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Direct Combination

KCl+conc.H2SO...

Question

$K C l + c o n c . H_{2} S O_{4} + K_{2} C r_{2} O_{7} Δ - \to (X) N a O H - --- \to (Y)$ ,
$(X)$ is reddish brown colored gas soluble in NaOH forming $(Y)$ . $(X)$ and $(Y)$ are:

A

C r_{2} O C l_{2}

,

N a_{2} C r O_{3}

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B

C r_{2} O_{2} C l_{2}

,

N a_{2} C r O_{3}

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C

C r O_{2} C l

,

N a_{2} C r O_{4}

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D

C r O_{2} C l_{2}

,

N a_{2} C r O_{4}

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Solution

The correct option is D $C r O_{2} C l_{2}$ , $N a_{2} C r O_{4}$
It is chromyl chloride test for confirmation of chloride ions.

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.

$4 K C l + K_{2} C r_{2} O_{7} + 6 H_{2} S O_{4}$ $⟶$ $2 K H S O_{4}$ + $2 C r O_{2} C l_{2} (c h r o m y l c h l o r i d e)$ + $4 N a H S O_{4} + 3 H_{2} O$

$C r O_{2} C l_{2} + 4 N a O H ⟶ N a_{2} C r O_{4} + 2 N a C l + 2 H_{2} O$

Option (D) is correct.

$(X)$ and $(Y)$ are $C r O_{2} C l_{2}$ and $N a_{2} C r O_{4}$ .

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