The bond angle in a molecule is primarily determined by the hybridization of the central atom and the number of lone pairs on it. According to VSEPR theory, lone pairs repel more strongly than bond pairs, leading to a reduction in bond angles.

Let's analyze the options:

• A) NH₃: Nitrogen is in Group 15, like arsenic. In NH₃, N is sp³ hybridized with one lone pair. The ideal tetrahedral angle is 109.5°, but due to the lone pair, the bond angle is reduced to approximately 107°. AsH₃ also has a central atom (As) from Group 15 with one lone pair. However, as we move down the group, the electronegativity of the central atom decreases, and the bond pair-bond pair repulsion decreases, leading to a smaller bond angle. Also, the tendency for hybridization decreases down the group, and the bond angle approaches 90° (p-orbital bonding). Therefore, the bond angle in AsH₃ (approx. 91.8°) is less than that in NH₃ (approx. 107°).
• B) H₂O: Oxygen is in Group 16. In H₂O, O is sp³ hybridized with two lone pairs. The ideal tetrahedral angle is 109.5°, but due to two lone pairs, the bond angle is significantly reduced to approximately 104.5°. The bond angle in AsH₃ (approx. 91.8°) is less than that in H₂O.
• C) BCl₃: Boron is in Group 13. In BCl₃, B is sp² hybridized with no lone pairs. The molecular geometry is trigonal planar, and the bond angle is exactly 120°. The bond angle in AsH₃ (approx. 91.8°) is less than that in BCl₃.

In all the given options (NH₃, H₂O, BCl₃), the bond angles are greater than the bond angle in AsH₃. Therefore, the bond angle in AsH₃ is not greater than any of the given options.

Correct Answer: (D)