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What is the predicted bond angle in a water molecule and how is it determined using VSEPR theory?

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In a water molecule  H2O , the central atom is oxygen, which is bonded to two hydrogen atoms and has two lone pairs of electrons. According to the Valence Shell Electron Pair Repulsion  VSEPR  theory, the electron pairs around the central atom will arrange themselves to minimize repulsion between them.In the case of water, there are four electron pairs around the oxygen atom: two bonding pairs  from the O-H bonds  and two lone pairs. The VSEPR theory predicts that these electron pairs will adopt a tetrahedral arrangement to minimize repulsion, which corresponds to an ideal bond angle of 109.5.However, the lone pairs of electrons occupy more space than the bonding pairs, causing greater repulsion between the lone pairs and the bonding pairs. As a result, the bonding pairs are pushed closer together, and the H-O-H bond angle is reduced from the ideal tetrahedral angle.The predicted bond angle in a water molecule is approximately 104.5, which is smaller than the ideal tetrahedral angle due to the presence of the lone pairs on the oxygen atom.
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