Question

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> An example of a disproportionation reaction is:

JEE MAIN 2019

• A. $2CuBr\to CuB{r}_2+Cu$
• B. $2KMn{O}_4\to K_{2}Mn{O}_4+Mn{O}_2+O_2$
• C. $2Mn{O}_4^{-}+10I^{-}+16H^{+}\to 2M{n}^{2+}+5I_2+8H_{2}O$
• D. $2NaBr+C{l}_2\to 2NaCl+B{r}_2$

Answer 1

The correct option is A $2CuBr\to CuB{r}_2+Cu$

$\text{(A)}2KMn{O}_4\to K_{2}Mn{O}_4+Mn{O}_2+O_2$

-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 $Mn{O}_2$ respectively. This indicates manganese is only getting reduced. And $O$ is getting oxidised from $-2$ to $0$. So, this reaction is not a disproportionation reaction.

$\text{(B)}2Mn{O}_4^{-}+10I^{-}+16H^{+}\to 2M{n}^{2+}+5I_2+8H_{2}O$
-In this reaction, manganese has $+7$ oxidation state in $Mn{O}_4^{-}$ and $+2$ oxidation state in the product side.
And $I$ is getting oxidised from $-1$ to $0$.
So, this reaction is not a disproportionation reaction.

$\text{(C)}2CuBr\to CuB{r}_2+Cu$
-In this reaction, copper is $+1$ in $CuBr$ and $+2$ oxidation state in $CuB{r}_2$and zero oxidation state 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.

$\text{(D)}2NaBr+C{l}_2\to 2NaCl+B{r}_2$
In this reaction, $Br$ is getting oxidised from $-1$ to $0$ and $Cl$ is getting reduced from $0$ to $-1$. Hence, this is not a disproportionation reaction.