In alcohols, carbon and oxygen atom are
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(sp^3\) and \(sp^3\) hybridised respectively
The hybridisation of an atom follows from the number of electron pairs around it (sigma bonds plus lone pairs). Four pairs need four equivalent orbitals, which means \( \mathrm{sp^{3}} \) hybridisation.
The carbon that carries the OH group in an alcohol forms four sigma bonds (to O and to H or C atoms) and has no multiple bond, so it is \( \mathrm{sp^{3}} \).
The oxygen forms two sigma bonds (to C and to H) and carries two lone pairs: four electron pairs, so it is also \( \mathrm{sp^{3}} \).
The C–O bond is formed by the overlap of an \( \mathrm{sp^{3}} \) orbital of carbon with an \( \mathrm{sp^{3}} \) orbital of oxygen.
Why the other options are incorrect
Option (2) \(sp^2\) and \(sp^2\) hybridised respectively — Incorrect. This describes a carbonyl group, C=O, as in aldehydes and ketones, where both atoms take part in a double bond.
Option (3) \(sp^3\) and \(sp^2\) hybridised respectively — Incorrect. The oxygen of an alcohol has four electron pairs and no double bond, so it is not \( \mathrm{sp^{2}} \).
Option (4) \(sp^2\) and \(sp^3\) hybridised respectively — Incorrect. This fits phenol, where the OH group is attached to an \( \mathrm{sp^{2}} \) carbon of the benzene ring, not an alcohol.
NCERT (2026-27 edition):
Structures of Functional Groups
