In alcohols, C-O-H bond angle is

The correct answer is Option (1) → Less than the tetrahedral bond angle (109° 28')

In alcohols the oxygen atom is sp³ hybridised: two of its four hybrid orbitals form the C–O and O–H bonds, and the other two hold lone pairs. A lone pair is held only by the oxygen nucleus and spreads out more than a bonding pair. The repulsion between the two lone pairs, and between lone pairs and bond pairs, pushes the two bonds closer together.

So the C–O–H angle is a little smaller than the regular tetrahedral angle; in methanol it is 108.9°, which is less than 109° 28′.

Why the other options are incorrect

• Option (2) Higher than the tetrahedral bond angle (109° 28') — Incorrect. A larger angle is found in ethers (C–O–C about 111.7°), where two bulky alkyl groups repel each other; the small H atom in alcohols does not do this.

• Option (3) “Exactly equal to tetrahedral bond angle i.e. 109° 28'” — Incorrect. The ideal angle is seen only when all four orbitals hold bond pairs, as in methane; lone pairs on oxygen make it smaller.

• Option (4) “Exactly equal to 90°” — Incorrect. An angle of 90° would mean bonding through pure p orbitals with no hybridisation, which is not the case for oxygen in alcohols.

NCERT (2026-27 edition):
Structures of Functional Groups – …

Note: The same/similar question has been asked in different years by NTA as per the details given below: