Question

In the formation of compounds bromine undergoes $s{p^3}{d^2}$ hybridization in.
A) Ground state X.
B) First excited state.
C) Second excited state.
D) Third excited state.

Answer 1

Hint:Hybridization is the concept in which atomic orbitals combine to make a new hybridized orbital which successively, influences molecular geometry and bonding properties.
We know that the electrons which are present at the outermost shell of an atom are called valence electrons and the valency of an electron is that the number of electrons during which atom accepts or donate to make a bond.

Complete step by step answer:We can write the ground state electronic configuration of Bromine as, \[1{s^{2{\text{ }}}}2{s^2}2{p^6}{\text{ }}3{s^2}3{p^6}4{s^{2{\text{ }}}}3{d^{10}}4{p^5}\].
We must remember that the atomic orbitals that participate in $s{p^3}{d^2}$ are \[4s,4p,4d\].
The excited state electronic configuration of Bromine is \[1{s^{2{\text{ }}}}2{s^2}2{p^6}{\text{ }}3{s^2}3{p^6}4{s^{2{\text{ }}}}3{d^{10}}4{p^3}4{d^2}\]
Although \[5s\] having lower energy than \[4d\] is the first excited state remains empty and electrons jump to the second excited state.

Therefore, the option C is correct.

Example: In the formation of compounds like \[Br{F_5}\] , bromine undergoes $s{p^3}{d^2}$hybridization in its second excited state. We know that the ground state electronic configuration of bromine is \[4{s^2}{\text{ }}4{p^5}\] and the electronic configuration of bromine in second excited state is \[4{s^1}4{p^3}{\text{ }}4{d^2}\]. This leads to the octahedral arrangement of lone pairs. The molecule features a square pyramid shape.

Note: Now we can discuss about the concept of ${\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}$ hybridization:
The ${\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}$ hybridization has 1s, 3p and 2d orbitals they undergo intermixing to form 6 identical ${\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}$ hybrid orbitals. These six orbitals are located at the corners of an octahedron. They are inclined at an angle of ${\text{9}}{{\text{0}}^{\text{o}}}$ to one another. The central atoms which have ${\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}$ hybrid orbitals forms the bonds with bond angle of ${\text{9}}{{\text{0}}^{\text{o}}}$.