MRI machines use superconducting magnets that only work near absolute zero. The technology relies on understanding how electrons fill energy levels in certain transition metals. Fireworks glow because electrons in metal salts jump to excited states then crash back down, releasing photons of specific colours — copper gives blue, strontium gives red, barium gives green. Every colour of light from every atom in the universe traces back to electron configuration. This lesson gives you the full quantum mechanical picture of how electrons are arranged in atoms — the foundation of everything from the periodic table trends to chemical bonding.
Core Content
| Subshell | l value | Number of orbitals | Max electrons | Notation example |
|---|---|---|---|---|
| s | 0 | 1 | 2 | 1s², 2s², 3s² |
| p | 1 | 3 (pₓ, p_y, pz) | 6 | 2p⁶, 3p⁴ |
| d | 2 | 5 | 10 | 3d¹⁰, 4d⁵ |
| f | 3 | 7 | 14 | 4f¹⁴ |
The key to electron configuration is knowing the filling order. The energy of subshells does not simply increase with shell number — the 4s subshell is lower in energy than 3d, so it fills first.
The diagonal rule (Madelung's rule) gives a systematic way to remember this: fill in diagonal rows from top-right to bottom-left through a grid of subshells. At HSC level, the key order to know precisely is up to at least 5p.
| Subshell | Max e⁻ | Running total | Periodic table location |
|---|---|---|---|
| 1s | 2 | 2 | Period 1 (H, He) |
| 2s | 2 | 4 | Period 2, Groups 1–2 |
| 2p | 6 | 10 | Period 2, Groups 13–18 |
| 3s | 2 | 12 | Period 3, Groups 1–2 |
| 3p | 6 | 18 | Period 3, Groups 13–18 |
| 4s | 2 | 20 | Period 4, Groups 1–2 |
| 3d | 10 | 30 | Period 4, Groups 3–12 (transition metals) |
| 4p | 6 | 36 | Period 4, Groups 13–18 |
| Rule | What it says | Practical implication |
|---|---|---|
| Aufbau | Fill lowest energy first | Always start at 1s and work up the energy ladder |
| Pauli | Max 2 electrons per orbital; must have opposite spins | Each orbital shows ↑↓ when full, never ↑↑ |
| Hund's | Fill orbitals of same energy singly before pairing | Nitrogen: 1s²2s²2p³ → three p orbitals each get 1e⁻ (not 2p² + empty) |
For elements beyond the first 10, writing the full configuration is tedious. The abbreviation uses the preceding noble gas in brackets, then the remaining electrons:
Chromium (Z=24) and copper (Z=29) have configurations that deviate from the expected pattern. This arises because half-filled and fully-filled d subshells have extra stability:
Worked Examples
Activities
1 Write the full subshell electron configuration for each: (a) Oxygen (Z=8), (b) Argon (Z=18), (c) Calcium (Z=20), (d) Vanadium (Z=23).
2 Write the abbreviated (noble gas core) configuration for: (a) Sodium (Z=11), (b) Bromine (Z=35), (c) Chromium (Z=24).
A Identify the error(s) in each student configuration and write the correct version: (a) Mg (Z=12): 1s²2s²2p⁴3s⁴, (b) Sc (Z=21): [Ar]4s²3d¹ → written as [Ar]3d³, (c) Cu (Z=29): [Ar]3d⁹4s².
Multiple Choice
1. Which is the correct electron configuration for sulfur (Z=16)?
2. The electron configuration [Ar]3d⁵4s¹ describes which element?
3. Which configuration violates Hund's rule for the 2p subshell in nitrogen (Z=7)?
4. An element has the abbreviated configuration [Kr]4d¹⁰5s²5p⁴. Its atomic number is:
5. Why does the 4s subshell fill before the 3d subshell?
Short Answer
6. Write the full electron configuration for each element and identify: the number of valence electrons, and the group and period the element belongs to. (a) Silicon (Z=14), (b) Manganese (Z=25). 4 MARKS
7. Explain why chromium (Z=24) has the configuration [Ar]3d⁵4s¹ rather than the expected [Ar]3d⁴4s². In your answer, refer to orbital stability and electron repulsion. 3 MARKS
1. (a) O (Z=8): 1s²2s²2p⁴. (b) Ar (Z=18): 1s²2s²2p⁶3s²3p⁶. (c) Ca (Z=20): 1s²2s²2p⁶3s²3p⁶4s². (d) V (Z=23): 1s²2s²2p⁶3s²3p⁶3d³4s² (or [Ar]3d³4s²).
2. (a) Na [Ne]3s¹. (b) Br: [Ar]3d¹⁰4s²4p⁵. (c) Cr: [Ar]3d⁵4s¹ (anomalous — half-filled d subshell).
A: (a) Mg (Z=12) error: 1s²2s²2p⁴3s⁴ — the 2p is not full (should be 2p⁶) and 3s holds max 2 (not 4). Correct: 1s²2s²2p⁶3s². (b) Sc error: [Ar]3d³ — this gives only 21 electrons but puts all 3 extra electrons in 3d. Sc should be [Ar]3d¹4s² (one 3d + two 4s; 4s fills before 3d). Correct: [Ar]3d¹4s². (c) Cu (Z=29) error: [Ar]3d⁹4s² — this is the expected but incorrect configuration. Cu is anomalous. Correct: [Ar]3d¹⁰4s¹ (fully-filled d is extra stable).
1. C — S: 1s²2s²2p⁶3s²3p⁴. Total = 2+2+6+2+4 = 16 ✓. Option A puts all 6 extra after 2p into 3s (wrong). B has 2p⁴ (not full). D puts 4 in 3s (max is 2).
2. B — [Ar]=18 electrons + 3d⁵(5) + 4s¹(1) = 24 = Cr. This is anomalous — Cr prefers the half-filled d subshell. Mn would be [Ar]3d⁵4s².
3. D — Hund's rule: fill each p orbital singly before doubling up. Having 2p_x² before 2p_y and 2p_z are filled violates Hund's rule. Options A, B, C all correctly represent one electron per p orbital.
4. A — Kr (Z=36) + 4d¹⁰(10) + 5s²(2) + 5p⁴(4) = 36+16 = 52. This is tellurium (Te).
5. C — The relative energies of 4s and 3d are affected by electron-electron repulsion and shielding. 4s penetrates close to the nucleus (lowering its energy relative to 3d) for lighter elements. B is wrong — 4s doesn't always fill before shell 3 subshells (3s and 3p fill before 4s).
Q6 (4 marks): (a) Si (Z=14): 1s²2s²2p⁶3s²3p² (1 mark). Valence electrons: 4 (3s²3p²). Group 14, Period 3 (1 mark). (b) Mn (Z=25): 1s²2s²2p⁶3s²3p⁶3d⁵4s² (or [Ar]3d⁵4s²) (1 mark). For transition metals, valence electrons include both 4s and 3d electrons: 3d⁵4s² = 7 valence electrons for bonding purposes. Group 7, Period 4 (1 mark).
Q7 (3 marks): The expected configuration [Ar]3d⁴4s² has 4 of the 5 d orbitals singly occupied and two electrons in 4s, with the 3d subshell half-full minus one (1 mark). The actual [Ar]3d⁵4s¹ promotes one 4s electron into the last empty 3d orbital, achieving a half-filled d subshell where all 5 d orbitals are each singly occupied (1 mark). This half-filled arrangement has extra stability due to exchange energy — electrons with parallel spins spread across the 5 d orbitals minimise electron-electron repulsion and maximise the number of favourable same-spin interactions. The energy gained from this extra stability outweighs the cost of promoting the 4s electron (1 mark).
Tick when you've finished all activities and checked your answers.