Question

In the formation of compounds bromine undergoes $s{p}^3{d}^2$ hybridisation in:

• A. ground state $X$
• B. 1st excited state
• C. 2nd excited state
• D. 3rd excited state

Answer 1

The correct option is C 2nd excited state

Electronic configuration of ground state of $Br$ is:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^5$

Atomic orbitals tha participate in $s{p}^3{d}^2$ are:

$4s,4p,4d$

Electronic configuration of $Br$: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{3}4d^2$

Although $5s$ having lower energy than $4d$ is the first excited state remains empty and electrons jump to second excited state i.e. $4d$.

Hence, option $C$ is correct.