Question

Sulphur reacts with chlorine in 1:2 ratio and forms X. Hydrolysis of X gives a sulphur compound Y. What is the hybridisation state of central atom in the anion of F?

• A. $ s{{p}^{3}} $
• B. sp
• C. $ s{{p}^{2}} $
• D. $ s{{p}^{3}}d $

Answer 1

The Correct Option is A

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}}. $