Concept: Bond angles are determined by the VSEPR theory, which predicts the geometry of molecules based on minimizing repulsion between electron pairs (both bonding and non-bonding) around the central atom. The number of electron domains (lone pairs + bonding pairs) dictates the electron geometry, which in turn influences the molecular geometry and bond angles. Lone pairs exert more repulsion than bonding pairs, leading to a reduction in bond angles.

Why (D) CO₂ is correct:

• A) H₂O: The central atom is oxygen. It has two bond pairs and two lone pairs. The electron geometry is tetrahedral, but the molecular geometry is bent. The ideal bond angle for a tetrahedral arrangement is 109.5°, but due to the repulsion from two lone pairs, the H-O-H bond angle is reduced to approximately 104.5°.
• B) NH₃: The central atom is nitrogen. It has three bond pairs and one lone pair. The electron geometry is tetrahedral, but the molecular geometry is trigonal pyramidal. The ideal bond angle is 109.5°, but due to the repulsion from one lone pair, the H-N-H bond angle is reduced to approximately 107°.
• C) SF₄: The central atom is sulfur. It has four bond pairs and one lone pair. The electron geometry is trigonal bipyramidal, but the molecular geometry is seesaw. The ideal bond angles for trigonal bipyramidal are 90° and 120°. Due to the lone pair, the bond angles are distorted from these ideal values, typically around 102° and 173°.
• D) CO₂: The central atom is carbon. It forms two double bonds with two oxygen atoms and has no lone pairs. According to VSEPR theory, there are two electron domains (two double bonds count as two domains). This leads to a linear electron geometry and molecular geometry. The O=C=O bond angle is exactly 180°.

Comparing the bond angles: H₂O (≈104.5°), NH₃ (≈107°), SF₄ (distorted 90° and 120°, largest ≈173°), CO₂ (180°). Therefore, CO₂ has the largest bond angle.

Correct Answer: (D)