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Q4.26(i):
Indicate the steps in the preparation of: (i) \(K_{2}Cr_{2}O_{7}\) from chromite ore.

Solution :

Step 1: Conversion of Chromite Ore to Sodium Chromate

The process begins with the fusion of powdered chromite ore ($FeCr_{2}O_{4}$) with sodium carbonate ($Na_{2}CO_{3}$) in the presence of excess air. This step facilitates the oxidation of the Chromium(III) ion to Chromium(VI).

Chemical Reaction:

$4FeCr_{2}O_{4} + 8Na_{2}CO_{3} + 7O_{2} \rightarrow 8Na_{2}CrO_{4} + 2Fe_{2}O_{3} + 8CO_{2}$

[Justification: The iron (II) is oxidized to iron (III) oxide, and the chromium (III) is oxidized to the chromate ion ($CrO_{4}^{2-}$), which is soluble in water.]

Step 2: Acidification of Sodium Chromate to Sodium Dichromate

The resulting yellow solution of sodium chromate ($Na_{2}CrO_{4}$) is filtered and acidified with concentrated sulfuric acid ($H_{2}SO_{4}$). This shifts the equilibrium to form the orange-colored sodium dichromate ($Na_{2}Cr_{2}O_{7}$).

Chemical Reaction:

$2Na_{2}CrO_{4} + 2H_{2}SO_{4} \rightarrow Na_{2}Cr_{2}O_{7} + 2Na_{2}SO_{4} + H_{2}O$

[Justification: Per Le Chatelier's Principle, increasing the concentration of $H^{+}$ ions (acidification) promotes the dimerization of chromate ions to dichromate ions.]

Step 3: Conversion to Potassium Dichromate

Sodium dichromate is highly soluble and difficult to crystallize. To obtain potassium dichromate ($K_{2}Cr_{2}O_{7}$), the solution is treated with potassium chloride ($KCl$). Potassium dichromate is significantly less soluble than sodium dichromate at low temperatures and precipitates out upon cooling.

Chemical Reaction:

$Na_{2}Cr_{2}O_{7} + 2KCl \rightarrow K_{2}Cr_{2}O_{7} + 2NaCl$

[Justification: Based on the difference in solubility products ($K_{sp}$), $K_{2}Cr_{2}O_{7}$ crystallizes preferentially from the solution as it cools, leaving $NaCl$ in the mother liquor.]

Visual Representation of the Structural Interconversion

Chromate ($CrO_{4}^{2-}$) (Yellow) $H^{+}$ Dichromate ($Cr_{2}O_{7}^{2-}$) (Orange) $OH^{-}$

Final Solution: The preparation of $K_{2}Cr_{2}O_{7}$ involves the fusion of chromite ore with $Na_{2}CO_{3}$ to form sodium chromate, subsequent acidification to produce sodium dichromate, and final double decomposition with $KCl$ to precipitate $K_{2}Cr_{2}O_{7}$.


More Questions from Class 12 Chemistry The d- and f-Block Elements Exercises


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