Question

The correct order of hybridization of the central atom in the following species $N{H_3},{\text{ }}{[PtC{l_4}]^{2 - }},{\text{ }}PC{l_5}\,and{\text{ }}\,BC{l_3}\,is\,:$
A. $ds{p^2},\,ds{p^3},\,s{p^2},s{p^3}$
B. $s{p^3},\,ds{p^2},\,s{p^3}d,s{p^2}$
C. $ds{p^2},\,s{p^2},\,s{p^3},ds{p^3}$
D. $ds{p^2},\,s{p^3},\,s{p^2},ds{p^3}$

Answer 1

Hybridization is the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of a new set of orbitals of equivalent energies and shape.

Complete step by step answer:
Hybridization in $N{H_3}$: In $N{H_3}$ molecule, $N$ is the central atom, the electronic configuration of $N = 1{s^2}2{s^2}2{p^3}$.
There are four $s{p^3}$ hybrid orbitals of nitrogen atom of ammonia which are formed by overlapping three half-filled orbitals of nitrogen atom with $s$ orbitals of $3$ hydrogen atoms. Geometry of ammonia is pyramidal or distorted tetrahedral due to the presence of lone pairs.

Hybridization in ${\left[ {PtC{l_4}} \right]^{2 - }}$
Atomic number of $Pt$ is $78,$ its electronic configuration is $5{d^9}6{s^1},$ the oxidation state of$Pt$in ${\left[ {PtC{l_4}} \right]^{2 - }}$ complex is $+ 2.$

The four-chlorine atom filled the empty orbitals. Therefore, the hybridization of ${\left[ {PtC{l_4}} \right]^{2 - }}{\text{is ds}}{{\text{p}}^2}$
Complex should be tetrahedral instead of square planar theoretically but the size of $Pt$ is large enough that it forms a strong bond with ligands. Due to which strong repulsion between the electron of $Pt$ and ligands takes place which results in strong crystal field splitting. The strong field splitting breaks the degeneracy of $d{x^2} - {y^2}{\text{ and d}}{{\text{z}}^2}$ orbitals. Hence stabilized the square planar arrangement more than tetrahedral thus, it should be square planar.
Hybridization of $PC{l_5} \to P$ is the central atom. Its electronic configuration is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^5}$. Phosphorus has $5$ valence electrons.

The five $s{p^3}d$ hybrid orbitals are singly occupied. These hybrid orbitals overlap with singly filled $3{p_z}$ atomic orbital of five chlorine atom to form five sigma bond $\left( {P - Cl} \right).$ Geometry of $PC{l_5}$ molecule is trigonal bipyramidal. Then the hybridization of $PC{l_5}$is$s{p^3}d$.
Hybridization of $BC{l_3}:$ Boron (B) is the central atom in $BC{l_3}$, its electronic configuration is $1{s^2}2{s^2}2{p^1}$. Boron has $3$ valence electrons in their outermost shell. The three $s{p^2}$ hybrid orbitals are singly occupied. These hybrid orbitals overlap with singly filled $3{p_z}$ orbital of three chlorine atom to from three sigma bond $\left( {B - Cl} \right).$ Geometry of $BC{l_3}$ is trigonal planar and hybridization is $s{p^2}.$

Hence, the correct option is (B).

Note: Hybridization is a theoretical concept which has been introduced to explain some structural properties, such as shape of molecules or equivalency of bonds, etc. which cannot be explained by simple theories of valency.