Topic 3 of 4

Writing electron configurations

Fill orbitals, account for exceptions and remove the correct electrons from ions.

A-Level 9476 (2026-2027)

Fill orbitals, then remove electrons from the outer shell

Use electron count, Pauli exclusion and Hund's rule; remember Cr, Cu and 4s removal.

Count electrons first. Fill lower-energy orbitals before higher-energy orbitals. Two electrons in one orbital have opposite spins (Pauli exclusion). In a set of equal-energy orbitals, put one electron into each with parallel spins before pairing (Hund's rule). Thus the three 2p electrons of nitrogen occupy three different p orbitals, rather than pairing in one immediately.

Configurations to reason from
Species	Electron count	Ground-state configuration
N	7	1s² 2s² 2p³
Cl	17	1s² 2s² 2p⁶ 3s² 3p⁵
Ca	20	[Ar]4s²
Cr	24	[Ar]3d⁵ 4s¹
Fe	26	[Ar]3d⁶ 4s²
Cu	29	[Ar]3d¹⁰ 4s¹
Fe²⁺	24	[Ar]3d⁶
Fe³⁺	23	[Ar]3d⁵
Cu²⁺	27	[Ar]3d⁹

Superscripts count electrons: 3d⁶ means six electrons distributed over five d orbitals. [Ar] abbreviates the 18-electron argon core. Chromium and copper have the ground-state exceptions shown above; a rigid filling diagram does not capture every balance of orbital energies and electron repulsions.

Worked example

An ion is not filled like a neutral atom with the same electron count

Write the configuration of Cr³⁺ (Z = 24).

Neutral chromium is [Ar]3d⁵ 4s¹.
Remove the 4s electron first, then two 3d electrons.
Check the total: 18 + 3 = 21 electrons, matching 24 - 3.

Answer

Cr³⁺ is [Ar]3d³. For these transition-metal cations, remove 4s electrons before 3d electrons.

Check your understandingFe²⁺ and Cr both have 24 electrons. Must their configurations be the same?Think it through, then reveal the answer

No. Their nuclear charges differ. Fe²⁺ is [Ar]3d⁶; Cr is [Ar]3d⁵ 4s¹. Count electrons and identify the actual species, rather than treating every 24-electron species as the same atom.