Answer : The hybridization of the bromine in [tex]BrO_2^{-}[/tex] is, [tex]sp^3[/tex]
Explanation :
Formula used :
[tex]\text{Number of electron pair}=\frac{1}{2}[V+N-C+A][/tex]
where,
V = number of valence electrons present in central atom
N = number of monovalent atoms bonded to central atom
C = charge of cation
A = charge of anion
Now we have to determine the hybridization of the bromine in [tex]BrO_2^{-}[/tex] molecule.
The given molecule is, [tex]BrO_2^{-}[/tex]
[tex]\text{Number of electrons}=\frac{1}{2}\times [7+1]=4[/tex]
The number of electron pair are 4 that means the hybridization will be [tex]sp^3[/tex] and the electronic geometry of the molecule will be tetrahedral.
But as there are 2 atoms around the central oxygen atom, the third and fourth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be bent or angular.
Hence, the hybridization of the bromine in [tex]BrO_2^{-}[/tex] is, [tex]sp^3[/tex]
