1. First ionisation energy
Ionisation energies are always positive (endothermic), because energy is needed to pull an electron away from the attraction of the nucleus. The units are kJ mol−1. Always include the (g) state symbols, even for elements that are normally solids.
2. Successive ionisation energies
Each successive ionisation removes one more electron from each ion:
| Ionisation | Equation |
|---|---|
| 1st | Mg(g) → Mg+(g) + e− |
| 2nd | Mg+(g) → Mg2+(g) + e− |
| 3rd | Mg2+(g) → Mg3+(g) + e− |
The second ionisation energy is not Mg(g) → Mg2+(g) + 2e−. That equation describes two ionisations added together. Check that charges balance: the left side's charge equals the right side's total.
3. What affects ionisation energy
Ionisation energy depends on how strongly the nucleus attracts the electron being removed. Three factors matter:
- Nuclear charge: more protons give a stronger attraction.
- Distance: an electron further from the nucleus (in a higher shell) is attracted less strongly.
- Shielding: inner-shell electrons repel the outer electrons, reducing the attraction they feel.
Explain trends using these factors and the attraction between the nucleus and the electron. "Full shells are stable" on its own does not explain anything.
4. Trend down Group 2
First ionisation energy decreases from Be to Ba.
Refer specifically to the outer electron.
5. Trend across Period 3
There is a general increase from Na to Ar.
The distance isn't exactly the same across a period: the radius gets smaller, which strengthens the attraction even more.
Diagram placeholder
First ionisation energy across Period 3
Labels to include:
- x-axis: element Na → Ar
- y-axis: first IE / kJ mol⁻¹
- Na lowest; Ar highest
- Drop from Mg to Al
- Drop from P to S
The graph shows a general rise from sodium (lowest) to argon (highest), with two small drops: between Mg and Al, and between P and S.
6. The Mg → Al and P → S drops
| Element | Configuration | Electron removed |
|---|---|---|
| Mg | [Ne] 3s2 | 3s |
| Al | [Ne] 3s2 3p1 | 3p (higher energy) |
| P | [Ne] 3s2 3p3 | 3p: three single electrons |
| S | [Ne] 3s2 3p4 | 3p: one orbital contains a pair |
The same patterns explain the drops in Period 2 from Be to B and from N to O. Helium has the highest first IE of all: its electron is in 1s, very close to the nucleus with no inner shielding, and it has more protons than hydrogen.
7. Why successive ionisation energies rise
Each electron is removed from an increasingly positive ion. The nucleus stays the same, but there are fewer electrons to repel each other and the ion gets smaller. This means the remaining electrons are attracted more strongly. The number of protons does not change.
8. Big jumps and electron shells
For a main-group atom, if the first large jump comes after k electrons have been removed, the element has k outer-shell electrons: that jump marks the change from removing outer-shell electrons to starting on an inner shell. Later jumps reflect deeper shells, so use only the first. The next electron comes from an inner shell, which is closer to the nucleus, lower in energy and less shielded, so it is attracted much more strongly.
Worked example (illustrative data)
| Ionisation | 1st | 2nd | 3rd | 4th |
|---|---|---|---|---|
| IE / kJ mol⁻¹ | 580 | 1820 | 2750 | 11 600 |
The large jump between the 3rd and 4th IEs means there are 3 outer electrons, so the element is in Group 13 (aluminium). The 4th electron is removed from the 2p sub-shell, which is closer to the nucleus and less shielded than 3s/3p. Note that this is Group 13, the old "Group 3", not the modern transition-metal Group 3.
Diagram placeholder
Successive ionisation energies with a shell jump
Labels to include:
- x-axis: number of electrons removed
- y-axis: IE / kJ mol⁻¹ (or log IE)
- Gradual rise within a shell
- Large jump where an inner shell is reached
9. Second ionisation energy across elements
This is different from successive IEs of a single element. When comparing the second IE of different elements, look at the configuration of the 1+ ion you are removing from:
- Na+ is [Ne]. Its second electron comes from 2p, an inner shell close to the nucleus, so sodium's second IE is unusually large.
- Mg+ is [Ne] 3s1. Its second electron comes from 3s, so magnesium's second IE is much lower than sodium's.
Work out each ion's configuration rather than just shifting the first-IE graph along by one element.
Diagram placeholder
Optional: second IE across Period 3
Labels to include:
- x-axis: element Na → Ar
- y-axis: second IE / kJ mol⁻¹
- Na as the peak
Only the shape is needed here: sodium's second IE is the highest, then it falls sharply at magnesium. This placeholder deliberately shows no data values.
10. Finesse practice
Original Finesse practice questions, not AQA past-paper questions.
Q1. Write the equation for the third ionisation energy of calcium.Show answer
Ca2+(g) → Ca3+(g) + e−
Q2. Explain why sulfur has a lower first ionisation energy than phosphorus.Show answer
Sulfur has a pair of electrons in one 3p orbital. The repulsion between them makes one easier to remove. Phosphorus's three 3p electrons are all unpaired.
Q3. A main-group element shows its first large jump between the sixth and seventh ionisation energies. What does this tell you?Show answer
It has 6 outer-shell electrons, so it is in Group 16. The 7th electron is in an inner shell that is closer to the nucleus and less shielded.
Q4. Why does first IE decrease from magnesium to calcium?Show answer
Calcium's outer electron is in the 4s sub-shell, which is further from the nucleus with more shielding. This outweighs the extra nuclear charge, so the attraction is weaker.
Q5. Why is a second ionisation energy higher than the first?Show answer
The second electron is removed from a positive ion. There are fewer electrons but the same number of protons, so the remaining electrons are attracted more strongly.
11. Atomic structure topic checklist
- Describe how the model of the atom developed, and interpret alpha-scattering evidence
- Give the relative mass, charge and location of protons, neutrons and electrons
- Calculate p, n and e for atoms and ions using A, Z and charge
- Define isotopes, and compare the particles in neutral isotopes
- Explain why isotopes have the same chemical properties
- Describe the four stages of TOF mass spectrometry, including both ionisation methods
- Use KE = ½mv² and v = d/t, with mass in kg per ion
- Define Ar and calculate it from spectra or abundances; find missing abundances
- Write configurations up to Kr, including Cr, Cu and ions
- Define first and successive IEs with correct equations and state symbols
- Explain the trends down Group 2 and across Period 3, including the drops
- Use jumps in successive IEs to work out an element's group
12. Sources
Sources and examiner guidance (reviewed 1 October 2026)
- Chemrevise — AQA 1.1 Atomic Structure revision guide (N. Goalby) — primary reference, pp8–10 (ionisation energies)
- AQA 7405 specification — 3.1.1 Atomic structure — 3.1.1.3 Ionisation energies
- AQA 7404/1 mark scheme, June 2022 — Q01.1–01.3, p11: Group 2 trend, third IE equation
- AQA 7404/1 examiner report, June 2022 — Q1, p3: refer to the outer electron, include gas states
- AQA 7404/1 mark scheme, June 2023 — Q01.1–01.3, p11: gas states; jump after the 6th electron
- AQA 7404/1 examiner report, June 2023 — Q1, p3: missing inner-shell reasoning
- AQA 7404/1 mark scheme, 2020 series (November archive; footer June 2020) — Q01.1, p11: Al higher-energy 3p vs 3s; S paired 3p repulsion; configurations alone insufficient
- AQA 7405/1 mark scheme, June 2023 — Q05.6, p19: successive IEs and attraction
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.
