The bond angle in a water molecule H2O is approximately 104.5 degrees. This bond angle can be predicted using the Valence Shell Electron Pair Repulsion VSEPR theory and the electron pair geometry of the molecule.According to the VSEPR theory, the electron pairs around the central atom will arrange themselves in a way that minimizes the repulsion between them. In the case of a water molecule, the central atom is oxygen, which has two lone pairs of electrons and two bonding pairs of electrons one for each hydrogen atom .The electron pair geometry of the water molecule is tetrahedral, as there are four electron pairs two lone pairs and two bonding pairs around the oxygen atom. However, the molecular geometry, which describes the arrangement of atoms in the molecule, is bent or angular due to the presence of the two lone pairs.The bond angle in a perfect tetrahedral geometry would be 109.5 degrees. However, in the case of a water molecule, the presence of the two lone pairs on the oxygen atom causes a greater repulsion between the lone pairs and the bonding pairs, pushing the hydrogen atoms closer together. This results in a smaller bond angle of approximately 104.5 degrees.