
The species that undergo(es) disproportionation reaction in alkaline medium are:
(A) \[C{{l}_{2}}\]
(B) \[Mn{{O}_{4}}^{2-}\]
(C) \[N{{O}_{2}}\]
(D) \[\,\,Cl{{O}_{4}}^{-}\]
Answer
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Hint: Try recalling the definition of redox reaction and how it occurs. We should have a basic understanding of oxidation and reduction.
Complete step-by-step answer:
>Reduction: It is a process in which an atom gains an electron and therefore decreases (or reduces its oxidation number). In other words, the positive character of the species is reduced.
>Oxidation: It is a process in which an atom loses an electron and therefore increases its oxidation number. In other words, the positive character of the species is increased.
So basically, Redox reactions are reactions in which one species is reduced while the other is oxidised in which the oxidation states of the species changes ultimately.
Example:
\[F{{e}^{2+}}\to \,F{{e}^{3+}}\,+\,{{e}^{-}}\]
\[C{{e}^{4+}}\,+\,{{e}^{-}}\,\to C{{e}^{3+}}\]
\[F{{e}^{2+}}\,+\,C{{e}^{4+}}\to \,F{{e}^{3+}}\,+\,\,C{{e}^{3+}}\]
> Disproportionate reaction is a special type of redox reaction in which oxidation and reduction of an element occurs simultaneously with an atom, element, molecule i.e. it gives electrons and accepts electrons at the same time to form different products.
> \[C{{l}_{2}}\] disproportionate in alkaline medium forming chloride ion \[C{{l}^{-}}\] and hypochlorite \[Cl{{O}^{-}}\] ion.
\[C{{l}_{2}}\,+\,2O{{H}^{-}}\,\to \,C{{l}^{-}}\,+\,Cl{{O}^{-}}\,+{{H}_{2}}O\]
> In \[Mn{{O}_{4}}^{2-}\], oxidation state of \[Mn\] is \[+6\] so it can give disproportionation reaction in acidic medium because \[Mn\] shows oxidation state of \[+2\,+3\]\[+4\]\[+6\,+7\].
> This reaction involves disproportionation of \[N{{O}_{2}}(+\,4\,state\,)\] into \[N{{O}_{2}}^{-}(+\,3\,state)\] and \[N{{O}_{3}}^{-}(+\,5\,state)\] in presence of water. Therefore, this reaction is an example of disproportionation redox reaction in aqueous medium.
\[2N{{O}_{2}}(g)\,+\,2O{{H}^{-}}(aq)\,\to \,N{{O}_{2}}^{-}(aq)\,+\,{{H}_{2\,}}O(l)\,+ N{{O}_{3}}^{-}\]
> In \[Cl{{O}_{4}}^{-}\] , the oxidation state of\[\,\,Cl\] is \[+7\] which is its highest oxidation state. It can be reduced so it cannot give a disproportionation reaction.
So, the correct answers are (a).
Note: Except \[\,\,Cl{{O}_{4}}^{-}\] rest of the ions and molecules \[C{{l}_{2}}\] , \[Mn{{O}_{4}}^{2-}\] , \[N{{O}_{2}}\] can undergoes disproportionation reaction but in alkaline medium only \[C{{l}_{2}}\] can show disproportionation reaction.
Complete step-by-step answer:
>Reduction: It is a process in which an atom gains an electron and therefore decreases (or reduces its oxidation number). In other words, the positive character of the species is reduced.
>Oxidation: It is a process in which an atom loses an electron and therefore increases its oxidation number. In other words, the positive character of the species is increased.
So basically, Redox reactions are reactions in which one species is reduced while the other is oxidised in which the oxidation states of the species changes ultimately.
Example:
\[F{{e}^{2+}}\to \,F{{e}^{3+}}\,+\,{{e}^{-}}\]
\[C{{e}^{4+}}\,+\,{{e}^{-}}\,\to C{{e}^{3+}}\]
\[F{{e}^{2+}}\,+\,C{{e}^{4+}}\to \,F{{e}^{3+}}\,+\,\,C{{e}^{3+}}\]
> Disproportionate reaction is a special type of redox reaction in which oxidation and reduction of an element occurs simultaneously with an atom, element, molecule i.e. it gives electrons and accepts electrons at the same time to form different products.
> \[C{{l}_{2}}\] disproportionate in alkaline medium forming chloride ion \[C{{l}^{-}}\] and hypochlorite \[Cl{{O}^{-}}\] ion.
\[C{{l}_{2}}\,+\,2O{{H}^{-}}\,\to \,C{{l}^{-}}\,+\,Cl{{O}^{-}}\,+{{H}_{2}}O\]
> In \[Mn{{O}_{4}}^{2-}\], oxidation state of \[Mn\] is \[+6\] so it can give disproportionation reaction in acidic medium because \[Mn\] shows oxidation state of \[+2\,+3\]\[+4\]\[+6\,+7\].
> This reaction involves disproportionation of \[N{{O}_{2}}(+\,4\,state\,)\] into \[N{{O}_{2}}^{-}(+\,3\,state)\] and \[N{{O}_{3}}^{-}(+\,5\,state)\] in presence of water. Therefore, this reaction is an example of disproportionation redox reaction in aqueous medium.
\[2N{{O}_{2}}(g)\,+\,2O{{H}^{-}}(aq)\,\to \,N{{O}_{2}}^{-}(aq)\,+\,{{H}_{2\,}}O(l)\,+ N{{O}_{3}}^{-}\]
> In \[Cl{{O}_{4}}^{-}\] , the oxidation state of\[\,\,Cl\] is \[+7\] which is its highest oxidation state. It can be reduced so it cannot give a disproportionation reaction.
So, the correct answers are (a).
Note: Except \[\,\,Cl{{O}_{4}}^{-}\] rest of the ions and molecules \[C{{l}_{2}}\] , \[Mn{{O}_{4}}^{2-}\] , \[N{{O}_{2}}\] can undergoes disproportionation reaction but in alkaline medium only \[C{{l}_{2}}\] can show disproportionation reaction.
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