Q.Although geometries of $NH_3$ and $H_2O$ molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.

Concept: VSEPR theory and lone-pair repulsion hierarchy.

Both molecules have four electron pairs around the central atom in a tetrahedral arrangement, but the number of lone pairs differs. Ammonia has one lone pair and three bond pairs (trigonal pyramidal shape), while water has two lone pairs and two bond pairs (bent shape).

The key lies in repulsion strength: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. In water, the two lone pairs exert stronger mutual repulsion, compressing the bonding pairs closer together and reducing the H–O–H angle more severely. Ammonia's single lone pair causes less compression of the bond pairs. …

The question asks us to explain a subtle but important observation: why does water, with the same basic shape as ammonia, end up with a smaller bond angle? The answer lies in understanding not just the geometry, but the forces that distort it.

Why VSEPR Theory matters here

VSEPR (Valence Shell Electron Pair Repulsion) theory tells us that electron pairs around a central atom arrange themselves to minimize repulsion. But not all electron pairs repel equally. Lone pairs, which are held close to the nucleus and occupy more angular space than bonding pairs, exert stronger repulsive forces. This differential repulsion is the key to understanding bond-angle compression.

Both $NH_3$ and $H_2O$ start from a tetrahedral electron-pair arrangement (four regions of electron density around the central atom), but the number of lone pairs differs, and that makes all the difference.

Step-by-step comparison

1. Count the electron pairs and predict the base geometry

For ammonia ($NH_3$):

• Nitrogen has 5 valence electrons
• Three are used in N–H bonds, leaving one lone pair
• Total: 3 bonding pairs + 1 lone pair = 4 electron pairs
• Electron-pair geometry: tetrahedral

For water ($H_2O$):

• Oxygen has 6 valence electrons
• Two are used in O–H bonds, leaving two lone pairs
• Total: 2 bonding pairs + 2 lone pairs = 4 electron pairs
• Electron-pair geometry: tetrahedral

In a perfect tetrahedron, all angles would be $109.5°$. But lone pairs change everything.

2. Understand the repulsion hierarchy

The strength of repulsion follows this order:

Lone pairs are closer to the nucleus and spread out more in space, so they push harder against neighboring electron pairs.

3. Apply the repulsion to ammonia

In $NH_3$, the single lone pair repels the three N–H bonding pairs more strongly than the bonding pairs repel each other. This pushes the three hydrogen atoms slightly closer together, compressing the H–N–H angle from the ideal $109.5°$ down to approximately $107°$.

The molecular geometry is trigonal pyramidal (three atoms at the base, lone pair at the apex).

4. Apply the repulsion to water

In $H_2O$, there are two lone pairs. These exert even stronger repulsion:

• The two lone pairs repel each other (the strongest repulsion)
• Each lone pair also repels both O–H bonding pairs

This double dose of lone-pair repulsion compresses the H–O–H bond angle much more severely, down to approximately $104.5°$.

The molecular geometry is bent or angular. …