Concept: The bond angle and dipole moment of a molecule are determined by its molecular geometry, hybridization, and the electronegativity difference between the bonded atoms, as well as the presence of lone pairs.

Why (C) is correct:

Water ($H_{2}O$) has a central oxygen atom bonded to two hydrogen atoms and two lone pairs of electrons. The oxygen atom is $s{p}^3$ hybridized. According to VSEPR theory, the four electron domains (two bonding pairs and two lone pairs) would ideally lead to a tetrahedral geometry with a bond angle of $109.5^{\circ }$. However, the lone pairs exert greater repulsion than bonding pairs, compressing the H-O-H bond angle. This repulsion reduces the bond angle from the ideal $109.5^{\circ }$ to approximately $104.5^{\circ }$.

Due to the significant electronegativity difference between oxygen and hydrogen, the O-H bonds are polar. The bent molecular geometry ensures that the individual bond dipoles do not cancel out, resulting in a net dipole moment. The experimental value for the dipole moment of water is approximately $1.84\mathrm{D}$.

Option Analysis:

• A) $109.5^{\circ },1.84\mathrm{D}$: $109.5^{\circ }$ is the ideal tetrahedral angle, not the actual bond angle in water due to lone pair repulsion.
• B) $107.5^{\circ },1.56\mathrm{D}$: $107.5^{\circ }$ is the bond angle of ammonia ($N{H}_3$), which has one lone pair and three bonding pairs. The dipole moment is also incorrect.
• D) $102.5^{\circ },1.56\mathrm{D}$: Both the bond angle and dipole moment values are incorrect for water.

Correct Answer: (C)