How will you account for 104.5^ bond angle in water? - Vidyadip

Question

How will you account for $104.5^{\circ}$ bond angle in water?

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Answer

In $\mathrm{H}_2 \mathrm{O}$ molecule the oxygen is $\mathrm{sp}^3$-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 ${ }^{\circ}$, but the actual angle is $104.5^{\circ}$. The lone pair-lone pair repulsions are greater than bond pair-bond pair repulsions. As a result, bond angle in water is reduced from $109.5^{\circ}$ to $104.5^{\circ}$.

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