Electron configurations

IB Chemistry HL· IB Chemistry HL Topic 1.3· 25 min read

1. Core Rules for Electron Filling★★☆☆☆⏱ 8 min

📘 Definition

Pauli Exclusion Principle

No two electrons in the same atom can have an identical set of four quantum numbers. This means each orbital can hold a maximum of two electrons with opposite spin.

Example:

A 1s orbital can hold 2 electrons with opposite spin, but cannot hold 3 electrons.

Three core rules govern how electrons fill orbitals in the ground state (lowest energy state) of an atom:

Aufbau Principle: Electrons fill lower energy orbitals before higher energy orbitals. Energy is determined by the $n + l$ rule: lower $n + l$ = lower energy; for equal $n + l$ , lower $n$ fills first.
Hund's Rule: When filling degenerate (equal energy) orbitals, electrons fill them singly first with parallel spins, before pairing up. This minimizes electron-electron repulsion.

📐 Worked Example

Draw the orbital filling diagram and write the electron configuration for a neutral nitrogen atom (atomic number = 7)

1
Calculate total number of electrons: for neutral N, $Z = 7$ , so 7 electrons total.
2
Apply filling order: 1s < 2s < 2p. Fill lower energy orbitals first: 2 electrons in 1s, 2 electrons in 2s, leaving 3 electrons for 2p.
3
Apply Hund's rule to the three degenerate 2p orbitals: place one unpaired electron in each orbital with parallel spin.
4
$1s^2 2s^2 2p^3$

2. Full and Condensed Electron Configurations★★☆☆☆⏱ 7 min

Full electron configurations list every occupied subshell with its electron count, ordered by principal quantum number $n$ . Condensed (noble gas core) configurations replace the electron configuration of the nearest preceding noble gas with its symbol in square brackets, only showing outer valence electrons.

📐 Worked Example

Write the full and condensed electron configuration for neutral calcium (Z = 20)

1
Neutral calcium has 20 electrons total. Fill orbitals in order of increasing energy: 1s → 2s → 2p → 3s → 3p → 4s.
2
$1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$
3
Find the nearest noble gas that comes before calcium: argon (Z = 18), which has the configuration $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$ .
4
$\text{Condensed configuration: } [Ar] 4s^2$

3. Electron Configurations of Ions★★★☆☆⏱ 5 min

For charged ions, the total number of electrons changes: add one electron for each negative charge, subtract one for each positive charge. A critical rule for transition metal cations: $n s$ electrons are lost before $(n - 1) d$ electrons when ions form.

📐 Worked Example

Write the condensed electron configuration for Fe³⁺ (iron(III) ion, Z = 26 for Fe)

1
First write the configuration for neutral iron, with 26 electrons.
2
$Fe: [Ar] 3d^6 4s^2$
3
Fe³⁺ has lost 3 electrons total. Remove 4s electrons first, as they are higher in energy once 3d starts filling.
4
Remove both 4s electrons, then one more electron from the 3d subshell.
5
$Fe^{3+}: [Ar] 3d^5$

✓ Quick check

Test your understanding of ion configurations:

What is the correct electron configuration of V²⁺ (vanadium(II), Z = 23 for V)?
- [Ar] 3d³
- [Ar] 4s² 3d¹
- [Ar] 3d¹ 4s²
- [Ar] 4s² 3d³
Reveal answer
[Ar] 3d³ —
Correct! 4s electrons are always lost before 3d when transition metals form cations.

4. Common Exceptions to Filling Rules★★★★☆⏱ 5 min

A small number of neutral atoms have unexpected electron configurations due to the extra stability of fully filled or half-filled d subshells. Symmetric electron distribution and low repulsion make these configurations lower energy than the expected Aufbau filling pattern.

📐 Worked Example

Explain why copper's electron configuration is not the expected $[A r] 4 s^{2} 3 d^{9}$

1
Following generic Aufbau filling rules, the expected configuration for copper (Z = 29) is $[A r] 4 s^{2} 3 d^{9}$ .
2
A fully filled 3d subshell ( $3 d^{10}$ ) has extra stability from symmetric electron distribution and reduced electron-electron repulsion.
3
One electron is promoted from the 4s orbital to the 3d orbital to achieve this lower energy state.
4
$\text{Actual ground state configuration: } [Ar] 4s^1 3d^{10}$

5. Common Pitfalls

Wrong move:

Writing 4s before 3d in the final electron configuration

Why:

Filling order is 4s before 3d, but exam markers require configurations ordered by principal quantum number $n$

Correct move:

Order subshells by increasing $n$ , so 3d always comes before 4s in the final configuration

Wrong move:

Removing 3d electrons before 4s when writing transition metal cation configurations

Why:

Once 3d starts filling, 4s electrons have higher energy and are lost first during ionization

Correct move:

Remove all $n s$ electrons first, then remove additional electrons from $(n - 1) d$ as needed

Wrong move:

Forgetting the exceptions for Cr and Cu, writing them as $4 s^{2} 3 d^{4}$ and $4 s^{2} 3 d^{9}$

Why:

Generic filling rules do not account for the extra stability of half-filled and fully filled d subshells

Correct move:

Memorize: Cr = $[A r] 4 s^{1} 3 d^{5}$ , Cu = $[A r] 4 s^{1} 3 d^{10}$

Wrong move:

Using atomic number (proton count) as electron count for charged ions

Why:

Cations have fewer electrons than protons, anions have more electrons than protons

Correct move:

Total electrons = atomic number minus positive charge, plus negative charge

Wrong move:

Pairing electrons early in degenerate orbitals to follow filling order

Why:

This violates Hund's rule and increases electron-electron repulsion

Correct move:

Always fill each degenerate orbital with one parallel spin electron before pairing any

6. Quick Reference Cheatsheet

Rule/Concept	Summary
Filling order	Lower $n + l$ first; lower $n$ for equal $n + l$
Final configuration order	Ordered by increasing principal quantum number $n$
Transition metal cation electron loss	$n s$ electrons lost before $(n - 1) d$ electrons
Common exceptions (4th period)	Cr: $[A r] 4 s^{1} 3 d^{5}$ ; Cu: $[A r] 4 s^{1} 3 d^{10}$
Condensed configuration	Use nearest preceding noble gas core in square brackets

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · 1
Write configuration for Cu²⁺ ion
2024 · 2
Explain Cr electron configuration exception
2023 · 1
Compare atom and iron ion configurations

Going deeper

syllabusOfficial IB Chemistry HL Subject GuideOfficial syllabus reference

What's Next

Electron configurations are the foundation for understanding all periodic trends, chemical bonding, and reactivity of elements. The arrangement of electrons directly determines an atom’s ionization energy, electronegativity, the number of bonds it can form, and its overall chemical behavior. Mastery of this sub-topic is required for nearly all subsequent topics in IB Chemistry, from explaining periodic trends to predicting molecular geometry and reaction behavior in organic chemistry. This knowledge also underpins the study of transition metal chemistry, explored later in the course.