How will you account for 104.5° bond angle in water?

1.	How will you account for 104.5° bond angle in water?
Answer» In H₂O molecule the oxygen is sp³ −hybridizedand thus, tetrahedral configuration comes into existence. Two positions are occupied by H atoms by forming sigma bonds with two hybrid orbitals and two positions are occupied by lone pairs. The expected bond angle should be109.5^∘, but the actual angle is 104.5^∘. The lone pair-lone pair repulsions are greater than bond pair-bond pair repulsions. As a result, bond angle in water is reduced from 109.5^o to 104.5^o.

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How will you account for 104.5° bond angle in water?

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