Topic 4 of 4
Ionisation energies as evidence
Explain an energy change and use large jumps to reconstruct an atom.
A-Level 9476 (2026-2027)
Explain the attraction to the electron being removed
Nuclear charge, distance, shielding and pairing compete.
First ionisation energy is the energy needed to remove one mole of electrons from one mole of gaseous atoms, forming one mole of gaseous singly positive ions: X(g) → X+(g) + e-. It is positive because energy is needed to separate an electron from the attraction of the nucleus. The gas state prevents lattice or intermolecular effects being included.
| Change | Typical effect on ionisation energy | Reason |
|---|---|---|
| Greater nuclear charge, similar shielding and distance | Increases | The electron is attracted more strongly. |
| An outer electron in a more distant shell | Decreases | Greater distance and extra inner-shell shielding weaken attraction. |
| More shielding between nucleus and electron | Decreases | Inner electrons reduce the nuclear attraction felt by the outer electron. |
| Electron removed from a higher-energy subshell | Often decreases | That electron is less tightly bound; compare B 2p with Be 2s. |
| Pairing in one orbital | Can decrease | Repulsion between paired electrons assists removal; compare O with N. |
Across a period, proton number rises while added electrons enter the same principal shell. Shielding does not increase enough to cancel the increased nuclear charge, so the general trend is increasing ionisation energy. Down a group, extra shells increase distance and shielding, usually outweighing the increased nuclear charge.
Worked example
Explain a dip, not just the general trend
Why is oxygen's first ionisation energy lower than nitrogen's despite oxygen having more protons?
- Both lose a 2p electron. Their inner-shell shielding is similar.
- Nitrogen has 2p3: one electron in each p orbital. Oxygen has 2p4, so one p orbital contains a pair.
- Removing a paired oxygen electron relieves additional repulsion. This effect outweighs the increased nuclear attraction in this comparison.
The dip is explained by paired-electron repulsion, not by claiming oxygen has fewer protons or a new principal shell.
A large jump reveals an inner shell
Count how many electrons are removed before the jump; use other evidence to determine the period.
Successive ionisation energies remove electrons one at a time from increasingly positive gaseous species. The second process is X+(g) → X2+(g) + e-. Values rise because the remaining electrons are attracted more strongly to the increasingly positive ion. A particularly large jump means the next electron is removed from a more tightly held inner shell.
Worked example
Reconstruct a period-3 atom
A main-group period-3 element has the first four ionisation energies 580, 1820, 2740 and 11600 kJ mol-1 (rounded teaching data). Deduce its outer configuration and group.
- The large jump is between the third and fourth removals: 11600/2740 is about 4.2, much larger than the previous ratio.
- Three electrons lie outside the inner core. For a period-3 main-group atom, they occupy 3s2 3p1.
- The group is 13. Combining period and group identifies aluminium; the complete configuration is [Ne]3s2 3p1.
Three outer electrons, group 13, period 3, [Ne]3s2 3p1. The period came from the question, not from the first jump alone.
For a main-group atom, one outer electron suggests group 1; two suggest group 2; three to eight suggest groups 13 to 18. Helium is the familiar two-electron exception in group 18. Do not apply this simple outer-count rule uncritically to transition elements, where 3d and 4s energies and occupations also matter.
- Find unusually large jumps
Compare the changes or ratios. On a logarithmic plot, read the axis before interpreting spacing.
- Count removals before each jump
This gives groups of electrons with substantially different binding energies.
- Combine with independent clues
Use proton number, total electron count or a stated period to identify the shell numbers.
- Check the proposed configuration
Its electron count and shell boundaries must explain the observed pattern.