Ncert Exercises & Solutions

When a chromite ore (A) is fused with sodium carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (B) is obtained. After treatment of this yellow solution with sulphuric acid, compound (C) can be crystallised from the solution. When compound (C) is treated with KCL, orange crystals of compound (D) crystallise out. Identify A to D and also explain the reactions.

NEETprep Answer: $K_{2} C r_{2} O_{7},$ is an orange compound. It is formed when

N a_{2} C r_{2} O_{7}

reacts with KCL.In acidic medium, yellow coloured

C r O_{4}^{2 -}

(chromate ion) changes into dichromate.

The given process is the preparation method of potassium dichromate from chromite ore.

A = F e C r_{2} O_{4}; B = N a_{2} C r O_{4}; C = N a_{2} C r_{2} O_{7}; D = K_{2} C r_{2} O_{7}

(i) 4 F e C r_{2} O_{4} + 8 N a_{2} C O_{3} + 7 O_{2} \to 8 N a_{2} C r O_{4} + 2 F e_{2} O_{3} + 8 C O_{2}

[A] [B]

(i i) 2 N a_{2} C r O_{4} + 2 H^{+} \to N a_{2} C r_{2} O_{7} + 2 N a^{+} + H_{2} O

(i i i) N a_{2} C r_{2} O_{7} + 2 K C L \to K_{2} C r_{2} O_{7} + 2 N a C L

[C] [D]

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