
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
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Hint: 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.
\[\begin{align}
& \overset{0}{\mathop{Cl}}\,\to \overset{+1}{\mathop{Cl}}\,+{{e}^{-}} \\
& \overset{0}{\mathop{Cl}}\,+{{e}^{-}}\to \overset{-1}{\mathop{Cl}}\, \\
\end{align}\]
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.
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.
\[\begin{align}
& \overset{0}{\mathop{Cl}}\,\to \overset{+1}{\mathop{Cl}}\,+{{e}^{-}} \\
& \overset{0}{\mathop{Cl}}\,+{{e}^{-}}\to \overset{-1}{\mathop{Cl}}\, \\
\end{align}\]
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.
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