Question

Which among the following is a disproportionation reaction?
[A]$C{{l}_{2}}+2O{{H}^{-}}\to C{{l}^{-}}+Cl{{O}^{-}}+{{H}_{2}}O$
[B]$2HCuC{{l}_{2}}\to Cu+C{{u}^{2+}}+4C{{l}^{-}}+2{{H}^{+}}$
[C]$HCHO+O{{H}^{-}}\to C{{H}_{3}}OH+HCO{{O}^{-}}$
[D]$MgC{{O}_{3}}\to MgO+C{{O}_{2}}$

Answer 1

If the same element is oxidised as well as reduced then it is called a disproportionation reaction. If the compound forms two products, one as a result of oxidation and other as a result of reduction, we can say it undergoes disproportionation.

Complete step by step answer:
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.
In simpler words, we can explain it as a reaction where the same element is both reduced and oxidised at the same time.
Let us discuss all the options one by one to find out whether it is a disproportionation reaction or not-
In option [A], we have $\overset{0}{\mathop{C{{l}_{2}}}}\,+2O{{H}^{-}}\to \overset{-1}{\mathop{C{{l}^{-}}}}\,+\overset{+1}{\mathop{Cl}}\,{{O}^{-}}+{{H}_{2}}O$
In this reaction, oxidation number of Cl is changed from 0 to both +1 and -1.
$\begin{aligned} & \overset{0}{\mathop{Cl}}\,\to \overset{+1}{\mathop{Cl}}\,+{{e}^{-}} \\\ & \overset{0}{\mathop{Cl}}\,+{{e}^{-}}\to \overset{-1}{\mathop{Cl}}\, \\\ \end{aligned}$
Here, since chlorine undergoes both oxidation and reduction therefore it is a disproportionation reaction.$C{{l}_{2}}$ is both oxidant and reductant.
In option [B] we have,$2H\overset{+1}{\mathop{Cu}}\,C{{l}_{2}}\to \overset{0}{\mathop{Cu}}\,+\overset{2+}{\mathop{C{{u}^{2+}}}}\,+4C{{l}^{-}}+2{{H}^{+}}$
In this reaction, Cu oxidised to +2 and reduced to 0. Therefore, this is also a disproportionation reaction.
In option[C] we have $H\overset{0}{\mathop{C}}\,HO+O{{H}^{-}}\to \overset{-2}{\mathop{C}}\,{{H}_{3}}OH+H\overset{3}{\mathop{C}}\,O{{O}^{-}}$
This is also a disproportionation reaction as C is both oxidised and reduced to -2 and +3.
Lastly, in option [D] we have, $\overset{+2}{\mathop{Mg}}\,\overset{+4}{\mathop{C}}\,{{O}_{3}}\to \overset{+2}{\mathop{Mg}}\,O+\overset{+4}{\mathop{C}}\,{{O}_{2}}$
It is not a disproportionation reaction, as there is no change in the oxidation states.

Therefore, options [A], [B], and [C] are disproportionation reactions.

Note: It is important to remember here that disproportionation reaction is itself a redox reaction. It is called disproportionation as the same species undergoes both oxidation and reduction. Gain of electron is reduction and loss of electron is oxidation.