MnO₄²⁻, in acidic medium, disproportionates to : (A) MnO₄⁻ and MnO (B) MnO₄⁻ and MnO₂ (C) Mn2O₇ and MnO₂ (D) Mn2O₇ and MnO

MnO₄²⁻, in acidic medium, disproportionates to :

Q. $MnO_4^{2-}$, in acidic medium, disproportionates to :

(A) $MnO_4^-$ and $MnO$

(B) $MnO_4^-$ and $MnO_2$

(C) $Mn_2O_7$ and $MnO_2$

(D) $Mn_2O_7$ and $MnO$

Correct Answer: B

Option (B) is correct

Explanation

The manganate ion ($MnO_4^{2-}$) is stable only in very strong alkaline solutions. In acidic, neutral, or even slightly alkaline media, it undergoes disproportionation. Disproportionation is a specific type of redox reaction where the same element is both oxidized and reduced.

The Reaction:

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

Oxidation State Analysis:

In $MnO_4^{2-}$, the oxidation state of Manganese (Mn) is +6.
In $MnO_4^-$ (Permanganate), the oxidation state of Mn is +7 (Oxidation).
In $MnO_2$ (Manganese dioxide), the oxidation state of Mn is +4 (Reduction).

Since Manganese goes from +6 to both a higher (+7) and a lower (+4) oxidation state, the products are $MnO_4^-$ and $MnO_2$.

Related Theory

Understanding Disproportionation

Disproportionation (or autoxidation) occurs when a species in an intermediate oxidation state is unstable relative to its higher and lower oxidation states. For Manganese, the +6 state in the Manganate ion is highly sensitive to the pH of the solution. While it remains green and stable in high concentrations of hydroxide ($OH^-$), the addition of even small amounts of acid causes it to turn purple (due to $MnO_4^-$) and form a brown precipitate (due to $MnO_2$).

Chemistry of Manganate vs. Permanganate

The transition between these two ions is a fundamental topic in d-block chemistry:

Manganate ($MnO_4^{2-}$): Green in color, tetrahedral geometry, paramagnetic (one unpaired electron).
Permanganate ($MnO_4^-$): Deep purple/magenta in color, tetrahedral geometry, diamagnetic (no unpaired electrons, though it shows temperature-independent paramagnetism).

The Role of pH

The stability of Manganese oxoanions is heavily pH-dependent. This is often summarized in a Latimer Diagram or Frost Diagram for Manganese. In acidic medium, the reduction potential for the transition from $MnO_4^{2-}$ to $MnO_4^-$ and $MnO_2$ is favorable, making the disproportionation spontaneous.

Preparation of Potassium Permanganate ($KMnO_4$)

This disproportionation reaction is actually a key step in the industrial preparation of $KMnO_4$. Pyrolusite ore ($MnO_2$) is fused with $KOH$ in the presence of an oxidizing agent like $KNO_3$ or air to give the green potassium manganate:

$$2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$$

The green $K_2MnO_4$ is then disproportionated using acids (like $H_2SO_4$ or $CO_2$) or oxidized electrolytically to produce the final purple $KMnO_4$.

Detailed Redox Half-Reactions

To understand the stoichiometry, we can look at the half-reactions in acidic medium:

Oxidation: $$MnO_4^{2-} \rightarrow MnO_4^- + e^-$$ Reduction: $$MnO_4^{2-} + 4H^+ + 2e^- \rightarrow MnO_2 + 2H_2O$$ Multiplying the oxidation half by 2 and adding them gives the net balanced equation seen in the explanation.

Shortcut for JEE/NEET

Whenever you see Manganese in the +6 state ($MnO_4^{2-}$) in any medium that is not strongly basic, immediately look for products in the +7 and +4 states. This "6 $\rightarrow$ 7 + 4" rule is a reliable way to solve these MCQ types quickly.

Common Student Errors

Confusing MnO with MnO2: Students often pick Option A because they assume the reduction goes to the +2 state ($MnO$). However, in this specific disproportionation, Mn stops at the +4 state ($MnO_2$).
Selecting Mn2O7: Option C and D include $Mn_2O_7$. While this is the anhydride of permanganic acid, it is an explosive oily liquid formed only when $KMnO_4$ reacts with cold concentrated $H_2SO_4$, not during simple disproportionation.
Color Confusion: Forgetting that Manganate is green and Permanganate is purple can lead to confusion in "observation-based" questions.

Structure and Bonding

Both $MnO_4^{2-}$ and $MnO_4^-$ involve $d^3s$ hybridization of the Manganese atom, resulting in a tetrahedral shape. The intense colors are not due to d-d transitions (as Mn is $d^1$ in manganate and $d^0$ in permanganate) but rather due to Charge Transfer (specifically, Ligand-to-Metal Charge Transfer or LMCT), where an electron momentarily shifts from the Oxygen p-orbitals to the Manganese d-orbitals.

Frequently Asked Questions (FAQs)

1. What is a disproportionation reaction? It is a redox reaction where the same reactant is simultaneously oxidized to a higher state and reduced to a lower state.

2. Why is manganate ion unstable in acidic medium? Thermodynamically, the +6 state of Manganese is less stable than the +7 and +4 states in acidic conditions, driving the system toward those products.

3. What is the color change observed in this reaction? A green solution ($MnO_4^{2-}$) turns into a purple solution ($MnO_4^-$) with a brown/black precipitate ($MnO_2$).

4. What is the oxidation state of Mn in MnO2? In $MnO_2$, Manganese is in the +4 oxidation state.

5. Can MnO4^2- be stabilized? Yes, it is stable in strongly alkaline solutions (pH > 13).

6. Is KMnO4 an oxidizing agent? Yes, it is one of the most powerful oxidizing agents used in titrations and organic synthesis.

7. What is the magnetic nature of Permanganate? It is essentially diamagnetic because Manganese is in the +7 state with a $d^0$ configuration.

8. What happens if you add cold conc. H2SO4 to KMnO4? It forms $Mn_2O_7$, a green, highly explosive covalent oil. This is a dangerous reaction.

9. How many moles of MnO4^- are produced from 3 moles of MnO4^2-? According to the balanced equation, 3 moles of manganate produce 2 moles of permanganate.

10. Is this reaction relevant for NEET? Extremely. d-block oxoanions and their redox behavior are high-weightage topics for both JEE and NEET.

Expert Contribution by: JEE NEET Experts

10+ years of experience in Inorganic Chemistry and JEE Advanced training.