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Question

For an sp³ molecule with one lone pair, how and why does the bond angle change compared to the ideal tetrahedral angle?

This question was previously asked in
RRB JE 2025 CBT 2 Mechanical and Allied Engg Question Paper English (2-Jul-2026) (Shift-1)
The correct answer is

The bond angle will decrease due to lone pair-bond pair repulsion

This question relates to the Valence Shell Electron Pair Repulsion (VSEPR) theory, which predicts molecular geometry based on minimising repulsion between electron pairs (both bonding and lone pairs) around a central atom.

For a central atom with sp³ hybridisation and four electron domains arranged with no lone pairs (e.g., methane, CH₄), the electron pairs arrange themselves symmetrically at the vertices of a regular tetrahedron, giving the ideal bond angle of 109.5°.

However, when one of those four hybrid orbitals holds a lone pair instead of a bonding pair (as in ammonia, NH₃), the geometry is distorted. VSEPR theory ranks the strength of repulsive interactions as:

lone pair–lone pair > lone pair–bond pair > bond pair–bond pair

This ordering arises because a lone pair is held only by the central atom's nucleus and is not "pulled" outward by a second bonded nucleus, so its electron cloud is more diffuse and occupies more angular space close to the central atom than a bonding pair does. Consequently, the lone pair pushes the neighbouring bond pairs closer together, compressing the angle between them.

For NH₃, which has three N–H bond pairs and one lone pair, this lone pair–bond pair repulsion squeezes the H–N–H bond angle down from the ideal 109.5° to approximately 107°. This confirms that the bond angle decreases due to lone pair–bond pair repulsion, which is the correct explanation.

The remaining options do not hold up: the angle does not increase, since a lone pair repels neighbouring bonds rather than allowing them to spread further apart; the geometry is not unchanged, precisely because introducing a lone pair breaks the perfect symmetry of the parent tetrahedral arrangement; and "increased nucleus attraction" is not the relevant mechanism at all — bond angle changes in VSEPR theory are explained through electron pair repulsion, not through nuclear attraction forces. As a further illustration of this trend, each additional lone pair compresses the angle further: water (H₂O), with two lone pairs, has an even smaller bond angle of about 104.5°.

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