A) chlorosulphonic acid
B) thionyl chloride
C) sulphuryl chloride
D) sulphurous acid
Correct Answer: C
Solution :
\[PC{{l}_{5}}\] reacts with cone \[{{H}_{2}}S{{O}_{4}}\]to give sulphuryl chloride by replacing its hydroxyl groups with chlorine atoms. \[\underset{\begin{smallmatrix} \,\,\,\,\,\,\,\,\,\,\,{{H}_{2}}S{{O}_{4}} \\ sulphuric\,acid \end{smallmatrix}}{\mathop{\underset{or}{\mathop{S{{O}_{2}}{{(OH)}_{2}}}}\,}}\,+2PC{{l}_{5}}\to \underset{\begin{smallmatrix} sulphuryl \\ chloride \end{smallmatrix}}{\mathop{S{{O}_{2}}C{{l}_{2}}}}\,+2POC{{l}_{3}}+2HCl\]
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