`MnO_(4)^(2-)` (`1` mole) in neutral aqueous medium is disproportionate to

Options

A. ` 2//3` mol of `MnO_4^(-)` and `1//3` mol of `MnO_2`
B. ` 1//3` mol of `MnO_4^(-)` and `2//3` mol of `MnO_2`
C. ` 1//3` mol of `Mn_(2)O_(7)` and `1//3` mol of `MnO_2`
D. ` 2//3` mol of `Mn_(2)O_(7)` and `1//3` mol of `MnO_2`

Answer

A

Explanation

` MnO_4^(2-)` is quite strongly oxdizing and is only stable in very strong alkali. In dilute alkali, water, or acidic solutions, it disproportionates :
` MnO_4^(2-) rarr MnO_4^(-) MnO_2`
Oxidation half-reaction:
`overset(+6)(M)nO_(4)^(2-) rarr overset(+7)(M)nO_(4)^(-) +e^(-)`
Reduction half-reaction:
`overset(+6)(M)nO_(4)^(2-)+2e^(-) rarr overset(+4)(M)nO_(2)`
Balance ionic charges by adding `H^(+)` ions:
`MnO_(4)^(2-) rarr MnO_(4)^(-)+e^(-)`
`MnO_(4)^(2-)+2e^(-)+4H^(+) rarr MnO_(2)`
Balance H and O by adding `H_(2)O` molecules
`MnO_(4)^(2-) rarr MnO_(4)^(-)+e^(-)`
`MnO_(4)^(2-)+2e^(-)+4H^(+) rarr MnO_(2)+2H_(2)O`
To equalize electron transfer, we multiply oxidation half-reaction by 2 and add them, cancelling electrons on both sides:
`3MnO_(4)^(2-) +4H^(+) rarr 3MnO_(4)^(-)+MnO_(2)+2H_(2)O`
Similarly we can balance it by means of `OH^(-)` ions:
`3MnO_(4)^(2-)+2H_(2)O rarr 2MnO_(4)^(-)+MnO_(2)+4OH^(-)`
Thus, 3 mol of `MnO_(4)^(2-)` yield 2 mol of `MnO_(4)^(-)` and 1 mole of `MnO_(2)`. Consequently, 1 mol of `MnO_(4)^(2-)` disproportionates to yield `2//3` mol of `MnO_(4)^(2-)` and `1//3` mol of `MnO_(2)`.
Note that in neutral aqueous meadium, we can balance the charge either using `H^(+)` ions or using `OH^(-)` ions.