Question

Hybridization of sulfur atom is in \[{\mathbf{S}}{{\mathbf{F}}_2}\]​, \[{\mathbf{S}}{{\mathbf{F}}_4}\] and \[{\mathbf{S}}{{\mathbf{F}}_6}\]​ respectively as:

A \[s{p^2}\],\[s{p^3}\],\[s{p^3}{d^2}\]
B \[s{p^3}\],\[s{p^3}\] ,\[s{p^3}{d^2}\]
C \[s{p^3}\],\[s{p^3}d\],\[s{p^3}{d^2}\]
D\[s{p^3}\],\[s{p^3}{d^2}\],\[{d^2}s{p^3}\]

Answer 1

Hint:Hybridization involves mixing and recasting of atomic orbitals of the same element. The orbitals involved in this process must have nearly the same energy. Only the atomic orbitals, not electrons undergo hybridization. Several hybrid orbitals produced several hybrid orbitals are involved in hybridization.

\[\begin{array}{*{20}{l}}
  {s{p^3},s{p^2},sp,ds{p^2},s{p^3}d,s{p^3}{d^2},s{p^3}{d^3}.} \\
  \;
\end{array}\]

Complete step-by-step answer:
(C) is the answer, \[s{p^3}\],\[s{p^3}d\],\[s{p^3}{d^2}\]
No. of electron pair= No. of atoms bonded to it \[ + \dfrac{1}{2}\] (Group number of central atom- Valency of central atom)

\[\; = 2 + 21\left( {6 - 2} \right){\text{ }} = {\text{ }}4\]
 Hybridisation = \[s{p^3}\]

\[{\mathbf{S}}{{\mathbf{F}}_4}\] ​\[ = 4 + 21\left( {6 - 4} \right){\text{ }} = {\text{ }}5\]
 Hybridization = \[s{p^3}d\]

\[{\mathbf{S}}{{\mathbf{F}}_6}\] \[ = {\text{ }}6 + 21\left( {6 - 6} \right){\text{ }} = {\text{ }}6\]
 Hybridization =\[s{p^3}{d^2}\] 

The two lone pairs of electrons give \[{\mathbf{S}}{{\mathbf{F}}_2}\] a bent shape. So, it is a bent molecule. The hybridization by the central Sulfur is \[s{p^3}\]

 When bonding takes place there is a formation of 4 single bonds in sulfur and it has 1 lone pair. Looking at this, we can say that the number of regions of electron density is 5. The middle S atom containing the 5 valence atomic orbitals are so, \[{\mathbf{S}}{{\mathbf{F}}_4}\] hybridized to form five \[s{p^3}d\] hybrid orbitals.
It Shows \[s{p^3}{d^2}\] hybridization. The structure is octahedral. Each \[s{p^3}{d^2}\] orbital overlaps with a \[2p\] orbital of fluorine to form the SF bond.

Note:Sigma Bond - The covalent bond formed by the coaxial overlap of atomic orbitals is called sigma bonding.

For Example: Methane molecule contains \[4{\text{ }}C - H\] sigma bonding.

Pi Bond - The covalent bond formed by the lateral overlap of atomic orbitals is called a pi bond. For example, the ethylene molecule contains 5 sigma bonding and 1 pi bonding in it. 

Sigma Bond and Pi Bond - Sigma bonds are formed by coaxial overlap whereas the pi bond is formed by lateral overlapping .Sigma being a strong bond whereas pi being weak due to electron density is high along the axis of the molecule in sigma bonding whereas it is zero in pi bonding.