Question

explain SP2 hybridization by taking bcl3 as an example

Answer 1

B(5) - 1s2 2s2 2p1

outermost shell e- = 3. thus, makes 3 sigma bonds with 3 cl atoms with no lone pairs and hence sp2 hybridisation corresponding to d 3 sigma bonds.

Answer 2

Explanation:

Formula used for calculating hybridization

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

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

For $BCl_3$

$\text{Number of electrons}=\frac{1}{2}[3+3-+0]=3$

The number of electrons are 3 that means the hybridization will be $sp^2$ and the electronic geometry of the molecule will be trigonal planar.

As there are no lone pair of electrons,  the molecular geometry also will be trigonal planar.

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