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Understanding how electrons are arranged in atoms is fundamental to chemistry. Electron configuration determines an element's chemical properties, the types of bonds it forms, and its position in the periodic table. In this lesson, we will build up electron configurations from first principles.
Electrons in atoms occupy energy levels (also called shells), numbered n = 1, 2, 3, 4, and so on. The higher the value of n, the further the electron is from the nucleus and the higher its energy.
Each energy level contains one or more subshells, labelled s, p, d, and f:
| Energy Level (n) | Subshells Available | Total Orbitals | Maximum Electrons |
|---|---|---|---|
| 1 | 1s | 1 | 2 |
| 2 | 2s, 2p | 1 + 3 = 4 | 8 |
| 3 | 3s, 3p, 3d | 1 + 3 + 5 = 9 | 18 |
| 4 | 4s, 4p, 4d, 4f | 1 + 3 + 5 + 7 = 16 | 32 |
Each subshell contains a specific number of orbitals, and each orbital holds a maximum of 2 electrons (with opposite spins):
| Subshell | Number of Orbitals | Maximum Electrons | Shape |
|---|---|---|---|
| s | 1 | 2 | Spherical |
| p | 3 | 6 | Dumbbell |
| d | 5 | 10 | Clover / complex |
| f | 7 | 14 | Very complex |
The general pattern: shell n contains n subshells and n-squared orbitals, holding a maximum of 2n-squared electrons.
Each type of orbital has a characteristic shape:
The shape of an orbital determines the spatial distribution of electron density around the nucleus. This has direct consequences:
Three rules govern how electrons fill orbitals:
Electrons fill orbitals in order of increasing energy. The filling order is:
1s -> 2s -> 2p -> 3s -> 3p -> 4s -> 3d -> 4p -> 5s -> 4d -> 5p -> 6s -> 4f -> 5d -> 6p
graph LR
A[1s] --> B[2s]
B --> C[2p]
C --> D[3s]
D --> E[3p]
E --> F[4s]
F --> G[3d]
G --> H[4p]
H --> I[5s]
I --> J[4d]
J --> K[5p]
K --> L[6s]
L --> M[4f]
M --> N[5d]
N --> O[6p]
Notice that 4s fills before 3d. This is because the 4s subshell is slightly lower in energy than 3d in neutral atoms. However, once 3d electrons are present, 4s becomes slightly higher in energy -- which is why when forming ions, 4s electrons are removed first.
A useful mnemonic: subshells fill in order of increasing (n + l), where n is the principal quantum number and l is the angular momentum quantum number (s=0, p=1, d=2, f=3). If two subshells have the same (n + l), the one with the lower n fills first.
| Subshell | n | l | n + l | Filling Order |
|---|---|---|---|---|
| 1s | 1 | 0 | 1 | 1st |
| 2s | 2 | 0 | 2 | 2nd |
| 2p | 2 | 1 | 3 | 3rd |
| 3s | 3 | 0 | 3 | 4th |
| 3p | 3 | 1 | 4 | 5th |
| 4s | 4 | 0 | 4 | 6th |
| 3d | 3 | 2 | 5 | 7th |
| 4p | 4 | 1 | 5 | 8th |
When filling orbitals of equal energy (degenerate orbitals, e.g., the three 2p orbitals), electrons occupy each orbital singly first, with parallel spins, before pairing up. This minimises electron-electron repulsion.
For example, the three electrons in nitrogen's 2p subshell occupy one each of the three 2p orbitals, all with the same spin: up, up, up
Why parallel spins? Electrons with parallel spins tend to stay further apart (due to the Pauli exclusion principle), reducing electron-electron repulsion and lowering the total energy. This is related to exchange energy -- the quantum mechanical stabilisation that occurs when electrons with parallel spins occupy different orbitals.
No two electrons in the same atom can have the same set of four quantum numbers. In practice, this means each orbital can hold a maximum of 2 electrons, which must have opposite spins (up-down).
Write the subshells in filling order with the number of electrons as a superscript:
| Element | Electrons | Configuration |
|---|---|---|
| Hydrogen | 1 | 1s1 |
| Helium | 2 | 1s2 |
| Lithium | 3 | 1s2 2s1 |
| Carbon | 6 | 1s2 2s2 2p2 |
| Neon | 10 | 1s2 2s2 2p6 |
| Sodium | 11 | 1s2 2s2 2p6 3s1 |
| Argon | 18 | 1s2 2s2 2p6 3s2 3p6 |
| Potassium | 19 | 1s2 2s2 2p6 3s2 3p6 4s1 |
| Calcium | 20 | 1s2 2s2 2p6 3s2 3p6 4s2 |
After calcium, the 3d subshell begins to fill:
| Element | Z | Configuration |
|---|---|---|
| Scandium | 21 | [Ar] 4s2 3d1 |
| Titanium | 22 | [Ar] 4s2 3d2 |
| Vanadium | 23 | [Ar] 4s2 3d3 |
| Chromium | 24 | [Ar] 4s1 3d5 (exception) |
| Manganese | 25 | [Ar] 4s2 3d5 |
| Iron | 26 | [Ar] 4s2 3d6 |
| Cobalt | 27 | [Ar] 4s2 3d7 |
| Nickel | 28 | [Ar] 4s2 3d8 |
| Copper | 29 | [Ar] 4s1 3d10 (exception) |
| Zinc | 30 | [Ar] 4s2 3d10 |
Two important exceptions occur in the first row of transition metals:
Chromium (24):
Copper (29):
In both cases, one electron is "promoted" from 4s to 3d. This produces either a half-filled (Cr: 3d5) or fully filled (Cu: 3d10) d subshell. These arrangements provide extra stability due to the symmetrical distribution of electrons and exchange energy effects.
When atoms form positive ions (cations), electrons are removed in a specific order:
For main group elements: Remove electrons from the highest energy level first.
For transition metals: Remove the 4s electrons before 3d electrons.
Misconception: "4s fills before 3d, so 3d electrons should be removed first when forming ions."
Correction: Although 4s fills before 3d in neutral atoms (because 4s is lower in energy in the absence of d electrons), once the 3d subshell is occupied, the 4s orbital is pushed to a higher energy level. In transition metal ions, 4s is higher in energy than 3d, so 4s electrons are removed first. Think of it this way: filling order is not the same as ionisation order.
When atoms form negative ions (anions), electrons are added:
Ions that have the same electron configuration are said to be isoelectronic. For example, Na+, Mg2+, Al3+, F-, O2-, and Ne all have the configuration 1s2 2s2 2p6 (10 electrons). Isoelectronic species have the same number of electrons but different nuclear charges, which affects their ionic radii.
Long configurations can be abbreviated using the noble gas core:
The periodic table is structured according to electron configurations:
| Block | Subshell Being Filled | Groups |
|---|---|---|
| s-block | s subshell | 1 and 2 |
| p-block | p subshell | 13-18 |
| d-block | d subshell | 3-12 |
| f-block | f subshell | Lanthanides / Actinides |
The group number tells you the number of outer-shell electrons:
The period number tells you the highest occupied principal energy level.
Edexcel 9CH0 specification Topic 1, sub-topic 1.2 covers the existence of s, p, d and f orbitals; the shapes of s and p orbitals; the relative energies of subshells in atoms up to Z = 36 (krypton); the rules for filling orbitals (aufbau, Hund, Pauli); the use of electron-in-box notation; and electron configurations of atoms and ions for the first 36 elements, including the anomalous configurations of chromium and copper and the order of removal of 4s before 3d for d-block ion formation (refer to the official Pearson Edexcel specification document for exact wording). The content is examined in Paper 1 (Advanced Inorganic and Physical Chemistry) and re-emerges in Paper 1 Topic 15 (Transition metals), where d-block ion configurations underpin colour, catalysis and complex-ion chemistry. Synoptic appearances in Paper 2 are limited but real — periodic trends in alkene reactivity, for example, trace back to s/p valence-electron counts.
Question (7 marks):
(a) Write the full electron configuration of (i) chromium (Cr, Z = 24), and (ii) the iron(III) ion, Fe³⁺ (Z of Fe = 26). (2) (b) Explain, in terms of orbital stability, why the configuration of chromium is not [Ar] 3d⁴ 4s². (3) (c) State and explain which 3d-block ion has electron configuration [Ar] 3d⁶. (2)
Solution with mark scheme:
(a) (i) B1 — Cr is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹. Common error: writing 3d⁴ 4s² for Cr (the "expected" but wrong configuration).
(ii) B1 — Fe³⁺ is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵. Common error: removing electrons from 3d before 4s. The 4s electrons are removed first on ionisation of d-block atoms, then 3d. So Fe is [Ar] 3d⁶ 4s²; lose 4s² first (→ Fe²⁺ = [Ar] 3d⁶); lose one more from 3d (→ Fe³⁺ = [Ar] 3d⁵).
(b) M1 — a half-filled 3d subshell (3d⁵) has lower energy than 3d⁴ because of the symmetric distribution of electron density and the favourable exchange energy among five parallel-spin electrons. M1 — promoting one 4s electron to 3d in chromium gives a half-filled 3d subshell and a half-filled 4s subshell simultaneously. A1 — the energy gain from achieving these two half-filled subshells outweighs the small energy cost of the 4s → 3d promotion. Common error: candidates write "3d⁵ is more stable" without explaining why (exchange energy / symmetric repulsion) and lose the M2 mark.
(c) B1 — [Ar] 3d⁶ corresponds to the loss of two electrons from iron (Fe = [Ar] 3d⁶ 4s², lose the 4s² to give [Ar] 3d⁶). B1 — therefore the ion is Fe²⁺.
Total: 7 marks (B2 + M2 A1 + B2).
Question (6 marks): Aluminium (Z = 13) and sulfur (Z = 16) are both period 3 elements.
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