In the following reaction:

MnO₄²⁻ → (acidic medium) → ?

Manganate ion undergoes disproportionation in acidic medium to form:

The correct answer is MnO₂, MnO₄⁻.

Step-by-Step Explanation:

1. Understanding Disproportionation:
A disproportionation reaction is a specific type of redox reaction in which a single species undergoes both oxidation (increase in oxidation state) and reduction (decrease in oxidation state) simultaneously to form two different products.

2. Analyzing the Manganate Ion:
The given starting material is the manganate ion, ${\mathrm{MnO}}_4^{2-}$.
Let us calculate the oxidation state of manganese (Mn) in ${\mathrm{MnO}}_4^{2-}$:
Let the oxidation state of Mn be $x$.
Since oxygen typically has an oxidation state of -2:
$x+\; 4(-2)\; =\; -2$
$x-\; 8\; =\; -2$
$x=\; +6$
So, manganese is in the +6 oxidation state in the manganate ion.

3. Disproportionation in Acidic Medium:
In an acidic medium, the manganate ion (${\mathrm{MnO}}_4^{2-}$, green in color) is unstable and undergoes a disproportionation reaction to form:
• Permanganate ion (${\mathrm{MnO}}_4^{1-}$, purple in color), where manganese is in the +7 oxidation state (oxidized product).
• Manganese dioxide (MnO₂, dark brown precipitate), where manganese is in the +4 oxidation state (reduced product).

4. Balanced Chemical Equation:
The balanced chemical equation for the disproportionation of manganate ion in acidic medium is:
$3{\mathrm{MnO}}_4^{2-}+\; 4$H⁺ $\to 2{\mathrm{MnO}}_4^{\text{-}}+$MnO₂ $+2$H₂O

Here, the oxidation state of Mn changes from +6 to +7 (oxidation in ${\mathrm{MnO}}_4^{\text{-}}$) and +4 (reduction in MnO₂).

Thus, the disproportionation of manganate ion in an acidic medium yields manganese dioxide (MnO₂) and permanganate ion (${\mathrm{MnO}}_4^{\text{-}}$).