The bond angle of water is 104.5^{circ} due t | vedclass

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Question

(A) In $H_2O$,the oxygen atom undergoes $sp^3$ hybridization,which would ideally result in a tetrahedral geometry with a bond angle of $109.5^{\circ}$

Vedclass · Accepted Answer

Repulsion between lone pair and bond pair

Vedclass · Answer

(A) In $H_2O$,the oxygen atom undergoes $sp^3$ hybridization,which would ideally result in a tetrahedral geometry with a bond angle of $109.5^{\circ}$.However,the presence of two lone pairs on the oxygen atom exerts strong repulsion on the two $O-H$ bond pairs.According to the $VSEPR$ theory,the order of repulsion is $lp-lp > lp-bp > bp-bp$.Due to the strong $lp-lp$ repulsion,the bond angle is compressed from the ideal $109.5^{\circ}$ to $104.5^{\circ}$.

The bond angle of water is $104.5^{\circ}$ due to

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