A) \[N{{a}_{2}}S\]
B) \[N{{a}_{2}}S{{O}_{4}}\]
C) \[N{{a}_{2}}{{S}_{4}}{{O}_{6}}\]
D) \[{{S}_{2}}\]
Correct Answer: C
Solution :
\[2N{{a}_{2}}{{S}_{2}}{{O}_{3}}+{{I}_{2}}\xrightarrow[{}]{{}}~\underset{Sod.\,terathionate}{\mathop{N{{a}_{2}}{{S}_{4}}{{O}_{6}}}}\,+2NaI\] \[\therefore \]The product is sodium tetrathionate.
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