How is bond angle influenced by the presence of lone pairs?

The presence of lone pairs can decrease the bond angle due to their greater repulsion compared to bonding pairs.

In a molecule, the shape and bond angles are determined by the repulsion between electron pairs in the valence shell of the central atom. This is explained by the Valence Shell Electron Pair Repulsion (VSEPR) theory. According to this theory, electron pairs will arrange themselves to minimise repulsion and maximise the distance between them.

Lone pairs of electrons, which are not involved in bonding, occupy more space in the electron cloud than bonding pairs. This is because they are held by only one nucleus and hence spread out more. As a result, they exert a greater repulsive force on the other electron pairs. This greater repulsion pushes the bonding pairs closer together, decreasing the bond angle.

For example, consider the molecules of water (H2O) and ammonia (NH3). In water, there are two bonding pairs (between the oxygen and hydrogen atoms) and two lone pairs on the oxygen atom. In ammonia, there are three bonding pairs (between the nitrogen and hydrogen atoms) and one lone pair on the nitrogen atom. According to the VSEPR theory, the bond angles should be 109.5 degrees for a tetrahedral arrangement. However, due to the greater repulsion from the lone pairs, the bond angle in water is 104.5 degrees and in ammonia it is 107 degrees.

In summary, the presence of lone pairs can significantly influence the bond angle in a molecule. The greater repulsion exerted by lone pairs pushes the bonding pairs closer together, resulting in a smaller bond angle. Understanding this concept is crucial for predicting the shape and properties of molecules.