The bond angle in a water molecule H2O is approximately 104.5 degrees. This bond angle can be predicted based on the Valence Shell Electron Pair Repulsion VSEPR theory and the hybridization of the atoms involved.According to the VSEPR theory, the electron pairs around a central atom repel each other and arrange themselves in a way that minimizes the repulsion. In a water molecule, the central atom is oxygen, which has two bonding pairs with the two hydrogen atoms and two lone pairs of electrons. The electron pair geometry around the oxygen atom can be described as tetrahedral, with the four electron pairs two bonding pairs and two lone pairs repelling each other and trying to be as far apart as possible.The hybridization of the oxygen atom in a water molecule is sp3, which means that the oxygen atom's orbitals are hybridized to form four new, equivalent orbitals one s orbital and three p orbitals combine to form four sp3 orbitals . These sp3 hybrid orbitals are arranged in a tetrahedral geometry.However, the bond angle in a water molecule is not the ideal tetrahedral angle of 109.5 degrees. This is because the lone pairs of electrons on the oxygen atom occupy more space and repel the bonding pairs more strongly than the bonding pairs repel each other. As a result, the bond angle between the two hydrogen atoms is slightly compressed to approximately 104.5 degrees.