The predicted bond angle for a water molecule H2O is approximately 104.5 degrees. This bond angle can be explained by considering the molecular geometry, electron pair repulsion, and hybridization of the central oxygen atom.1. Molecular Geometry: The molecular geometry of a water molecule is bent or V-shaped. This is because the central oxygen atom is bonded to two hydrogen atoms and has two lone pairs of electrons. The presence of these lone pairs causes the molecule to adopt a bent geometry.2. Electron Pair Repulsion: According to the Valence Shell Electron Pair Repulsion VSEPR theory, electron pairs around a central atom will arrange themselves in a way that minimizes repulsion between them. In the case of water, there are four electron pairs around the central oxygen atom two bonding pairs and two lone pairs . These electron pairs will arrange themselves in a tetrahedral geometry to minimize repulsion. However, since we only consider the positions of the hydrogen atoms when describing the molecular geometry, the shape appears bent.3. Hybridization: The central oxygen atom in a water molecule is sp3 hybridized. This means that the s and p orbitals of the oxygen atom have combined to form four equivalent hybrid orbitals, which are directed towards the corners of a tetrahedron. The bond angle in a perfect tetrahedron is 109.5 degrees. However, in a water molecule, the bond angle is slightly smaller due to the greater repulsion between the lone pairs compared to the bonding pairs.The combination of the bent molecular geometry, electron pair repulsion, and sp3 hybridization of the central oxygen atom results in a bond angle of approximately 104.5 degrees in a water molecule. This bond angle allows for the minimization of electron pair repulsion and contributes to the unique properties of water, such as its high boiling point and strong hydrogen bonding.