The C-O-H bond angle in alcohols is slightly less than tetrahedral angle 109°28' due to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → lone pair - lone pair repulsion
In an alcohol the oxygen atom is \(sp^3\) hybridised. Two of its four hybrid orbitals form the C–O and O–H bonds and the other two hold lone pairs. Electron pairs repel each other in the order:
\( \displaystyle \text{lone pair--lone pair} \gt \text{lone pair--bond pair} \gt \text{bond pair--bond pair} \)
The repulsion between the two unshared electron pairs of oxygen is the strongest. It pushes the lone pairs apart and squeezes the two bond pairs closer together, so the C–O–H angle becomes slightly less than the tetrahedral angle of 109°28′ (about 108.9° in methanol).
Why the other options are incorrect
Option (1) bond pair - lone pair repulsion — Incorrect. This repulsion is also present, but it is weaker than the lone pair–lone pair repulsion, which is the main reason the angle closes.
Option (3) bond pair-bond pair repulsion — Incorrect. This is the weakest of the three; on its own it would keep the four pairs at the ideal 109°28′, not reduce the angle.
Option (4) Electronegativity of oxygen atom — Incorrect. Electronegativity makes the C–O and O–H bonds polar; it does not explain why the bond angle is smaller than tetrahedral.
NCERT (2026-27 edition):
Structures of Functional Groups – “In alcohols, the oxygen of the –OH group is attached to carbon by a sigma (σ) bond formed by the overlap of a sp3 hybridised orbital of carbon with a sp3 hybridised orbital of oxygen. … The bond angle in alcohols is slightly less than the tetrahedral angle (109°-28′). It is due to the repulsion between the unshared electron pairs of oxygen.”