Electron Configuration: Subshell Notation
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
Sodium (Na, Z = 11) and potassium (K, Z = 19) are both in Group 1 and have similar chemical properties — they both react violently with water and form +1 ions. Yet sodium has 11 electrons and potassium has 19. What do their electron configurations have in common that explains this similarity?
Key facts
- The four subshell types (s, p, d, f), their orbital counts (1, 3, 5, 7) and max electrons (2, 6, 10, 14)
- The shell capacity rule: 2n² electrons per shell
- The Aufbau filling order up to at least 5p (memorise 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p)
Concepts
- Why 4s fills before 3d (lower energy) yet is written after 3d in shell order
- How the Aufbau, Pauli and Hund's rules constrain electron arrangement
- Why Cr and Cu adopt anomalous configurations (extra stability of half-filled or full d⁵/d¹⁰)
Skills
- Write the full and noble-gas-abbreviated configuration for any element up to Z ≈ 36
- Draw orbital box notation showing arrows that obey Pauli and Hund's rule
- Use a configuration to deduce an element's group, period and block
| 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 |
Subshells: s (1 orbital, max 2e⁻); p (3 orbitals, max 6e⁻); d (5 orbitals, max 10e⁻). Each orbital holds at most 2 electrons with opposite spins (Pauli exclusion principle). Shell n holds 2n² electrons max: shell 1 → 2, shell 2 → 8, shell 3 → 18.
Pause — copy the highlighted subshell capacity table into your book before moving on.
Quick check: what is the maximum number of electrons that can occupy the 3d subshell?
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.
1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f → 5d → 6p → 7s → 5f → 6d → 7p
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 |
We just saw the capacity of each subshell type. That raises a question: in what order do electrons fill these subshells, and why does the sequence not always follow increasing shell number? This card answers it → electrons always fill the lowest available energy subshell first (Aufbau); 4s is lower in energy than 3d, so it fills first.
Aufbau filling order: 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p. 4s fills before 3d because 4s is lower in energy. When writing electron configurations, list subshells in increasing shell number order (not filling order). Exception: in transition-metal ions, 4s electrons are removed before 3d.
Add the highlighted Aufbau order to your notes before the check below.
Two truths and a lie: two of these statements about the Aufbau filling order are true, one is the lie. Pick the lie.
Noble Gas (Abbreviated) Configurations
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:
Na (Z=11): Full: 1s²2s²2p⁶3s¹ → Abbreviated: [Ne]3s¹
Ca (Z=20): Full: 1s²2s²2p⁶3s²3p⁶4s² → Abbreviated: [Ar]4s²
Fe (Z=26): Full: 1s²2s²2p⁶3s²3p⁶3d⁶4s² → Abbreviated: [Ar]3d⁶4s²
Anomalous Configurations: Cr and Cu
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:
Cu (Z=29): Expected: [Ar]3d⁹4s² — Actual: [Ar]3d¹⁰4s¹ (fully filled d subshell = extra stability)
We just saw the Aufbau filling order. That raises a question: when multiple orbitals of the same energy are available, how do electrons distribute themselves, and are there any exceptions to the standard filling order? This card answers it → Hund's rule governs same-energy orbitals; Cr and Cu are important exceptions due to extra stability of half-filled and fully-filled d subshells.
Three rules for electron configuration: Aufbau — fill lowest energy first. Pauli — max 2 electrons per orbital, opposite spins (↑↓, never ↑↑). Hund's — for degenerate orbitals, fill each singly before pairing (e.g. 2p³ = ↑ ↑ ↑). Noble-gas notation: [Ar]3d⁶4s² for Fe. Anomalies: Cr = [Ar]3d⁵4s¹ and Cu = [Ar]3d¹⁰4s¹ (half-filled/fully-filled d are extra stable).
Pause — write the highlighted three rules into your book.
Match each rule to what it says about how electrons fill orbitals.
- Aufbau principle
- Pauli exclusion principle
- Hund's rule
- Noble-gas notation
- A single orbital can hold at most 2 electrons, and they must have opposite spins (↑↓)
- Replace the inner electrons with the preceding noble gas in square brackets, e.g. Ca = [Ar]4s²
- Electrons fill the lowest-energy subshell first, then move up the energy ladder
- Within a subshell, electrons singly occupy each orbital before any pairing occurs (e.g. 2p³ → ↑ ↑ ↑)
1. Which is the correct electron configuration for sulfur (Z=16)?
We just saw the three configuration rules and key anomalies. That raises a question: how do you efficiently decode an electron configuration in a multiple-choice question to determine period, group, and block? This card answers it → highest shell number → period; highest shell electron count → group for main-group elements; always verify by summing superscripts to Z.
For multiple choice on electron configurations: superscripts must add to Z (always check). Valence electrons = electrons in the highest-numbered shell for main-group elements. For d-block (e.g. Mn, Z=25 = [Ar]3d⁵4s²): Group = ns + (n−1)d electrons → Group 7. Common trap: 3p holds 6 electrons — never assign more than capacity.
Add the highlighted verification rules to your notes before the check below.
Quick check: which is the correct ground-state electron configuration for sulfur (Z=16)?
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
We just saw how to use configurations to find period and group. That raises a question: how do you write and verify electron configurations in exam answers, especially for exceptions and ions? This card answers it → follow the rules, apply noble-gas notation for efficiency, and handle Cr/Cu exceptions by explaining the extra stability of special d-subshell fillings.
For short-answer configuration questions: write in subshell notation (1s²2s²2p⁶…), use noble-gas shorthand for long configurations, and always verify total = Z. For Cr and Cu exceptions: explain that a half-filled (3d⁵) or fully-filled (3d¹⁰) subshell has extra stability from minimised electron–electron repulsion and maximised exchange energy. For ions: remove 4s electrons first, then 3d.
Pause — copy the highlighted exam strategy into your book before moving on.
Two truths and a lie: pick the false statement about chromium's anomalous configuration.
Worked examples · reveal as you go
Write the full subshell electron configuration and the abbreviated (noble gas core) configuration for: (a) Phosphorus (Z=15) and (b) Iron (Z=26)
Fill in order: 1s → 2s → 2p → 3s → 3p
1s² (2 e⁻; total = 2)
2s² (2 e⁻; total = 4)
2p⁶ (6 e⁻; total = 10)
3s² (2 e⁻; total = 12)
3p³ (3 e⁻ remain; total = 15) ✓
Valence electrons: 3s²3p³ → 5 valence e⁻ → Group 15 ✓
Abbreviated: Preceding noble gas is Ne (Z=10): [Ne]3s²3p³
Fill in order: 1s → 2s → 2p → 3s → 3p → 4s → 3d
1s²2s²2p⁶3s²3p⁶ (18 e⁻ = [Ar] core)
4s² (2 more; total = 20)
3d⁶ (6 more; total = 26) ✓
(Write 3d before 4s in notation even though 4s filled first)
Abbreviated: [Ar]3d⁶4s²
A student writes the electron configuration of chlorine (Z=17) as: 1s²2s²2p⁵3s²3p⁶. Identify all errors and write the correct configuration
Student's config: 1s²2s²2p⁵3s²3p⁶
Total = 2+2+5+2+6 = 17 ✓
Aufbau says: 2p must fill completely before 3s starts
2p holds max 6 electrons
Student has: 2p⁵ (incomplete) and 3p⁶ (full)
But if 3p is full, and we only have 17 total...something is wrong
Cl is Group 17 → should have 5 valence electrons in the p subshell, so 3p⁵
That means 2p must be COMPLETE: 2p⁶
Error identified: Student moved one electron from 2p to 3p too early
Should be 2p⁶ and 3p⁵, not 2p⁵ and 3p⁶
Correct: 1s²2s²2p⁶3s²3p⁵
Check: 2+2+6+2+5 = 17 ✓
Abbreviated: [Ne]3s²3p⁵
Valence electrons: 3s²3p⁵ = 7 → Group 17 (halogens) ✓
Predict the full electron configuration of iron (Fe, Z=26). Write the answer in subshell notation, e.g. 1s²2s²2p⁶...
How close was your prediction?
Common errors · the 3 traps that cost marks
Misconception to fix
Wrong: You should fill 3d before 4s when writing electron configurations.
Misconception to fix
Right: The 4s subshell fills before 3d due to lower energy, but when forming transition metal ions, 4s electrons are removed before 3d electrons. The Aufbau principle gives the filling order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, etc.
Removing 3d electrons first when forming transition-metal ions
When iron forms Fe²⁺, students often remove two 3d electrons because 3d appears written after 4s in shell order. In reality the 4s electrons are removed first — once the subshells are populated, 4s sits at a higher energy than 3d, so 4s electrons leave first. Fe = [Ar]3d⁶4s² but Fe²⁺ = [Ar]3d⁶ (not [Ar]3d⁴4s²).
Fix: For any transition-metal ion, always remove the highest-n s electrons first, then any required d electrons.
Quick-fire practice · 5 reps +2 XP per reveal
How many electrons can the 3d subshell hold in total?
Write the full ground-state electron configuration of phosphorus (Z = 15).
Write the noble-gas-abbreviated configuration for iron (Z = 26).
Draw the orbital box notation for the 2p subshell of nitrogen (Z = 7) and state which rule it illustrates.
Write the configuration of the Fe²⁺ ion, and justify which electrons were removed.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Pick your answer, then rate your confidence — that tells the system what to drill next.
A student writes the ground-state electron configuration of nickel (Ni, Z=28). One line contains an error — click it.
- Z = 28, so 28 electrons to place using Aufbau order
- 1s² 2s² 2p⁶ 3s² 3p⁶ accounts for the first 18 electrons (argon core)
- Remaining 10 electrons → 3d⁶ 4s⁴
- Ni has 2 valence electrons in the 4s subshell, consistent with Group 10 of the d-block
Q1. 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).
Q2. 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.
📖 Comprehensive answers (click to reveal)
Activity 1
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).
️ Activity 2
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).
❓ Multiple Choice
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).
Short Answer Model Answers
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).
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