Question

Which of the following statements regarding the structure of ${\text{SOC}}{{\text{l}}_2}$ is not correct?
A. The sulphur is ${\text{s}}{{\text{p}}^3}$ hybridised and it has a tetrahedral shape.
B. The sulphur is ${\text{s}}{{\text{p}}^3}$ hybridised and it has a trigonal pyramidal shape.
C. The oxygen – sulphur bond is ${{p\pi }} - {{d\pi }}$ bond.
D. It contains one lone pair of electrons in the ${\text{s}}{{\text{p}}^3}$ hybrid orbital of sulphur.

Answer 1

The central atom in ${\text{SOC}}{{\text{l}}_2}$ is sulphur. Oxygen is always attached with a double bond. One lone pair will be present on the sulphur after bonding, so the shape is not regular. Backbonding occurs when there is a vacant d-orbital in the central atom and the atom attached to it is electron rich.

Complete step-by-step answer: Sulphur is group number 16 element. The atomic number of sulphur is 16 and it belongs to the third period. It has 6 valence electrons in its valence shell. Chlorine is a monovalent species that is always attached by a single bong with a central element. Oxygen is a bivalent species that attaches via double bong with the central atom. Hence 4 electrons will be used in the bond formation and one lone pair will be there. Now let us first calculate the hybridisation using steric number method. The formula for calculation of steric number is as follow:
${\text{S}}{\text{.No}} = \dfrac{1}{2}\left[ {{\text{L}}{\text{.P}} + {\text{B}}{\text{.P}} - {\text{C}} + {\text{A}}} \right]$
Here S.No is steric number value, L.P is number of lone pair of electrons on central atom, B.P is number of bond pair of electrons, C is charge on cation and A is charge on anion.
As we can in our case there is 1 lone pair of electrons and 2 bond pairs of electrons because only monovalents bonded atoms are considered while calculating the steric number. The species given is neutral hence the charge on cation and anion would not be considered. The value of steric number will come out to be:
${\text{S}}{\text{.No}} = \dfrac{1}{2}\left[ {6 + 2} \right] = 4$
The hybridisation with steric number 4 is ${\text{s}}{{\text{p}}^3}$ that is tetrahedral geometry and due to presence of one lone pair the shape will be trigonal pyramidal and not tetrahedral. The lone pairs are never included while stating the shapes of molecules.

Hence, statement A is incorrect. The correct option is option A.

Note: The molecule given that is ${\text{SOC}}{{\text{l}}_2}$ is thionyl chloride. It is often confused with the compound named sulphuryl chloride whose formula is ${\text{S}}{{\text{O}}_2}{\text{C}}{{\text{l}}_2}$ .