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 concept here is oxidation state control by reaction conditions. Manganese can exist in many oxidation states, and the product formed depends on the strength of the oxidising agent and the medium (acidic vs. alkaline). In a strongly alkaline fusion with a moderate oxidant (O₂), manganese(IV) in MnO₂ is oxidised to manganese(VI), not to manganese(VII). The deep green colour of the manganate(VI) ion is a classic identifying feature.

Let’s walk through the reasoning step by step.

1. Identify the starting material and conditions.

   We begin with $\mathrm{MnO}_2$, where manganese is in the +4 oxidation state. It is fused (heated strongly) with KOH (a strong base) in the presence of oxygen gas ($\mathrm{O}_2$). The fusion creates a molten, highly alkaline environment.

2. Determine the possible oxidation products.

   In alkaline conditions, manganese can be oxidised to:

   • $\mathrm{Mn}_2\mathrm{O}_3$ (Mn in +3) — but this is less oxidised than the starting +4, so it would require reduction, not oxidation. Not possible here.
   • $\mathrm{MnO}_2$ (Mn in +4) — unchanged, but the reaction explicitly includes O₂, so oxidation is expected.
   • $\mathrm{K}_2\mathrm{MnO}_4$ (manganate, Mn in +6) — a dark green compound.
   • $\mathrm{KMnO}_4$ (permanganate, Mn in +7) — a purple compound.

3. Recall the classic fusion reaction.

   The standard laboratory preparation of potassium manganate is exactly this:

The product is potassium manganate(VI), which is dark green. This is a well-known reaction in inorganic chemistry.

4. Why not KMnO₄? …

Concept & Intuition

The chemistry here hinges on the acid–base equilibrium between dichromate ($\mathrm{Cr}_2\mathrm{O}_7^{2-}$) and chromate ($\mathrm{CrO}_4^{2-}$). In acidic solution, dichromate is stable; in basic solution, it shifts to chromate. Heating with aqueous NaOH provides a strongly basic environment, so the reaction is simply a deprotonation/condensation reversal — no redox occurs, and no insoluble hydroxide forms because chromate is soluble.

1. Recognize the species involved

   Potassium dichromate ($\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$) in water gives $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ ions. Aqueous NaOH supplies $\mathrm{OH}^-$ ions, making the solution basic.

2. Write the equilibrium

   The dichromate–chromate equilibrium is:

In basic solution, $\mathrm{OH}^-$ consumes $\mathrm{H}^+$, pulling the equilibrium to the right.

3. Apply Le Chatelier’s principle

   Adding $\mathrm{OH}^-$ removes $\mathrm{H}^+$ (forming water), so the system shifts to produce more $\mathrm{CrO}_4^{2-}$. Heating accelerates the reaction but does not change the product.

4. Check the options

   • (A) $\mathrm{Cr}(\mathrm{OH})_2$: This would require reduction of Cr(VI) to Cr(II), which does not occur here (no reducing agent). …

Step 1 — The concept: oxidant is reduced, reductant is oxidised

Half the options here are traps built on confusing these two words. Fix them first:

• Oxidising agent — accepts electrons ⇒ its own oxidation number falls ⇒ it is reduced.
• Reducing agent — donates electrons ⇒ its own oxidation number rises ⇒ it is oxidised.

So an option saying "$\ce{MnO4-}$ is oxidised to $\ce{Mn^{2+}}$" is self-contradictory on its face: $+7 \to +2$ is a fall, which is reduction. That immediately kills options (A) and (C).

Step 2 — What must $\ce{MnO4-}$ do?

Oxidation number of Mn in $\ce{MnO4-}$:

$+7$ is manganese's highest possible oxidation state — it has no electrons left to lose, so it cannot be oxidised. It can only gain electrons. In acidic medium (as stated) the standard 5-electron reduction is:

Mn: $\;+7 \to +2$. Reduced. ✓ (The purple colour discharges to the almost-colourless $\ce{Mn^{2+}}$ — the basis of permanganate titrations being self-indicating.)

Step 3 — What must $\ce{H2S}$ do?

Oxidation number of S in $\ce{H2S}$:

$-2$ is sulphur's lowest oxidation state — it has a full octet and no electrons left to gain, so it cannot be reduced. It can only lose electrons. So option (C)'s "$\ce{H2S}$ is reduced to $\ce{SO2}$" is doubly absurd: it is not reduction, and $-2 \to +4$ is a rise.

With permanganate, sulphide is oxidised to free sulphur:

Step 4 — Balance the overall reaction

Multiply the Mn half-reaction by 2 (10 e⁻) and the S half-reaction by 5 (10 e⁻) so the electrons cancel: …

Z is described as orange crystals used as an oxidising agent in acid medium - this is potassium dichromate, $\text{K}_2\text{Cr}_2\text{O}_7$. Reading the standard industrial preparation backwards:

Roasting with $\text{Na}_2\text{CO}_3$ / $\text{O}_2$: the chromite ore is fused in air, oxidising Cr(III) to Cr(VI) as sodium chromate (X):

Acidification ($\text{H}^+$): yellow chromate converts to orange dichromate (Y):

$2\,\text{Na}_2\text{CrO}_4 + 2\,\text{H}^+ \rightarrow \text{Na}_2\text{Cr}_2\text{O}_7 + 2\,\text{Na}^+ + \text{H}_2\text{O}$ …

The key to this question lies in understanding the nature of the ore and the method of extraction that is actually used in industry. Copper pyrites, also known as chalcopyrite, is a mixed sulphide of copper and iron. For sulphide ores, the common route is roasting followed by reduction. But here's the twist: copper extraction from chalcopyrite does not use coke or a separate reducing agent. Instead, it exploits a clever chemical cycle.

The process is called auto reduction (or sometimes "self-reduction"). Here's why it works:

1. Partial roasting: The concentrated ore is first roasted in a limited supply of air. This converts some of the copper sulphide ($Cu_2S$) into copper oxide ($Cu_2O$), while the iron sulphide ($FeS$) is largely converted to iron oxide ($FeO$), which is then removed as slag with silica.

2. The auto reduction step: The remaining $Cu_2S$ (which was not roasted) now reacts with the freshly formed $Cu_2O$ in the absence of air. This is a redox reaction where the sulphide acts as the reducing agent for the oxide:

Notice: no coke, no external reducing agent. The copper sulphide itself provides the reduction. This is the "auto" part — the ore reduces itself.

3. Why not the other options?
   • (B) Thermal decomposition: This works for ores like carbonates (e.g., $ZnCO_3$) or hydroxides, which break down on heating. Sulphides like $CuFeS_2$ do not simply decompose to metal; they need a chemical reaction. …

The key here is to connect each observation to a specific chemical property. The problem gives three clues: a white precipitate with $BaCl_2$, that precipitate dissolves in dilute HCl to produce a gas, and that gas decolourises acidified $KMnO_4$. Each clue narrows down the possibilities.

Let’s walk through it step by step.

1. White precipitate with $BaCl_2$

   Barium chloride ($BaCl_2$) is a common test for sulfate ($SO_4^{2-}$) and sulfite ($SO_3^{2-}$) ions. Both form white precipitates:

   • $Ba^{2+} + SO_4^{2-} \rightarrow BaSO_4$ (white, insoluble in dilute acids)
   • $Ba^{2+} + SO_3^{2-} \rightarrow BaSO_3$ (white, but soluble in dilute acids) So the precipitate could be either $BaSO_4$ or $BaSO_3$. We need the next clue to decide.

2. Precipitate dissolves in dilute HCl to produce a gas

   This is the decisive step.

   • $BaSO_4$ is insoluble even in dilute HCl — it does not dissolve. So if the precipitate were $BaSO_4$, it would not produce any gas.
   • $BaSO_3$, on the other hand, reacts with dilute HCl:

 The gas produced is sulfur dioxide ($SO_2$).  

Therefore, the precipitate must be $BaSO_3$, and the gas (B) is $SO_2$.

3. Gas (B) decolourises acidified $KMnO_4$ Acidified potassium permanganate ($KMnO_4$) is a strong oxidising agent. $SO_2$ is a reducing agent and readily reduces $MnO_4^-$ (purple) to $Mn^{2+}$ (colourless): …

The key idea: an amphoteric oxide sits in the middle of the acidity scale — it can act as an acid toward a strong base and as a base toward a strong acid. For transition metal oxides, the acidity increases as the oxidation state of the metal increases. A low oxidation state gives a basic oxide (like CrO, where Cr is +2), a very high oxidation state gives an acidic oxide (like Mn$_2$O$_7$, where Mn is +7), and intermediate oxidation states give amphoteric behaviour.

Let’s check each oxide in the options.

1. $V_2O_5$ — Vanadium is in the +5 oxidation state. This is a moderately high state, and indeed $V_2O_5$ is well-known as an amphoteric oxide. It dissolves in strong acids to give vanadyl salts (e.g., $VO_2^+$) and in strong bases to give vanadate ions ($VO_4^{3-}$). So it passes the test.

2. $Cr_2O_3$ — Chromium is in the +3 oxidation state. This is the classic amphoteric oxide of chromium. It reacts with acids to give $Cr^{3+}$ salts and with bases to give chromite ions ($CrO_2^-$). So it also passes.

3. $Mn_2O_7$ — Manganese is in the +7 oxidation state. This is a very high oxidation state, making the oxide strongly acidic. It reacts with water to give permanganic acid ($HMnO_4$) and does not behave as a base. So it is not amphoteric.

4. $CrO$ — Chromium is in the +2 oxidation state. This is a low oxidation state, so the oxide is basic. It reacts with acids to give $Cr^{2+}$ salts but does not react with bases. Not amphoteric.

5. $V_2O_4$ — Vanadium is in the +4 oxidation state. This oxide is actually amphoteric as well (it reacts with both acids and bases), but the question asks for a pair. Let’s see the options.

Now examine each option:

• (A) $V_2O_5$, $Cr_2O_3$ — Both are amphoteric. This looks correct.
• (B) $Mn_2O_7$, $Cr_2O_3$ — $Mn_2O_7$ is acidic, not amphoteric. So wrong.
• (C) $CrO$, $V_2O_5$ — $CrO$ is basic, not amphoteric. So wrong. …