Answer : The hybridization of the central atom in the tribromide anion is, [tex]sp^3d[/tex]
Explanation :
The given molecule is, [tex]Br^-_3[/tex]
Hybridization is given by the formula:
[tex]H=\frac{1}{2}(V+X-C+A)[/tex]
where,
V = number of valence electrons in central atom
X = number of monovalent atoms around the central atom
C = positive charge on cation
A = negative charge on anion
Bromine is the central atom which has 7 valence electrons.
Number of monovalent atoms = 2
Negative charge = 1
Now put all this values in the given formula, we get
[tex]H=\frac{1}{2}(7+2-0+1)=5[/tex]
If the value of 'H' is '5' then the hybridization is [tex]sp^3d[/tex]
Number of bond pair = 2
Number of lone pair = 5 - 2 = 3
Therefore, the hybridization of the central atom in the tribromide anion is, [tex]sp^3d[/tex]
The geometry is shown below.
