UNIT 2 CHEMISTRY · PEARSON AAQ · A1.3.3
Ionisation Energy Trends
FREE CHEMISTRY REVISION · TOPIC 9
Explain the general trend—and the dips.
Connect first ionisation energy to nuclear attraction, shielding, shell distance and subshell occupancy, then use those ideas to interpret real trend graphs.
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · A1.3.3
What you will learn
- Explain the general fall in first ionisation energy down a group.
- Explain the general rise across a period using nuclear charge and shielding.
- Interpret the Mg → Al and P → S exceptions, with equivalent Period 2 examples.
- Read trend graphs without confusing them with successive ionisation-energy graphs.
- Build comparisons that state which electron is being removed and why its attraction differs.
A1.3.3 specifies nuclear charge, number of shells, shielding and subshells as factors affecting ionisation-energy trends down groups and across periods.
1. Compare the same process for different elements
The first ionisation energy is the energy needed to remove one electron from each atom in one mole of gaseous atoms, forming gaseous 1+ ions. For every element on a first-ionisation trend graph, the starting species is a neutral gaseous atom:
X(g) → X⁺(g) + e⁻
Across a period or down a group, compare this first removal for different elements. A successive-ionisation graph instead compares first, second, third and later removals from one element. Read the horizontal axis before explaining a pattern.
Ionisation requires an energy input to overcome attraction between the negative electron and the positive nucleus. A more strongly bound electron generally requires a higher ionisation energy.
2. Four factors that control attraction
| Factor | Meaning | Effect, with other factors comparable |
|---|---|---|
| Nuclear charge | Positive charge of the nucleus; determined by proton number Z. | More protons increase electron–nucleus attraction. |
| Number of occupied shells | The principal shell occupied by the electron being removed. | A higher outer shell generally puts the electron farther from the nucleus and makes removal easier. |
| Shielding | Other electrons reduce the net nuclear attraction experienced by an electron. | More effective shielding reduces attraction. Inner-shell electrons shield strongly. |
| Subshell and occupancy | s, p or d environment and whether an orbital contains a pair. | Subshell energy, penetration and electron repulsion can produce departures from the simple trend. |
Effective nuclear attraction refers to the attraction experienced after accounting for other electrons. It is not identical to nuclear charge. Two atoms can have different proton numbers but the atom with more protons can still have a lower first ionisation energy if its outer electron is much farther away and more shielded.
Electron shielding is an electrostatic effect, not a physical wall. Electrons in the same shell also interact and shield imperfectly; “shielding is exactly constant across a period” is too strong. Across a main-group period, the inner-shell shielding is broadly similar and added same-shell electrons do not fully offset the extra nuclear charge.
3. Down a group: the outer electron is less strongly held
First ionisation energy generally decreases down a group. The outer electrons occupy higher principal shells, so the electron removed is farther from the nucleus. There are more inner electrons and more shielding.
Although nuclear charge increases, the increased distance and shielding outweigh its effect on the outer electron. The outer electron is therefore less strongly attracted and requires less energy to remove.
Down a group: more occupied shells → greater outer-electron distance and shielding → weaker attraction → lower first ionisation energy.
Group 1 atoms all have an outer ns¹ electron, but that electron is in 2s for Li, 3s for Na, 4s for K, 5s for Rb and 6s for Cs. This graph does not show five successive removals from lithium.
Use “generally” when discussing the entire periodic table. Detailed d- and f-block effects and heavy-element effects can complicate simple trends. The main-group comparisons above isolate the shell and shielding explanation.
4. Across a period: increasing attraction, with exceptions
First ionisation energy generally increases across a main-group period. Proton number rises while electrons are added to the same principal shell. Inner-shell shielding is broadly similar, so the effective nuclear attraction increases. Atomic radius generally decreases, and more energy is needed to remove an outer electron.
Across a period: greater nuclear charge + broadly similar inner shielding + same outer principal shell → stronger attraction → generally higher first ionisation energy.
| Na | Mg | Al | Si | P | S | Cl | Ar |
|---|---|---|---|---|---|---|---|
| 496 | 738 | 578 | 787 | 1012 | 1000 | 1251 | 1521 |
The broad increase is the trend. The Mg/Al and P/S comparisons require an additional subshell or pairing explanation. A graph that simply rises at every point would lose important electronic-structure evidence.
5. The Mg → Al dip: a different subshell
Magnesium is [Ne] 3s², while aluminium is [Ne] 3s² 3p¹. Mg loses a 3s electron; Al loses its single 3p electron. Both are in principal shell n = 3, but they are not in the same subshell.
The 3p electron is higher in energy and less penetrating than a 3s electron: it is more effectively shielded and less strongly bound. This effect outweighs the increase in nuclear charge from Mg to Al, so aluminium’s first ionisation energy is lower.
The equivalent drop in Period 2 is Be → B: Be loses a 2s electron, while B loses a higher-energy 2p electron. Do not explain Mg → Al by saying aluminium has an extra principal shell: it does not.
6. The P → S dip: repulsion within a paired orbital
Phosphorus is [Ne] 3s² 3p³. Its three 3p electrons occupy separate orbitals with parallel spins. Sulfur is [Ne] 3s² 3p⁴: one 3p orbital contains a pair.
Repulsion between the two electrons in sulfur’s paired orbital makes one of them easier to remove. This pairing effect outweighs the additional nuclear charge for this comparison, producing the small dip from P to S.
Both electrons being compared are removed from 3p; the distinction is pairing within an orbital, not a new principal shell or a switch from s to p. The analogous Period 2 drop is N → O (2p³ → 2p⁴).
7. Worked examples: build precise comparisons
Example 1 · Why is potassium lower than sodium?
- Na is [Ne] 3s¹; K is [Ar] 4s¹.
- K has more protons, but its first electron is removed from a higher shell.
- The 4s electron is farther from the nucleus and more shielded by inner electrons.
- These effects outweigh the greater nuclear charge, so K needs less energy: about 419 versus 496 kJ mol⁻¹.
Write the comparison explicitly: “Potassium’s outer electron is less strongly attracted than sodium’s,” rather than only “potassium is bigger.”
Example 2 · Explain the overall Na → Cl rise
- Both atoms lose an electron from principal shell n = 3.
- Proton number increases from 11 to 17; both retain a [Ne] inner core.
- Added same-shell electrons do not fully shield the extra nuclear charge.
- Stronger attraction and a generally smaller radius make Cl’s first removal harder: about 1251 compared with 496 kJ mol⁻¹.
This is an overall comparison, not a claim that every intermediate point rises.
Example 3 · Explain and quantify Mg → Al
- Mg loses a 3s electron; Al loses a 3p electron.
- The 3p electron is higher in energy, less penetrating and more shielded than 3s.
- It is easier to remove despite Al’s greater nuclear charge.
- The rounded decrease is 738 − 578 = 160 kJ mol⁻¹.
The numerical difference describes the dip; the subshell comparison explains it.
Example 4 · Explain the small P → S drop
- P has three singly occupied 3p orbitals; S has a 3p pair.
- Remove one electron from sulfur’s paired orbital.
- Repulsion in the pair reduces the energy needed relative to P’s singly occupied 3p orbitals.
- This produces a small decrease: 1012 − 1000 = 12 kJ mol⁻¹ using the rounded graph values.
Do not use the s-to-p explanation here: both removals involve p electrons.
Example 5 · Identify the graph and choose an explanation
A graph plots first ionisation energy for Na, Mg, Al, Si, P, S, Cl and Ar.
- The horizontal axis lists different elements, so this is a periodic trend, not successive removal.
- Identify the broad rise and the two dips.
- Use increasing effective nuclear attraction for the general rise.
- Use 3s/3p occupancy for Mg → Al and paired-electron repulsion for P → S.
A single sentence such as “more protons means higher energy” cannot explain the entire graph.
8. Common misconceptions and exam tips
- “More protons always means higher first ionisation energy.” Compare distance, shielding and subshell occupancy too.
- “Down a group, nuclear charge decreases.” It increases; increased shielding and distance dominate the outer-electron comparison.
- “Every point rises across a period.” Explain the s-to-p and paired-p exceptions.
- “Al has more shells than Mg.” Both have occupied shells n = 1, 2 and 3.
- “P → S is an s-to-p change.” Both lose 3p electrons; pairing is the key difference.
- “A first-ionisation graph shows electrons removed one by one from a single atom.” Check whether the x axis lists elements or ionisation stages.
Exam method: identify the electron → compare nuclear charge → compare shell distance and shielding → check subshell and pairing → state attraction and energy consequence.
Use named species and comparative language. Explain which effects dominate rather than listing factors without connecting them to the result.
Your revision summary
Down groups, extra shells and shielding generally lower first ionisation energy. Across main-group periods, increasing nuclear charge and broadly similar inner shielding generally raise it. Higher-energy p occupancy explains Mg → Al and Be → B; paired-p repulsion explains P → S and N → O.
Graph values are rounded to whole kJ mol⁻¹ from NIST first-ionisation energies, using 1 eV per particle ≈ 96.485 kJ mol⁻¹. NIST periodic table and ionisation-energy data (PDF). Graphs and orbital diagrams are original BTECRevision representations.
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