Question

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

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

Answer 1

The correct option is C $2^{nd}$ excited state

In the formation of compounds (such as $Br{F}_5$), bromine undergoes $s{p}^3{d}^2$ hybridization in $2^{nd}$ excited state. The ground state electronic configuration of bromine is $4s^{2}4p^5$. The electronic configuration of bromine in second excited state is $4s^{1}4p^{3}4d^2$. This results in the octahedral arrangement of lone pairs. The molecule has a square pyramid shape. It contains one lone pair and five bond pairs.