Q.In both water and dimethyl ether (CH3—O—CH3), oxygen atom is central atom, and has the same hybridisation, yet they have different bond angles. Which one has greater bond angle? Give reason.

The key idea is VSEPR Theory: bond angles are determined by the repulsion between electron pairs around the central atom, and lone pairs repel more strongly than bonding pairs.

1. In water ($H_2O$), oxygen has 2 bond pairs and 2 lone pairs — a tetrahedral electron-pair geometry. The lone-pair–lone-pair repulsion compresses the $H-O-H$ angle to about 104.5°.
2. In dimethyl ether ($CH_3-O-CH_3$), oxygen again has 2 bond pairs and 2 lone pairs, so the same tetrahedral arrangement. However, the two bonding pairs are attached to bulky methyl groups ($-CH_3$). …

The key to this question lies in VSEPR (Valence Shell Electron Pair Repulsion) Theory. This theory states that electron pairs around a central atom arrange themselves to minimise repulsion. But not all repulsions are equal — the strength of repulsion follows a clear hierarchy: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair.

In both water ($H_2O$) and dimethyl ether ($CH_3-O-CH_3$), the central oxygen atom has the same electronic geometry. Let's see why.

1. Count the electron pairs around oxygen.

   Oxygen has 6 valence electrons. In water, it forms two single bonds with hydrogen atoms, using 2 electrons. The remaining 4 electrons form two lone pairs. That gives 2 bond pairs + 2 lone pairs — a total of 4 electron domains.

   In dimethyl ether, oxygen forms two single bonds with carbon atoms (from the methyl groups). Again, 2 electrons are used in bonding, leaving 4 as two lone pairs. So the electron domain count is identical: 2 bond pairs + 2 lone pairs.

2. Determine the hybridisation.

   Four electron domains around a central atom always correspond to $sp^3$ hybridisation. The ideal bond angle for a perfect tetrahedron is $109.5^\circ$. But the presence of lone pairs distorts this angle because lone pairs occupy more space and repel more strongly than bond pairs.

3. Compare the bond pair repulsions.

   In water, the two bond pairs are attached to small hydrogen atoms. The repulsion between the two O–H bond pairs is relatively modest. The dominant repulsion comes from the two lone pairs pushing the bond pairs closer together, resulting in the well-known bond angle of $104.5^\circ$.

   In dimethyl ether, the bond pairs are attached to methyl groups ($-CH_3$). These are much bulkier than hydrogen atoms. The bond pair–bond pair repulsion between the two O–C bonds is significantly stronger because the electron clouds of the methyl groups are larger and more polarisable. This increased repulsion pushes the two bond pairs further apart, opening the bond angle.

4. The result. …