
Physics IV course sequence · Interactive lesson
The periodic table from shell structure
The table on the wall is a picture of a filling order, drawn to scale. This lesson rebuilds its shape from two facts — how many electrons a subshell holds, and which subshell fills next — then tests how far the trends that shape predicts can actually be trusted.
UPIV-11.05 · Lesson 5 of 9 in this unit. Position is not a completion score.Build the model
Row lengths are a filling order, not a shell capacity.
The key idea
The length of a period is the total capacity of the subshells that fill between one ns level and the np level closing the row — 2(lₘₐₓ + 1)² electrons, where lₘₐₓ is the largest orbital angular-momentum quantum number entering by that row.
Example. Period 4 runs 4s, 3d, 4p, so lₘₐₓ = 2 and its length is 2(2 + 1)² = 18, potassium to krypton — even though the n = 4 shell holds 2 × 4² = 32 and the n = 3 shell is not full until copper.
The complete physical model and assumptions
The periodic table is not a separate law of nature; it is the bookkeeping of a one-electron model, and its shape follows from two ingredients. Antisymmetry caps a subshell at 2(2l + 1) electrons, so a block can only be 2, 6, 10 or 14 wide. Screening then decides which subshell fills next: in a many-electron atom the Coulomb degeneracy in l is gone, penetrating low-l orbitals lie deeper than their principal quantum number suggests, and the empirical n + l rule summarises the result. Put the two together and a row is one subshell of each l from 0 up to lₘₐₓ, holding 2(lₘₐₓ + 1)² elements — 2, 8, 8, 18, 18, 32, 32 — with lₘₐₓ advancing only every second period, which is why every length after the first is used twice before a new block enters. The same two ingredients set the trends inside a row: the effective charge climbs while n stays fixed, so radii shrink and ionisation energies rise, and the places where that rise stumbles mark a change of subshell or the first forced spin pairing. The cost is exactness. The filling order is fitted rather than derived and already fails at chromium; the orbital energies the picture rests on are not ionisation energies, because ionising an atom lets the survivors relax inward; and in heavy atoms the s electrons move at an appreciable fraction of c, shifting levels by amounts the table's geometry cannot see.
1Where the four block widths come from
An orbital is fixed by n, l and mₗ, and an electron in it by one further label, mₛ = ±½. For a given l there are 2l + 1 values of mₗ, so a subshell holds Nₗ = 2(2l + 1) electrons: 2 for s, 6 for p, 10 for d, 14 for f. Nothing in the Schrödinger equation forbids a third 2s electron — antisymmetry does. Write the state as a Slater determinant of one-electron spin-orbitals and it vanishes identically the moment two columns repeat, so each set of labels is used once and once only. Sum over l = 0 to n − 1 and a principal shell holds 2n²: 2, 8, 18, 32. Those are the only capacities on offer, which is why the periodic table has exactly four block widths and not, say, a block of seven. Below Z = 118 only l = 0, 1, 2 and 3 are ever occupied in a ground state, so s, p, d and f exhaust the vocabulary the table is written in.
2A period is a Madelung diagonal, not a shell
A row opens at ns and closes at np, and what fills between them is set by increasing n + l, lower n first on a tie. Period 4 therefore reads 4s (n + l = 4), then 3d (5), then 4p (also 5, but 4 > 3, so 3d goes first): 2 + 10 + 6 = 18 elements, potassium to krypton. Period 6 reads 6s (6), 4f (7), 5d (7), 6p (7): 2 + 14 + 10 + 6 = 32, caesium to radon. Now look at what those lists have in common. Each is exactly one subshell of every l from 0 up to some lₘₐₓ, drawn from whichever n the ordering happens to supply. So the length is Lₖ = Σ 2(2l + 1) over l = 0 to lₘₐₓ, which is 2(lₘₐₓ + 1)², with lₘₐₓ = ⌊k/2⌋. Feeding k = 1 to 7 gives 2, 8, 8, 18, 18, 32, 32. The repetition is not a coincidence to be memorised: lₘₐₓ advances only on every second period, because each l first appears at n = l + 1, on the diagonal n + l = 2l + 1, so successive new l values sit two diagonals apart while each period advances by one. The formula also predicts forward — period 8 would have lₘₐₓ = 4, an 18-wide g block, and 2 × 5² = 50 elements.
3Why a new block waits: penetration beats n
In hydrogen 3s, 3p and 3d are degenerate and the n + l rule would be meaningless. In a screened field they are not. A low-l radial function has inner lobes that reach inside the core, where the electron sees close to the bare nuclear charge; a high-l orbital is held out by the centrifugal term ℏ²l(l + 1)/2mr², sits outside the screening cloud, and sees something near Z minus the core count. At potassium the 4s orbital, despite its larger principal number, is bound more tightly than 3d, and the nineteenth electron goes into 4s. That single competition is the whole reason the d block is inset by one period and the f block by two. It is also a competition barely won: across the 3d series the 4s and 3d levels sit within about an electronvolt of each other, which is why the order already fails at chromium ([Ar]3d⁵4s¹) and copper ([Ar]3d¹⁰4s¹), and why once d electrons are present the 3d level drops below 4s, so transition-metal cations lose their 4s electrons first — Fe²⁺ is [Ar]3d⁶, not [Ar]3d⁴4s².
4Reading a row: rising Zeff against a fixed n
Two terms compete in every trend. Across a period Z rises by one per step, but the new electron joins a shell already occupied, where Slater's rules charge it only 0.35 of a unit of screening against its neighbours. So Zeff climbs by roughly 0.65 per element — 1.30 at lithium to 5.85 at neon — while n stays at 2. The orbital contracts and binds harder: covalent radii fall from 128 pm at lithium to 57 pm at fluorine, and the first ionisation energy rises from 5.39 eV to 21.56 eV, a factor 4.0. Down a group the other term wins. A new shell adds a radial node and puts the valence electron outside a much larger core, and the extra protons are screened almost completely — 0.85 per n − 1 electron, 1.00 per deeper one. Lithium, sodium and potassium have radii 128, 166 and 203 pm and ionisation energies 5.39, 5.14 and 4.34 eV. Keep the model honest about magnitudes, though: Zeff rises by a factor 4.5 across period 2 while the measured radius falls by only about 2.2, so r ≈ n*²a₀/Zeff gets the direction and the mechanism right and the size wrong.
5Every break in the trend names a physical step
The rise across a row is not monotonic, and the two kinds of stumble have different sizes and different causes. The first is a change of subshell. Beryllium is 2s² and boron is 2s²2p¹, so boron's outermost electron leaves a 2p level that penetrates the core less than 2s and is screened by the filled 2s² pair: the first ionisation energy falls from 9.32 eV to 8.30 eV. Magnesium to aluminium repeats it, 7.65 eV down to 5.99 eV. The second is the first forced spin pairing. Nitrogen is 2p³ with one electron in each of the three 2p orbitals and their spins parallel; oxygen is 2p⁴ and must put a second electron into one of them, and that pairing repulsion is refunded when an electron leaves: 14.53 eV falls to 13.62 eV. Phosphorus to sulfur repeats it, 10.49 eV to 10.36 eV. Compare the scales in period 3: 1.66 eV for the subshell step against 0.13 eV for the pairing step, a ratio of about 13. That both breaks recur in period after period is the evidence that block structure is a real feature of the level scheme.
6Orbital energies are not ionisation energies
All of this rests on orbital energies, and Koopmans' theorem tempts you to read them straight off as ionisation energies, IEᵢ ≈ −εᵢ. The identification assumes the other N − 1 orbitals stay exactly where they were. They do not: they contract around the hole, lowering the ion's energy and therefore the true ionisation energy, so −ε is an overestimate, and the deeper the hole the larger the relaxation. For neon, −ε(2p) = 23.13 eV against a measured 21.565 eV, an error of 1.57 eV; −ε(1s) = 891.8 eV against 870.2 eV, an error of 21.6 eV. Correlation pushes the other way, since the neutral correlates more strongly than the cation, and the partial cancellation is why valence errors stay at a few per cent. Relativity moves the levels again in heavy atoms: gold's 6s contracts by about 18%, and francium's first ionisation energy, 4.07 eV, is higher than caesium's 3.89 eV, reversing the group-1 trend the geometry alone would predict. The lanthanide contraction leaves hafnium the same size as zirconium, 32 protons later. The table's combinatorics are exact; its energetics are not.
Key equations
Read the units, assumptions and limits before substituting.
Subshell capacity
Nₗ = 2(2l + 1)s 2, p 6, d 10, f 14
The only place antisymmetry enters the table's geometry: a block can be 2, 6, 10 or 14 wide and nothing else.
l is the orbital quantum number; 2l + 1 counts the mₗ values and the 2 counts mₛ = ±½. Both dimensionless.Madelung (n + l) filling order
fill by increasing n + llower n first on a tie
Sets the d block one period below its own n and the f block two, which is why the table is inset rather than rectangular.
n and l dimensionless. 4s (n + l = 4) precedes 3d (5); 3d precedes 4p on the tie; 6s (6) precedes 4f (7).Period length
Lₖ = 2(lₘₐₓ + 1)², with lₘₐₓ = ⌊k/2⌋
lₘₐₓ advances only on every second period, so every row length after the first is used twice before the next block joins.
k is the period number; Lₖ is the sum of 2(2l + 1) for l = 0 to lₘₐₓ. Gives 2, 8, 8, 18, 18, 32, 32.Screened valence charge (Slater)
Zeff = Z − Sr ≈ n*² a₀ / Zeff
Li 2s gives Zeff = 1.30 and Na 3s gives 2.20, so r(Na)/r(Li) ≈ (9/2.20)/(4/1.30) = 1.33 against a measured 1.30.
Slater: S = 0.35 per same-shell electron, 0.85 per n − 1 electron, 1.00 per deeper one. n* is the effective quantum number (n* = n up to n = 3) and a₀ = 52.9 pm is the Bohr radius, so r comes out in pm.Koopmans' identification
IEᵢ ≈ −εᵢ (Hartree–Fock)
Neon 2p: −ε = 23.1 eV against a measured 21.6 eV. The 1.5 eV gap is the relaxation of the surviving electrons, partly cancelled by correlation.
εᵢ is the Hartree–Fock orbital energy in eV; the equality holds only if the other N − 1 orbitals stay frozen.Relativistic s contraction
v/c ≈ Zαγ = (1 − (Zα)²)⁻¹⁄²r ∝ 1/γ
Gold, Z = 79: v/c = 0.58 and γ = 1.22, an 18% 6s contraction — enough to pull the 5d-to-6s gap into the visible.
α = 1/137.04 and Z is the nuclear charge. The estimate is for a 1s electron and carries down to every s orbital.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Start at period 6: a 32-element row with an f block. Now drag the shell slider — 2n² lands on the row's end only at n = 4, never at n = 6. Then set period 3 against n = 3 and watch the dashed line overshoot by ten, the 3d subshell that period 3 never fills.
Live results
- HIGHEST l IN THE ROW
- 3
- PERIOD LENGTH
- 32 elements
- SHELL CAPACITY 2n²
- 72 electrons
- EXCESS OVER THE ROW
- 40 electrons
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. Period k: 6; Compare with shell n: 6
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The commonest error is to read the period lengths off the shell capacity 2n². It returns 2, 8, 18, 32 — all four of which really are period lengths, which is exactly why the mistake survives — but from period 3 onward it files them in the wrong rows. Period 3 would have to hold 18, and it holds 8: argon is 1s²2s²2p⁶3s²3p⁶, with all five 3d orbitals empty, and potassium's next electron goes into 4s rather than 3d because 4s has n + l = 4 against 3d's 5. The n = 3 shell is not completed until copper, [Ar]3d¹⁰4s¹, eleven elements into period 4. What sets a row is the capacity of a Madelung diagonal — one subshell of each l up to lₘₐₓ — not the capacity of a principal shell. The numerical coincidence is real but misleading: Lₖ = 2(lₘₐₓ + 1)² is 2m² for m = lₘₐₓ + 1, so period 6's 32 is the capacity of shell 4, not of shell 6.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Use the n + l rule to list the subshells that fill in period 5 and in period 6, add their capacities, and then predict the length of period 8 and the width of the new block it would introduce.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- A period opens at ks and closes at kp. For k = 5 the n + l order between them is 5s (n + l = 5), 4d (6), 5p (6, and 4 < 5 so 4d goes first); for k = 6 it is 6s (6), 4f (7), 5d (7), 6p (7).
- Capacities are Nₗ = 2(2l + 1): s 2, p 6, d 10, f 14. Period 5 is 2 + 10 + 6 = 18 elements; period 6 is 2 + 14 + 10 + 6 = 32.
- Both rows carry exactly one subshell of each l from 0 to lₘₐₓ, so L = 2(lₘₐₓ + 1)². Period 5 has lₘₐₓ = 2, giving 2 × 3² = 18 ✓; period 6 has lₘₐₓ = 3, giving 2 × 4² = 32 ✓.
- Period 8 has lₘₐₓ = ⌊8/2⌋ = 4, so a g block enters with capacity 2(2 × 4 + 1) = 18, and L₈ = 2 × 5² = 50. Written out, the row is 8s, 5g, 6f, 7d, 8p = 2 + 18 + 14 + 10 + 6 = 50.
Answer. Period 5 holds 18 elements and period 6 holds 32. Period 8 would hold 50, opening an 18-wide g block — the first row in which l = 4 appears.
MediumExample 2
First ionisation energies across period 3, in eV: Na 5.14, Mg 7.65, Al 5.99, Si 8.15, P 10.49, S 10.36, Cl 12.97, Ar 15.76. Find the two places where the value falls instead of rising, size each drop, and give the reason for each.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- The row rises overall from 5.14 eV to 15.76 eV, a factor 15.76/5.14 = 3.07, because Z climbs by seven while every added electron stays in n = 3 and screens its neighbours only weakly.
- First fall, Mg to Al: 7.65 → 5.99 eV, a drop of 1.66 eV, or 1.66/7.65 = 21.7% of the magnesium value. Magnesium loses a 3s electron; aluminium loses a 3p electron, which lies above 3s because it penetrates the core less and is screened by the filled 3s² pair.
- Second fall, P to S: 10.49 → 10.36 eV, a drop of 0.13 eV, or 1.2%. Phosphorus is 3p³ with one electron in each 3p orbital and spins parallel; sulfur is 3p⁴ and must double-occupy one of them, and that pairing repulsion is refunded when the electron leaves.
- The two mechanisms differ by an order of magnitude: 1.66/0.13 = 13. Changing subshell moves a whole level; forcing a pair only pays a repulsion penalty inside one.
Answer. Mg → Al falls 1.66 eV (21.7%) at the 3s-to-3p subshell change; P → S falls 0.13 eV (1.2%) at the first forced 3p pairing. Both are wobbles on a rise of a factor 3.07 across the row.
HardExample 3
Hartree–Fock orbital energies for neon are ε(1s) = −32.772 Ha, ε(2s) = −1.930 Ha and ε(2p) = −0.850 Ha, with 1 Ha = 27.211 eV. The measured binding energies are 870.2 eV, 48.5 eV and 21.565 eV. Convert, find the Koopmans error at each level, and say what the pattern of errors is measuring.
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.
- Convert to eV: −ε(1s) = 32.772 × 27.211 = 891.8 eV; −ε(2s) = 1.930 × 27.211 = 52.52 eV; −ε(2p) = 0.850 × 27.211 = 23.13 eV.
- Errors, Koopmans minus measured: 891.8 − 870.2 = 21.6 eV, or 2.5%; 52.52 − 48.5 = 4.0 eV, or 8.3%; 23.13 − 21.565 = 1.57 eV, or 7.3%.
- Every error carries the same sign. Koopmans freezes the surviving nine orbitals, but in reality they contract around the hole, lowering the ion's energy and therefore the true ionisation energy — so the frozen-orbital value must be too large.
- The error tracks the depth of the hole: 1.57 eV for a 2p vacancy against 21.6 eV for a 1s vacancy, because a core hole leaves the outer electrons a far larger unscreened charge to relax towards.
- Correlation pushes the other way — the ten-electron neutral correlates more strongly than the nine-electron ion, which raises the true ionisation energy — and the partial cancellation is why the valence errors stay near 7% rather than running away.
Answer. −ε overestimates every binding energy: by 1.57 eV (7.3%) at 2p, 4.0 eV (8.3%) at 2s and 21.6 eV (2.5%) at 1s. The pattern measures orbital relaxation, which grows with the depth of the hole and is partly offset by correlation.
Quick check
Select an answer and explain it before checking.
Review & continue
The length of a period is the total capacity of the subshells that fill between one ns level and the np level closing the row — 2(lₘₐₓ + 1)² electrons, where lₘₐₓ is the largest orbital angular-momentum quantum number entering by that row.
- 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.