Q.Indicate the steps in the preparation of:

Concept: Inorganic Synthesis — extraction and conversion of ores into pure compounds via oxidation and fusion.

(i) $K_2Cr_2O_7$ from chromite ore ($FeCr_2O_4$)

1. Fusion with alkali: Chromite is fused with $Na_2CO_3$ in air to form sodium chromate: $4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$
2. Leaching & acidification: The yellow solution of $Na_2CrO_4$ is leached, then acidified with $H_2SO_4$ to give $Na_2Cr_2O_7$: $2Na_2CrO_4 + H_2SO_4 \rightarrow Na_2Cr_2O_7 + Na_2SO_4 + H_2O$
3. Conversion to potassium salt: $Na_2Cr_2O_7$ is treated with KCl to precipitate $K_2Cr_2O_7$ (less soluble): $Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 + 2NaCl$

(ii) $KMnO_4$ from pyrolusite ore ($MnO_2$)

1. Fusion with alkali: $MnO_2$ is fused with KOH and an oxidising agent (KNO$_3$ or air) to form green $K_2MnO_4$: $2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$ …

(i) Preparation of $K_2Cr_2O_7$ from chromite ore

Concept and intuition: Chromite ore ($FeCr_2O_4$) contains chromium in the +3 oxidation state. To make dichromate (where Cr is +6), we need to oxidise it. The trick is to first convert the insoluble ore into a water-soluble chromate salt by fusing it with an alkali (like $Na_2CO_3$) in the presence of air (oxygen). This step also separates chromium from iron, which remains as insoluble oxide. Once we have sodium chromate in solution, we acidify it to get sodium dichromate, and then use a double displacement reaction with KCl to precipitate the less soluble potassium dichromate.

Step-by-step:

1. Fusion of chromite ore: The finely powdered ore is mixed with sodium carbonate ($Na_2CO_3$) and a small amount of quicklime ($CaO$, to keep the mass porous) and roasted in a reverberatory furnace in a current of air. The reaction is:

Here, $FeCr_2O_4$ is oxidised: $Fe^{2+} \rightarrow Fe^{3+}$ (as $Fe_2O_3$) and $Cr^{3+} \rightarrow Cr^{6+}$ (as $Na_2CrO_4$). The sodium chromate is water-soluble, while $Fe_2O_3$ and any unreacted ore settle out.

2. Leaching and filtration: The fused mass is cooled and treated with hot water. Sodium chromate dissolves, leaving behind a red-brown residue of ferric oxide. The solution is filtered to get a clear yellow solution of $Na_2CrO_4$.

3. Conversion to dichromate: The sodium chromate solution is acidified with concentrated sulphuric acid. In acidic medium, chromate ($CrO_4^{2-}$) converts to dichromate ($Cr_2O_7^{2-}$):

The colour changes from yellow (chromate) to orange (dichromate). This step is driven by the equilibrium shift: in acidic solution, dichromate is the stable form.

4. Double decomposition with KCl: To get the potassium salt, we add a calculated amount of potassium chloride ($KCl$) to the concentrated solution of sodium dichromate. Because $K_2Cr_2O_7$ is less soluble than $Na_2Cr_2O_7$ at lower temperatures, it crystallises out:

The orange crystals of $K_2Cr_2O_7$ are filtered, washed with a little cold water, and dried.

(ii) Preparation of $KMnO_4$ from pyrolusite ore

Concept and intuition: Pyrolusite ($MnO_2$) contains manganese in the +4 state. To get permanganate ($MnO_4^-$), where Mn is +7, we need a strong oxidation. The classic route is to first fuse $MnO_2$ with an alkali (KOH) and an oxidising agent (like $KNO_3$ or air) to form potassium manganate ($K_2MnO_4$, green, Mn +6). Then, this manganate is oxidised further to permanganate (purple, Mn +7) — either by electrolytic oxidation or by chemical oxidation (e.g., with chlorine or ozone). The electrolytic method is preferred industrially because it's cleaner and gives high purity.

Step-by-step:

1. Fusion of pyrolusite: Finely powdered $MnO_2$ is mixed with potassium hydroxide ($KOH$) and an oxidising agent like potassium nitrate ($KNO_3$) or potassium chlorate ($KClO_3$), and heated strongly in a furnace. The reaction is:

(If $KNO_3$ is used: $MnO_2 + 2KOH + KNO_3 \rightarrow K_2MnO_4 + KNO_2 + H_2O$)

The green mass of potassium manganate is formed. This step oxidises Mn from +4 to +6.

2. Leaching: The fused green mass is cooled and treated with a small amount of water (or dilute KOH) to dissolve $K_2MnO_4$. The solution is filtered to remove any insoluble impurities. The filtrate is a deep green solution of potassium manganate. …

Preparation of $K_2Cr_2O_7$ from Chromite Ore

Method: Alkaline oxidative fusion followed by acidification.

Steps:

1. Fusion (oxidation): Finely powdered chromite ore ($FeCr_2O_4$) is mixed with sodium carbonate ($Na_2CO_3$) and lime ($CaO$) or soda ash, then heated in a reverberatory furnace in the presence of air.

2. Leaching:

   The fused mass is treated with hot water. Sodium chromate ($Na_2CrO_4$) dissolves, while iron oxide ($Fe_2O_3$) remains insoluble and is filtered off.

3. Conversion to dichromate:

   The yellow solution of $Na_2CrO_4$ is acidified with concentrated $H_2SO_4$ to convert it to sodium dichromate:

4. Double decomposition: Hot concentrated solution of $Na_2Cr_2O_7$ is treated with KCl (or $K_2CO_3$). Due to lower solubility of $K_2Cr_2O_7$, it crystallizes out:

5. Crystallization: Orange-red crystals of $K_2Cr_2O_7$ are separated by filtration and dried.

Preparation of $KMnO_4$ from Pyrolusite Ore

Method: Alkaline oxidative fusion followed by electrolytic oxidation.

Steps:

1. Fusion (first oxidation): Finely powdered pyrolusite ($MnO_2$) is fused with KOH (or $K_2CO_3$) and an oxidizing agent like KNO$_3$ (or air) in a furnace.

(Potassium manganate — green mass)

2. Leaching:

   The green fused mass is treated with water. $K_2MnO_4$ dissolves, giving a green solution.

3. Oxidation to permanganate:

   The green solution is oxidized to purple $KMnO_4$ by one of two methods:

   • Method A (Electrolytic oxidation): The solution is electrolyzed using Ni electrodes. At the anode: $\text{MnO}_4^{2-} \rightarrow \text{MnO}_4^- + e^-$ …

Here are the common mistakes students make when writing the preparation of $K_2Cr_2O_7$ and $KMnO_4$, along with the correct conceptual approach to avoid each.

(i) Preparation of $K_2Cr_2O_7$ from Chromite Ore ($FeCr_2O_4$)

Mistake 1: Writing the wrong formula for chromite ore

• Wrong: $FeO \cdot CrO_3$ or $Cr_2O_3$ alone.
• Correct: Chromite is $FeCr_2O_4$ (or $FeO \cdot Cr_2O_3$).
• Why it matters: The oxidation state of Cr in ore is +3, not +6. You must start with the correct reactant to balance the redox steps.

Mistake 2: Forgetting the fusion step with air/oxygen

• Wrong: Directly leaching the ore with acid or water.
• Correct: The ore is fused with $Na_2CO_3$ (or $NaOH$) in the presence of air (oxygen) at high temperature.
• Why: Cr(III) is insoluble and must be oxidized to soluble Cr(VI) ($Na_2CrO_4$). Without air, no oxidation occurs.
• Avoid: Always write: “Fused with $Na_2CO_3$ in excess of air”.

Mistake 3: Confusing the yellow chromate ($CrO_4^{2-}$) with orange dichromate ($Cr_2O_7^{2-}$)

• Wrong: Writing $K_2Cr_2O_7$ forms directly after fusion.
• Correct: The fusion gives sodium chromate ($Na_2CrO_4$, yellow). This is then acidified to convert to sodium dichromate ($Na_2Cr_2O_7$), and finally treated with KCl to precipitate $K_2Cr_2O_7$.
• Avoid: Remember the sequence: $FeCr_2O_4 \xrightarrow{Na_2CO_3/air} Na_2CrO_4 \xrightarrow{H^+} Na_2Cr_2O_7 \xrightarrow{KCl} K_2Cr_2O_7 \downarrow$

Mistake 4: Omitting the purification step (recrystallization)

• Wrong: Stopping at the precipitation step.
• Correct: $K_2Cr_2O_7$ is less soluble in cold water, so it is recrystallized to obtain pure orange-red crystals.
• Avoid: Mention “recrystallization from hot water” as the final step.

(ii) Preparation of $KMnO_4$ from Pyrolusite Ore ($MnO_2$)

Mistake 1: Using the wrong ore formula

• Wrong: $Mn_2O_3$ or $Mn_3O_4$.
• Correct: Pyrolusite is $MnO_2$ (Mn in +4 state).
• Why it matters: The oxidation state determines the amount of oxidant needed.

Mistake 2: Skipping the alkaline fusion step

• Wrong: Directly treating $MnO_2$ with acid or $KOH$ alone.
• Correct: $MnO_2$ is fused with $KOH$ (or $K_2CO_3$) in the presence of air (or an oxidant like $KNO_3$) to form potassium manganate ($K_2MnO_4$, green).
• Avoid: Always write: “Fused with KOH and an oxidizing agent (air/KNO$_3$)”.

Mistake 3: Confusing manganate ($MnO_4^{2-}$) with permanganate ($MnO_4^-$)

• Wrong: Writing $KMnO_4$ forms directly from fusion.
• Correct: Fusion gives $K_2MnO_4$ (green). This is then disproportionated in acidic/neutral medium to give $KMnO_4$ (purple) and $MnO_2$.
• Avoid: Remember the color change: green ($Mn^{VI}$) → purple ($Mn^{VII}$) + brown ($Mn^{IV}$).

Mistake 4: Writing the wrong disproportionation reaction …