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# In ammonia, the bond angle of $H-N-H$ is ______.A. $106.7{}^\circ$B. $104.5{}^\circ$C. $109.5{}^\circ$D. $120{}^\circ$

Last updated date: 19th Jun 2024
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Hint: Think about the geometry of the ammonia molecule according to the hybridization of the nitrogen atom. Consider the standard angle known for that type of geometry and then take into consideration the number of lone pairs and bond pairs present.

Complete step by step solution:
We know that the number of valence electrons in the nitrogen atom is 7. The electronic configuration of nitrogen is $1{{s}^{2}}2{{s}^{2}}2{{p}^{3}}$. So, the ground state and hybridized state of the nitrogen atom will be:
- Ground state

- Hybridized state

So, here we can see that the second orbital undergoes $s{{p}^{3}}$ hybridization and forms four degenerate (having same energy) orbitals. One of those orbitals will be occupied by a lone pair and the other three will be occupied by bond pairs of electrons. We know that the atoms that undergo $s{{p}^{3}}$ hybridization have a tetrahedral geometry, and the standard angle for tetrahedral geometry is $109.5{}^\circ$. But here, we need to consider the fact that not all the orbitals are occupied by bond pair interactions. One of the orbitals is occupied by the lone pair. We know that the repulsion between two lone pairs is the most followed by the repulsion between a lone pair and a bond pair and the repulsion between two bond pairs is the least. So, $\text{lone pair-lone pair}>\text{lone pair-bond pair}>\text{bond pair-bond pair}$.
The standard angle between two orbitals in tetrahedral geometry will be reduced due to the presence of this lone pair. The angle will be less than $109.5{}^\circ$. The structure of the ammonia molecule will be:

Here, we can see that the angle of the bond $H-N-H$ is less than the standard angle.

Hence, the answer to this question is ‘A. $106.7{}^\circ$’.

Note: Remember that the methane molecule has the standard tetrahedral geometry so it will exhibit the standard angle, the presence of a lone pair changes the angle in case of ammonia. Considering the water molecule too, oxygen has 2 lone pairs so the angle between $H-O-H$ will be even lesser than $106.7{}^\circ$.