Preparation of K₂Cr₂O₇ from chromite ore — complete steps

Question

Describe the preparation of potassium dichromate ( $\text{K}_2\text{Cr}_2\text{O}_7$ ) from chromite ore ( $\text{FeCr}_2\text{O}_4$ ). Write the balanced equations for each step.

(NCERT Class 12, Chapter 8 — commonly asked in boards for 5 marks)

Solution — Step by Step

Chromite ore is fused with sodium carbonate in the presence of air (oxidation):

4\text{FeCr}_2\text{O}_4 + 8\text{Na}_2\text{CO}_3 + 7\text{O}_2 \rightarrow 8\text{Na}_2\text{CrO}_4 + 2\text{Fe}_2\text{O}_3 + 8\text{CO}_2

Cr(III) in the ore is oxidised to Cr(VI) in sodium chromate (yellow). The iron forms insoluble $\text{Fe}_2\text{O}_3$ .

The fused mass is extracted with water. Sodium chromate ( $\text{Na}_2\text{CrO}_4$ ) dissolves in water, while $\text{Fe}_2\text{O}_3$ remains as an insoluble residue. Filtration separates them.

The filtrate contains yellow $\text{Na}_2\text{CrO}_4$ solution.

The yellow sodium chromate solution is acidified with sulphuric acid:

2\text{Na}_2\text{CrO}_4 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{Cr}_2\text{O}_7 + \text{Na}_2\text{SO}_4 + \text{H}_2\text{O}

The yellow chromate ( $\text{CrO}_4^{2-}$ ) converts to orange dichromate ( $\text{Cr}_2\text{O}_7^{2-}$ ). This is the chromate-dichromate equilibrium shifted by acid.

Sodium dichromate is treated with potassium chloride:

\text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{KCl} \rightarrow \text{K}_2\text{Cr}_2\text{O}_7 + 2\text{NaCl}

$\text{K}_2\text{Cr}_2\text{O}_7$ is less soluble than $\text{Na}_2\text{Cr}_2\text{O}_7$ and crystallises out on cooling. It is collected by filtration and dried.

Why This Works

The entire process relies on the chemistry of chromium’s oxidation states. In the ore, chromium is in the +3 state. Fusion with alkali and oxygen oxidises it to +6 (chromate), making it water-soluble. Acidification converts the tetrahedral $\text{CrO}_4^{2-}$ to the bridged $\text{Cr}_2\text{O}_7^{2-}$ (dichromate). The final metathesis step exploits the lower solubility of the potassium salt.

The chromate-dichromate equilibrium is pH-dependent:

2\text{CrO}_4^{2-} + 2\text{H}^+ \rightleftharpoons \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O}

In base: yellow chromate dominates. In acid: orange dichromate dominates. This colour change is itself a common exam question.

Alternative Method

In some industrial processes, chromite ore is directly oxidised with concentrated $\text{H}_2\text{SO}_4$ and an oxidising agent (like $\text{HNO}_3$ ). But the alkali fusion method described above is the standard NCERT method and is what examiners expect in board exams.

For CBSE boards, this is typically a 5-mark question. Write each step with its balanced equation, name the products, and mention the colour changes (yellow chromate to orange dichromate). The examiners specifically look for balanced equations — losing marks here is avoidable.

Common Mistake

Students often skip the intermediate $\text{Na}_2\text{Cr}_2\text{O}_7$ step and write a direct reaction to $\text{K}_2\text{Cr}_2\text{O}_7$ . The process goes through sodium dichromate first — you cannot directly get $\text{K}_2\text{Cr}_2\text{O}_7$ from the chromate. Also, do not confuse chromate ( $\text{CrO}_4^{2-}$ , yellow, in base) with dichromate ( $\text{Cr}_2\text{O}_7^{2-}$ , orange, in acid).

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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