Question

\( \mathrm{MnO_4^{2-}} \) in acidic medium disproportionates to :

• A. \( \mathrm{Mn_2O_7} \) and \( \mathrm{MnO_2} \)
• B. \( \mathrm{MnO_4^-} \) and \( \mathrm{MnO} \)
• C. \( \mathrm{MnO_4^-} \) and \( \mathrm{MnO_2} \)
• D. \( \mathrm{Mn_2O_7} \) and \( \mathrm{MnO} \)

Hint:

Remember the oxidation states of manganese: \[ \mathrm{MnO_4^-} \rightarrow +7 \] \[ \mathrm{MnO_4^{2-}} \rightarrow +6 \] \[ \mathrm{MnO_2} \rightarrow +4 \] Manganate ion (\( \mathrm{MnO_4^{2-}} \)) is unstable in acidic medium and disproportionates into permanganate and manganese dioxide.

Answer 1

The Correct Option is C

Concept: Disproportionation reaction is a reaction in which the same species undergoes:

• Oxidation
• Reduction

simultaneously. In manganate ion: \[ \mathrm{MnO_4^{2-}} \] oxidation state of manganese is: \[ x + 4(-2) = -2 \] \[ x - 8 = -2 \] \[ x = +6 \] Thus, manganese is in: \[ +6 \] oxidation state. In acidic medium, manganate ion is unstable and undergoes disproportionation.

Step 1: Understanding the oxidation and reduction processes. During disproportionation: \[ \mathrm{Mn^{+6}} \] undergoes:

• Oxidation to \(+7\)
• Reduction to \(+4\)

The corresponding compounds are: For manganese in \(+7\) oxidation state: \[ \mathrm{MnO_4^-} \] For manganese in \(+4\) oxidation state: \[ \mathrm{MnO_2} \] Hence: \[ \mathrm{MnO_4^{2-} \rightarrow MnO_4^- + MnO_2} \]

Step 2: Writing the balanced disproportionation reaction. The balanced reaction in acidic medium is: \[ 3\mathrm{MnO_4^{2-}} + 4\mathrm{H^+} \rightarrow 2\mathrm{MnO_4^-} + \mathrm{MnO_2} + 2\mathrm{H_2O} \] This clearly shows the formation of: \[ \mathrm{MnO_4^-} \] and \[ \mathrm{MnO_2} \]

Step 3: Checking all options carefully. Option (1): \[ \mathrm{Mn_2O_7} \] is manganese(VII) oxide and is not formed in this disproportionation. Hence incorrect. Option (2): \[ \mathrm{MnO} \] contains manganese in \(+2\) oxidation state, not formed here. Hence incorrect. Option (3): \[ \mathrm{MnO_4^- \ and \ MnO_2} \] This matches the actual disproportionation products. Hence correct. Option (4): Again contains incorrect products. Thus, the correct answer is: \[ \boxed{(3)\ \mathrm{MnO_4^- \ and \ MnO_2}} \]