Electron Configuration Evidence Revision Notes | A Level AQA Chemistry

1. Why Ionisation Energy Is the Key to Unlocking Electron Structure

You cannot see inside an atom. Yet chemists in the early twentieth century worked out the precise arrangement of electrons — not by looking, but by measuring the energy needed to remove them. That measurement is ionisation energy, and it is one of the most powerful pieces of evidence for the shell and sub-shell model you need to know for AQA.

The core idea is this: electrons that are harder to remove are closer to the nucleus or in a filled sub-shell. Electrons that are easier to remove are further away or in a less stable arrangement. Every dip, rise, or sudden jump in ionisation energy data is telling you something precise about electron configuration.

Notice the equation uses gaseous atoms. This is a standard condition you must state in an exam — the process must start with isolated gaseous atoms, not a solid or liquid. Examiners deduct marks when students omit state symbols or write the equation for a solid.

2. Electron Configurations: Shells, Sub-Shells, and the Rules

Before you can interpret ionisation energy data, you need to be confident writing electron configurations. The AQA specification requires you to write configurations for atoms and ions up to Z = 36 (krypton) using s, p, and d notation.

Electrons fill sub-shells in order of increasing energy: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p. The 4s sub-shell fills before the 3d sub-shell because it is at a lower energy level in neutral atoms. This ordering is the source of one of the most common errors at A-level.

• s sub-shells: 1 orbital, maximum 2 electrons
• p sub-shells: 3 orbitals, maximum 6 electrons
• d sub-shells: 5 orbitals, maximum 10 electrons
• Electrons occupy orbitals singly before pairing (Hund's rule)
• The 4s sub-shell fills before 3d in neutral atoms

Worked example: Writing Electron Configurations for Atoms up to Z = 36

Write the full electron configurations for: (a) chlorine (Z = 17), (b) iron (Z = 26), (c) copper (Z = 29).

1. 1s² 2s² 2p⁶ 3s² 3p⁵ — Cl has 17 electrons. After filling 1s (2), 2s (2), 2p (6), 3s (2), there are 5 electrons left for 3p. The 3p sub-shell can hold 6, so it is one short of full.
2. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² — Fe has 26 electrons. After filling up to 3p (18 electrons total), 4s fills next (2 electrons, total 20), then 3d takes the remaining 6 electrons. The configuration is written in shell order (3d before 4s), which is the standard convention. AQA accepts both shell order (3d⁶ 4s²) and filling order (4s² 3d⁶).
3. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ — Cu (Z = 29) has an anomalous configuration. A fully filled 3d sub-shell (3d¹⁰) combined with a half-filled 4s (4s¹) is more stable than the expected 3d⁹ 4s². This is a well-known exception you must memorise.
4. Cl: 2+2+6+2+5 = 17 ✓ Fe: 2+2+6+2+6+6+2 = 26 ✓ Cu: 2+2+6+2+6+10+1 = 29 ✓ — Always sum the superscripts to verify the total equals the atomic number. This is a quick self-check that takes five seconds and prevents dropped marks.

Answer: Cl: 1s² 2s² 2p⁶ 3s² 3p⁵ | Fe: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² | Cu: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹

Ions require one extra step: add electrons for negative ions, remove them for positive ions. Crucially, when transition metal atoms form positive ions, electrons are removed from the 4s sub-shell first, not from 3d. This is because once the 3d sub-shell begins to fill, 3d electrons become lower in energy than 4s electrons in the resulting ion.

Worked example: Writing Electron Configurations for Ions

Write the electron configurations for: (a) Fe²⁺, (b) Fe³⁺, (c) Cl⁻.

1. Fe: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶ — Always write the neutral atom configuration first, then adjust for charge.
2. Fe²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ — Remove both 4s electrons. The 3d electrons are not touched for Fe²⁺. Total electrons: 2+2+6+2+6+6 = 24 = 26 − 2 ✓
3. Fe³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ — After removing both 4s electrons, one 3d electron is also removed. A half-filled 3d⁵ is particularly stable. Total electrons: 2+2+6+2+6+5 = 23 = 26 − 3 ✓
4. Cl⁻: 1s² 2s² 2p⁶ 3s² 3p⁶ — Chlorine gains one electron to complete its 3p sub-shell. Total electrons: 2+2+6+2+6 = 18 ✓ — this is isoelectronic with argon.

Answer: Fe²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ | Fe³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ | Cl⁻: 1s² 2s² 2p⁶ 3s² 3p⁶

3. Successive Ionisation Energies: Evidence for Shell Structure

Now that you can write electron configurations, you can use them to interpret ionisation energy data. Successive ionisation energies — the energies to remove the 1st, 2nd, 3rd electron and so on from the same atom — are the clearest evidence for the existence of distinct electron shells.

The pattern is straightforward: as you remove more electrons from an atom, the remaining electrons experience a greater effective nuclear charge (the same nucleus, but fewer electrons shielding it). So each successive ionisation energy is generally larger than the previous one. The dramatic part comes when you cross a shell boundary.

Look at the sodium data. There is a large jump between the 1st and 2nd ionisation energies — from 496 to 4562 kJ mol⁻¹. This tells you that the 1st electron is in an outer shell (n = 3), while the 2nd electron is in a much closer shell (n = 2) with far less shielding and much stronger attraction to the nucleus. There is a second enormous jump between the 9th and 10th ionisation energies, indicating the transition from shell 2 to shell 1.

This is how you use successive ionisation energy data to identify which group an element belongs to. The position of the first large jump tells you how many electrons are in the outer shell — and therefore the group number. For sodium, the jump after the 1st electron confirms it is in Group 1.

4. First Ionisation Energies Across Period 3: Evidence for Sub-Shells

Successive ionisation energies reveal shell boundaries. First ionisation energies across a period reveal something finer: the existence of sub-shells within those shells. The trend across Period 3 (Na to Ar) is broadly increasing, but with two notable dips that are the key evidence for sub-shell structure.

The general increase from Na to Ar makes sense: across a period, nuclear charge increases while electrons are added to the same shell (n = 3), so shielding stays roughly constant and the effective nuclear charge experienced by outer electrons increases. Higher effective nuclear charge means more energy is needed to remove an electron.

Now look at the two dips. The first is at aluminium (Al), which has a lower first ionisation energy than magnesium (Mg) despite having a higher nuclear charge. The second is at sulphur (S), which has a lower first ionisation energy than phosphorus (P). Each dip is direct evidence for sub-shell structure.

4a. The Mg–Al Dip: Evidence for the 3p Sub-Shell

Magnesium has the configuration [Ne] 3s². Its outer electron is in the 3s sub-shell. Aluminium has the configuration [Ne] 3s² 3p¹. Its outer electron is in the 3p sub-shell, which is at a higher energy than 3s and is also slightly shielded by the 3s electrons. The combination of higher energy and additional shielding means the 3p electron in Al is easier to remove than the 3s electron in Mg, despite Al having a greater nuclear charge.

This dip is the experimental evidence that the 3p sub-shell exists and is at a higher energy than the 3s sub-shell. Without sub-shells, you would expect a smooth increase from Na to Ar — the dip at Al tells you the model of a single n = 3 shell is insufficient.

4b. The P–S Dip: Evidence for Electron Pairing in 3p Orbitals

Phosphorus has the configuration [Ne] 3s² 3p³. By Hund's rule, each of the three 3p orbitals contains one electron — none are paired. Sulphur has the configuration [Ne] 3s² 3p⁴, meaning one 3p orbital now contains a paired electron. That paired electron experiences extra repulsion from its orbital partner, making it easier to remove than the unpaired 3p electrons in phosphorus.

The dip at sulphur is therefore evidence that electrons occupy orbitals singly before pairing, and that electron–electron repulsion within a shared orbital lowers the ionisation energy. This is direct experimental support for Hund's rule and the existence of three separate 3p orbitals.

5. First Ionisation Energies Down Group 2: Evidence for Increasing Shell Number

Moving from Period 3 to a group trend, the first ionisation energies of Group 2 elements (Be to Ba) decrease steadily as you go down the group. This trend provides evidence that successive elements have their outer electrons in shells that are progressively further from the nucleus.

As you descend Group 2, each element has its outer 2 electrons in a shell with a higher principal quantum number (n). These outer electrons are further from the nucleus and are shielded by an increasing number of inner electron shells. The combined effect — greater distance and greater shielding — reduces the effective nuclear charge experienced by the outer electrons, so less energy is needed to remove them.

The fact that the decrease is broadly gradual rather than showing sudden jumps is itself significant. It confirms that each successive Group 2 element has its outer electrons in a new, higher principal quantum shell. However, the decrease is not perfectly uniform — the step from Ca to Sr is smaller in part because the intervening 3d transition elements add extra shielding, making the pattern slightly less regular. Overall, the trend still provides strong experimental support for the shell model: each period corresponds to filling a new principal quantum shell.

6. Putting It Together: Interpreting Ionisation Energy Data in Exam Questions

Exam questions on this topic typically ask you to either explain a trend or identify an element from data. Both require the same underlying reasoning chain: nuclear charge, distance, shielding, effective nuclear charge, and energy required to remove an electron. Here is how to structure a high-scoring response.

Worked example: Exam-Style: Explaining a Trend in First Ionisation Energy

Explain why the first ionisation energy of aluminium (Al) is lower than that of magnesium (Mg). [3 marks]

1. Mg: 1s² 2s² 2p⁶ 3s² Al: 1s² 2s² 2p⁶ 3s² 3p¹ — Establishing the configurations shows the examiner you know which sub-shell the outer electron occupies in each case. [B1]
2. Mg outer electron: 3s sub-shell Al outer electron: 3p sub-shell — The 3p sub-shell is at a higher energy than the 3s sub-shell. [M1]
3. The 3p electron in Al is also shielded by the 3s² electrons, in addition to all inner shells. — Extra shielding reduces the effective nuclear charge experienced by the 3p electron. [M1]
4. Therefore, less energy is required to remove the outer electron from Al than from Mg, despite Al having a higher nuclear charge. — Linking back to the question with a clear conclusion secures the final mark. [A1]

Answer: Al has a lower first IE than Mg because Al's outer electron is in the 3p sub-shell (higher energy, more shielded by 3s electrons), whereas Mg's outer electron is in the 3s sub-shell. The 3p electron experiences a lower effective nuclear charge and requires less energy to remove.

The same reasoning structure applies to the P–S dip: identify the configurations, note the paired electron in sulphur's 3p sub-shell, explain the electron–electron repulsion, and conclude that this lowers the energy needed for removal. Practise writing this out in full sentences — the marks are for the reasoning, not just the conclusion.

7. Exam Technique and Common Pitfalls

The most reliable way to lose marks on this topic is to give incomplete explanations. AQA mark schemes for ionisation energy questions almost always require three linked ideas: the location of the electron (which shell or sub-shell), the reason it is easier or harder to remove (distance, shielding, repulsion), and the consequence for ionisation energy. Missing any one of these typically costs a mark.

For electron configuration questions, the two most penalised errors are: writing 3d before 4s when filling (the 4s fills first in neutral atoms), and removing 3d electrons before 4s electrons when forming transition metal cations (4s is always removed first). These are tested repeatedly in AQA papers and are worth drilling until they are automatic.