A) \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}\] is oxidised
B) \[Cu{{I}_{2}}\] is formed
C) \[C{{u}_{2}}{{I}_{2}}\]is formed
D) Evolved \[{{I}_{2}}\] is reduced
Correct Answer: B
Solution :
\[2\overset{\,\,-1}{\mathop{KI}}\,+2CuS{{O}_{4}}\to {{\overset{\text{O}}{\mathop{I}}\,}_{2}}+C{{u}_{2}}{{I}_{2}}+2{{K}_{2}}S{{O}_{4}}\] \[{{\overset{\text{O}}{\mathop{I}}\,}_{2}}+2N{{a}_{2}}\overset{+2}{\mathop{{{S}_{2}}}}\,{{O}_{3}}\to \overset{+2.5}{\mathop{N{{a}_{2}}}}\,{{S}_{4}}{{O}_{6}}+\overset{\,\,\,\,\,-1}{\mathop{2NaI}}\,\]
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