Question

What is the hybridization of $SnC{l_2}$ ?

Answer 1

Hint : In order to determine the hybridization of the compound, first we need to determine the number of valence electrons present in $Sn$ . Hybridization is defined as the process of mixing two atomic orbitals with the same energy levels to give a degenerated new type of orbitals. Depending on the types of orbitals involved in mixing, hybridization can be classified as $sp,s{p^2},s{p^3},s{p^3}{d^2},s{p^3}{d^3}$ .

Complete Step By Step Answer:
In $SnC{l_2}$ , the central atom $Sn$ has $4$ valence. $2$ electrons are present as a lone pair of electrons and two electrons are shared with chlorine atoms to form two tin chloride bonds. Thus, there are two bond pairs of electrons and one lone pair of electrons. Hence, the steric number is $3$ .The steric number is related to hybridization. Three atomic orbitals of $Sn$ that is one $s$ and two $p$ orbitals undergo hybridization to form three degenerate hybrid orbitals. Degenerate means that all the three orbitals have the same energy. Out of the three hybrid orbitals, one contains a lone pair of electrons and two contain a bond pair of electrons. Hence, we can say that $SnC{l_2}$ undergoes $s{p^2}$ hybridization.
The electron pair geometry is trigonal planar and the molecular geometry is angular or bent.

Note :
The $s{p^2}$ hybridization is defined as the hybridization in which one s and two $p$ orbitals combine to form a new hybrid orbital. All the three hybrid orbitals remain in one plane and make an angle of ${120^0}$ with one another. All the compounds of boron and all the compounds of carbon containing a carbon-carbon double bond