The predicted bond angle in a water molecule H2O is approximately 104.5 degrees. This bond angle is determined using the Valence Shell Electron Pair Repulsion VSEPR theory. According to the VSEPR theory, the electron pairs around the central atom repel each other and arrange themselves in a way that minimizes this repulsion. In a water molecule, the central atom is oxygen, which has two lone pairs of electrons and two bonding pairs of electrons shared with the two hydrogen atoms .The VSEPR notation for water is AX2E2, where A is the central atom oxygen , X represents the bonded atoms hydrogen , and E represents the lone pairs of electrons. The electron pairs arrange themselves in a tetrahedral geometry, but since we only consider the positions of the hydrogen atoms, the molecular shape is bent or angular. The bond angle in a perfect tetrahedron is 109.5 degrees, but the presence of the lone pairs on the oxygen atom causes a slight distortion, reducing the H-O-H bond angle to approximately 104.5 degrees.In comparison, the bond angle in an ammonia molecule NH3 is approximately 107.3 degrees. Ammonia has a similar structure to water, with one lone pair of electrons and three bonding pairs of electrons shared with the three hydrogen atoms around the central nitrogen atom. The VSEPR notation for ammonia is AX3E1. The electron pairs also arrange themselves in a tetrahedral geometry, but the molecular shape is trigonal pyramidal due to the presence of the lone pair. The lone pair on the nitrogen atom causes a smaller distortion compared to water, resulting in a bond angle of approximately 107.3 degrees.In summary, the bond angle in a water molecule is approximately 104.5 degrees, while the bond angle in an ammonia molecule is approximately 107.3 degrees. The difference in bond angles is due to the different number of lone pairs on the central atoms and their repulsion effects on the bonding pairs of electrons.