Final answer:
The HOH bond angle in water is 104.5 degrees due to the bent shape caused by the repulsive interactions between the non-bonding and bonding electron pairs around the oxygen atom, leading to a smaller angle than the ideal tetrahedral geometry.
Step-by-step explanation:
The value of the HOH bond angle in a molecule of H2O is 104.5 degrees. This is due to the bent or V-shaped geometry of the water molecule. While the oxygen atom's valence shell contains four pairs of electrons, which would typically be arranged in a tetrahedral geometry with bond angles of 109.5 degrees, the presence of two non-bonding pairs, or lone pairs, on the oxygen causes stronger repulsion against the bonding pairs. This repulsion pushes the hydrogen atoms closer together, resulting in a smaller H-O-H bond angle of 104.5 degrees. Quantum-mechanical calculations and the valence bond theory, which includes hybridization of the oxygen atom's orbitals, provide an explanation for the observed bond angle in water that deviates from the ideal tetrahedral angle.