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

Answer: (A)

$s{{p}^{3}}$

Answer 2

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 $s{{p}^{3}}.$