Did you know? Sodium’s second ionisation energy is more than nine times its first. The first removes an outer 3s electron; the second must remove a much more tightly held electron from the inner n=2 shell.
This original Hougang JC1 H1 Chemistry tutorial follows SEAB 2027 syllabus 8873 Atomic Structure. It interprets sodium, magnesium and aluminium successive-ionisation energies, explaining shells, shielding, ionic charge, electronic configurations and the position of the large energy jump.
EDUKATESG · HOUGANG · JC1 H1 CHEMISTRY
12 chapters · Ionisation table · Transfer checks · FAQs · Official 2027 syllabus
Full chapter index
- 1. Why the first and second sodium ionisations differ so much
- 2. Write the gas-phase ionisation equations
- 3. Use the orbital configuration to locate each electron
- 4. Explain distance, shielding and nuclear attraction
- 5. A noble-gas configuration does not mean infinite energy
- 6. Read a large jump as evidence of one outer electron
- 7. Compare magnesium and locate its different jump
- 8. Compare aluminium and find the jump after three electrons
- 9. Do not confuse periodic first-ionisation trends with jump positions
- 10. Interpret fictional data without knowing the element
- 11. Distinguish ionisation energy from lattice and hydration energy
- 12. Repair the recurring explanations before an exam
1. Why the first and second sodium ionisations differ so much
The SEAB Chemistry Data Booklet lists sodium’s first ionisation energy at about 494 kJ mol⁻¹ and its second at about 4560 kJ mol⁻¹. Both quantities describe electron removal from gaseous species, but the second requires more than nine times as much energy as the first. This sharp contrast is evidence that the second electron is in a much more strongly attracted region of the atom.
Sodium has eleven protons and eleven electrons when neutral. Its single outer 3s electron is removed first, producing Na⁺ and leaving a filled n=2 electron shell. The second removal must reach into that inner shell. Its greater proximity to the same eleven-proton nucleus and lower shielding make it much harder to extract.
The 2027 H1 Chemistry 8873 syllabus expressly includes deducing electronic configurations and Periodic Table positions from successive-ionisation data. The best answer identifies the shell transition, rather than simply saying the ion has become stable.
2. Write the gas-phase ionisation equations
First ionisation is Na(g)→Na⁺(g)+e⁻. The starting species is a neutral gaseous atom, and the energy refers to removing one mole of electrons from one mole of gaseous sodium atoms. It is not the reaction of sodium metal with water or the dissolution of sodium chloride.
Second ionisation is Na⁺(g)→Na²⁺(g)+e⁻. This starts with a singly charged gaseous ion, not a neutral atom. Writing Na(g)→Na²⁺(g)+2e⁻ is a different two-electron overall transformation and must not be labelled the second ionisation enthalpy on its own.
Students should label all starting and resulting charges, then check conservation of electrical charge. The correct definitions are essential to explaining why remaining electrons become harder to remove.
3. Use the orbital configuration to locate each electron
Neutral sodium has the ground-state configuration 1s²2s²2p⁶3s¹. The outermost occupied level is n=3 with just one electron. The first ionisation removes that 3s electron and leaves Na⁺ with 1s²2s²2p⁶.
That Na⁺ electron arrangement matches neutral neon’s arrangement, but sodium remains sodium because it still has eleven protons, while neon has ten. Isoelectronic does not mean identical chemical identity or identical ionisation energy.
The next electron must be removed from the 2p portion of Na⁺’s n=2 core. When students mark the orbital supplying each electron, the large energetic jump becomes physically understandable.
4. Explain distance, shielding and nuclear attraction
The n=3 valence electron of neutral sodium is, on average, farther from the nucleus than the filled n=2 core. The inner electrons also shield the valence electron from some of the full positive nuclear charge, reducing its effective attraction to the centre.
After the 3s electron leaves, the next available electron occupies n=2. That electron has a more compact distribution and is shielded by fewer electrons. The eleven-proton nucleus therefore attracts it strongly. The initial species is also a positive ion rather than a neutral atom.
Use all three points where appropriate: a different starting ionic charge, a smaller mean electron–nucleus distance and less screening for the core electron. The phrase stable octet by itself does not show the physical cause.
5. A noble-gas configuration does not mean infinite energy
Na⁺ is stable in ordinary ionic chemistry because it has a filled-shell arrangement and forms favourable electrostatic interactions in compounds. However, the second ionisation energy is finite. With sufficient energy in an appropriate atomic process, Na⁺ can lose another electron.
Students sometimes write that the second electron cannot be removed because sodium has become neon. Both halves are incorrect. Na⁺ is not neon, and the existence of a measured second ionisation energy establishes that the process is possible even though costly.
The correct conclusion is that the second electron is much more tightly held, not that the periodic table forbids its removal. Precision matters when interpreting the difference between a common chemical oxidation state and a high-energy ionic species.
6. Read a large jump as evidence of one outer electron
Successive-ionisation tables show the energy required for each additional electron removal. For main-group atoms, a particularly large rise often comes after the last outer-shell electron has been removed and before the first core-shell electron is taken.
Sodium has a large jump between IE₁ and IE₂. Therefore the data support one electron in its outermost principal shell, consistent with a Group 1 element. This conclusion is based on the jump’s location, not on an arbitrary numerical threshold for the ratio of adjacent energies.
A small rise from one successive ionisation to the next can occur because the ion grows more positively charged. A very large jump usually signals a more fundamental change in the electron shell from which the next electron must be taken.
7. Compare magnesium and locate its different jump
Magnesium has ground-state configuration 1s²2s²2p⁶3s². It contains two valence electrons in n=3. Removing one gives Mg⁺, and removing the second gives Mg²⁺. Only the third ionisation reaches into the n=2 core.
The SEAB booklet reports approximate IE₁=736, IE₂=1450 and IE₃=7740 kJ mol⁻¹ for magnesium. The major jump therefore follows the second removal, unlike sodium’s jump after the first.
Students should not say that magnesium’s second ionisation is easy or that the first two values are equal. Both cost energy; it is their relative scale and the much larger third value that reveal the outer-shell electron count.
8. Compare aluminium and find the jump after three electrons
Aluminium has electron configuration 1s²2s²2p⁶3s²3p¹. Its third principal shell contains three valence electrons. Successive removal of these electrons gives Al⁺, Al²⁺ and then Al³⁺, leaving the compact neon-like electron configuration.
Approximate booklet values are 577, 1820, 2740 and 11600 kJ mol⁻¹ for successive aluminium ionisations. The fourth ionisation is especially demanding because it removes an electron from the now-exposed n=2 core.
Compare the pattern with sodium and magnesium: major jumps after one, two and three electrons respectively. The structural evidence is the number of less costly removals before the pronounced shell change.
9. Do not confuse periodic first-ionisation trends with jump positions
Across Period 3, first ionisation energy generally increases with increasing effective nuclear attraction, but there are local dips due to subshell energies and electron pairing. For example, aluminium’s first ionisation can be lower than magnesium’s because aluminium’s electron is removed from the higher-energy 3p subshell.
That local first-ionisation anomaly does not imply aluminium has fewer outer electrons. Its successive-ionisation sequence still shows three comparatively accessible third-shell electrons before a major fourth-ionisation jump.
Students should therefore separate two questions: how the first ionisation changes across different neutral elements, and what a large jump within one element’s successive energies reveals about its electron arrangement. The correct explanations overlap but are not identical.
10. Interpret fictional data without knowing the element
A fictional set has successive energies 600, 1250, 7100 and 9800 kJ mol⁻¹. The large rise between the second and third implies two outer-shell electrons. Under a simple representative-element model, a Group 2-like configuration is reasonable.
Another fictional set has 520, 4900, 7300 and 9800 kJ mol⁻¹. The first-to-second jump indicates one readily removed valence electron and a Group 1-like pattern. The exact species cannot be identified solely from a rough invented table without additional information.
The tutor should ask learners to circle the exceptional jump, name the number of outer electrons and explain why the subsequent removal reaches a more tightly bound shell. A Group label alone is not a complete chemical justification.
11. Distinguish ionisation energy from lattice and hydration energy
Ionisation energies are defined for gaseous atoms and ions losing electrons. Lattice energy refers to forming or separating an ionic crystal from gaseous ions under a specified convention. Hydration enthalpy refers to interactions when gaseous ions become hydrated by water.
These processes can all occur within broader energy cycles, but they are not synonymous. Dissolving a sodium chloride crystal does not simply repeat Na(g)→Na⁺(g)+e⁻: the sodium ion already exists in the crystal before it dissolves.
A student who uses first ionisation energy as though it were the direct enthalpy of salt dissolution will generally get a wrong chemical calculation. Every quoted energy must be tied to its correct reaction equation and physical states.
12. Repair the recurring explanations before an exam
One incorrect answer says sodium’s second ionisation is larger because it gains protons. The nucleus retains eleven protons. Another says Na⁺ has become neon; it shares neon’s electron configuration but has a different nucleus.
A third explanation says only that Na⁺ has a stable octet, with no mention of distance and shielding. A fourth writes the second ionisation as a two-electron event from neutral sodium. A fifth classifies magnesium as Group 1 by comparing only its first ionisation with sodium’s.
The repair sequence is to write the electron configuration, cross out one electron at each ionisation, label the shell change, then examine the energy table. A strong solution explains the actual physical reason for the exceptionally large jump.
Worked evidence: successive-ionisation data
The sodium, magnesium and aluminium values are rounded from the official SEAB Chemistry Data Booklet. Fictional examples are labelled and are not actual elemental measurements.
| Ionisation or set | Equation or inputs | Energy/data | Interpretation |
|---|---|---|---|
| Sodium IE₁ | Na(g)→Na⁺(g)+e⁻ | 494 kJ mol⁻¹ | 3s outer electron |
| Sodium IE₂ | Na⁺(g)→Na²⁺(g)+e⁻ | 4560 kJ mol⁻¹ | 2p core electron; large jump |
| Magnesium IE₁ | Mg(g)→Mg⁺(g)+e⁻ | 736 kJ mol⁻¹ | First 3s electron |
| Magnesium IE₂ | Mg⁺(g)→Mg²⁺(g)+e⁻ | 1450 kJ mol⁻¹ | Second 3s electron |
| Magnesium IE₃ | Mg²⁺(g)→Mg³⁺(g)+e⁻ | 7740 kJ mol⁻¹ | Core-shell jump |
| Aluminium IE₁ | Al(g)→Al⁺(g)+e⁻ | 577 kJ mol⁻¹ | 3p electron |
| Aluminium IE₃ | Al²⁺(g)→Al³⁺(g)+e⁻ | 2740 kJ mol⁻¹ | Third outer-shell electron |
| Aluminium IE₄ | Al³⁺(g)→Al⁴⁺(g)+e⁻ | 11600 kJ mol⁻¹ | Core-shell jump |
| Fictional X | 600,1250,7100,9800 | Jump after second | Two outer electrons inferred |
| Fictional Y | 520,4900,7300,9800 | Jump after first | One outer electron inferred |
Four independent student exercises
- Explain why Na IE₁≈494 and IE₂≈4560 differ, using the 3s and 2p electron arrangements.
- Use Mg IE₁≈736, IE₂≈1450 and IE₃≈7740 to infer two outer-shell electrons.
- Use aluminium’s large jump from IE₃≈2740 to IE₄≈11600 to infer three third-shell electrons.
- Given fictional 600,1250,7100,9800, identify the shell-boundary jump and explain what Group is plausible.
Tutor preparation checklist
- Confirm SEAB 2027 H1 8873 Section 1 Atomic Structure.
- Write each gas-phase ionisation equation with the correct ion charge.
- Mark sodium’s 3s¹ electron and its inner-shell 2p⁶ core.
- Explain shielding and average electron–nucleus distance.
- Distinguish isoelectronic Na⁺ and neutral neon.
- Use the large jump position to infer valence-electron count.
- Contrast Na, Mg and Al without inventing an exact fixed jump ratio.
- Do not confuse first-ionisation periodic anomalies with successive-energy jump positions.
- Separate ionisation energy from dissolution and hydration.
- Finish with an unfamiliar energy table and a written shell inference.
Frequently asked questions
Why is sodium’s second ionisation so large?
Its next electron comes from the much more tightly held n=2 core, not the original 3s shell.
What is Na’s ground-state configuration?
1s²2s²2p⁶3s¹.
Is Na⁺ a neon atom?
No. It has neon’s electron configuration but retains eleven protons.
Where is the largest jump for magnesium?
Between the second and third ionisations.
Where is the large jump for aluminium?
Between the third and fourth ionisations.
Does a large ionisation value mean removal is impossible?
No. It means the positive energy requirement is large, not infinite.
Does ionisation energy describe dissolving NaCl?
No. It describes electron removal from gaseous species.
Does H1 syllabus 8873 require interpreting successive ionisation?
Yes, including shell configurations and Periodic Table position.
Related atomic-structure guides
- JC1 H2 Chemistry | First-Ionisation Energy Trends and Exceptions
- Hougang Chemistry Tuition and Tutor Library
Bring the student’s examination year, current atomic-structure worksheet and an attempted successive-ionisation inference to a tutoring consultation. Confirm venue, teacher, class size, fees and availability directly.
Official 2027 syllabus and data context
- SEAB 2027 H1 Chemistry 8873, Section 1 Atomic Structure
- Official SEAB 2027 A-Level syllabuses and Chemistry Data Booklet
- Immutable eduKateSG Clementi small-group teaching reference
- Hougang Chemistry Tuition and Tutor Library
The 2027 H1 syllabus requires successive-ionisation-energy interpretation. The real ionisation values above are rounded from the official Chemistry Data Booklet; fictional values are exercise-only. The shell explanation is an introductory quantum-structure model and does not imply that electrons follow classical planetary trajectories.