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