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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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
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.
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.
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⁻
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.
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 electrons from the highest n first: ns before (n-1)d for transition metals.
Added electrons continue the ordinary aufbau filling sequence.
Species with identical electron counts and configurations, such as Na⁺, Ne and F⁻.
Main-group metals lose their valence electrons, and nonmetals gain enough to complete the p subshell.
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.
0 of 3 steps revealed.
Which electrons are removed first when Fe becomes Fe²⁺, and what is the resulting configuration?
0 of 3 steps revealed.
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.
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.
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.
No practice questions are available for this topic yet. You can still practice the whole unit.
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.
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