Electron configurations
Electron configurations
Filling order and electron configurations
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Start by counting the electrons. A neutral atom has one electron for every proton, so its total equals its atomic number.
- Follow the filling order shown in the diagram.
- Use the capacities s², p⁶ and d¹⁰ as you fill each subshell.
- Stop when all the atom's electrons have been placed.
- Add the superscripts as a final check: their total must equal Z.
In a configuration, the number names the shell, the letter names the subshell and the superscript counts its electrons. [Ar] is shorthand for the complete 18-electron argon core.
The 4s and 3d energies are close and depend on occupancy. Use the stated filling order without assuming one fixed energy ordering applies to every atom and ion.
Titanium, Z = 22
Write the full electron configuration and the noble-gas shorthand configuration of a neutral titanium atom. Titanium has atomic number 22.
Worked answer
- A neutral Ti atom has 22 electrons.
- Fill 18 core electrons, then 4s² and 3d².
- Check: 18 + 2 + 2 = 22.
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
- Write the subshell filling order from 1s to 4p. [2]
- Which subshell fills immediately before 3d, and which immediately after it? [2]
- Explain why 4s fills before 3d in the building-up order. [1]
Total: 5 marks
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- 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p. [2]
- Before: 4s; after: 4p. [2]
- For building up these neutral-atom configurations, 4s is filled before 3d because it is initially lower in energy. [1]
- Write the full configuration of Ca, Z = 20, and check its electron total. [2]
- Write the full configuration of Kr, Z = 36, and check its electron total. [2]
Total: 4 marks
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- Ca: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² [1]; total 20 [1].
- Kr: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ [1]; total 36 [1].
Atomic numbers: K 19, Ca 20, Zn 30, Kr 36.
Each student configuration contains an error.
| Species | Student configuration |
|---|---|
| K | [Ar] 3d¹ |
| Ca | [Ar] 3d² |
| Zn | [Ar] 3d⁸ 4s⁴ |
| Kr | [Ar] 3d¹⁰ 4s² 4p⁵ |
- Write the correct configuration for each. [4]
- For each correction, state the rule broken. [4]
Total: 8 marks
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| Species | Correct configuration [1 each] | Rule [1 each] |
|---|---|---|
| K | [Ar] 4s¹ | 4s fills before 3d. |
| Ca | [Ar] 4s² | 4s fills before 3d. |
| Zn | [Ar] 3d¹⁰ 4s² | An s subshell holds at most two electrons. |
| Kr | [Ar] 3d¹⁰ 4s² 4p⁶ | Neutral Kr must have 36 electrons. |
A student writes selenium as [Ar] 3d¹⁰ 4s² 4p⁴ and says this proves that 3d fills before 4s.
- State the filling sequence immediately after 3p. [1]
- Write selenium’s full configuration. [2]
- Explain why the written order does not prove the filling order. [2]
- Check its electron total. [1]
Total: 6 marks
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- 4s → 3d → 4p. [1]
- 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁴. [2]
- Writing 3d before 4s is a convention [1]. The writing order does not change the building-up sequence, in which 4s fills before 3d [1].
- 2 + 2 + 6 + 2 + 6 + 2 + 10 + 4 = 34. [1]
The chromium and copper exceptions
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Chromium and copper are the two exceptions to remember when using the simple filling order. Their measured configurations contain 4s¹, rather than the 4s² you might first predict.
Chromium has a half-filled 3d⁵ subshell; copper has a filled 3d¹⁰ subshell. The 3d and 4s energies are close, and these observed arrangements have lower total energy than the simple predictions.
Use Cr, Z = 24, and Cu, Z = 29, to check the total electrons. Learn their observed configurations without turning the half-filled/full-filled explanation into a rule for every element.
| Atom | Simple prediction | Observed configuration |
|---|---|---|
| Cr | [Ar] 3d⁴ 4s² | [Ar] 3d⁵ 4s¹ |
| Cu | [Ar] 3d⁹ 4s² | [Ar] 3d¹⁰ 4s¹ |
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
For each atom, write the simple predicted configuration and the observed configuration using [Ar].
- Cr, Z = 24. [2]
- Cu, Z = 29. [2]
Total: 4 marks
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- Predicted [Ar] 3d⁴ 4s² [1]; observed [Ar] 3d⁵ 4s¹ [1].
- Predicted [Ar] 3d⁹ 4s² [1]; observed [Ar] 3d¹⁰ 4s¹ [1].
Neutral atom Y contains 29 electrons.
Atomic numbers: Ni 28, Cu 29, Zn 30.
- Write the simple predicted configuration using [Ar]. [2]
- Identify Y and write its observed configuration. [2]
- Represent the 3d and 4s electrons of the observed atom in boxes. [3]
Total: 7 marks
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- [Ar] 3d⁹ 4s². [2]
- Copper [1]: [Ar] 3d¹⁰ 4s¹ [1].
Subshell Orbital 1 Orbital 2 Orbital 3 Orbital 4 Orbital 5 3d ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ Correct 3d occupancy and paired spins [2]; one 4s electron [1].Subshell Orbital 1 4s ↑
Electron configurations of s- and p-block ions
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Required scope: s- and p-block ions of elements up to Z = 36. Transition-metal ion configurations are outside this requirement.
- Write the neutral atom's electron configuration.
- Use the charge to work out how many electrons it has lost or gained.
- For a positive ion, remove electrons from the highest occupied principal shell, taking p before s within that shell.
- For a negative ion, add electrons to the available outer subshell.
- Check the final electron total.
For Ga³⁺, remove the 4p electron and both 4s electrons, leaving [Ar] 3d¹⁰. Do not treat the ion as a new neutral atom and blindly refill its electrons.
| Ion | Neutral atom configuration | Ion configuration | Electron count |
|---|---|---|---|
| O²⁻ (Z = 8) | O: 1s² 2s² 2p⁴ | [Ne] | 10 |
| Na⁺ (Z = 11) | Na: [Ne] 3s¹ | [Ne] | 10 |
| Ca²⁺ (Z = 20) | Ca: [Ar] 4s² | [Ar] | 18 |
| Br⁻ (Z = 35) | Br: [Ar] 3d¹⁰ 4s² 4p⁵ | [Kr] | 36 |
A p-block ion with an occupied d subshell
A neutral gallium atom has electron configuration [Ar] 3d¹⁰ 4s² 4p¹ and atomic number 31. Write the electron configuration of Ga⁺ and explain which electron is removed.
Worked answer
- Ga⁺ has 31 − 1 = 30 electrons.
- Remove the outer 4p electron.
- Keep 3d¹⁰ and 4s²; check 18 + 10 + 2 = 30.
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Atomic numbers: Al 13, S 16, Br 35.
Calculate the electron count and write the noble-gas configuration for each ion.
- Al³⁺. [2]
- S²⁻. [2]
- Br⁻. [2]
Total: 6 marks
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- 13 − 3 = 10 electrons [1]; [Ne] [1].
- 16 + 2 = 18 electrons [1]; [Ar] [1].
- 35 + 1 = 36 electrons [1]; [Kr] [1].
Gallium has Z = 31.
- Write the configuration of Ga using [Ar]. [2]
- Calculate the electron count of Ga³⁺. [1]
- Write its configuration using [Ar]. [1]
- Identify the three electrons removed. [2]
Total: 6 marks
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- Ga: [Ar] 3d¹⁰ 4s² 4p¹. [2]
- 31 − 3 = 28. [1]
- Ga³⁺: [Ar] 3d¹⁰. [1]
- Remove 4p¹ [1] and then 4s² [1].
A 2− ion of a p-block element has 18 electrons.
Atomic numbers: Si 14, P 15, S 16, Cl 17, Ar 18.
- Calculate the neutral atom’s Z and identify it. [2]
- Write the ion’s full configuration. [2]
- Draw the ion’s 3p box diagram. [2]
Total: 6 marks
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- Z = 18 − 2 = 16 [1]; sulfur [1].
- 1s² 2s² 2p⁶ 3s² 3p⁶. [2]
- Three filled orbitals [1], opposite spins in each pair [1].
Subshell Orbital 1 Orbital 2 Orbital 3 3p⁶ ↑↓ ↑↓ ↑↓
A²⁻, B⁻, C and D⁺ each contain 18 electrons. A–D are consecutive elements in increasing atomic number.
Atomic numbers: Si 14, P 15, S 16, Cl 17, Ar 18, K 19, Ca 20.
- Identify A, B, C and D. [4]
- Write their shared configuration. [1]
Total: 5 marks
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- A = S; B = Cl; C = Ar; D = K. [4]
- [Ar]. [1]
Gallium has Z = 31; germanium has Z = 32.
- Write each neutral configuration using [Ar]. [2]
- Calculate the electron counts of Ga³⁺ and Ge⁴⁺. [2]
- Write each ion’s configuration. [2]
- Explain why these ions have the same configuration. [1]
Total: 7 marks
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- Ga: [Ar] 3d¹⁰ 4s² 4p¹; Ge: [Ar] 3d¹⁰ 4s² 4p². [2]
- Ga³⁺: 31 − 3 = 28; Ge⁴⁺: 32 − 4 = 28. [2]
- Both are [Ar] 3d¹⁰. [2]
- Both lose their occupied outer 4p and 4s electrons and retain the same 28-electron arrangement. [1]
Putting electron structure together
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These questions connect several electron-structure skills. Keep returning to the electron count so you do not lose track of the atom or ion.
- Use atomic number to count electrons in a neutral atom.
- If an ion is given, use its charge to relate its electron count to the neutral atom.
- Use the supplied atomic numbers to identify the element.
- Write the configuration, then translate it into orbital boxes if asked. Paired electrons need opposite spins.
Show each step. A correct total alone does not prove the electrons occupy the correct subshells.
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Atom X has configuration [Ar] 3d¹⁰ 4s² 4p³.
Atomic numbers: Ge 32, As 33, Se 34, Br 35, Kr 36.
- Calculate Z and identify X. [2]
- Give the electron count and configuration of X³⁻. [2]
- Draw X’s 4p³ box diagram. [2]
Total: 6 marks
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- 18 + 10 + 2 + 3 = 33 [1]; arsenic, As [1].
- 36 electrons [1]; [Kr] [1].
- One electron in each orbital [1], parallel spins [1].
Subshell Orbital 1 Orbital 2 Orbital 3 4p³ ↑ ↑ ↑
An atom has configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁴.
Atomic numbers: Ge 32, As 33, Se 34, Br 35, Kr 36.
- Deduce Z and identify the element. [1]
- Write the configuration using [Ar]. [1]
- Give the count and configuration of its 2− ion. [2]
- Draw the neutral atom’s 4p box diagram. [2]
Total: 6 marks
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- Z = 34: selenium, Se. [1]
- [Ar] 3d¹⁰ 4s² 4p⁴. [1]
- 36 electrons [1]; [Kr] [1].
- One pair and two singles [1], correct spins [1].
Subshell Orbital 1 Orbital 2 Orbital 3 4p⁴ ↑↓ ↑ ↑
Stretch: combine particle counting with configurations. X⁻ contains 35 protons, 44 neutrons and 36 electrons.
Atomic numbers: As 33, Se 34, Br 35, Kr 36.
- Identify the element and write the full nuclide symbol, with mass number, atomic number and charge. [2]
- Write the full ion configuration and check its total. [2]
- Write the neutral atom using [Ar]. [2]
- Draw the 4p diagrams of the neutral atom and the ion. [4]
Total: 10 marks
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- Bromine [1]: ⁷⁹₃₅Br⁻, since 35 + 44 = 79 [1].
- 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ [1]; total 36 [1].
- [Ar] 3d¹⁰ 4s² 4p⁵. [2]
Subshell Orbital 1 Orbital 2 Orbital 3 Br, 4p⁵ ↑↓ ↑↓ ↑ Two marks per correctly occupied diagram with correct spins. [4]Subshell Orbital 1 Orbital 2 Orbital 3 Br⁻, 4p⁶ ↑↓ ↑↓ ↑↓