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

We assume you know nothing except: an atom has a positive centre (the nucleus) and electrons live in "rooms" called orbitals. Everything else we build.

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Step 1 — What is an orbital, and what are $n$ and $l$?

WHAT. Every orbital carries two address-labels. The first, $n$, is a whole number $1,2,3,\dots$ that tells you which shell — roughly how far out from the nucleus the electron lives. The second, $l$, is a whole number $0,1,2,3$ that tells you the shape of the room. We give the shapes letter-names:

$l=\underset{s}{\underbrace{0}},\underset{p}{\underbrace{1}},\underset{d}{\underbrace{2}},\underset{f}{\underbrace{3}}$

WHY these two. We need $n$ because distance from the nucleus decides attraction; we need $l$ because — as Step 3 shows — the shape decides how deep the electron dips toward the centre. One number is not enough to rank energy. (These labels come from Quantum Numbers (n, l, m, s).)

PICTURE. Look at the figure: concentric rings are shells (bigger $n$ = wider ring). Inside each ring, the little shape-icons are the $l$-flavours. The one accent (red) marks the innermost, smallest-$n$, roundest-$l$ orbital — the "cheapest" room, where filling begins.

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Step 2 — Why "fill by $n$ alone" is tempting, and where it snaps

WHAT. The lazy guess: farther out = higher energy, so just order rooms by $n$: $1s,2s,2p,3s,3p,3d,4s,\dots$

WHY it feels right. More distance really does mean weaker pull from the nucleus, and weaker pull means higher energy. For the first few shells this guess is even correct.

WHAT IT LOOKS LIKE. In the figure, the black staircase is the naive "by-$n$" energy ladder. It rises smoothly — until we reach the red step, where the real energy of 3d (top, faint) sits above 4s even though $3<4$. The staircase crosses itself. That single red crossing is the entire reason we need a better rule.

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Step 3 — The hidden physics: penetration decides the tie

WHAT. Why can a $4s$ electron (far shell) sit lower than a $3d$ electron (nearer shell)? Because "far out on average" is not the same as "never comes close." A round $s$ orbital has an inner lobe that dives right through the shielding electrons and touches the nucleus. A $d$ orbital is pushed outward and almost never gets that close.

WHY it matters. Diving close = feeling more of the raw positive charge = getting pulled down in energy. This is penetration, and it partly cancels the distance penalty. (Full story: Penetration and Shielding.)

WHAT IT LOOKS LIKE. The figure plots, along a line out from the nucleus, how much of each orbital's presence sits at each distance. The red curve (4s) has a sharp spike very near the nucleus — its penetrating inner lobe. The black curve (3d) has no such spike; it stays out. That red spike is worth enough energy to drop 4s below 3d.

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Step 4 — Package both effects into one number: $K=n+l$

WHAT. We invent a single ranking number that grows when either $n$ grows or $l$ grows:

$K=\underset{\text{distance cost}}{\underbrace{n}}+\underset{\text{shape/penetration cost}}{\underbrace{l}}$

Each symbol is doing a job: $n$ charges you for living far out; $l$ charges you for a spread-out, poorly-penetrating shape. Add the two bills and you get $K$, the total "energy price tag."

WHY addition (and not, say, $n\times l$). Multiplication would collapse to zero whenever $l=0$ (all $s$ orbitals would tie at zero — nonsense, since $1s$ and $5s$ differ hugely). Addition keeps the $n$-cost alive even when $l=0$, and still lets $l$ nudge the ranking. It is the simplest formula that respects both bullet points from Step 3.

WHAT IT LOOKS LIKE. In the figure each orbital is a dot on a grid: horizontal axis $n$, vertical axis $l$. Its price $K=n+l$ is just the diagonal it sits on. The red diagonal line marks one constant-$K$ family — every dot it touches costs the same. Cheaper diagonals are toward the bottom-left.

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Step 5 — Rule 2: breaking a tie, and why smaller $n$ wins

WHAT. Several orbitals can share one $K$. Take $K=5$: it is hit by $3d(3+2)$, $4p(4+1)$, $5s(5+0)$. Among these, we fill smaller $n$ first: $3d<4p<5s$.

WHY. Along one diagonal, the cheaper $n$-value means the electron's shell is genuinely more inside. Being inside, it feels the nucleus more directly (less outer material to shield it), so its true energy is a hair lower. The tie-break is not arbitrary — it re-uses the Step 3 penetration logic within a tie.

WHAT IT LOOKS LIKE. The figure is the same $n$–$l$ grid, but now we walk the red constant-$K$ diagonal from its bottom-right end toward the top-left. The arrow shows the reading direction: large $n$ down to small $n$ we fill last-to-first, i.e. we start at the smallest-$n$ corner. Watch the order light up: $3d$, then $4p$, then $5s$.

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Step 6 — Assembling the full order by sweeping diagonals

WHAT. Now just do Rule 1 across diagonals and Rule 2 within each. Sweep $K=1,2,3,4,5,\dots$; inside each $K$ list orbitals by rising $n$. Out pops:

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

Each "$<$" means "fills before." Read left to right = pour electrons in that order.

WHY this is the whole rule. Nothing else is needed. Rule 1 orders the diagonals; Rule 2 orders inside them; together they generate the sequence with no diagram to memorize.

WHAT IT LOOKS LIKE. The figure overlays all the constant-$K$ diagonals on the $n$–$l$ grid as parallel lines, numbered by $K$. Follow them bottom-left to top-right; the red path threads through the dots in exactly the filling order. This red thread is the famous "Madelung staircase." (The same order literally builds the Periodic Table Blocks (s, p, d, f).)

Recall Self-test: 5s vs 4d

Which fills first? ::: 5s has $K=5$, 4d has $K=6$; smaller $K$ wins, so 5s first.

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Step 7 — Edge & degenerate cases you must not trip on

WHAT. Three corner cases that break naive habits.

(a) The $l=0$ floor. For any $s$ orbital $l=0$, so $K=n$. Here — and only here — ranking by $K$ and ranking by $n$ agree. Do not overgeneralize this coincidence to $p,d,f$.

(b) Filling ≠ emptying. Once filled, $4s$ ($n=4$) is higher $n$ than $3d$ ($n=3$). On ionization we pull the highest-$n$ electron first, so $4s$ leaves before $3d$: ${\text{Fe}}^{2+}=[\text{Ar}]3d^6$, not $[\text{Ar}]3d^{4}4s^2$. Filling uses $n+l$; removal uses $n$ — two different questions. (See Ionization and Electron Removal Order.)

(c) Aufbau is a forecast, not a law. For a few atoms a half-filled or full $d$ shell is extra stable, so an electron hops: $\text{Cr}=[\text{Ar}]3d^{5}4s^1$, $\text{Cu}=[\text{Ar}]3d^{10}4s^1$. Aufbau gives the starting guess; real energetics can override it. (See Electron Configuration Exceptions (Cr, Cu). How electrons spread within a subshell is Hund's Rule; how many fit is set by Pauli Exclusion Principle: $2(2l+1)$.)

WHAT IT LOOKS LIKE. Left panel: filling arrow points into $3d$ after $4s$. Right panel: the red ionization arrow points out of $4s$ first. Same two boxes, opposite arrows — the picture of "first in is NOT last out."

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The one-picture summary

WHAT. Everything at once: the $n$–$l$ grid, all constant-$K$ diagonals, and the single red thread that walks the orbitals in ground-state filling order — with the 4s→3d crossover circled as the payoff.

Recall Feynman retelling — the whole walkthrough in plain words

Every room in the atom has two tags: how far out it is ($n$) and how spread-out its shape is ($l$). Both cost energy, so we glue them into one price tag $K=n+l$ (we add because even the roundest room still charges rent for distance). Electrons are cheapskates: they grab the lowest-$K$ room first. If two rooms cost the same, they take the one nearer the middle (smaller $n$), because a round room that dips close to the nucleus gets a discount — that's penetration, and it's exactly why 4s (price 4) beats 3d (price 5) even though 3d has the lower row number. Draw a grid of rooms, sweep diagonal lines of equal price from cheap to dear, and the red thread you trace is the filling order. Two footnotes: emptying an atom is the reverse question (pull the outermost, highest-$n$ electron first, so 4s leaves before 3d), and a couple of atoms (Cr, Cu) cheat for a tidy half/full shell. That's the entire rule — two numbers, one sum, one tie-break.

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Connections

• Parent · Aufbau / Madelung
• Quantum Numbers (n, l, m, s) — where $n$ and $l$ come from.
• Penetration and Shielding — the physics under Steps 3 & 5.
• Pauli Exclusion Principle — subshell capacity $2(2l+1)$.
• Hund's Rule — arrangement within a subshell.
• Electron Configuration Exceptions (Cr, Cu) — Step 7(c).
• Ionization and Electron Removal Order — Step 7(b).
• Periodic Table Blocks (s, p, d, f) — the order builds the table.