Question

Which out $S{O_4^{2 - }}$, $S{F_4}$ and $S{F_2}$ does not undergo $s{p^3}$ - hybridization?
A. \[S{O_4^{2 - }}\]
B. $S{F_2}$ and $S{O_4^{2 - }}$
C. $S{F_2}$
D. $S{F_4}$

Answer 1

Hint: We know that orbital hybridisation is defined as the mixing of two atomic orbitals. The result of orbital hybridization is the formation of new orbits which have different energy and shape with respect to parent orbits. The condition for hybridization is that the atomic orbital should be at the same energy which takes part in hybridization. Half full or full both types of orbitals can take part in hybridization.

Complete answer:
$s{p^3}$ - hybridization – Hybrid orbitals are the result of the mixing of atomic orbitals . In this type of hybridization one ‘s’ and 3 ‘p’ orbitals which belong to the same shell of an atom takes part. Then four equal orbits of the same energy level are formed. These four orbitals are directed towards the corners of a regular tetrahedron and have an angle of ${109^0}28'$. $s{p^3}$ hybrid orbitals have $25\% $ s character and $75\% $ p character.
$S{F_2}$ have two lone pairs of electron which gives $S{F_2}$ to a bent shape and the hybridization by the central atom sulphur is $s{p^3}$.$S{O_4^{2 - }}$ has four oxygen atom which are arranged in a tetrahedral shape. In $S{O_4^{2 - }}$ there is minor mixing of d orbital but it is negligible. So $S{O_4^{2 - }}$ also have $s{p^3}$ hybridization.
In $S{F_4}$ molecule there are five electron pairs and one is lone pair among these five electron pairs which lead to trigonal bi pyramidal geometry .So in $S{F_4}$ molecule $s{p^3}d$ is found. Now we see that there is only one molecule which does not go $s{p^3}$ hybridization is $S{F_4}$,so option D is correct.

Note: We have approached this problem by finding out each molecule's hybridization and we come to know that only $S{F_4}$ molecules have different hybridizations. We should find out the number of lone pairs of electrons as it is helpful in finding the hybridization of molecules.