Explain: The geometry of $NH_3$ molecule is trigonal pyramidal.

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(N/A) This geometry is explained by the $sp^3$ hybridization of $N$ in $NH_3$ and the $VSEPR$ theory.
In $NH_3$,the valence shell electronic configuration of Nitrogen is $2s^2 2p_x^1 2p_y^1 2p_z^1$. It undergoes $sp^3$ hybridization to form four $sp^3$ hybrid orbitals,three of which contain unpaired electrons and one contains a lone pair of electrons.
These three hybrid orbitals overlap with $1s$ orbitals of three hydrogen atoms to form three $N-H$ sigma bonds. One $sp^3$ orbital remains occupied by a non-bonding lone pair.
According to $VSEPR$ theory,the force of repulsion between a lone pair and a bond pair is greater than the force of repulsion between two bond pairs.
Due to this lone pair-bond pair repulsion,the molecule gets distorted from the ideal tetrahedral geometry,and the bond angle is reduced from $109.5^{\circ}$ to $107^{\circ}$. Thus,the geometry of the $NH_3$ molecule becomes trigonal pyramidal.

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