Worked example 1
Write the ground-state configuration of Cr³⁺ (Cr, Z = 24).
Try it first: Write neutral chromium first, and remember it is one of the two exceptions.
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Periodic TrendsIon configurationsContent level: Core 20 min
Electron Configurations of Ions
What you'll be able to do: Write the electron configuration of any common cation or anion, including transition-metal ions.
Best after: Electron Configurations and Orbital Filling
Introduction
Atoms become ions by gaining or losing electrons, and which electrons move is not always the ones you added last. This lesson fixes the removal order so transition-metal ions stop being guesswork.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Write configurations for main-group cations and anions
- Apply the ns-before-(n-1)d removal rule to transition-metal ions
- Predict the common ion charge from a neutral configuration
- Explain why the removal order differs from the filling order
Lesson
Anions: keep filling
Adding electrons continues the normal aufbau order. Chlorine, [Ne] 3s² 3p⁵, gains one electron to become Cl⁻ with [Ne] 3s² 3p⁶, matching argon. Most main-group anions end up isoelectronic with the nearest noble gas.
Main-group cations: empty the valence shell
Magnesium, [Ne] 3s², loses both 3s electrons to give Mg²⁺ = [Ne]. Aluminium loses three to reach [Ne]. The number of electrons lost is normally the group''s valence count, which is why group 1 forms 1+ and group 2 forms 2+ ions.
Mg: [Ne] 3s² → Mg²⁺: [Ne] + 2 e⁻
Transition metals: 4s out first
The 4s orbital fills before 3d in a neutral atom, but once 3d is occupied the 3d electrons drop below 4s in energy. Ionization therefore removes 4s electrons first. Iron, [Ar] 4s² 3d⁶, becomes Fe²⁺ = [Ar] 3d⁶ and Fe³⁺ = [Ar] 3d⁵.
Fill 4s first, remove 4s first. The two rules are not contradictory: adding a nuclear-charge-shielding d population changes the relative energies.
Exceptions carry through to their ions
Copper is [Ar] 4s¹ 3d¹⁰, so Cu²⁺ removes the single 4s electron and one 3d electron to give [Ar] 3d⁹. Chromium is [Ar] 4s¹ 3d⁵, so Cr³⁺ is [Ar] 3d³. Always start from the correct neutral configuration before removing anything.
Remove from the highest n shell, then check the charge
For a first-row transition-metal cation, write the neutral atom first and remove 4s electrons before 3d electrons. Cobalt begins [Ar] 4s² 3d⁷; Co³⁺ removes two 4s electrons and one 3d electron, giving [Ar] 3d⁶. This may look opposite to the 4s-before-3d filling order, but the relative orbital energies change as electrons are added and the atom becomes an ion. Always finish with an electron-count check: Co has 27 protons, so Co³⁺ must have 24 electrons; the argon core plus six d electrons gives 24.
Separate filling order from removal order
Neutral transition metals fill 4s before 3d, but their cations lose 4s electrons before 3d electrons. Write the neutral atom first and remove the outermost occupied shell for a cation; do not reverse the order just because the written configuration puts 3d beside 4s. Finally, count electrons to verify the charge.
For a positive transition-metal ion, inspect 4s before removing 3d electrons.
Key ideas
Rule
Removal order
Remove electrons from the highest n first: ns before (n-1)d for transition metals.
Rule
Addition order
Added electrons continue the ordinary aufbau filling sequence.
Definition
Isoelectronic
Species with identical electron counts and configurations, such as Na⁺, Ne and F⁻.
Key concept
Charge from the group
Main-group metals lose their valence electrons, and nonmetals gain enough to complete the p subshell.
Worked examples
Worked example 2
Which electrons are removed first when Fe becomes Fe²⁺, and what is the resulting configuration?
0 of 3 steps revealed.
Worked example 3
Find the ground-state configuration of Co³⁺ (Co, Z = 27).
Try it first: Start from neutral Co, then take electrons from 4s before 3d.
0 of 3 steps revealed.
Worked example 4
Find the condensed configuration of Cu⁺ and check its electron count.
Try it first: Start with neutral copper's observed exception.
0 of 3 steps revealed.
Common mistakes
Removing 3d electrons before 4s electrons.
Why it's wrong: After the d subshell starts filling, 3d sits lower in energy than 4s, so 4s is the outermost and least tightly held.
Check instead: Delete the ns electrons first every time, then take from (n-1)d if more are needed.
Building an ion configuration from scratch by counting electrons in the aufbau order.
Why it's wrong: For Fe²⁺ that method gives [Ar] 4s² 3d⁴, which is not the observed ground state.
Check instead: Write the neutral atom, then remove electrons from it.
Forgetting the Cr and Cu exceptions when forming their ions.
Why it's wrong: Starting from [Ar] 4s² 3d⁹ for copper gives the wrong Cu²⁺ result.
Check instead: Recall Cu = [Ar] 4s¹ 3d¹⁰, so Cu²⁺ = [Ar] 3d⁹.
After the lesson · Question bank
Independent practice
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What you should now know
Anions add electrons to the next available orbital, main-group cations empty the valence shell, and transition-metal cations lose their ns electrons before any (n-1)d electrons.
- Anions continue the aufbau filling order
- Main-group cations empty their valence shell
- Transition metals lose ns electrons before (n-1)d
- Always build ions from the correct neutral configuration
Sources and further reading
- Chemistry 2e, Section 6.4: Electronic Structure of Atoms (Electron Configurations)Flowers, Theopold, Langley, Robinson · OpenStax, Rice University · Chapter 6.4View source
Access for free at openstax.org License