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How will your account for 104.5^@ bond angle in water ? |
Answer» Solution :In water OXYGEN undergoes `sp^3` HYBRIDISATION so H-O-H bond angle should have been `109^@28.` but In water, the oxygen atomis SURROUNDED by two shared pairs two lone pairs ELECTRONS. H And ACCORDING to VSEPR theory, lone pair - lone pair repulsions are stronger than bond pair-bond pair repulsions. As a result, the HOH bond angle in water slightly decreases from the regular bond angle of `109^@ 28.` to `104.5^@`. 
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