Answer: The boron atom of [tex]BBr_3[/tex] has three hybrid orbitals
Explanation: Hybridization of the molecule can be calculated by:
[tex]Hyb=\frac{1}{2}(V+H-C+A)[/tex] .....(1)
where,
V = number of valence electrons in the central metal atom
H = number of monovalent atoms
C = cationic charge
A = anionic charge
Here, electronic configuration of B = [tex]1s^22s^22p^1[/tex]
Number of valence electrons = 3
Number of monovalent atoms = 3
Cationic charge = 0
Anionic charge = 0
Putting these values in equation 1, we get
[tex]Hyb=\frac{1}{2}(3+3-0+0)=3[/tex]
As the Hybridization is 3, which means that boron has 3 hybrid orbitals.