In the following reaction,
\[
{MnO4^{2-} \xrightarrow{H^+} \, ?}
\]
Manganate ion undergoes disproportionation to form:
Show Hint
Important manganate reactions:
Manganate ion (\(+6\)) is unstable in acidic medium
It disproportionates into permanganate (\(+7\)) and manganese dioxide (\(+4\))
This is a classic redox disproportionation reaction
The given reaction is the disproportionation of the manganate ion (\( \text{MnO}_4^{2-} \)) in acidic medium (\( \text{H}^+ \)). In a disproportionation reaction, a single substance is simultaneously reduced and oxidized to form two different products.
Let's analyze the reaction:
For the manganate ion (\( \text{MnO}_4^{2-} \)), manganese is in the +6 oxidation state.
During the reaction, this can be both reduced to a lower oxidation state and oxidized to a higher oxidation state. Possible products could be:
Manganese(IV) oxide (\( \text{MnO}_2 \)) where manganese is in the +4 oxidation state - this represents reduction.
Permanganate ion (\( \text{MnO}_4^- \)) where manganese is in the +7 oxidation state - this represents oxidation.
Given these possible products, we can conclude that the manganate ion disproportionates into \( \text{MnO}_2 \) and \( \text{MnO}_4^- \).
Therefore, the correct option is \(({MnO2,\ MnO4^-})\).
Let's reason why the other options are incorrect:
\(({MnO,\ MnO2})\) - In this pair, MnO corresponds to Mn in +2 oxidation state which usually does not form in such reactions.
\(({MnO2,\ MnO})\) - Again, MnO represents an uncommon and unlikely state for manganese in this context.
\(({MnO4^-,\ MnO})\) - The permanganate ion is correctly there, but MnO is not typically formed in these disproportionation reactions.
Tip: In an acidic disproportionation reaction involving manganese, always check the standard oxidation states that can feasibly be generated from the intermediate state of Mn(+6).
