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

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(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.

Explore More

Similar Questions

In the $OF_2$ molecule,what are the number of bonding electron pairs and lone pairs of electrons present,respectively?

The correct shape and $I-I-I$ bond angle respectively in $I_{3}^{-}$ ion are :-

The number of lone pairs of electrons on the central $I$ atom in $I_{3}^{-}$ is $.......$

The correct order of bond angles in the following compounds/species is:

Which set of species is arranged in order of increasing $O-N-O$ bond angle?

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo