A) \[2PC{{l}_{5}}+{{H}_{2}}S{{O}_{4}}\to 2POC{{l}_{3}}+2HCl+S{{O}_{2}}C{{l}_{2}}\]
B) \[2NaOH+{{H}_{2}}S{{O}_{4}}\to N{{a}_{2}}S{{O}_{4}}+2{{H}_{2}}O\]
C) \[NaCl+{{H}_{2}}S{{O}_{4}}\to NaHS{{O}_{4}}\text{+}HCl\]
D) \[2HI+{{H}_{2}}S{{O}_{4}}\to {{I}_{2}}+S{{O}_{2}}+2{{H}_{2}}O\]
Correct Answer: D
Solution :
[d] \[2HI+{{H}_{2}}S{{O}_{4}}\to {{I}_{2}}+S{{O}_{2}}+2{{H}_{2}}O\] (HI is oxidised to \[{{I}_{2}}\] (\[{{H}_{2}}S{{O}_{4}}\] is reduced to\[S{{O}_{2}}\])
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