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. \[\underset{Sulphuric\,acid}{\mathop{\underset{{{H}_{2}}S{{O}_{4}}}{\mathop{\underset{or}{\mathop{S{{O}_{2}}{{(OH)}_{2}}}}\,}}\,}}\,+2PC{{l}_{5}}\xrightarrow{{}}\underset{chloride}{\mathop{\underset{Sulphuryl}{\mathop{S{{O}_{2}}C{{l}_{2}}}}\,}}\,\]
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