Explain why the bond angle in ammonia (NH₃) is approximately 107°, which is less than the bond angle of 109.5° found in methane (CH₄).

IB DP Chemistry Standard Level (2023 syllabus) — S2.2 The covalent model · Explain · 4 marks · View as Markdown

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

Model answer (4 marks)

The VSEPR principle states that electron pairs around a central atom arrange to minimise repulsion.
NH₃ has four electron domains: three N–H bonds and one lone pair, giving a tetrahedral electron geometry.
A lone pair exerts a greater repulsive force than a bonding pair because it is more concentrated around the nucleus.
This stronger lone‑pair repulsion pushes the three N–H bonds closer together, reducing the H–N–H angle from the ideal 109.5° to about 107°.

Examiner tips

  • State the VSEPR principle first; mention the four electron domains in NH₃; explain lone‑pair repulsion; give the resulting angle.
  • Use the exact angle 107° and the comparison to 109.5°; keep the answer concise and to the point.

Common mistakes

  • Failing to mention the lone pair or the tetrahedral electron geometry; giving the wrong angle (e.g. 109.5°) or not explaining the reduction; using vague terms like ‘more repulsive’ without specifying lone pair vs bonding pair.

Mark scheme (4 marks)

  1. Electron pairs (electron domains) arrange themselves to minimise repulsion / VSEPR principle stated
  2. NH₃ has three bonding pairs and one lone pair (tetrahedral electron geometry / four electron domains around nitrogen)
  3. A lone pair exerts greater repulsion than a bonding pair because it is closer to / more concentrated around the central atom
  4. The greater lone pair repulsion pushes the three N–H bonds closer together, reducing the H–N–H bond angle below 109.5° to approximately 107°

Key terms in this question

bond angle

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