Question

The hybrid state of B in $BF_{4}^{-}$ :
A. $s{{p}^{2}}$
B. sp
C. $s{{p}^{3}}$
D. Not specific

Answer 1

Hint: To calculate the number of hybrid orbitals or hybridization of the central atom there is a formula.
The formula to calculate hybridization on the central atom is as follows.
\[H=\dfrac{1}{2}[G+M-C+A]\]
Where H = Hybridization of the central atom
G = number of valence electrons
M = number of monovalent atoms
C = the charge on the cation
A = the charge on the anion

Complete step by step answer:
- In the question it is asked to find the hybridization of the Boron in $BF_{4}^{-}$ .
- The electronic configuration of the boron is $1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}$ .
- We have to substitute all the values in the below formula to get the hybridization of the Boron in $BF_{4}^{-}$
\[H=\dfrac{1}{2}[G+M-C+A]\]
G = number of valence electrons in boron = 3 (from electronic configuration)
M = number of monovalent atoms = 4 (fluorine atoms)
C = the charge on the cation = 0
A = the charge on the anion = 1 (negative charge in $BF_{4}^{-}$ )
Then
\[\begin{align}
  & H=\dfrac{1}{2}[G+M-C+A] \\
 & H=\dfrac{1}{2}[3+4+1] \\
 & H=4 \\
\end{align}\]
- H = 4 means there are four hybrid orbitals involved in hybridization.
- Therefore the hybridization of the boron in $BF_{4}^{-}$ is $s{{p}^{3}}$ .

- So, the correct option is C.

Note: The shape of the $BF_{4}^{-}$ is tetrahedral and in the molecule there are four bond pairs and zero lone pairs are present. By using VSEPR theory we can find the structure and hybridization of the molecules. In $BF_{4}^{-}$ there are no lone pair electrons present then repulsion between the electrons and bonds is also very less.