Concept: VSEPR Theory and Bond Angles

The Valence Shell Electron Pair Repulsion (VSEPR) theory predicts the geometry of molecules based on the repulsion between electron pairs (both bonding and non-bonding) in the valence shell of the central atom. Lone pair-lone pair repulsion is greater than lone pair-bond pair repulsion, which is greater than bond pair-bond pair repulsion. These repulsions influence the bond angles in a molecule.

Why (C) is correct:

1. Methane (CH\u2084): The central carbon atom in CH\u2084 has four bond pairs and no lone pairs. According to VSEPR theory, these four electron pairs arrange themselves tetrahedrally, resulting in an ideal H-C-H bond angle of $109.5^{\circ }$.

2. Water (H\u2082O): The central oxygen atom in H\u2082O has two bond pairs (with hydrogen atoms) and two lone pairs of electrons. The total number of electron domains is four, suggesting a tetrahedral electron geometry. However, the presence of two lone pairs causes greater repulsion than bond pairs. Lone pair-lone pair repulsion > lone pair-bond pair repulsion > bond pair-bond pair repulsion. This stronger repulsion from the lone pairs compresses the H-O-H bond angle from the ideal tetrahedral angle of $109.5^{\circ }$.

3. Result: Due to the increased repulsion from the two lone pairs on the oxygen atom, the H-O-H bond angle in water is reduced to approximately $104.5^{\circ }$, which is less than $109.5^{\circ }$.

Option Analysis:

• A) remain the same: Incorrect. The presence of lone pairs significantly alters bond angles.
• B) increase: Incorrect. Lone pair repulsion compresses bond angles, it does not increase them.
• C) decrease: Correct. The two lone pairs on oxygen exert greater repulsion, pushing the bonding pairs closer and decreasing the H-O-H bond angle.
• D) become $180^{\circ }$: Incorrect. A $180^{\circ }$ bond angle would imply a linear geometry, which is not possible for a molecule with two lone pairs and two bond pairs around the central atom.

Correct Answer: (C)