Compound (D) is

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

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P b C l_{2}

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P b I

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P b I_{2}

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Solution

The correct option is D $P b I_{2}$
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 passes, 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. The reaction

P b (N O_{3})_{2} + 2 N a I \to 2 N a N O_{3} + P b I_{2} ↓

is classified as:

Q. Complete the reaction :

P b (N O_{3})_{2} + 2 H C l \to P b C l_{2} + 2

Q. How many atoms are present in the given reaction?

{P b ({N O}_{3})}_{2} + 2 K I \to P b I_{2} + 2 K N O_{3}

Q. Two solutions of following compounds having given volume are mixed.

[P b (N O_{3})_{2}]_{1} = 3.0 \times 10^{- 2} M

[N a C l]_{2} = 8.0 \times 10^{- 2} M

V_{1} = 0.15 L

V_{2} = 300 m L

K_{s p} (P b C l_{2}) = 2.4 \times 10^{- 4}

What will happen to

P b C l_{2}

Given the reaction:-

$P b {(N O_{3})}_{2} (a q) + 2 K I \to P b I_{2} (s) + 2 K N O_{3} (a q)$ What is the mass of $P b I_{2}$ that will precipitate if 10.2 g of $P b {(N O_{3})}_{2}$ is mixed with 5.73 g of $K I$ in a sufficient quantity of $H_{2} O$ ?

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