Aufbau principle — order of filling (Madelung rule, n + l): Worked examples

Before we start, three tools we will reuse. Each is defined the moment it appears, in plain words:

The master order (built entirely from $K$ , no diagram to memorise):

$1 s < 2 s < 2 p < 3 s < 3 p < 4 s < 3 d < 4 p < 5 s < 4 d < 5 p < 6 s < 4 f < 5 d < 6 p < 7 s$

Figure — Aufbau principle — order of filling (Madelung rule, n + l)

The staircase above is the rule drawn out: every diagonal line links orbitals with the same $K$ . Sweep them from top ( $K = 1$ ) downward and read off the sequence.

The scenario matrix

Here is every distinct kind of case this topic can throw at you. Each worked example below is tagged with the cell it fills.

Cell	Scenario class	What is tricky about it	Example
A	Straight neutral atom	just count in $K$ -order	Ex 1 (Fe)
B	The famous $4 s$ / $3 d$ crossover boundary	$K$ tie & near-tie region	Ex 2 (Ca vs Sc)
C	Tie-break, equal $K$	Rule 2, smaller $n$ first	Ex 3 ( $3 d$ vs $4 p$ vs $5 s$ )
D	Cation — electron removal	removal order $\neq =$ filling order	Ex 4 (Fe²⁺)
E	Anion — electrons added	keep filling in $K$ -order	Ex 5 (S²⁻)
F	Aufbau exception	half/full subshell stability	Ex 6 (Cr, Cu)
G	Degenerate / zero input	$Z = 0$ , $Z = 1$ , hydrogen	Ex 7
H	Real-world word problem	translate story → $Z$ → config	Ex 8 (colour of gold-ish metal)
I	Exam-style twist	reverse question: config → element	Ex 9

The worked examples

Worked example Example 2 — Cell B: the crossover boundary (Ca

$Z = 20$ → Sc $Z = 21$ ) Forecast: Calcium is $Z = 20$ , Scandium $Z = 21$ . Where does that one extra electron of Sc go — into $4 s$ or $3 d$ ?

Step 1 — Fill Ca ( $Z = 20$ ). Why this step? Ca is the last element before $3 d$ starts, so it's our reference floor. $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2}$ → count $= 20$ ✅. $4 s$ is now full.

Step 2 — Add one more electron for Sc. Why this step? $4 s$ ( $K = 4$ ) is already full; the next-lowest $K$ is $3 d$ ( $K = 5$ ). The 21st electron goes into $3 d$ : $[Ar] 3 d^{1} 4 s^{2}$ .

Step 3 — Read the figure. Look at the red arrow in the diagram: it shows the energy of $3 d$ dropping below $4 s$ once electrons occupy it — but for filling order we still used $K$ , which correctly put $4 s$ first. Why this step? This is the subtle bit: $4 s$ fills first, yet in the built atom $3 d$ ends up lower. That is why on ionisation (Ex 4) $4 s$ leaves first. See Penetration and Shielding.

Verify: Sc config electron count $= 18 (Ar) + 1 + 2 = 21$ ✅.

Worked example Example 7 — Cell G: degenerate / zero inputs

Forecast: What is the "configuration" of (a) a bare nucleus with $Z = 0$ electrons, and (b) hydrogen $Z = 1$ ? Does Madelung even matter for one electron?

Step 1 — Case $Z = 0$ (no electrons). Why this step? Always check the empty edge case — a rule must not break at zero. Zero electrons → configuration is empty (written as nothing / $[]$ ). Total electrons $= 0$ ✅.

Step 2 — Case $Z = 1$ (hydrogen). Why this step? With one electron there is no shielding, so $K$ -ranking and pure- $n$ ranking agree — the crossover ( $4 s$ below $3 d$ ) does not happen in H. $H = 1 s^{1} .$ The single electron takes the smallest- $K$ orbital, $1 s$ ( $K = 1$ ).

Verify: Electron count $= 1$ ✅. (Note: in H alone, $3 d$ actually lies below $4 s$ because there is nothing to shield — Madelung's crossover is a multi-electron effect.)

Worked example Example 8 — Cell H: real-world word problem

Forecast: A jeweller has a soft, yellow, corrosion-resistant metal used in circuits. Its atomic number is 29. Predict its ground-state configuration and say why one $4 s$ electron seems "missing."

Step 1 — Identify the element. Why this step? Translate the story into a number. $Z = 29$ → copper (Cu). The "yellow, conductive, corrosion-resistant" clues match Cu.

Step 2 — Build the configuration. Why this step? Cu is an exception (Cell F), so use the $3 d^{10}$ rule. $Cu = [Ar] 3 d^{10} 4 s^{1} .$

Step 3 — Explain the "missing" $4 s$ electron. Why this step? The word problem asks why. A full $3 d^{10}$ shell is so stable that one $4 s$ electron migrates into $3 d$ , leaving $4 s^{1}$ — this loosely-held $4 s^{1}$ electron is exactly what makes copper such a good conductor.

Verify: $18 + 10 + 1 = 29$ ✅.

Connections

Parent topic
Quantum Numbers (n, l, m, s) — where $n$ and $l$ come from.
Penetration and Shielding — why the $4 s$ / $3 d$ crossover is real (Ex 2).
Pauli Exclusion Principle — the $2 (2 l + 1)$ capacities we counted with.
Hund's Rule — how electrons spread within a chosen subshell.
Electron Configuration Exceptions (Cr, Cu) — Cell F (Ex 6, Ex 8).
Ionization and Electron Removal Order — Cell D (Ex 4).
Periodic Table Blocks (s, p, d, f) — Cell I (Ex 9).