Question

N atom in $$N{H}_4^{+}$$ ion involves the hybridization:
(A) $$sp$$
(B) $$s{p}^2$$
(C) $$s{p}^3$$
(D) $$s{p}^{3}d$$

Answer 1

Hint: We can find the hybridization and shape of the molecules using VSEPR (Valence Shell Electron Pair Repulsion) theory. According to VSEPR theory each atom present in a molecule achieves geometry that reduces the repulsions between electrons present in the valence shell of that particular atom.

Complete step by step answer:
The given molecule in the question is $$N{H}_4^{+}$$ (Ammonium ion).
The formation of Ammonium ion is as follows.
$$N{H}_3+H^{+}\to N{H}_4^{+}$$
We can represent the above equation in the form of structure as follows.
 
According to VSEPR theory we can find the hybridization of atoms in a molecule by the summation of the number of lone pairs of electrons and the number of sigma bonds.
Ammonia reacts with hydrogen ions and forms ammonium cation as the product by donating a lone pair of electrons.
In the structure of the ammonium cation we can say that nitrogen atom has four sigma bonds with four hydrogen atoms.
The ammonium does not contain any lone pair of electrons in its structure.
Means Nitrogen atom in Ammonium ion has only four bonding orbitals or 4 sigma bonds.
Therefore the hybridization of Nitrogen (N) in ammonium ion is $$s{p}^3$$.

So, the correct option is C.

Note: The hybridization of nitrogen atom in ammonia is also $$s{p}^3$$. Because the nitrogen present in ammonia molecules has three sigma bonds and one lone pair of electrons.
Therefore total number orbitals = Bonding orbitals + lone pair of electrons
=3+1
= 4
So, the hybridization of nitrogen in ammonia is $$s{p}^3$$