A) \[s{{p}^{3}}\]
B) sp
C) \[s{{p}^{2}}\]
D) \[s{{p}^{3}}d\]
Correct Answer: A
Solution :
When sulphur reacts with chlorine in 1:2 ratio then \[\text{SC}{{\text{l}}_{\text{4}}}\] (sulphur tetra chloride) is obtained which on hydrolysis gives sulphurous acid \[\text{(}{{\text{H}}_{2}}\text{S}{{\text{O}}_{3}}\text{)}\text{.}\] So, the compound X is \[\text{SC}{{\text{l}}_{\text{4}}}\]and Y is \[{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{3}}}\text{.}\] \[S+2C{{l}_{2}}\xrightarrow{{}}\underset{(X)}{\mathop{SC{{l}_{4}}}}\,\] \[SC{{l}_{4}}+4{{H}_{2}}O\xrightarrow{{}}\underset{(unstable)}{\mathop{S{{(OH)}_{4}}}}\,+4HCl\] \[S{{(OH)}_{4}}\xrightarrow{{}}\underset{\begin{smallmatrix} \text{Sulphrous}\,\text{acid} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{(}\gamma \text{)} \end{smallmatrix}}{\mathop{{{H}_{2}}S{{O}_{3}}}}\,+{{H}_{2}}O\] The hybridization of sulphur in \[\text{SO}_{3}^{2-}\]is \[s{{p}^{3}}.\]
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