Q.A violet compound of manganese (A) decomposes on heating to liberate oxygen and compounds (B) and (C) of manganese are formed. Compound (C) reacts with KOH in the presence of potassium nitrate to give compound (B). On heating compound (C) with conc. $H_2SO_4$ and NaCl, chlorine gas is liberated and a compound (D) of manganese along with other products is formed. Identify compounds A to D and also explain the reactions involved.

The key idea is the thermal decomposition of potassium permanganate and the redox chemistry of manganese in different oxidation states.

Step 1 – Identify A and its decomposition.

The violet compound is potassium permanganate, $\text{KMnO}_4$ (A). On heating, it decomposes:

Thus B is potassium manganate ($\text{K}_2\text{MnO}_4$, green) and C is manganese dioxide ($\text{MnO}_2$, black).

Step 2 – Reaction of C with KOH and KNO$_3$.

$\text{MnO}_2$ (C) is oxidised by KNO$_3$ in molten KOH to give $\text{K}_2\text{MnO}_4$ (B):

(Here KNO$_3$ is the oxidising agent, itself reduced to KNO$_2$.)

Step 3 – Reaction of C with conc. H$_2$SO$_4$ and NaCl. …

This is a classic inorganic synthesis sequence that tests your understanding of manganese chemistry — specifically the interconversion between different oxidation states of manganese under different conditions. The violet colour of compound A is the first big clue: only one common manganese compound is intensely violet — potassium permanganate ($\text{KMnO}_4$), where Mn is in the +7 oxidation state.

Let’s work through each clue systematically.

1. Compound A is violet and decomposes on heating to give oxygen, plus compounds B and C. The only violet manganese compound you need to know is $\text{KMnO}_4$. On heating, it decomposes:

Here, $\text{K}_2\text{MnO}_4$ (potassium manganate, green) is compound B, and $\text{MnO}_2$ (manganese dioxide, black/brown) is compound C. Oxygen gas is liberated.

2. Compound C reacts with KOH in the presence of potassium nitrate to give compound B. Compound C is $\text{MnO}_2$. Here, $\text{KNO}_3$ acts as an oxidising agent in a fused alkaline medium. The reaction is:

Mn in $\text{MnO}_2$ (+4) is oxidised to +6 in $\text{K}_2\text{MnO}_4$ (compound B). This confirms B is potassium manganate.

3. On heating compound C with conc. $\text{H}_2\text{SO}_4$ and NaCl, chlorine gas is liberated and a compound D of manganese is formed. This is a two-step process in one pot. First, conc. $\text{H}_2\text{SO}_4$ reacts with NaCl to produce HCl gas:

Then, $\text{MnO}_2$ (C) oxidises this HCl to chlorine gas — the classic laboratory preparation of $\text{Cl}_2$ — itself being reduced to $\text{Mn}^{2+}$:

$\text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + \text{Cl}_2 + 2\text{H}_2\text{O}$ …

Method: Reaction Pathway Deduction via Oxidation States & Known Manganese Chemistry

This method uses the characteristic oxidation states and colours of manganese compounds, combined with step-by-step logical deduction from the given reactions.

Step 1: Identify compound A from colour and decomposition

• Observation: Violet compound of manganese → most likely potassium permanganate ($KMnO_4$) — deep violet, common in inorganic synthesis.
• Reaction given: A decomposes on heating to give $O_2$ + B + C.

Known fact: $KMnO_4$ on heating decomposes as:

• B = $K_2MnO_4$ (potassium manganate, green)
• C = $MnO_2$ (manganese dioxide, black/brown)

✓ A = $KMnO_4$

Step 2: Verify reaction of C with KOH + KNO₃ to give B

• Given: C + KOH + KNO₃ → B
• C = $MnO_2$, B = $K_2MnO_4$

Reaction: $MnO_2$ is oxidised to manganate in alkaline medium using an oxidising agent like KNO₃.

✓ Confirms B and C.

Step 3: Identify D from reaction of C with conc. $H_2SO_4$ + NaCl

• Given: C + conc. $H_2SO_4$ + NaCl → $Cl_2$ + D + other products
• C = $MnO_2$

Known reaction: $MnO_2$ with conc. $H_2SO_4$ and NaCl produces chlorine gas (used in lab preparation of $Cl_2$).

• D = $MnCl_2$ (manganese(II) chloride, pale pink)

✓ D is $MnCl_2$.

Final Answer

| Compound | Name | Formula |

|----------|------|---------| …

Here is a breakdown of the common mistakes students make with this classic inorganic synthesis problem, and how to avoid them.

The Core Concept (The "Why")

This question tests your knowledge of the oxidation states of Manganese (Mn) and how they change under different conditions (heating, acidic/alkaline media). The key is to track the color and the oxidation state of Mn.

• Mn in +7 (Violet): Permanganate ($MnO_4^-$), e.g., $KMnO_4$.
• Mn in +6 (Green): Manganate ($MnO_4^{2-}$), e.g., $K_2MnO_4$.
• Mn in +4 (Black/Brown): Manganese dioxide ($MnO_2$).
• Mn in +2 (Pale Pink/Colorless): Manganous salts ($Mn^{2+}$), e.g., $MnCl_2$.

The reactions are a cycle of oxidation and reduction.

Common Mistake #1: Misidentifying Compound (A)

The Mistake: Students often guess $MnO_2$ or $Mn_2O_7$ as the starting violet compound. $MnO_2$ is black/brown, not violet. $Mn_2O_7$ is a greenish-brown oily liquid, not a solid, and is violently explosive.

The Correct Approach:

• Clue: "A violet compound of manganese."
• Knowledge: The only common violet manganese compound is Potassium Permanganate ($KMnO_4$) . The color comes from the permanganate ion ($MnO_4^-$).
• Result: A = $KMnO_4$ .

Common Mistake #2: Forgetting the Products of $KMnO_4$ Decomposition

The Mistake: Writing the decomposition of $KMnO_4$ as:

This is incorrect because it doesn't account for the stable potassium manganate ($K_2MnO_4$) that also forms.

The Correct Approach:

• Clue: Heating $KMnO_4$ liberates $O_2$ and forms two manganese compounds (B and C).
• Knowledge: The thermal decomposition of $KMnO_4$ is a classic reaction. It produces a green compound ($K_2MnO_4$) and a black compound ($MnO_2$).
• Reaction:

• Result: B = $K_2MnO_4$ (Potassium manganate, green), C = $MnO_2$ (Manganese dioxide, black).

Common Mistake #3: Confusing the Reagents for Converting C to B

The Mistake: Thinking that $MnO_2$ (C) reacts directly with KOH to give $K_2MnO_4$ (B). This is wrong. $MnO_2$ is already in the +4 state, and $K_2MnO_4$ has Mn in the +6 state. You need an oxidizing agent to increase the oxidation state.

The Correct Approach:

• Clue: "Compound (C) reacts with KOH in the presence of potassium nitrate to give compound (B)."
• Knowledge: $KNO_3$ is the oxidizing agent. It oxidizes $MnO_2$ (Mn+4) to $K_2MnO_4$ (Mn+6). The reaction is a fusion (heating with a solid).
• Reaction:

• Key Point: The $KNO_3$ is essential. Without it, $MnO_2$ and KOH won't form $K_2MnO_4$.

Common Mistake #4: Writing the Wrong Product for the Chlorine Evolution Reaction

The Mistake: Writing the reaction of $MnO_2$ (C) with conc. $H_2SO_4$ and NaCl as producing $MnSO_4$ and $Cl_2$, but forgetting the water or writing the wrong manganese compound (D).

The Correct Approach: …