$\mathrm{MnO_4^{2-}}$, in acidic medium, disproportionates to :

Question

Accepted Answer

D. 1. **Find the oxidation state of Mn in $\mathrm{MnO_4^{2-}}$.**

Let the oxidation state of Mn be $\mathit{x}$.

$$
   \mathit{x}+4(-2)=-2
   $$

$$
   \mathit{x}-8=-2
   $$

$$
   \mathit{x}=+6
   $$

Thus, manganese is in the **$+6$ oxidation state**.

2. **Understand disproportionation.**

A disproportionation reaction is one in which the same element in a single oxidation state undergoes both **oxidation** and **reduction** simultaneously.

Here, Mn in the $+6$ state changes into:

* **$+7$** by oxidation, forming $\mathrm{MnO_4^-}$ (permanganate).
   * **$+4$** by reduction, forming $\mathrm{MnO_2}$.

3. **Balanced reaction in acidic medium.**

$$
   3\mathrm{MnO_4^{2-}}+4\mathrm{H^+}
   \rightarrow
   2\mathrm{MnO_4^-}+\mathrm{MnO_2}+2\mathrm{H_2O}
   $$

4. **Why the other options are incorrect.**

* **Option A:** MnO contains Mn in the $+2$ oxidation state and is not formed in this disproportionation.
   * **Option C:** $\mathrm{Mn_2O_7}$ is not produced during the reaction.
   * **Option D:** Neither $\mathrm{Mn_2O_7}$ nor MnO are the products of this disproportionation.

5. Therefore, in acidic medium, $\mathrm{MnO_4^{2-}}$ disproportionates to **$\mathrm{MnO_4^-}$ and $\mathrm{MnO_2}$**.

**Answer: D**

MnO42−\mathrm{MnO_4^{2-}}MnO42−​, in acidic medium, disproportionates to :

$\mathrm{MnO_4^{2-}}$ , in acidic medium, disproportionates to :