Q.Give reasons for the following:

Concept: Bonding and Hybridisation Effects

(a) Bond angle in alcohol is slightly less than the tetrahedral angle ($109.5^\circ$).

The oxygen atom in an alcohol ($R-OH$) is $sp^3$ hybridised. Two of the four $sp^3$ orbitals form sigma bonds with C and H, while the other two hold lone pairs. Lone pairs exert greater repulsion than bond pairs (lone pair–bond pair repulsion > bond pair–bond pair repulsion). This pushes the C–O–H bond angle slightly inward, reducing it from the ideal $109.5^\circ$ to about $108.9^\circ$.

(b) C–OH bond length in $CH_3OH$ is slightly more than in phenol. …

Why This Approach Works

Before diving into numbers, let's build the intuition. Both parts of this question hinge on the electronic structure around oxygen — specifically, how lone pairs and resonance affect geometry and bond strength.

Part (a) is about bond angles. You know the tetrahedral angle is $109.5^\circ$, found in perfect $sp^3$ hybrids like methane. But in water, the H–O–H angle is only $104.5^\circ$. Why? Lone pairs repel more strongly than bonding pairs. Alcohols sit between water and methane: the oxygen has two lone pairs and two sigma bonds, so the angle should be less than tetrahedral but greater than water's angle (since the alkyl group is bulkier and slightly different in electronegativity).

Part (b) is about bond lengths. A shorter bond means stronger bonding — often due to multiple bond character. Phenol's oxygen can donate its lone pair into the aromatic ring, creating resonance structures with C–O double bond character. Methanol has no such resonance. So the phenol C–O bond should be shorter (stronger) than methanol's.

Let's verify with actual data.

Step-by-Step Reasoning

1. Bond angle in alcohols: why less than $109.5^\circ$

The oxygen atom in an alcohol (say methanol, $CH_3OH$) is $sp^3$ hybridized. It forms two sigma bonds (one to C, one to H) and holds two lone pairs. According to VSEPR theory, the electron pairs — both bonding and non-bonding — arrange themselves to minimize repulsion.

The key point: lone pair–lone pair repulsion > lone pair–bond pair repulsion > bond pair–bond pair repulsion. The two lone pairs on oxygen push the bonding pairs closer together, compressing the C–O–H bond angle.

In methanol, the experimental C–O–H angle is about $108.9^\circ$ (some sources give $108.5^\circ$). Compare:

• Tetrahedral angle: $109.5^\circ$
• Water H–O–H: $104.5^\circ$
• Methanol C–O–H: $\approx 108.9^\circ$

So the alcohol angle is indeed slightly less than tetrahedral, but not as small as water's. Why the difference? The methyl group is bulkier than a hydrogen atom, so it pushes the C–O–H angle open a bit compared to water. Still, the lone pair repulsion dominates, keeping it below $109.5^\circ$.

2. C–OH bond length: methanol vs phenol

Now consider the bond between carbon and oxygen in the hydroxyl group.

In methanol ($CH_3OH$): The C–O bond is a pure single bond. Oxygen is $sp^3$ hybridized, carbon is $sp^3$ hybridized. There is no possibility of resonance — the lone pairs on oxygen have no adjacent pi system to delocalize into. The bond length is typical for a C–O single bond: about $1.43 \, \text{Å}$. …