The hybridization of any atom in a molecule can be calculated by the summation of the number of sigma bond and lone pair of electrons. Now, ammonia forms ammonium cation by the donation of lone pair on nitrogen atom . So ,in ammonium cation the ’N’ -atom formed 4 sigma bond with the four ‘H’-atom . Therefore, the hybridization of NH4+ is sp3 .

Hybridization of NH₄

To determine the hybridization of NH₄ (ammonium ion), we need to consider the central atom, which is nitrogen (N). The Lewis structure of NH₄ shows that nitrogen is bonded to four hydrogen atoms (H).

Lewis structure of NH₄:
H
|
H - N - H
|
H

Valence electrons:
Nitrogen (N) has five valence electrons, and hydrogen (H) has one valence electron each.
So, the total valence electrons in NH₄ can be calculated as follows:
(5 from N) + (4 from H) = 9

Distribution of electrons:
To distribute the electrons around the central atom (N), we start by pairing each hydrogen atom with one of the valence electrons of nitrogen. This leaves one unpaired electron on nitrogen, as shown below:

H
|
H - N - H
|
H

Drawing resonance structures:
Since nitrogen has one unpaired electron, it can form multiple resonance structures. In this case, we can draw three resonance structures by moving the unpaired electron to different hydrogen atoms.

Resonance structures of NH₄:
H H H
| | |
H - N - H H - N - H H - N - H
| | |
H H H

Hybridization:
The hybridization of an atom is determined by the number of sigma bonds and lone pairs around it. In the case of NH₄, nitrogen is bonded to four hydrogen atoms (four sigma bonds) and has no lone pairs.

Counting sigma bonds and lone pairs:
Number of sigma bonds = 4
Number of lone pairs = 0

Hybridization formula:
Hybridization = (Number of sigma bonds) + (Number of lone pairs)

Calculating hybridization for NH₄:
Hybridization = 4 + 0 = 4

Conclusion:
The hybridization of NH₄ is sp³, which means that the nitrogen atom in NH₄ forms four sigma bonds using four hybrid orbitals.