
The nitrogen atoms in \[N{H_3},N{H_2}^ - \] and $N{H_4}^ + $ are all surrounded by eight electrons. When these species are arranged in the increasing order of bond angle, correct order is
(A) $N{H_3},N{H_2}^ - ,N{H_4}^ + $
(B) $N{H_4}^ + ,N{H_2}^ - ,N{H_3}$
(C) $N{H_3},N{H_4}^ + ,N{H_2}^ - $
(D) $N{H_2}^ - ,N{H_3},N{H_4}^ + $
Answer
233.1k+ views
Hint: As the total number of shared electrons are mentioned with this the geometry of the molecules can be determined easily.
The bond angle will be affected by the lone pairs.
Complete step by step solution:
-The central atom in all the three structures is the Nitrogen atom.
-The structure of any compound depends on the number of electron pairs shared.
-Electron pairs consist of bond pairs as well as lone pairs.
-The geometry of the compound changes as the number of lone pairs increases.
-In the question, it is mentioned that eight electrons are shared in each structure which means 4 electron pairs. So accordingly the structure will be Tetrahedral.
-Bond angle is the angle between two bonds originating from the same central atom.
-Bond angle changes as the number of lone pairs increases.
-Structure of $N{H_4}^ + $

-In$N{H_4}^ + $ there are 4 bond pairs and zero lone pairs so the structure and geometry will be the same as the tetrahedral. The bond angle is 109⁰.
-Structure of $N{H_3}$

-In$N{H_3}$ there are 3 bond pairs and one lone pair so though the structure is tetrahedral but the geometry is trigonal pyramidal. The geometry is changed due to lone pairs. The bond angle is less than 109⁰ due to repulsion between lone pairs and bond pairs.
-Structure of $N{H_2}^ - $

-In $N{H_2}^ - $there are 2 bond pairs and 2 lone pairs so though the structure is tetrahedral but the geometry is bent. The geometry is again changed due to lone pairs. The bond angle is less than 109⁰ due to repulsion between two lone pairs along with that lone pair and bond pairs.
So the order for bond angle is $N{H_2}^ - ,N{H_3},N{H_4}^ + $
That means option (D) is correct.
Note: -The decreasing order of repulsion for the two types of electron pairs are
Lone pair-lone pairlone pair-bond pair bond pair-bond pair.
-Structure and geometry will not be the same always.
The bond angle will be affected by the lone pairs.
Complete step by step solution:
-The central atom in all the three structures is the Nitrogen atom.
-The structure of any compound depends on the number of electron pairs shared.
-Electron pairs consist of bond pairs as well as lone pairs.
-The geometry of the compound changes as the number of lone pairs increases.
-In the question, it is mentioned that eight electrons are shared in each structure which means 4 electron pairs. So accordingly the structure will be Tetrahedral.
-Bond angle is the angle between two bonds originating from the same central atom.
-Bond angle changes as the number of lone pairs increases.
-Structure of $N{H_4}^ + $

-In$N{H_4}^ + $ there are 4 bond pairs and zero lone pairs so the structure and geometry will be the same as the tetrahedral. The bond angle is 109⁰.
-Structure of $N{H_3}$

-In$N{H_3}$ there are 3 bond pairs and one lone pair so though the structure is tetrahedral but the geometry is trigonal pyramidal. The geometry is changed due to lone pairs. The bond angle is less than 109⁰ due to repulsion between lone pairs and bond pairs.
-Structure of $N{H_2}^ - $

-In $N{H_2}^ - $there are 2 bond pairs and 2 lone pairs so though the structure is tetrahedral but the geometry is bent. The geometry is again changed due to lone pairs. The bond angle is less than 109⁰ due to repulsion between two lone pairs along with that lone pair and bond pairs.
So the order for bond angle is $N{H_2}^ - ,N{H_3},N{H_4}^ + $
That means option (D) is correct.
Note: -The decreasing order of repulsion for the two types of electron pairs are
Lone pair-lone pairlone pair-bond pair bond pair-bond pair.
-Structure and geometry will not be the same always.
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