Q.When an oxide of manganese (A) is fused with KOH in the presence of an oxidising agent and dissolved in water, it gives a dark green solution of compound (B). Compound (B) disproportionates in neutral or acidic solution to give purple compound (C). An alkaline solution of compound (C) oxidises potassium iodide solution to a compound (D) and compound (A) is also formed. Identify compounds A to D and also explain the reactions involved.

Concept: Inorganic Synthesis — Manganese in variable oxidation states

Step 1 — Identify A and B

The oxide of manganese that reacts with KOH and an oxidising agent (like KNO₃ or air) to give a dark green solution is MnO₂ (A). Fused with KOH, it forms potassium manganate(VI), K₂MnO₄ (B), which is dark green in water.

Step 2 — Disproportionation of B to C

In neutral or acidic solution, manganate(VI) disproportionates:

$3 \text{MnO}_4^{2-} + 4 \text{H}^+ \rightarrow 2 \text{MnO}_4^- + \text{MnO}_2 + 2 \text{H}_2\text{O}$

The purple product is potassium permanganate, KMnO₄ (C). …

The entire problem revolves around the variable oxidation states of manganese — from +4 in MnO₂ to +6 in manganate, then +7 in permanganate, and back down. Each colour change is a direct signal of the oxidation state shift. Let’s walk through it step by step.

1. Identifying compound A

   The starting material is an “oxide of manganese”. The most common stable oxide is manganese dioxide, MnO₂. It is a brown-black solid, widely used in dry cells and as an oxidising agent. When fused with KOH and an oxidising agent (like KNO₃ or air), MnO₂ is oxidised from Mn(IV) to Mn(VI).

   So A = MnO₂.

2. Formation of compound B — the dark green solution

   The fusion reaction:

(KNO₃ can replace O₂ as the oxidiser.)

The product is potassium manganate, K₂MnO₄, which dissolves in water to give a dark green solution. The colour comes from the manganate ion, MnO₄²⁻, where Mn is in the +6 state.

So B = K₂MnO₄.

3. Disproportionation of B to give purple compound C In neutral or acidic solution, the manganate ion is unstable and disproportionates:

One Mn(VI) is oxidised to Mn(VII) (purple permanganate) and another is reduced to Mn(IV) (brown MnO₂). The purple colour is unmistakable.

So C = KMnO₄ (or simply the permanganate ion, MnO₄⁻).

4. Reaction of C with KI in alkaline medium The problem says: “An alkaline solution of compound C oxidises potassium iodide solution to a compound D and compound A is also formed.” …

Here is the clear solution method for this problem.

Method: Systematic Deduction via Oxidation States & Colour

This method uses the characteristic colours and oxidation states of manganese compounds, combined with the logic of redox reactions (disproportionation and oxidation), to identify each species step-by-step.

Step 1: Identify Compound (A) from the starting material.

• The problem begins with "an oxide of manganese (A)". The most common and stable manganese oxide is manganese dioxide ($MnO_2$).
• Result: A = $MnO_2$ (Manganese dioxide, brown-black solid).

Step 2: Identify Compound (B) from the fusion reaction.

• Reaction: $MnO_2$ is fused with KOH in the presence of an oxidising agent (like $KNO_3$ or air). This oxidises Mn(IV) to Mn(VI).
• Colour clue: The product gives a dark green solution in water. The only common manganese compound with a dark green colour is potassium manganate ($K_2MnO_4$).
• Result: B = $K_2MnO_4$ (Potassium manganate, dark green).
• Reaction: $2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$

Step 3: Identify Compound (C) from disproportionation.

• Reaction: Compound B ($K_2MnO_4$) disproportionates in neutral or acidic solution. Mn(VI) is unstable and converts to Mn(VII) and Mn(IV).
• Colour clue: The product is purple. The only common purple manganese compound is potassium permanganate ($KMnO_4$).
• Result: C = $KMnO_4$ (Potassium permanganate, purple).
• Disproportionation reaction:
  • In neutral solution: $3K_2MnO_4 + 2H_2O \rightarrow 2KMnO_4 + MnO_2 + 4KOH$
  • In acidic solution: $2K_2MnO_4 + 2H_2SO_4 \rightarrow 2KMnO_4 + MnO_2 + 2K_2SO_4 + 2H_2O$
  • (Note: $MnO_2$ is also formed here, which is compound A). …

Here are the common mistakes students make on this classic inorganic synthesis problem, along with the correct conceptual approach to avoid them.

Mistake 1: Misidentifying Compound (A) — The Starting Oxide

• The Mistake: Students often guess $\text{MnO}_2$ (manganese dioxide) for (A) because it is the most common oxide of manganese. While $\text{MnO}_2$ can be used, the problem specifies the oxide is fused with KOH in the presence of an oxidising agent. This implies (A) is in a lower oxidation state so it can be oxidised to the dark green compound.
• Why it’s wrong: If (A) were $\text{MnO}_2$ (Mn in +4 state), fusing it with KOH and an oxidant (like air or $\text{KNO}_3$) would still work, but the problem’s wording “an oxide of manganese (A) is fused…” often points to the most common lower oxide that undergoes this specific fusion.
• The Correct Approach: Recall the fusion of $\text{MnO}_2$ with KOH and an oxidant gives potassium manganate ($\text{K}_2\text{MnO}_4$). However, the problem’s phrasing “an oxide of manganese (A)” is ambiguous. The most common exam answer is that (A) is $\text{MnO}_2$ (manganese dioxide). The fusion reaction is:

**Avoid this mistake by:** Remembering that the fusion of any manganese oxide (usually $\text{MnO}_2$) with KOH in air/oxidant yields the manganate ion ($\text{MnO}_4^{2-}$), which is dark green.

Mistake 2: Confusing Compound (B) — The Dark Green Solution

• The Mistake: Thinking the dark green solution is potassium permanganate ($\text{KMnO}_4$), which is purple.
• Why it’s wrong: Permanganate ($\text{MnO}_4^-$) is purple. The dark green colour is characteristic of the manganate ion ($\text{MnO}_4^{2-}$).
• The Correct Approach: Compound (B) is potassium manganate ($\text{K}_2\text{MnO}_4$). The colour comes from the $\text{MnO}_4^{2-}$ ion. Remember the colour code:
  • Green → Manganate ($\text{MnO}_4^{2-}$, +6 oxidation state)
  • Purple → Permanganate ($\text{MnO}_4^-$, +7 oxidation state)

Mistake 3: Forgetting the Disproportionation Condition

• The Mistake: Stating that compound (B) disproportionates in any solution, or writing the wrong products.
• Why it’s wrong: The disproportionation of manganate is pH-dependent. In strongly alkaline solution, $\text{MnO}_4^{2-}$ is stable. It only disproportionates when the solution is made neutral or acidic.
• The Correct Approach: The reaction is:

- **Purple product:** $\text{MnO}_4^-$ (compound C)
- **Brown precipitate:** $\text{MnO}_2$ (which is actually compound A again!)
**Avoid this mistake by:** Always linking disproportionation of manganate to the addition of acid or $\text{CO}_2$ (which makes the solution acidic via carbonic acid).

Mistake 4: Misidentifying Compound (D) — The Product of Iodide Oxidation

• The Mistake: Saying compound (D) is $\text{I}_2$ (iodine), by analogy with permanganate's reaction with iodide in acidic medium.
• Why it’s wrong: The question specifies an alkaline solution of compound C. In alkaline medium, iodide is oxidised all the way to iodate ($\text{IO}_3^-$) — any $\text{I}_2$ formed does not survive and is itself further oxidised by the excess $\text{OH}^-$/permanganate present.
• The Correct Approach: The reaction of alkaline $\text{KMnO}_4$ (compound C) with KI: $2\text{KMnO}_4 + \text{KI} + \text{H}_2\text{O} \rightarrow 2\text{MnO}_2 + \text{KIO}_3 + 2\text{KOH}$ …