A solution of `KMnO_4` is reduced to various products depending upon its pH. At pH lt 7 it is reduced to a colourless solution (A), at pH = 7 it forms a brown precipitate (B) and at pH gt 7 it gives a green solution ( C), (A),(B) and( C) are

Options

A. (A)-`Mn^(2+) , (B)-`MnO_2` , (C ) -`MnO_4^(2-)`
B. `(A)-MnO_2 , (B)-Mn^(2+) , (C ) - MnO_4^(2-)`
C. `(A)-Mn^(2+) , (B ) -MnO_4^(2-) , (C ) - MnO_2`
D. `(A)-MnO_4^(2-) , (B) - Mn^(2+) , (C ) - MnO_2`

Answer

A

Explanation

At pH lt 7 , in acidic medium
`MnO_4^(-) + 8H^(+) + 5e^(-) to underset"(Colourless)"(Mn^(2+)) + 4H_2O`
At pH=7 , in neutral medium
`MnO_4^(-) + 2H_2O + 3e^(-) to underset"(Brown precipitate )"(MnO_2 )+ 4OH^(-)`
At pH gt 7 , in alkaline medium
`MnO_4^(-) + e^(-) to underset"(green)"(MnO_4^(2-))`