Thus, the compound (D) is $P b I_{2}$ .

State whether the given statement is true or false.

True

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False

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Solution

The correct option is A True
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 form 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 lead 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.

$2 P b (N O_{3})_{2} h e a t - - \to 2 P b O + 4 N O_{2} + O_{2}$

Hence, the given statement is $true$

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