The hybridization in \[{\rm{B}}{{\rm{F}}_{\rm{3}}}\] molecule is
A) \[sp\]
B) \[s{p^2}\]
C) \[s{p^3}\]
D) \[s{p^3}d\]

Hint: We know that hybridization is the process of mixing of orbitals of similar nature to give a new set of orbitals that have different shapes and energies. Here, we have to use a formula to know the hybridization of the molecule.

Formula Used:The formula of hybridization is,
\[H = \dfrac{{V + X - C + A}}{2}\]
Here, V is for valence electrons, X is the count of monovalent atoms that are bonded with the central atom, C is for cationic charge and A is for anionic charge.

Complete step by step solution:If the value of H is 4, that means, hybridization of the compound is \[s{p^3}\].
If the value of H is 3, that means, hybridization of the compound is \[s{p^2}\].
And if the value of H is 2,that means, the hybridization of the compound is \[sp\] .
Let’s check the hybridization of \[{\rm{B}}{{\rm{F}}_{\rm{3}}}\] molecule.
The valence electrons of B are 3 and the count of monovalent atoms in the compound is 3.
So,
\[H = \dfrac{{3 + 3}}{2} = 3\]
Therefore, the \[{\rm{B}}{{\rm{F}}_{\rm{3}}}\]molecule has the hybridization of \[s{p^2}\] .

Hence, option B is right.

Note:Another way to check hybridization of a compound is to check the count of groups surrounding the central atom. The surrounding groups are atoms that form bonds with the central atom and the lone pairs on the central atom. In \[{\rm{B}}{{\rm{F}}_{\rm{3}}}\], only three fluorine atoms form bonds with the boron atom and no lone pair is present on the Boron atom. So, the count of groups is three and therefore, the hybridization is \[s{p^2}\] .