| I. \[2Rb+2{{H}_{2}}O\xrightarrow{{}}2RbOH+{{H}_{2}}\] |
| II. \[2Cu{{I}_{2}}\xrightarrow{{}}2CuI+{{I}_{2}}\] |
| III. \[N{{H}_{4}}Br+KOH\xrightarrow[{}]{{}}KBr+N{{H}_{3}}+{{H}_{2}}O\] |
| IV. \[4KCN+Fe{{(CN)}_{2}}\xrightarrow{{}}{{K}_{4}}[Fe{{(CN)}_{6}}]\] |
A) I and II
B) I and III
C) I, III and IV
D) III and IV
Correct Answer: D
Solution :
I. \[Rb\xrightarrow{{}}R{{b}^{+}}+{{e}^{-}}\] (Oxidation) \[{{H}_{2}}O+{{e}^{-}}\xrightarrow{{}}\overset{\bigcirc -}{\mathop{OH}}\,+\frac{1}{2}{{H}_{2}}\] (Reduction) Hence, redox reaction. II. \[{{e}^{-}}+C{{u}^{2+}}\xrightarrow{{}}C{{u}^{+}}\] (Reduction) \[2{{I}^{\bigcirc -}}\xrightarrow{{}}{{I}_{2}}+2{{e}^{-}}\] (Oxidation) Hence, redox reaction. III. \[\underset{x\,=\,\,-\,3}{\mathop{\overset{\oplus }{\mathop{N{{H}_{4}}}}\,}}\,\xrightarrow{{}}\underset{x\,=\,\,-\,3}{\mathop{N{{H}_{3}}}}\,\] (No change in oxidation number) Hence, reaction is not redox. IV. No change in the oxidation number of either \[F{{e}^{2+}}\] or \[C{{N}^{\bigcirc -}}\]in both reactant and product, hence, not redox reaction.
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