
When $\text{N}{{\text{H}}_{\text{3}}}$ is treated with HCl, $\text{H -- N -- H}$ bond angle:
A. increases
B. decreases
C. remains same
D. depends upon temperature
Answer
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Hint: The increase or decrease of the bond angle of a compound is determined by the s character. By s character we mean the s hybridisation. So, to determine the increase or decrease of the bond angle, we need to find the hybridisation of the compound.
Complete step-by-step answer:
We know that $\text{N}{{\text{H}}_{\text{3}}}$ has a bond angle of $\text{107 }\!\!{}^\circ\!\!\text{ }$. This bond angle is less as compared to the tetrahedral angle, owing to the increase in the repulsion to bond pairs. This happens due to the presence of lone pairs of nitrogen.
Now on reacting with HCl, $\text{N}{{\text{H}}_{\text{3}}}$ results in the formation of $\text{N}{{\text{H}}_{\text{4}}}\text{Cl}$. The reaction is given below:
$\text{N}{{\text{H}}_{\text{3}}}\text{ }+\text{ HCl}\xrightarrow{{}}\text{ N}{{\text{H}}_{\text{4}}}\text{Cl}$
In this case, $\text{N}{{\text{H}}_{\text{4}}}^{\text{+}}$ has no longer any lone pair at nitrogen. Hence, the ammonium ion is perfectly tetrahedral, just like the $\text{C}{{\text{H}}_{\text{4}}}$. So, now the bond angle is $\text{109}\text{.5 }\!\!{}^\circ\!\!\text{ }$.
Therefore, we can say that when $\text{N}{{\text{H}}_{\text{3}}}$ is treated with HCl, $\text{H -- N -- H}$ the bond angle increases.
So, the correct answer is Option A.
Both $\text{N}{{\text{H}}_{\text{3}}}$ and $\text{N}{{\text{H}}_{\text{4}}}^{\text{+}}$ are $\text{s}{{\text{p}}^{\text{3}}}$ hybridized. Here is the structure of both of the structures.
Note: Angle between two adjacent bonds at an atom in a molecule made up of three or more atoms is known as the bond angle. Bond angle is affected by the presence of a lone pair of electrons at the central atom.
The electronegativity of elements decreases as we move down the group. The lone pair - bond pair repulsion is stronger as compared to the lone pair - lone pair or bond pair - bond pair interaction. But as electronegativity decreases, the repulsion also decreases which results in a decrease in bond angle.
Complete step-by-step answer:
We know that $\text{N}{{\text{H}}_{\text{3}}}$ has a bond angle of $\text{107 }\!\!{}^\circ\!\!\text{ }$. This bond angle is less as compared to the tetrahedral angle, owing to the increase in the repulsion to bond pairs. This happens due to the presence of lone pairs of nitrogen.
Now on reacting with HCl, $\text{N}{{\text{H}}_{\text{3}}}$ results in the formation of $\text{N}{{\text{H}}_{\text{4}}}\text{Cl}$. The reaction is given below:
$\text{N}{{\text{H}}_{\text{3}}}\text{ }+\text{ HCl}\xrightarrow{{}}\text{ N}{{\text{H}}_{\text{4}}}\text{Cl}$
In this case, $\text{N}{{\text{H}}_{\text{4}}}^{\text{+}}$ has no longer any lone pair at nitrogen. Hence, the ammonium ion is perfectly tetrahedral, just like the $\text{C}{{\text{H}}_{\text{4}}}$. So, now the bond angle is $\text{109}\text{.5 }\!\!{}^\circ\!\!\text{ }$.
Therefore, we can say that when $\text{N}{{\text{H}}_{\text{3}}}$ is treated with HCl, $\text{H -- N -- H}$ the bond angle increases.
So, the correct answer is Option A.
Both $\text{N}{{\text{H}}_{\text{3}}}$ and $\text{N}{{\text{H}}_{\text{4}}}^{\text{+}}$ are $\text{s}{{\text{p}}^{\text{3}}}$ hybridized. Here is the structure of both of the structures.
Note: Angle between two adjacent bonds at an atom in a molecule made up of three or more atoms is known as the bond angle. Bond angle is affected by the presence of a lone pair of electrons at the central atom.
The electronegativity of elements decreases as we move down the group. The lone pair - bond pair repulsion is stronger as compared to the lone pair - lone pair or bond pair - bond pair interaction. But as electronegativity decreases, the repulsion also decreases which results in a decrease in bond angle.
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