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Precipitation Reaction

The gas liber...

Question

The gas liberated on heating a mixture of two salts with $N a O H$ , give a reddish brown precipitate with an alkaline solution of $K_{2} H g l_{4}$ . The aqueous solution of the mixture on treatment with $B a C l_{2}$ gives a white precipitate which is sparingly soluble in concentrated $H C l$ . On heating the mixture with $K_{2} C r_{2} O_{7}$ and concentrated $H_{2} S O_{4}$ , red vapour $(A)$ are produced. The aqueous solution of the mixture gives a deep blue ppt. $(B)$ with Potassium ferricyanide solution. Identify the radicals.

A

(A) - C r O_{2} C l_{2},

(B) - F e_{3} [F e (C N)_{6}]_{2}

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B

(A) - C r O_{2} C l_{2},

(B) - F e_{2} [F e (C N)_{6}]_{2}

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C

(A) - C r O_{2} C l,

(B) - F e_{3} [F e (C N)_{6}]_{2}

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D

(A) - C r O_{2} C l,

(B) - F e_{2} [F e (C N)_{6}]_{2}

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Solution

The correct option is A $(A) - C r O_{2} C l_{2},$ $(B) - F e_{3} [F e (C N)_{6}]_{2}$
Option (A) is correct.
(A) and (B) are $C r O_{2} C l_{2}$ and $F e_{3} [F e (C N)_{6}]_{2}$ .
Mix a little amount of salt and an equal amount of solid potassium dichromate ( $K_{2} C r_{2} O_{7}$ ) in a test tube and add $c o n c . H_{2} S O_{4}$ to it. Heat the test tube and pass the evolved gas through sodium hydroxide solution. If a yellow solution is obtained, divide the solution into two parts. Acidify the first part with acetic acid and then add lead acetate solution. Formation of a yellow precipitate of lead chromate confirms the presence of chloride ions in the salt. This test is called chromyl chloride test.
$4 N a C l + K_{2} C r_{2} O_{7} + 6 H_{2} S O_{4} ⟶ 2 K H S O_{4} + 2 C r O_{2} C l_{2} + 4 N a H S O_{4} + 3 H_{2} O$
$C r O_{2} C l_{2} + 4 N a O H ⟶ N a_{2} C r O_{4} + 2 N a C l + 2 H_{2} O$
Another cation is $F e^{3 +}$ .
$F e^{3 +} + K_{3} [F e (C N)_{6}] ⟶ F e_{3} [F e (C N)_{6}]_{2} + K C l$

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