The big idea: An electron configuration is a map of how an atom's electrons are spread across its sub-shells (1s, 2s, 2p, 3s, …).
Electrons fill those sub-shells in a fixed way, set by three rules. Get the rules right and you can write the configuration of any atom or ion in the first few rows of the periodic table.
- Aufbau — fill the lowest energy first. - Pauli — at most 2 electrons per orbital, with opposite spins. - Hund — fill orbitals singly first, then pair.
The vocabulary: - Sub-shell — a set of orbitals of the same type and energy: s (1 orbital), p (3 orbitals), d (5 orbitals). - Orbital — a region holding up to 2 electrons. Drawn as a box. - Ground state — the lowest-energy arrangement, the one these rules give. Anything else is an excited state. - The superscript (e.g. the 4 in 2p⁴) is the number of electrons in that sub-shell.
| Rule | What it says | In plain words |
|---|---|---|
| Aufbau | fill the lowest-energy sub-shell first | build up: 1s, then 2s, then 2p, … |
| Pauli exclusion | each orbital holds at most 2 electrons, with opposite spins | max ↑↓ per box |
| Hund's rule | fill each orbital in a sub-shell singly before pairing, spins parallel | ↑ ↑ ↑ before ↑↓ ↑ ↑ |
Drawing electrons in boxes makes the three rules visible. Each box is an orbital; an arrow is an electron; the direction of the arrow is its spin. Fill from the bottom up (Aufbau).
The filling order to memorise: The sub-shell energy order across the first four rows is:
1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p
Note the swap: 4s fills before 3d. The maximum electrons are 2 (s), 6 (p) and 10 (d), matching the number of boxes × 2.
Nitrogen (1s² 2s² 2p³): the three 2p electrons go singly into separate boxes, all spins parallel — that's Hund's rule.
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Why nitrogen looks like that: Nitrogen has 7 electrons: 1s² 2s² 2p³. The three 2p electrons could crowd into one or two boxes — but Hund's rule says spread them out singly with parallel spins first. So the 2p sub-shell is ↑ ↑ ↑, not ↑↓ ↑.
Oxygen (1s² 2s² 2p⁴): the 4th 2p electron must now pair up — one box holds ↑↓ while the other two stay singly filled.
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Then pairing begins: Oxygen has one more electron than nitrogen (1s² 2s² 2p⁴). The 2p sub-shell is full of singly-filled boxes, so the extra electron must now pair up (Pauli: opposite spin) — giving ↑↓ ↑ ↑.
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A full configuration lists every sub-shell. A condensed (or core) configuration replaces the inner electrons with the previous noble gas in square brackets, then lists only the outer electrons — quicker and clearer.
| Species | Full configuration | Condensed configuration |
|---|---|---|
| Na (Z = 11) | 1s² 2s² 2p⁶ 3s¹ | [Ne] 3s¹ |
| S (Z = 16) | 1s² 2s² 2p⁶ 3s² 3p⁴ | [Ne] 3s² 3p⁴ |
| Ca (Z = 20) | 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² | [Ar] 4s² |
| Fe (Z = 26) | 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² | [Ar] 3d⁶ 4s² |
Ions — add or remove from the outermost shell: To write an ion configuration, start from the atom, then:
- Negative ion (gained electrons) → add electrons to the next available sub-shell. e.g. O → O²⁻ adds 2 to give 1s² 2s² 2p⁶. - Positive ion (lost electrons) → remove electrons from the highest occupied main shell (largest n) first.
For transition metals this matters: the 4s electrons leave before the 3d electrons, even though 4s filled first.
The 4s-out-first trap: Iron is [Ar] 3d⁶ 4s². To make Fe²⁺, remove the two 4s electrons (highest n), not two 3d — giving [Ar] 3d⁶. Removing 3d by mistake is the classic lost mark.
Fe²⁺ (1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶): the two electrons are removed from 4s first, leaving a part-filled 3d sub-shell.
Interactive diagram
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The two exceptions: chromium and copper: A half-full or full d sub-shell is extra stable, so two elements break the simple Aufbau order:
- Chromium is [Ar] 3d⁵ 4s¹ (not 3d⁴ 4s²) — one electron promotes to give a half-full 3d⁵. - Copper is [Ar] 3d¹⁰ 4s¹ (not 3d⁹ 4s²) — to give a full 3d¹⁰.
These two are worth memorising; the IB expects them.
How this is tested: S1.3.3 shows up two ways.
- Paper 1A (MCQ): pick the correct orbital diagram for an atom (a Hund check), or the right condensed configuration — copper and chromium are favourite exceptions to spot. - Paper 2: a state or deduce question — write the full configuration of an atom, or deduce the configuration of an ion.
The deduce-an-ion question is the one that separates marks: you must remove electrons from the highest main shell first (4s before 3d).
Scoring the deduce-the-ion mark: Write the neutral atom first, then strike out the electrons you remove. For a transition-metal cation, take the 4s electrons out before any 3d — that is the marking point examiners look for.
IB-style question — deduce an ion configuration
Cobalt has atomic number 27. Deduce the full electron configuration of the cobalt(II) ion, Co²⁺. [2]
How to score the marks
- Mark 1 — the neutral atom. Cobalt (Z = 27) fills by Aufbau, with 4s before 3d:
- Mark 2 — remove 2 electrons from the highest main shell (4s) first, not from 3d, to form the 2+ ion:
Final answer
Co²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷ (the two 4s electrons are removed first).