Question

Which of the following is the disproportionation reaction?
(A) $ C{{l}_{2\left( g \right)}}+2OH_{\left( aq \right)}^{-}\to ClO_{\left( aq \right)}^{-}+Cl_{\left( aq \right)}^{-}+{{H}_{2}}{{O}_{\left( l \right)}} $
(B) $ C{{l}_{2\left( g \right)}}+2I_{\left( aq \right)}^{-}\to 2Cl_{\left( aq \right)}^{-}+I_{2\left( s \right)}^{{}} $
(C) $ 2F{{e}_{(s)}}+3{{H}_{2}}{{O}_{(l)}}\xrightarrow{\Delta }F{{e}_{2}}{{O}_{3(s)}}+3{{H}_{2(g)}} $
(D) $ 2{{H}_{2}}{{O}_{\left( l \right)}}+2F2(g)\to 4H{{F}_{(aq)}}+{{O}_{2(g)}} $

Answer 1

Hint: We know that the disproportionation reaction is a redox reaction where the oxidation and reduction both take place from the same element/species and form two or more different products.

Complete Step By Step Answer:
As we know, the reaction in which one reactant gets oxidized and the same reactant gets reduced is known as a disproportionation reaction. The disproportionation reaction is also known as the dismutation reaction. The disproportionation reaction is also known as the dismutation reaction.
The same element is oxidized and reduced in the disproportionation reaction. We can explain it as a reaction where the same element is both reduced and oxidized at the same time. In general, the oxidation number of hydrogen is $ +1 $ while oxygen has $ -2 $ .
In option A, we have $ \overset{0}{\mathop{C{{l}_{2\left( g \right)}}}}\,+2OH_{\left( aq \right)}^{-}\to \overset{+1}{\mathop{ClO_{\left( aq \right)}^{-}}}\,+\overset{-1}{\mathop{Cl_{\left( aq \right)}^{-}}}\,+{{H}_{2}}{{O}_{\left( l \right)}} $ here oxidation number of chlorine is changed from zero to both $ +1,-1 $ which means chlorine gets oxidized and reduced. Thus, it’s a disproportionate reaction.
In option B we have reaction: $ \overset{0}{\mathop{C{{l}_{2\left( g \right)}}}}\,+\overset{-1}{\mathop{2I_{\left( aq \right)}^{-}}}\,\to \overset{-1}{\mathop{2Cl_{\left( aq \right)}^{-}}}\,+\overset{0}{\mathop{I_{2\left( s \right)}^{{}}}}\, $ here oxidation number of chlorine and iodine is changed from $ 0,-1 $ to $ -1,0 $ respectively which means chlorine gets reduced and iodine gets oxidized.
In option C we have reaction: $ \overset{0}{\mathop{2F{{e}_{(s)}}}}\,+\overset{+1}{\mathop{3{{H}_{2}}{{O}_{(l)}}}}\,\xrightarrow{\Delta }\overset{+3}{\mathop{F{{e}_{2}}{{O}_{3(s)}}}}\,+\overset{0}{\mathop{3{{H}_{2(g)}}}}\, $ here oxidation number of Iron and hydrogen is changed from $ 0,+1 $ to $ +3,0 $ respectively which means hydrogen gets reduced and iron gets oxidized.
In option D we have reaction: $ 2\overset{+1}{\mathop{{{H}_{2}}}}\,\overset{-2}{\mathop{{{O}_{\left( l \right)}}}}\,+\overset{0}{\mathop{2{{F}_{2(g)}}}}\,\to 4\overset{+1}{\mathop{H}}\,\overset{-1}{\mathop{{{F}_{(aq)}}}}\,+\overset{0}{\mathop{{{O}_{2(g)}}}}\, $ here oxidation number of iron and oxygen is changed from $ 0,-2 $ to both $ -1,0 $ respectively which means iron gets reduced and oxygen gets oxidized.
Therefore, the correct answer is option A.

Note:
Remember that the easiest way to identify the disproportionation is that the central atom in the reactant forms two compounds in the product. Further, even the sum of change of oxidation state in products is equal to the oxidation state of the reactant.