Question

The hybridization of $NH_4^ + $ ion is:
A. $sp$
B. $s{p^2}$
C. $s{p^3}$
D. $s{p^3}d$

Answer 1

Hint: In chemistry, the orbital hybridization is the concept of mixing atomic orbitals of different energies into new hybrid orbitals of same energy levels but with different energies than the previous individual atomic ones, different in shapes, etc., suitable for the pairing of electrons to form chemical bonds in valence bond theory.

Complete step by step answer:
Orbitals are a model representation of the behavior of electrons within molecules. The orbitals are space inside the atom where there is a maximum probability of finding an electron. In the case of simple hybridization, this approximation is based on atomic orbitals, similar to those obtained for the hydrogen atom, the only neutral atom for which the Schrödinger equation can be solved exactly. In heavier atoms, such as carbon, nitrogen, and oxygen, the atomic orbitals used are the \[2s\] and \[2p\] orbitals, similar to excited state orbitals for hydrogen. Hybrid orbitals are assumed to be mixtures of atomic orbitals, superimposed on each other in various proportions.
In the case of an ammonium ion ($NH_4^ + $), the hybridization can be determined by the general formula:
Number of hybrid orbitals = $\dfrac{1}{2}(N + B - C)$
Where, \[N\] is the number of valence electrons in the central atom, $B$ is the number of atoms surrounding it and $C$ is the total charge on the atom.
As per the given ion, $N = 5$
$B = 4$
$C = 1$
Thus, substituting the values and solving, we have:
Number of hybrid orbitals = $\dfrac{1}{2}(5 + 4 - 1) = 4$
This means that the total number of atomic orbitals that are needed to constitute 4 hybrid orbitals is = 4.
Hence, the hybridization of ammonium ion ($NH_4^ + $) = $s{p^3}$

So, the correct answer is “Option A”.

Note:
As ammonium ion is $s{p^3}$hybridized, this means that it is tetrahedral in structure and all the bonds are substituted to each other at equal angles. The possible structure of the ammonium ion is as follows: