A) \[N{{a}_{2}}S+N{{a}_{2}}S{{O}_{3}}+C{{l}_{2}}\]
B) \[Na{{}_{2}}S+S{{O}_{2}}\]
C) \[N{{a}_{2}}S{{O}_{3}}+S\]
D) \[N{{a}_{2}}S+N{{a}_{2}}S{{O}_{3}}+\]
Correct Answer: D
Solution :
\[N{{a}_{2}}S+{{I}_{2}}+N{{a}_{2}}S{{O}_{3}}\to N{{a}_{2}}{{S}_{2}}{{O}_{3}}+2NaI\]
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