Question

Which of the following reactions are disproportion reactions ?
A) $2C{u}^{+}\to C{u}^{2+}+C{u}^0$
B) $3Mn{O}_4^{2-}+4H^{+}\to 2Mn{O}_4^{-}+Mn{O}_2+2H_{2}O$
C) $2KMn{O}_4\stackrel{\mathrm{\Delta }}{\to }{K}_{2}Mn{O}_4+Mn{O}_2+O_2$
D) $2Mn{O}_4^{-}+3M{n}^{2+}+2H_{2}O\to 5Mn{O}_2+4H^{+}$
a.) (A) and (B) only
b.) (A), (B) and (C)
c.) (A), (C) and (D)
d.) (A) and (D) only

Answer 1

Hint:. The disproportionation reaction is also called the dismutation reaction. In such a reaction, one element in the reactants form two compounds in the products, one with higher oxidation state and the other with lower oxidation state.

Complete step by step answer:
First, we will know about the disproportionation reactions and then we will be able to find the correct answer.
The disproportionation reaction can be defined as the redox reaction in which one compound of the intermediate oxidation state is converted into two compounds one with lower oxidation state and the other one with higher oxidation state.
In general, the disproportion reaction can be written as -
$2X\to X^{n+}+X^{n-}$

Now, let us see the examples given to us and find out which is an example of disproportionation reaction.
We will see the oxidation state of the central atom in each reaction and then see how it changes in products.
The first reaction is -
$2C{u}^{+}\to C{u}^{2+}+C{u}^0$
In this reaction, the copper in the reactant is in +1 oxidation state and in products, it is in +2 and zero oxidation states. So, it is a disproportionation reaction.

The second reaction is
$3Mn{O}_4^{2-}+4H^{+}\to 2Mn{O}_4^{-}+Mn{O}_2+2H_{2}O$
Here, the Mn is in +6 oxidation state in $Mn{O}_4^{2-}$ and +7 oxidation state in $Mn{O}_4^{-}$ and +4 in $Mn{O}_2$. So, this is also a disproportionation reaction.

The third reaction is
$2KMn{O}_4\stackrel{\mathrm{\Delta }}{\to }{K}_{2}Mn{O}_4+Mn{O}_2+O_2$
In this Mn is in +7 oxidation state in $KMn{O}_4$ in reactant while in products it is in +6 oxidation state in $K_{2}Mn{O}_4$ and +4 oxidation state in $Mn{O}_2$. Here, the oxidation state decreased in both the products. This is not a disproportionation reaction.

The fourth reaction is
$2Mn{O}_4^{-}+3M{n}^{2+}+2H_{2}O\to 5Mn{O}_2+4H^{+}$
In this Mn is in +7 oxidation state in $Mn{O}_4^{-}$ and +2 oxidation state in $M{n}^{2+}$ in reactant while in products it is in +4 oxidation state in $Mn{O}_2$. This is not a disproportionation reaction.
Thus, we can say that A) and B) both are disproportionation reactions.
So, the correct answer is “Option A”.

Note: 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.