Question

Hybridization and shape of [BF4]-

Answer 1

Answer:

sp3

Explanation:

octet of all F and B complete.No lone pair...

Answer 2

Answer: The hybridization and shape of $[BF_4]^-$ is $sp^3d$ and tetrahederal respectively.

Explanation:

To calculate the hybridization of a compound, we use the equation:

$\text{Number of electrons}=\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

Now we have to determine the hybridization of the $[BF_4]^-$ molecule

$\text{Number of electrons}=\frac{1}{2}\times [5+4+1]=5$

Bond pair electrons = 4

Lone pair electrons = 5 - 4 = 1

The number of electrons are 5 that means the hybridization will be $sp^3d$ and the electronic geometry of the molecule will be trigonal bipyramidal

But as there are four atoms around the central boron atom, the fifth position will be occupied by lone pair of electrons. Hence, the molecular geometry will be tetrahederal.

Hence, the hybridization and shape of $[BF_4]^-$ is $sp^3d$ and tetrahederal respectively.