Question

Which one of the following is an example of disproportionation reaction?

• A. $3\text{MnO}_4^{2-}$+ $4H^+$ $\rightarrow$ $\text{2MnO}_4^{-}$ + $MnO_2$ + $2H_2O$
• B. $\text{MnO}_4^{-}$ + $4H^+$ +$4e^-$ →$$ MnO_2 + $2H_2O$
• C. $10I^-$ + $\text{2MnO}_4^{-}$ +16H^+$$→\ $$2Mn^{2+} + $8H_2O$ +$5l_2$
• D. $\text{8MnO}_4^{-}$ +$\text{3S{2}O}_3^{2-}$ +$H_2O$ $→$ $8MnO_2$ + $\text{6SO}_4^{2-}$+ $\text{2OH}^{-}$

Answer 1

Answer: (A)

$3\text{MnO}_4^{2-}$+ $4H^+$ $\rightarrow$ $\text{2MnO}_4^{-}$ + $MnO_2$ + $2H_2O$

Answer 2

\stackrel{+6}{M}$$\text{nO}_4^{-2} $→$ \stackrel{+7}{M}$$\text{nO}_4^{-}

\stackrel{+6}{M}$$\text{nO}_4^{-2} $→$ \stackrel{+4}{M}$$\text{nO}_2

$\text{MnO}_4^{-2}$ is an intermediate oxidation state and is converted into compounds having higher and lower oxidation states.

So, the correct answer is (A): $3\text{MnO}_4^{2-}$+ $4H^+$ + $\text{2MnO}_4^{-}$ + $MnO_2$ + $2H_2O$