Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

The hybrid state of B in BF4 :
A. sp2
B. sp
C. sp3
D. Not specific


Answer
VerifiedVerified
490.8k+ views
like imagedislike image
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=12[G+MC+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 BF4 .
- The electronic configuration of the boron is 1s22s22p1 .
- We have to substitute all the values in the below formula to get the hybridization of the Boron in BF4
H=12[G+MC+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 BF4 )
Then
H=12[G+MC+A]H=12[3+4+1]H=4
- H = 4 means there are four hybrid orbitals involved in hybridization.
- Therefore the hybridization of the boron in BF4 is sp3 .

- So, the correct option is C.

Note: The shape of the BF4 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 BF4 there are no lone pair electrons present then repulsion between the electrons and bonds is also very less.
Latest Vedantu courses for you
Grade 10 | MAHARASHTRABOARD | SCHOOL | English
Vedantu 10 Maharashtra Pro Lite (2025-26)
calendar iconAcademic year 2025-26
language iconENGLISH
book iconUnlimited access till final school exam
tick
School Full course for MAHARASHTRABOARD students
PhysicsPhysics
BiologyBiology
ChemistryChemistry
MathsMaths
₹36,600 (9% Off)
₹33,300 per year
Select and buy