3d 1.697 eV higher; Sc²⁺ binds 3d by 24.742 eV with 4s 3.152 eV higher. Find n* = n − δ for 4s and 3d in each, locate where the order flips, and check the limit ζ → ∞.
- Write each binding energy as Eb = (13.606 eV)ζ²/n*², so n* = ζ√(13.606 eV/Eb). The ratio of two energies is dimensionless, so n* is a pure number, and a smaller n* means a more deeply bound level.
- K, ζ = 1: 4s has Eb = 4.341 eV, n* = 1.770, δ = 2.230. 3d has Eb = 4.341 − 2.670 = 1.671 eV, n* = 2.853, δ = 0.147.
- Ca⁺, ζ = 2: 4s has n* = 2√(13.606/11.872) = 2.141, δ = 1.859. 3d has Eb = 11.872 − 1.697 = 10.175 eV, n* = 2.313, δ = 0.687.
- Sc²⁺, ζ = 3: 3d has n* = 3√(13.606/24.742) = 2.225, δ = 0.775. 4s has Eb = 24.742 − 3.152 = 21.590 eV, n* = 2.382, δ = 1.618.
- n*(3d) − n*(4s) runs +1.083, +0.172, −0.157: 4s is lower for ζ = 1 and 2, and 3d for ζ = 3, so the order flips between Ca⁺ and Sc²⁺. The 3d defect grows more than fivefold as the orbital contracts into the core, while the 4s defect shrinks.
- Limit ζ → ∞: the core becomes a small correction, n* → n, and 3d must end below 4s by ζ²(13.606 eV)(1/9 − 1/16) = 0.66ζ² eV, hydrogen's order. That is the order the measured 3dⁿ ground states of Sc²⁺ to Zn²⁺ already follow.
Answern*(4s) = 1.770, 2.141, 2.382 and n*(3d) = 2.853, 2.313, 2.225 for ζ = 1, 2, 3: 3d drops below 4s between Ca⁺ and Sc²⁺, and at large ζ the order tends to hydrogen's, 3d lower by 0.66ζ² eV.