Question
Explain geometry of $H_2O$ molecule according to $\text{VSEPR}$ theory.
✓
Answer
In $H_2O$ molecule, the central atom oxygen has six electrons in its valence shell. On bond formation with two hydrogen atoms, there are $8$ electrons in the valence shell of oxygen. Out of these, two pairs are bond pairs and two are lone pairs.
Due to lone pair – lone pair repulsion, the lone pairs are pushed towards the bond pairs and bond pair – lone pair repulsions become stronger thereby reducing the $H-O-H$ bond angle from $109^\circ 28′ \ to \ 104^\circ 35′$ and the geometry of the molecule becomes angular $($bent$).$
Diagram:
Due to lone pair – lone pair repulsion, the lone pairs are pushed towards the bond pairs and bond pair – lone pair repulsions become stronger thereby reducing the $H-O-H$ bond angle from $109^\circ 28′ \ to \ 104^\circ 35′$ and the geometry of the molecule becomes angular $($bent$).$
Diagram:
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