Question

A solution of $KMn{O}_4$ is reduced to various products depending upon its pH. At pH < 7 it is reduced to a colourless solution (A), at pH = 7 it forms a brown precipitate (B) and at pH > 7 it gives a green solution (C).
(A), (B) and (C) are :

• A. $\left(A\right)-M{n}^{2+},\left(B\right)-Mn{O}_2\left(C\right)-Mn{O}_4^{2-}$
• B. $\left(A\right)-Mn{O}_2,\left(B\right)-M{n}^{2+},\left(C\right)-Mn{O}_4^{2-}$
• C. $\left(A\right)-M{n}^{2+},\left(B\right)-Mn{O}_4^{2-},\left(C\right)-Mn{O}_2$
• D. $\left(A\right)-Mn{O}_4^{2-},\left(B\right)-M{n}^{2+},\left(C\right)-Mn{O}_2$

Answer 1

Answer: (A)

$\left(A\right)-M{n}^{2+},\left(B\right)-Mn{O}_2\left(C\right)-Mn{O}_4^{2-}$

A solution of ${KMnO}_4$ is reduced to various products depending upon its $pH$.

$\to$ At $pH<7$, it is reduced to a colourless solution ${Mn}^{2+}$

$\to$ At $pH=7$, it forms a brown precipitate ${MnO}_2$

$\to$ At $pH>7$, it gives a green solution ${MnO}_4^{2-}$