Lesson 3 of 3
Reading energy data
Explain attraction and infer the arrangement.
A-Level 8873, revised syllabus (2026-2027)
I can explain ionisation energies.Syllabus
Syllabus 8873, 1(i). Explain ionisation energies.
I can infer configurations from successive energies.Syllabus
Syllabus 8873, 1(j). Infer configurations from successive energies.
I can connect successive energies to periodic position.Syllabus
Syllabus 8873, 1(k). Connect successive energies to periodic position.
Explain how strongly an electron is held
Balance nuclear attraction, shielding, distance and electron pairing.
Make a guess. You are not marked.
First ionisation energy generally rises across Period 3. How does sulfur compare with phosphorus?
Show the answer
Sulfur is lower, as it loses an electron from a paired 3p orbital
Sulfur loses an electron from a 3p orbital that holds a pair. Repulsion between the paired electrons makes that electron easier to remove.
The first ionisation energy is the energy needed to remove one electron from each atom in one mole of gaseous atoms, forming one mole of gaseous 1+ ions. For X: X(g) → X+(g) + e-. Values are positive and usually quoted in kJ mol-1: energy is supplied to overcome attraction to the nucleus.
| Factor | Effect when other factors are similar | How to use it |
|---|---|---|
| Greater nuclear charge | Stronger attraction, higher ionisation energy | Compare proton number, but also check shielding. |
| Greater shielding by inner electrons | Weaker attraction to the outer electron | A new shell often matters more than extra protons. |
| Greater electron-nucleus distance | Weaker attraction, lower ionisation energy | Name the shell from which the electron leaves. |
| Higher-energy subshell or electron pairing | Often makes an electron easier to remove | Identify the specific subshell and paired orbital. |
Across Period 3, first ionisation energy generally increases: nuclear charge rises while added electrons enter the same principal shell, so shielding changes relatively little. Two dips need more precise reasoning. Al loses a higher-energy 3p electron, whereas Mg loses 3s; Al is easier to ionise. S loses an electron from a paired 3p orbital, whereas P has three singly occupied 3p orbitals; pair repulsion makes the S electron easier to remove.
Down a group, the outer electron occupies a higher shell farther from the nucleus and is more shielded. These effects usually outweigh the greater nuclear charge, so first ionisation energy decreases. Say which effects dominate; merely saying "there are more electrons" does not explain the attraction.
Check your understanding
First ionisation energy dips from Mg to Al, even though Al has more protons. Why?
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Al loses a 3p electron, which is higher in energy and more shielded than Mg's 3s.
The 3p electron is further from the nucleus on average and shielded by the 3s electrons, so it is easier to remove. This outweighs the extra nuclear charge.
Find the jump that reveals the outer electrons
A large step means removal has reached an inner shell.
Successive ionisation energies refer to removing electrons one at a time from increasingly positive gaseous ions. The second process is X+(g) → X2+(g) + e-. Values rise because the remaining electrons are held more strongly. A particularly large jump indicates that the next electron comes from a lower, inner shell much closer to the nucleus.
| Electron removed | 1st | 2nd | 3rd | 4th |
|---|---|---|---|---|
| Energy / kJ mol-1 | 740 | 1450 | 7730 | 10500 |
Worked exampleLocate a group without overclaiming the element
Use the data to infer the outer-electron count. Then write the configuration if Z = 12.
Try it first. Then show the worked stepsHide the worked steps
- Compare steps: 740 to 1450 is an increase, but 1450 to 7730 is the pronounced jump.
- Two electrons are removed before reaching an inner shell, so the atom has two outer electrons.
- For a main-group element this suggests Group 2. The four energies alone do not establish the period.
- With Z = 12 supplied, place 12 electrons: 1s2 2s2 2p6 3s2.
Two outer electrons; Group 2. With the extra information Z = 12, the atom is magnesium in Period 3.
| Electron removed | Ionisation energy / kJ mol-1 |
|---|---|
| 1 | 496 |
| 2 | 4,560 |
| 3 | 6,910 |
| 4 | 9,540 |
| 5 | 13,400 |
| 6 | 16,600 |
| 7 | 20,100 |
| 8 | 25,500 |
| 9 | 28,900 |
| 10 | 141,000 |
| 11 | 159,000 |
Worked exampleRecover the whole configuration from complete data
All eleven successive ionisation energies of a neutral atom are shown. Deduce its shell arrangement, configuration, group and period.
Try it first. Then show the worked stepsHide the worked steps
- All electrons have been removed after the eleventh step, so the neutral atom contained eleven electrons. Its proton number is therefore 11; this comes from the completeness of the data, not from memorising a numerical energy.
- The large jumps are from the 1st to the 2nd energy and from the 9th to the 10th. Separate the removals into 1 | 8 | 2 electrons. These groups are removed from the outside in.
- Reverse their order to write the occupied shells from the nucleus outwards: 2, 8, 1. Three occupied principal shells place the atom in Period 3; one outer electron places it in Group 1.
- Fill the subshells: the first two electrons occupy 1s, the next eight occupy 2s and 2p, and the last occupies 3s. The smaller steps within a shell do not represent additional principal shells.
The atom is Na: 1s2 2s2 2p6 3s1, Group 1, Period 3. If only the first few energies had been supplied, that full conclusion would need additional evidence.
Check your understanding
A main-group atom has a large jump between its third and fourth ionisation energies. What can you infer?
Show the answer
It has three outer electrons, so it is in Group 13.
Three electrons come off before the next one must come from an inner shell, closer to the nucleus. The data do not tell you the period on their own.