Explain why the bond angle in a water molecule (H₂O) is approximately 104.5°, which is less than the ideal tetrahedral angle of 109.5°.

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 central oxygen atom has four electron domains (two O–H bonds and two lone pairs), giving a tetrahedral electron geometry. Lone pairs occupy more angular space and repel more strongly than bonding pairs. The two lone pairs on oxygen therefore repel the two O–H bonding pairs more strongly than bonding pairs repel each other. As a result the H–O–H bond angle is compressed below the ideal tetrahedral angle of 109.5° to about 104.5°.

Examiner tips

  • Use the term ‘electron geometry’ to show understanding of VSEPR. Include the idea that lone pairs repel more strongly than bonding pairs. Mention the compression of the bond angle to 104.5°.
  • Keep the answer concise – 4 marks can be earned with 3–4 short sentences. Use correct UK spelling (repel, geometry).

Common mistakes

  • Confusing the bond angle with the electron‑pair angle; students sometimes give 109.5° instead of 104.5°.
  • Failing to mention that lone pairs repel more strongly than bonding pairs, or omitting the reason for the angle compression.

Mark scheme (4 marks)

  1. The central oxygen atom has four electron domains (two bonding pairs and two lone pairs), giving a tetrahedral electron geometry.
  2. Lone pairs occupy greater angular space / repel more strongly than bonding pairs.
  3. The two lone pairs on oxygen repel the two O–H bonding pairs more strongly than bonding pairs repel each other.
  4. The H–O–H bond angle is compressed below 109.5° to approximately 104.5° as a result of this increased lone pair repulsion.

Key terms in this question

bond angle · tetrahedral

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