Question

An example of a disproportionation reaction is:
(A) $2KMn{{O}_{4}}\to {{K}_{2}}Mn{{O}_{4}}+Mn{{O}_{2}}+{{O}_{2}}$
(B) $2MnO_{4}^{-}+10{{I}^{-}}+16{{H}^{+}}\to 2M{{n}^{2+}}+5{{I}_{2}}+8{{H}_{2}}O$
(C) $2CuBr\to CuB{{r}_{2}}+Cu$
(D) $2NaBr+C{{l}_{2}}\to 2NaCl+B{{r}_{2}}$

Answer 1

Hint: Recollect what disproportionation reaction is. A Disproportionate word indicates a substance having a different proportion. Think what different proportions will mean in chemical reaction context. Write down each of the reactions and write down the oxidation states of metal in both reactant and product side to get the answer.

Complete step by step solution:
-A disproportionation reaction is a reaction in which a molecule having different atoms whose major element having a specific oxidation state gets oxidized and reduced in two separate half-reactions, yielding two other products containing the same major element having different oxidation states.
-Let’s start by writing the oxidation states of metals in each reaction.
(A) $2KMn{{O}_{4}}\to {{K}_{2}}Mn{{O}_{4}}+Mn{{O}_{2}}+{{O}_{2}}$
-In this reaction, manganese has +7 oxidation state in $KMn{{O}_{4}}$ and +6 and +4 oxidation states in ${{K}_{2}}Mn{{O}_{4}}$ and $Mn{{O}_{2}}$. This indicates manganese is only getting reduced. So, this reaction is not a disproportionation reaction.
(B) $2MnO_{4}^{-}+10{{I}^{-}}+16{{H}^{+}}\to 2M{{n}^{2+}}+5{{I}_{2}}+8{{H}_{2}}O$
-In this reaction, manganese has +7 oxidation state in $MnO_{4}^{-}$ and +2 oxidation state in the product side. Thus, in this reaction as well manganese is getting reduced. So, this reaction is not a disproportionation reaction.
(C) $2CuBr\to CuB{{r}_{2}}+Cu$
-In this reaction, copper is +1 in $CuBr$ and has +2 oxidation state in $CuB{{r}_{2}}$ and zero oxidation state in elemental form. This implies, in this reaction copper is getting both oxidized as well as reduced. Therefore, this reaction is an example of a disproportionation reaction.
(D) $2NaBr+C{{l}_{2}}\to 2NaCl+B{{r}_{2}}$
-In this reaction, sodium is +1 in both the reactant side as well as the product side. This is just an example of a displacement reaction.

Therefore an example is disproportionation reaction is $2CuBr\to CuB{{r}_{2}}+Cu$
Therefore, the answer is option (C).

Note: Remember in disproportionation reaction, anyone element having a specific oxidation state present in the reactant compound will get both oxidized as well as reduced to form two kinds of products containing different oxidation states. If in a reaction, an element in a reactant forms two products having different oxidation states but is only oxidized or only reduced then it is not an example of a disproportionation reaction.