Electron configurations
Electron configurations
Filling sub-shells and writing configurations
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An electron configuration is a record of where an atom's electrons go. Put electrons into the available subshells in increasing energy, following the arrows in the diagram.
- In 2p⁴, the 2 tells you the shell, p tells you the subshell and 4 tells you how many electrons are in it.
- Use the s, p and d capacities of 2, 6 and 10. Check that the superscripts add to the atom's total electron number.
- Within a set of equal-energy orbitals, put one electron in each orbital before pairing. Two electrons in one orbital must have opposite spins.
Bromine, Z = 35
Write the full electron configuration and the noble-gas shorthand configuration of a neutral bromine atom. Bromine has atomic number 35.
Worked answer
- A neutral atom has 35 electrons.
- Fill in order, respecting s², p⁶ and d¹⁰ capacities.
- The first 18 electrons are the [Ar] core.
- Check: 18 + 10 + 2 + 5 = 35.
Shorthand: [Ar] 3d¹⁰ 4s² 4p⁵.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
- Put 3p, 4s, 3d and 4p in filling order. [1]
- State the maximum number of electrons in 3d. [1]
- State the principal energy level number of an electron in 4p. [1]
Total: 3 marks
Show answer & marking points
- 3p → 4s → 3d → 4p. [1]
- 10. [1]
- n = 4. [1]
Atomic numbers: P = 15, Ca = 20, Se = 34.
- Write the full electron configuration of phosphorus. [1]
- Write the full electron configuration of calcium. [1]
- Write the full electron configuration of selenium. [1]
- Write selenium using [Ar] shorthand. [1]
Total: 4 marks
Show answer & marking points
- P: 1s² 2s² 2p⁶ 3s² 3p³. [1]
- Ca: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s². [1]
- Se: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁴. [1]
- Se: [Ar] 3d¹⁰ 4s² 4p⁴. [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.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
Atomic numbers: Cr = 24, Cu = 29.
- Write chromium using [Ar] shorthand. [1]
- Write copper using [Ar] shorthand. [1]
- Describe the 3d occupancy in each. [2]
Total: 4 marks
Show answer & marking points
- Cr: [Ar] 3d⁵ 4s¹. [1]
- Cu: [Ar] 3d¹⁰ 4s¹. [1]
- Chromium: half-filled 3d [1]. Copper: full 3d [1].
Electron configurations of ions
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Start with the neutral atom's configuration. The charge then tells you how many electrons to remove or add. A positive ion has lost electrons; a negative ion has gained them.
When removing electrons, start with the highest occupied principal shell. For transition-metal ions, remove the 4s electrons before the 3d electrons. For example, losing two electrons from 4s² leaves 4s empty.
Check the final electron total against atomic number and charge. Counting is a useful check, but remember that removing electrons from an atom is not the same as refilling a new neutral atom from scratch.
| Ion | Neutral atom configuration | Ion configuration | Electrons removed or added |
|---|---|---|---|
| O²⁻ | O: 1s² 2s² 2p⁴ | 1s² 2s² 2p⁶ | Gain 2 electrons |
| Ca²⁺ | Ca: [Ar] 4s² | [Ar] | Remove 4s² |
| Fe²⁺ | Fe: [Ar] 3d⁶ 4s² | [Ar] 3d⁶ | Remove both 4s electrons |
| Fe³⁺ | Fe: [Ar] 3d⁶ 4s² | [Ar] 3d⁵ | Remove both 4s electrons and one 3d electron |
Work backwards from an ion
X²⁺ has configuration [Ar] 3d⁷. Identify X and reconstruct its neutral configuration. Atomic numbers: Fe = 26, Co = 27, Ni = 28.
Worked answer
- X²⁺ has 18 + 7 = 25 electrons.
- The neutral atom has 27 electrons, so Z = 27: cobalt.
- Reverse ion formation by returning two electrons to 4s.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
Atomic numbers: O = 8, Ca = 20, Fe = 26.
- Write the electron configuration of O²⁻. [1]
- Write the electron configuration of Ca²⁺. [1]
- Write the electron configuration of Fe²⁺. [1]
- Write the electron configuration of Fe³⁺. [1]
- Which sub-shell loses electrons first when iron forms a positive ion? [1]
Total: 5 marks
Show answer & marking points
- O²⁻: 1s² 2s² 2p⁶. [1]
- Ca²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶, or [Ar]. [1]
- Fe²⁺: [Ar] 3d⁶. [1]
- Fe³⁺: [Ar] 3d⁵. [1]
- 4s is emptied before 3d. [1]
X³⁺ has configuration [Ar] 3d². Relevant atomic numbers: Sc = 21, Ti = 22, V = 23, Cr = 24.
- How many electrons are in X³⁺? [1]
- Identify X. [1]
- Write the neutral atom’s configuration. [1]
- Explain how you reconstructed it. [1]
Total: 4 marks
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- 18 + 2 = 20 electrons. [1]
- Vanadium: 20 + 3 = 23 electrons in the neutral atom. [1]
- V: [Ar] 3d³ 4s². [1]
- Reverse the 3+ charge by returning one electron to 3d and two to 4s. [1]
Y²⁺ has configuration [Ar] 3d⁵. Y has Z ≤ 36. Relevant atomic numbers: Cr = 24, Mn = 25, Fe = 26, Co = 27.
- How many electrons are in Y²⁺? [1]
- Determine the proton number and identity of Y. [2]
- Write the neutral Y configuration. [1]
- Write the Y³⁺ configuration. [1]
- Which sub-shell loses electrons first? [1]
- A student says Y is iron because Fe³⁺ is also [Ar] 3d⁵. Explain the error. [2]
Total: 8 marks
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- 23. [1]
- Z = 23 + 2 = 25 [1], manganese [1].
- Mn: [Ar] 3d⁵ 4s². [1]
- Mn³⁺: [Ar] 3d⁴. [1]
- 4s before 3d. [1]
- Charge must be used as well as electron count [1]. Fe³⁺ is [Ar] 3d⁵, but Fe²⁺ is [Ar] 3d⁶, so Y²⁺ cannot be Fe²⁺ [1].
Electron configuration and the Periodic Table
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The Periodic Table helps you predict electron structure. An element's block tells you the type of subshell being filled across that part of the table.
Look for the differentiating electron: the electron that distinguishes this element's configuration from the preceding element's. Do not simply choose the subshell written last on the page.
In the table below, n means the outer-shell principal energy level. Electron configurations up to Z = 36 use s, p and d subshells. You should also recognise where the f block sits.
| Block | Position | Typical ending |
|---|---|---|
| s | Groups 1 and 2, plus helium | ns¹ or ns² |
| p | Groups 3–7 and 0, except helium | np¹ to np⁶ |
| d | Central block | (n − 1)d¹ to (n − 1)d¹⁰ |
| f | Two rows below the main table | (n − 2)f¹ to (n − 2)f¹⁴ |
Identify the block
An atom has configuration [Ar] 3d⁵ 4s². State its block.
Worked answer
The differentiating electron enters 3d, even though 4s² may be written last.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
Element E has proton number 34 and forms E²⁻.
| Element | Ge | As | Se | Br | Kr |
|---|---|---|---|---|---|
| Atomic number | 32 | 33 | 34 | 35 | 36 |
- Identify E. [1]
- Write the full configuration of E. [1]
- Write the full configuration of E²⁻. [1]
- State the number of electrons in E²⁻. [1]
- Final electrons enter 3s, 3p, 3d and 4f in four different atoms. State their blocks. [4]
- Explain how a block is named. [1]
Total: 9 marks
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- Selenium, Se. [1]
- Se: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁴. [1]
- Se²⁻: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶. [1]
- 36 electrons. [1]
- s, p, d and f blocks respectively. [4]
- The type of sub-shell receiving the differentiating electron gives the block name. [1]