A) \[N{{a}_{2}}S\]
B) \[HCl\]
C) \[N{{a}_{2}}{{S}_{4}}{{O}_{6}}\]
D) \[N{{a}_{2}}S{{O}_{3}}\]
Correct Answer: C
Solution :
\[2N{{a}_{2}}{{S}_{2}}{{O}_{3}}+{{I}_{2}}\to N{{a}_{2}}{{S}_{4}}{{O}_{6}}+2NaI\].
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