Q.Assertion (A) : The C - O - H bond angle in alcohols is slightly less than the tetrahedral angle. Reason (R) : This is due to the repulsive interaction between the two lone electron pairs on oxygen. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.

Concept: Bond angle deviation due to lone pair repulsion (VSEPR theory).

Reasoning:

1. In alcohols, the oxygen atom in the C–O–H group has two bond pairs and two lone pairs, giving a bent (angular) geometry around oxygen.
2. According to VSEPR theory, lone pair–lone pair repulsion is stronger than bond pair–bond pair repulsion. This compresses the C–O–H bond angle below the ideal tetrahedral angle of $109.5^\circ$. …

Why this question tests your understanding of VSEPR theory

The geometry around oxygen in an alcohol (R–O–H) is not simply tetrahedral — it’s a distorted tetrahedron. Oxygen has four electron domains: two bonding pairs (C–O and O–H) and two lone pairs. According to VSEPR theory, lone pairs occupy more space than bonding pairs because they are held closer to the nucleus and experience less nuclear pull. This creates stronger repulsion between lone pairs, which pushes the bonding pairs closer together.

The tetrahedral angle is 109.5°, but the C–O–H angle in methanol, for example, is about 108.9°. That slight compression is exactly what the Reason describes.

Step-by-step reasoning

1. Identify the electron geometry around oxygen

   Oxygen in an alcohol has 6 valence electrons. Two are used in sigma bonds (one with carbon, one with hydrogen), leaving four electrons as two lone pairs. That gives four electron domains — a tetrahedral arrangement in terms of electron pairs.

2. Apply VSEPR: lone pairs repel more than bonding pairs

   The repulsion order is:

The two lone pairs on oxygen push each other apart, and in doing so, they also push the two bonding pairs (C–O and O–H) closer together. This reduces the bond angle from the ideal 109.5°.

3. Check the actual experimental value

   In methanol (CH₃OH), the C–O–H angle is measured as ~108.9°. In ethanol, it’s similar (~108.5°). Both are slightly less than 109.5°, confirming the Assertion.

4. Evaluate the Reason …