To determine the molecular geometry and bond angles of a water molecule H2O using the VSEPR theory, follow these steps:1. Determine the central atom: In H2O, oxygen O is the central atom.2. Count the valence electrons: Oxygen has 6 valence electrons, and each hydrogen atom has 1 valence electron. So, the total number of valence electrons in H2O is 6 + 1 + 1 = 8.3. Determine the electron pair geometry: Oxygen has two bonded atoms hydrogen atoms and two lone pairs of electrons. This gives a total of 4 electron groups around the central atom. According to the VSEPR theory, 4 electron groups result in a tetrahedral electron pair geometry.4. Determine the molecular geometry: Since there are two bonded atoms and two lone pairs around the central atom, the molecular geometry of H2O is bent or angular.5. Determine the bond angles: In a tetrahedral electron pair geometry, the ideal bond angle is 109.5. However, due to the presence of lone pairs, the bond angle in H2O is slightly less than 109.5. The actual bond angle in a water molecule is approximately 104.5.In summary, the molecular geometry of a water molecule H2O is bent or angular, with a bond angle of approximately 104.5, according to the VSEPR theory.