# Atomic Structure and Electron Configuration
At A-Level, we move beyond the simple 2, 8, 8 model and explore the quantum mechanical description of electron arrangement. Understanding subshells, orbitals, and ionisation energies is essential for explaining periodic trends, chemical bonding, and spectroscopy.
1. Subshells and Orbitals
Electrons are arranged in energy levels (shells), which are further divided into subshells, and each subshell contains orbitals.
The Four Types of Subshell
| Subshell | Number of Orbitals | Max Electrons |
|---|---|---|
| s | 1 | 2 |
| p | 3 | 6 |
| d | 5 | 10 |
| f | 7 | 14 |
Each orbital holds a maximum of 2 electrons with opposite spins (Pauli exclusion principle).
Energy Level and Subshell Structure
| Shell (n) | Subshells Available | Total Electrons |
|---|---|---|
| 1 | 1s | 2 |
| 2 | 2s, 2p | 8 |
| 3 | 3s, 3p, 3d | 18 |
| 4 | 4s, 4p, 4d, 4f | 32 |
Filling Order
Subshells fill in order of increasing energy (not always by shell number):
Key point: 4s fills before 3d because 4s is lower in energy.
2. Rules for Filling Orbitals
Aufbau Principle
Electrons fill orbitals starting from the lowest energy level available.
Pauli Exclusion Principle
Each orbital holds a maximum of 2 electrons with opposite spins (↑↓).
Hund's Rule
Electrons occupy orbitals in a subshell singly first (with parallel spins) before pairing up.
For example, the 2p subshell with 3 electrons: ↑ ↑ ↑ (one in each orbital) With 4 electrons: ↑↓ ↑ ↑ (pair in first, single in remaining two)
3. Writing Electron Configurations
Notation
Full notation shows each subshell and the number of electrons:
| Element | Z | Configuration |
|---|---|---|
| H | 1 | |
| He | 2 | |
| Li | 3 | |
| C | 6 | |
| Ne | 10 | |
| Na | 11 | |
| Ar | 18 | |
| K | 19 | |
| Ca | 20 | |
| Sc | 21 | |
| Fe | 26 | |
| Zn | 30 |
Exceptions
Chromium (Cr, Z=24): Expected , actual:
Copper (Cu, Z=29): Expected , actual:
Reason: Half-filled and fully-filled d subshells are particularly stable.
Ion Configurations
When forming ions, 4s electrons are lost before 3d electrons.
- Fe:
- Fe²⁺: (loses 2 × 4s electrons)
- Fe³⁺: (loses 2 × 4s + 1 × 3d)
4. Ionisation Energy
The first ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms:
The second ionisation energy:
Successive ionisation energies always increase because:
- Each electron is removed from an increasingly positive ion
- Greater nuclear attraction must be overcome
Factors Affecting Ionisation Energy
- Nuclear charge: More protons → stronger attraction → higher IE
- Atomic radius: Further from nucleus → weaker attraction → lower IE
- Shielding: More inner electrons → more shielding → lower IE
- Subshell: Paired electrons are slightly easier to remove (electron-electron repulsion)
Trends
Across a period: IE generally increases (more protons, same shielding)
Down a group: IE decreases (more shells, greater distance, more shielding)
Anomalies in Period 3
-
Be → B drop: B has ; the 2p electron is easier to remove than a 2s electron (2p is higher energy, less penetrating)
-
N → O drop: O has ; the extra electron in the 2p is paired, and electron-electron repulsion makes it easier to remove
5. Successive Ionisation Energies
A graph of successive IEs for an element shows jumps that indicate the electron shell structure.
For sodium ():
- 1st IE: relatively low (3s¹ electron)
- Big jump to 2nd IE (removing from 2p⁶ — inner shell, closer to nucleus)
- IEs 2–9 increase gradually (removing 2s²2p⁶)
- Another big jump at 10th IE (removing from 1s²)
The pattern of jumps tells you the number of electrons in each shell.
6. Emission Spectra
When atoms are excited (heated or given electrical energy), electrons jump to higher energy levels. When they fall back, they emit photons of specific energies.
where J·s, m/s
The line spectrum of an element (discrete lines at specific wavelengths) provides evidence for quantised energy levels.
Worked Example: Electron Configuration
Question: Write the electron configuration of V (Z=23) and V³⁺.
- V:
- V³⁺: Remove 2 × 4s then 1 × 3d:
Worked Example: Explaining IE Anomaly
Question: Explain why the first ionisation energy of oxygen is lower than that of nitrogen.
Nitrogen has configuration (half-filled 2p — extra stability). Oxygen has — the fourth 2p electron must be paired in an orbital already containing an electron. The electron-electron repulsion in this pair makes the electron easier to remove, so the IE of oxygen is lower.
Worked Example: Successive IEs
Question: The successive IEs (kJ/mol) of element X are: 578, 1817, 2745, 11578, 14831... Deduce which group X is in.
There is a large jump between the 3rd and 4th IEs. This means 3 electrons are relatively easy to remove (outer shell), then the 4th is much harder (inner shell). Element X has 3 outer electrons → Group 3.
8. Practice Questions
- Write the full electron configuration of Mn (Z=25) and Mn²⁺.
- Explain why 4s fills before 3d but 4s empties before 3d when forming ions.
- Sketch how first ionisation energy varies across Period 3 (Na to Ar). Explain the general trend and any anomalies.
- The successive IEs of an element are: 496, 4562, 6912, 9544, 13353... Identify the group.
- Explain why chromium has the configuration rather than .
Want to check your answers and get step-by-step solutions?
9. Exam Tips
- Always write configurations using the subshell notation ( etc.)
- For ions: remove 4s electrons before 3d electrons
- Know the Cr and Cu exceptions and be able to explain them
- In IE questions, identify where the big jumps are
- For anomalies across a period, mention subshell energy (s vs p) and electron pairing
Summary
- Electrons occupy subshells (s, p, d, f) within energy levels
- Filling order:
- Aufbau, Pauli, and Hund's rule govern filling
- Ionisation energy increases across a period (with anomalies) and decreases down a group
- Successive IE jumps reveal electron shell structure
- Cr and Cu are exceptions ( and )
