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Question: In the formation of compounds bromine undergoes \(s{p^3}{d^2}\) hybridization in. A) Ground state ...

In the formation of compounds bromine undergoes sp3d2s{p^3}{d^2} hybridization in.
A) Ground state X.
B) First excited state.
C) Second excited state.
D) Third excited state.

Explanation

Solution

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, 1s2 2s22p6 3s23p64s2 3d104p51{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 sp3d2s{p^3}{d^2} are 4s,4p,4d4s,4p,4d.
The excited state electronic configuration of Bromine is 1s2 2s22p6 3s23p64s2 3d104p34d21{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 5s5s having lower energy than 4d4d 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 BrF5Br{F_5} , bromine undergoes sp3d2s{p^3}{d^2}hybridization in its second excited state. We know that the ground state electronic configuration of bromine is 4s2 4p54{s^2}{\text{ }}4{p^5} and the electronic configuration of bromine in second excited state is 4s14p3 4d24{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 sp3d2{\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}} hybridization:
The sp3d2{\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}} hybridization has 1s, 3p and 2d orbitals they undergo intermixing to form 6 identical sp3d2{\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 90o{\text{9}}{{\text{0}}^{\text{o}}} to one another. The central atoms which have sp3d2{\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}} hybrid orbitals forms the bonds with bond angle of 90o{\text{9}}{{\text{0}}^{\text{o}}}.