Question

Given,$B{F_3} + {F^ - } \to BF_4^ - $
What is the hybridization state of $B$ in $B{F_3}$ and $BF_4^ - $ ?
$(A)s{p^2},s{p^3}$
$(B)s{p^3},s{p^3}$
$(C)s{p^2},s{p^2}$
$(D)s{p^3},s{p^3}d$

Answer 1

Hint: Hybridization can be defined as the concept in which mixing of two or more orbitals lead to the formation of new types of degenerate orbitals. Hybridization of the molecule can be calculated using the number of electrons in the valence shell of the central atom and the charge present on the molecule by applying a formula which we will see in the solution.

Complete answer:
We know that hybridization can be defined as the concept in which mixing of two or more orbitals lead to the formation of new types of degenerate orbitals. The new orbitals formed have different properties and are degenerate in nature (have the same energy level).
We can calculate the hybridisation state of inorganic compounds using a formula which is given below:
$Hybridization = \dfrac{1}{2}\left[ {V + M + A - C} \right]$
Where, V is the number of valence electrons on the central atom.
M is the number of monovalent species.
A stands for anion and C stands for cation.
The value of hybridization comes can be written as:
$2 - sp$
$3 - s{p^2}$
$4 - s{p^3}$
And so on.
For $B{F_3}$
Put the values in above formula:
$Hybridization = \dfrac{1}{2}\left[ {3 + 3 + 0 - 0} \right]$
$Hybridization = 3$
Therefore hybridization of $B$ in $B{F_3}$ is $s{p^2}.$
For $BF_4^ - $
Put the values in above formula:
$Hybridization = \dfrac{1}{2}\left[ {3 + 4 + 1 - 0} \right]$
$Hybridization = 4$
Therefore hybridization of $B$ in $BF_4^ - $ is $s{p^3}.$
So, the correct answer is “Option A”.

Note:
 The orbitals which belong to the same atom or ion can be hybridized. The total number of hybrid orbitals are always equal to the number of orbitals participating in the hybridization process. All the hybrid orbitals have the same shape and energy (they are degenerate in nature). Hybrid orbitals form more stable bonds as compared to pure orbitals.