Concept: The bond angle in hydrides of Group 15 elements is influenced by the electronegativity of the central atom and the size of the central atom, which affects the extent of p-orbital involvement in bonding and the repulsion between bond pairs.

Why (C) is correct:

Phosphine (PH₃) has a pyramidal geometry due to the presence of one lone pair and three bond pairs around the central phosphorus atom. According to VSEPR theory, the ideal bond angle for a molecule with three bond pairs and one lone pair would be less than the tetrahedral angle (109.5°) due to lone pair-bond pair repulsion. However, as we move down Group 15 from Nitrogen to Phosphorus, the electronegativity of the central atom decreases, and its size increases. This leads to a decrease in the s-character of the hybrid orbitals involved in bonding and an increase in the p-character. In PH₃, the bonding is primarily due to the overlap of almost pure p-orbitals of phosphorus with the s-orbitals of hydrogen. Pure p-orbitals are oriented at 90° to each other. Therefore, the bond angle in PH₃ is close to 90° (specifically, it is around 93.5°).

Option Analysis:

• A) Close to 109.28°: This is the ideal tetrahedral angle. While PH₃ has a tetrahedral electron geometry, the lone pair reduces the bond angle significantly, and the bonding involves more p-character than in NH₃.
• B) Same as NH₃: The bond angle in NH₃ is approximately 107°. Due to the higher electronegativity of nitrogen, there is more s-character in the hybrid orbitals, leading to a larger bond angle compared to PH₃.
• D) Close to 120°: This angle is characteristic of trigonal planar geometry (e.g., BF₃), which is not the geometry of PH₃.

Correct Answer: (C)