Explain why the first ionisation energy of oxygen is lower than that of nitrogen, despite oxygen having a greater nuclear charge.

IB DP Chemistry Higher Level (2023 syllabus) — S1.3 Electron configurations · Explain · 4 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

Model answer (4 marks)

1. Nitrogen’s 2p subshell is half‑filled (one electron in each of the three 2p orbitals), giving a particularly stable configuration.
2. In oxygen the 2p subshell contains four electrons; two occupy the same orbital as a paired pair.
3. The paired electrons experience electron–electron repulsion, which raises the energy of the 2p electrons.
4. This repulsion makes it easier to remove one of the paired electrons, so the first ionisation energy of O is lower than that of N, despite O’s greater nuclear charge.

Examiner tips

  • Use the term ‘half‑filled subshell’ for nitrogen’s stability. Mention electron–electron repulsion in oxygen. Explain that the repulsion outweighs the increased nuclear charge. Keep the answer to four short points, matching the marks.

Common mistakes

  • Confusing ‘paired electrons’ with ‘unpaired’. Forgetting to mention the half‑filled subshell. Saying the nuclear charge is higher without explaining the repulsion effect.

Mark scheme (4 marks)

  1. Nitrogen has a half-filled 2p subshell (one electron in each 2p orbital), which is a particularly stable arrangement.
  2. In oxygen, two electrons must occupy the same 2p orbital, resulting in electron–electron repulsion between the paired electrons.
  3. This repulsion makes it easier to remove one of the paired electrons from oxygen, lowering its first ionisation energy relative to nitrogen.
  4. The effect of electron–electron repulsion in oxygen outweighs the effect of the increased nuclear charge, so oxygen's first ionisation energy is lower than nitrogen's.

Key terms in this question

first ionisation energy · nuclear charge

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