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Example · Example 2

Q.Describe the shape of the ammonia molecule, NH3\text{NH}_3, stating the hybridisation of nitrogen and the approximate H-N-H\text{H-N-H} bond angle. Why is this angle smaller than the ideal tetrahedral angle of 109.5∘109.5^\circ?

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In NH3\text{NH}_3, nitrogen is sp3sp^3 hybridised: three of its four hybrid orbitals form σ\sigma bonds to the three hydrogen atoms, and the fourth holds the lone pair. This gives a trigonal pyramidal molecular shape, with nitrogen at the apex above a triangular base of three hydrogens. The measured H-N-H\text{H-N-H} bond angle is about 107.8∘107.8^\circ, noticeably smaller than the ideal tetrahedral angle of 109.5∘109.5^\circ expected for four exactly equivalent electron domains. The reason is that a lone pair, held by only one nucleus rather than shared between two, occupies more angular space around the central atom than a bonding pair does; by VSEPR theory, lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, so the lone pair pushes the three N-H bonding pairs closer together than a perfect tetrahedral arrangement would place them, compressing the angle. [!ANSWER] Ammonia is trigonal pyramidal (sp3sp^3 nitrogen, one lone pair), with an H-N-H bond angle of about 107.8∘107.8^\circ, smaller than 109.5∘109.5^\circ because lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion.

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