An aqueous solution of salt (A), gives white crystalline precipitate (B) with $N a C l$ 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 $N a I$ and cooling. The compound (A) does not give any gas with dilute $H C l$ but liberates reddish brown gas on heating.

Compound (A) 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 A $P b (N O_{3})_{2}$
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 sparingly 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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