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.
Use N(electrons) = Z − q, where q is the signed charge in units of the elementary charge. For S²⁻, Z = 16 and q = −2, so N = 18; for Al³⁺, N = 13 − 3 = 10. A positive ion has fewer electrons, not more protons. Ion formation leaves the element's proton count and identity unchanged. After counting, write the occupied subshells and verify the sum; a familiar noble-gas label is a result to check, not a substitute for counting.
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.
Removing electrons from a main-group atom starts at the highest occupied principal shell and, within that shell, the higher-energy occupied subshell. Aluminum is [Ne] 3s² 3p¹: Al⁺ loses 3p¹, Al²⁺ loses one 3s electron, and Al³⁺ is [Ne]. Chlorine instead gains one electron to complete 3p⁶. Group patterns predict common monatomic charges such as Na⁺, Mg²⁺ and Cl⁻, but not every element forms a stable ion by simply pursuing a noble-gas count. In particular, transition metals commonly retain partly filled d subshells.
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.
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.
A configuration describes a specified electron count; it does not establish that an ion is common or stable in every environment. Fe²⁺ and Fe³⁺ both occur even though neither has a noble-gas configuration. Removing electrons beyond an outer shell becomes much more costly, which helps explain why Na⁺ is common and Na²⁺ is not a usual aqueous ion.
Practice the two checks independently. O²⁻ and Mg²⁺ each have ten electrons and both are [Ne], but their nuclei contain eight and twelve protons respectively. They are isoelectronic, not the same species. Cu⁺ and Cu²⁺ are [Ar] 3d¹⁰ and [Ar] 3d⁹: after the single 4s electron is removed, the next removal must come from 3d. Never subtract two 4s electrons from copper's one-electron 4s population.
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?
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Find the ground-state configuration of Co³⁺ (Co, Z = 27).
Try it first: Start from neutral Co, then take electrons from 4s before 3d.
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Find the condensed configuration of Cu⁺ and check its electron count.
Try it first: Start with neutral copper's observed exception.
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| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Removing 3d electrons before 4s electrons. | After the d subshell starts filling, 3d sits lower in energy than 4s, so 4s is the outermost and least tightly held. | 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. | For Fe²⁺ that method gives [Ar] 4s² 3d⁴, which is not the observed ground state. | Write the neutral atom, then remove electrons from it. |
| Forgetting the Cr and Cu exceptions when forming their ions. | Starting from [Ar] 4s² 3d⁹ for copper gives the wrong Cu²⁺ result. | Recall Cu = [Ar] 4s¹ 3d¹⁰, so Cu²⁺ = [Ar] 3d⁹. |
Scroll horizontally if needed to read all columns.
After the lesson · Question bank
Work through new problems on your own. The lesson's quick checks and worked examples stay above if you need to review a step.
35 topic questions across 3 difficulty levels.
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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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