1.3.4Additional lessons
Writing Electron Configurations — IB Chemistry SL Revision Notes & Topical Questions
5 min read 5 parts Lesson 10 of 98
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Section 1 of 5Three Rules
The same three principles decide how electrons are arranged.
Aufbau principle
- Aufbau principle Electrons fill the lowest energy sublevel available first, then move to higher ones.
Pauli exclusion
- Pauli exclusion An orbital can hold at most two electrons, and these must have opposite spins.
Hund's rule
- Hund's rule Orbitals of the same sublevel are occupied singly with parallel spins before any orbital receives a second electron.
Condensed form
- Condensed form The inner electrons can be replaced by the symbol of the previous noble gas in square brackets.
Section 2 of 5Writing a Configuration
Follow the order of filling and count the electrons.
1
Find the number of electrons from the atomic number of the neutral atom.
2
Fill sublevels in order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, making sure that the total adds up to the atomic number.
3
Write the superscript for the number of electrons in each sublevel, such as 2p⁴ for four electrons.
4
Check the sum of the superscripts equals the atomic number of the atom, adjusted for charge in an ion.
Section 3 of 5Exceptions and Ions
Exceptions and Ions
Chromium
- Chromium Has the configuration [Ar] 4s¹ 3d⁵, not 4s² 3d⁴, because a half-filled d sublevel is more stable.
Copper
- Copper Has the configuration [Ar] 4s¹ 3d¹⁰, because a completely filled d sublevel is more stable.
Positive ions of d-block
- Positive ions of d-block Lose electrons from the 4s sublevel first, then from 3d, even though 4s fills first.
Negative ions
- Negative ions Gain electrons into the next available orbital, so the chloride ion has the argon configuration.
| Species | Configuration |
|---|---|
| O (Z = 8) | 1s² 2s² 2p⁴ |
| Cl (Z = 17) | [Ne] 3s² 3p⁵ |
| Fe (Z = 26) | [Ar] 4s² 3d⁶ |
| Fe³⁺ | [Ar] 3d⁵ |
Section 4 of 5Worked Example
Question Write the full electron configuration of the Fe³⁺ ion (Z = 26).
- Neutral Fe has 26 electrons: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶.
- Fe³⁺ has lost three electrons, giving 23 in total.
- The two 4s electrons are lost first, then one 3d electron: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵.
Answer 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵, or [Ar] 3d⁵
Section 5 of 5Exam Checklist & Tips
1.3.4 – Key Points for the Exam
- Aufbau principle Electrons occupy the lowest energy orbitals available first.
- Hund's rule Electrons occupy degenerate orbitals singly before pairing.
- Pauli exclusion principle An orbital holds at most two electrons, and these must have opposite spins.
- Condensed configuration A configuration that uses the previous noble gas symbol in brackets for the inner electrons.
Common Mistakes
- Removing 3d electrons before 4s when forming transition metal cations.
- Writing chromium as [Ar] 4s² 3d⁴ and copper as [Ar] 4s² 3d⁹.
- Pairing electrons in a p sublevel before each orbital has one electron.
- Writing a configuration whose electrons do not add up to the atomic number, less any ion charge.
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