An aqueous solution of salt (A), gives white crystalline precipitate (B) with NaCl solution. The filtrate gives a black precipitate (C) when $H_{2} S$ is passed through it. Compound (B) is dissolves in hot water and the solution gives a yellow precipitate (D) on treating with NaI and cooling. The compound (A) does not give any gas with dilute HCl but liberates reddish brown gas on heating.

Compound (B) is:

P b (N O_{3})_{2}

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P b C l_{2}

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P b S

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P b I_{2}

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Solution

The correct option is C $P b C l_{2}$
Since the crystalline compound (B) dissolves in hot water and gives a yellow precipitate with $N a I$ , it should be lead chloride, $P b C l_{2}$ and the solution (A) consists a lead salt.
$P b C l_{2} + 2 N a I \to P b I_{2} + 2 N a C l$
(B, white ppt.) (D, yellow ppt)

The compound (A) does not give any gas with dilute $H C l$ but liberates a reddish brown gas on heating, it should be lead nitrate, $P b (N O_{3})_{2}$ .
$2 P b (N O_{3})_{2} \to 2 P b O + 2 N O_{2} + O_{2}$
(A) (reddish brown gas)

Lead chloride is sparangly soluble in water. When $H_{2} S$ is passed, it gives a black precipitate of lead sulphide, $P b S$ .
$P b C l_{2} + H_{2} S \to P b S + 2 H C l$
(C, black ppt)

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Similar questions

Q. An aqueous solution of salt (A) gives a white crystalline precipitate (B) with NaCl solution. The filtrate gives a black precipitate (C) when

H_{2} S

is passed through it. Compound (B) dissolves in hot water and the solution gives yellow precipitate (D) on treatment with potassium iodine and on cooling. The compound (A) does not give any gas with dilute HCl but liberates a reddish brown gas on heating. Compounds (A) to (D) are identified as:

(A) :

P b (N O_{3})_{3}

, (B) :

P b C l_{2}

P b S

, (D) :

P b I_{2}

If true enter 1, else enter 0.

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