Explain why the first ionisation energy of phosphorus is greater than that of sulfur, even though sulfur has a higher nuclear charge.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Model answer (4 marks)
Phosphorus has the configuration 1s²2s²2p⁶3s²3p³, giving a half‑filled 3p sub‑level with one electron in each of the three 3p orbitals. This arrangement is stabilised by exchange energy and a symmetrical distribution of electrons. Sulphur has 1s²2s²2p⁶3s²3p⁴, so one of the 3p orbitals contains a paired set of electrons. The electron–electron repulsion between the paired electrons in sulphur makes it easier to remove an electron, lowering its first ionisation energy even though its nuclear charge is higher.
Examiner tips
- Mention the half‑filled 3p sub‑level of P and its extra stability
- Explain the paired electrons in S and the resulting repulsion
- Show the link between repulsion and lower ionisation energy
Common mistakes
- Confusing the effect of nuclear charge with electron configuration
- Forgetting to mention the exchange energy or electron repulsion
- Using the wrong symbols for sulphur (S vs. S)
Mark scheme (4 marks)
- Phosphorus has a half-filled 3p sub-level (3p³) with one electron in each 3p orbital
- This half-filled arrangement confers extra stability due to the exchange energy / symmetrical electron distribution
- Sulfur has configuration [Ne]3s²3p⁴, so one 3p orbital contains a pair of electrons
- Electron–electron repulsion between the paired electrons in sulfur makes it easier to remove one electron, lowering the first ionisation energy relative to phosphorus
Key terms in this question
Related
- All IB DP Chemistry Higher Level (2023 syllabus) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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