Question

An example of a disproportionation reaction is:

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

Answer 1

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

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

-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. 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.

Thus, in this reaction as well manganese is getting reduced. 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, sodium is $+1$ in both the reactant side as well as the product side. This is just an example of a displacement reaction not a disproportionation reaction.