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) : $P b C l_{2}$ (C) : $P b S$ , (D) : $P b I_{2}$
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Solution

Lead nitrate (A) reacts with NaCl to form a white precipitate of lead chloride (B).
$P b (N O_{3})_{2} + 2 N a C l \to P b C l_{2} w h i t e p p t B + 2 N a N O_{3}$
Lead(II) ions react with hydrogen sulphide to from black ppt (C) of lead sulphide.
$P b^{2 +} + H_{2} S \to P b S b l a c k, C + 2 H^{+}$
Lead chloride reacts with KI to form yellow ppt D of lad iodide.
$P b C l_{2} + 2 K I \to P b I_{2} y e l l o w p p t, D + 2 K C l$
Lead nitrate on heating forms lead oxide and liberates a reddish brown gas which is nitrogen dioxide.
$2 P b (N O_{3})_{2} h e a t - - \to 2 P b O + 4 N O_{2} + O_{2}$

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