How will you account for 104.5^(@) bond angle in water ?

1.	How will you account for 104.5^(@) bond angle in water ?
Answer» Solution :In water, O undergoes `sp^(3)`-hybridization and hence the `angleHOH` should have been `109^(@)-28'`. In `H_(2)O`, the oxygen atom is SURROUNDING by two SHARED pairs and two lone pairs of electrons. But acconding to VSEPR THEORY, lone pair-lone pair repulsions are stronger than bondpair-BOND pair repulsions. As a result, the HOH bond angle in water slightly decreases from the regular tetrahedral angle of `109^(@)-28'` to `104.5^(@)` .

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