Question

In the formation of compounds, bromine undergoes $\text{s}{{\text{p}}^{3}}{{\text{d}}^{2}}$ hybridisation in:
(A) ground state
(B) 1st excited state
(C) 2nd excited state
(D) 3rd excited state

Answer 1

For this problem, firstly we have to write the electronic configuration of the bromine in its ground state after which we have to excite the electrons so that it can make required bonds. After which we can write the hybridisation of the excited state of the bromine and choose the correct answer.

Complete step by step solution:
- In the given question, we have to explain the state of the bromine in the hybridisation of $\text{s}{{\text{p}}^{3}}{{\text{d}}^{2}}$ that whether it is in the ground state or the excited state.
- As we know that the s subshell has only one orbital, p subshell has three orbitals and d subshell has five orbitals.
- So, firstly we will write the electronic configuration of the bromine element whose atomic number is 35 i.e.
$\text{1}{{\text{s}}^{2}}\text{ 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{6}}\text{ 3}{{\text{s}}^{2}}\text{ 3}{{\text{p}}^{6}}\text{ 4}{{\text{s}}^{2}}\text{ 3}{{\text{d}}^{10}}\,4{{\text{p}}^{5}}$
- Now, in hybridisation, $\text{s}{{\text{p}}^{3}}{{\text{d}}^{2}}$ the orbitals that participate are 4s, 4p and 4d. As the energy of the 4d shell is lower than that of the 5s that's why the electron will excite into the 4d shell.
- Then the electronic configuration of the first excited state of the bromine will be:
$\text{1}{{\text{s}}^{2}}\text{ 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{6}}\text{ 3}{{\text{s}}^{2}}\text{ 3}{{\text{p}}^{6}}\text{ 4}{{\text{s}}^{2}}\text{ 3}{{\text{d}}^{10}}\,4{{\text{p}}^{4}}\text{ 4}{{\text{d}}^{1}}$
- Here, the hybridisation of the bromine will be $\text{s}{{\text{p}}^{2}}{{\text{d}}^{1}}$
- Now, the second electron will excite into the 4d shell and the electronic configuration becomes:
$\text{1}{{\text{s}}^{2}}\text{ 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{6}}\text{ 3}{{\text{s}}^{2}}\text{ 3}{{\text{p}}^{6}}\text{ 4}{{\text{s}}^{2}}\text{ 3}{{\text{d}}^{10}}\,4{{\text{p}}^{3}}\text{ 4}{{\text{d}}^{2}}$
- And here the hybridisation of the bromine atom will be $\text{s}{{\text{p}}^{3}}{{\text{d}}^{2}}$.

Therefore, option (C) is the correct answer.

Note: Compounds such as $\text{Br}{{\text{F}}_{5}}$ show the hybridisation of $\text{s}{{\text{p}}^{3}}{{\text{d}}^{2}}$ and the shape of the molecule will be square pyramidal. Hybridisation means the mixing of the orbitals of the atoms to form new hybrid orbitals.