Question
Explain geometry of $NH_3$ molecule according to $\text{VSEPR}$ theory.
✓
Answer
In $NH_3$ molecule, the central atom nitrogen has five electrons in its valence shell. On bond formation with three hydrogen atoms, there are $8$ electrons in the valence shell of nitrogen. Out of these, three pairs are bond pairs $(N – H$ covalent bonds$)$ and one forms lone pair. The expected geometry is tetrahedral and bond angle is $109^\circ 28′.$
There are two types of repulsions between the electron pairs: Lone pair – bond pair and bond pair – bond pair
The lone pair – bond pair repulsions are stronger and the bonded pairs are pushed inwards. Thus, reducing the bond angle to $107^\circ 18′$ and shape of the molecule becomes trigonal pyramidal.
Diagram:
There are two types of repulsions between the electron pairs: Lone pair – bond pair and bond pair – bond pair
The lone pair – bond pair repulsions are stronger and the bonded pairs are pushed inwards. Thus, reducing the bond angle to $107^\circ 18′$ and shape of the molecule becomes trigonal pyramidal.
Diagram:
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