Three separate samples of a solution of a single salt gave these results. One formed a white precipitate with excess ammonia solution, one formed a white precipitate with dil. $N a C l$ solution and one formed a black precipitate with $H_{2} S$ .
The salt could be :

A g N O_{3}

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

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H g (N O_{3})_{2}

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M n S O_{4}

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Solution

The correct option is B $P b (N O_{3})_{2}$
The salt could be $P b (N O_{3})_{2}$

It formed a white precipitate with excess ammonia solution:

$P b (N O_{3})_{2} + 2 N H_{4} O H \to P b (O H)_{2} ↓ (white ppt) + 2 N H_{4} N O_{3}$

It formed a white precipitate with dil. $N a C l$ solution:

$P b (N O_{3})_{2} + 2 N a C l \to P b C l_{2} ↓ (white ppt) + 2 N a N O_{3}$

It formed a black precipitate with $H_{2} S$ :
$P b (N O_{3})_{2} + H 2 S \to P b S ↓ (black ppt) + 2 H N O_{3}$

So, the correct option is $B$

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A white precipitate will be formed if we add common salt solution to :

(a) $B a (N O_{3})_{2}$ solution (b) $K N O_{3}$ solution (c) $A g N O_{3}$ solution (d) $M g (N O_{3})_{2}$ solution

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

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