Question

In the following reaction, \[ {MnO4^{2-} \xrightarrow{H^+} \, ?} \] Manganate ion undergoes disproportionation to form:

• A. \( {MnO2,\ MnO4^-} \)
• B. \( {MnO,\ MnO2} \)
• C. \( {MnO2,\ MnO} \)
• D. \( {MnO4^-,\ MnO} \)

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

Answer 1

The Correct Option is A

Step 1: Oxidation state of Mn in manganate ion: \[ {MnO4^{2-}} \Rightarrow \text{Mn oxidation state} = +6 \]
Step 2: In acidic medium, manganate ion undergoes disproportionation, where Mn is simultaneously oxidised and reduced. \[ 3{MnO4^{2-}} + 4{H^+} \rightarrow 2{MnO4^-} + {MnO2} + 2{H2O} \]
Step 3:
Mn is oxidised from \(+6\) to \(+7\) in \( {MnO4^-} \)
Mn is reduced from \(+6\) to \(+4\) in \( {MnO2} \) Thus, the products are: \[ {MnO2 \; and \; MnO4^-} \]