MnO4^2- undergoes disproportionate reaction in acidic medium but MnO4^- does not. Why?
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Manganese (VI) is usually in the form of the MnO42- ion. This ion disproportionates in acid solution but not in base. What is going on here?
The Mn(VI) state is not stable under acid conditions because it CAN disproportionate (notice that this is NOT a half-equation – it is a reaction):
3MnO4(2-) + 4H+ –> MnO2 + 2MnO4(-) + 2H2O
The Mn(VI) state cannot do this under alkaline conditions, and the alkaline disproportionation reaction (theoretical) would be:
3MnO4(2-) + 2H2O –> MnO2 + 2MnO4(-) + 4OH-
This is now in direct competition with the reverse reaction (as all reactions are) which we know DOES occur.
The Mn(VI) state is not stable under acid conditions because it CAN disproportionate (notice that this is NOT a half-equation – it is a reaction):
3MnO4(2-) + 4H+ –> MnO2 + 2MnO4(-) + 2H2O
The Mn(VI) state cannot do this under alkaline conditions, and the alkaline disproportionation reaction (theoretical) would be:
3MnO4(2-) + 2H2O –> MnO2 + 2MnO4(-) + 4OH-
This is now in direct competition with the reverse reaction (as all reactions are) which we know DOES occur.
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Answer:
Since, +2 oxidation state is more stable as compared to +3 oxidation state due to stable half filled electronic configuration, therefore Mn(III) disproportionates to Mn(II).
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