A) \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}\] is oxidized
B) \[Cu{{I}_{2}}\] is formed
C) \[C{{u}_{2}}{{I}_{2}}\] is formed
D) Evolved \[{{I}_{2}}\] is reduced
Correct Answer: B
Solution :
[b] \[4\overset{-1}{\mathop{KI}}\,+2CuS{{O}_{4}}\to \overset{0}{\mathop{{{I}_{2}}}}\,+C{{u}_{2}}{{I}_{2}}+2{{K}_{2}}S{{O}_{4}}\] \[{{\overset{0}{\mathop{I}}\,}_{2}}+2\overset{2+}{\mathop{N{{a}_{2}}{{S}_{2}}{{O}_{3}}}}\,\to \overset{-2.5}{\mathop{N{{a}_{2}}{{S}_{4}}{{O}_{6}}}}\,+2\overset{-1}{\mathop{NaI}}\,\] In this \[Cu{{I}_{2}}\] is not formed.
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