Question

An aqueous solution of a salt (A) gives white precipitate (B) with sodium chloride solution. The filtrate gives a black precipitate. (C) When $H_{2}S$ is passed into it. Compound (B) dissolves in hot water and the solution gives a yellow precipitate (D) on treatment with sodium iodide. The compound (A) does not give any gas with dilute HCl but liberates a reddish brown gas on heating. Identify the compound (A) to (D).

• A. $A=Pb(N{O}_3{)}_{2}B=PbC{l}_{2}C=PbSD=Pb{l}_2$
• B. $A=PbS{O}_{3}B=PbC{l}_{2}C=PbClD=Pb{O}_2$
• C. $A=Pb{l}_{2}B=PbOC=Pb{H}_{4}D=Pb(N{O}_3{)}_2$
• D. $A=PbSB=PbC{l}_{2}C=Pb(N{O}_3{)}_{2}D=Pb{l}_2$

Answer 1

The correct option is A

$A=Pb(N{O}_3{)}_{2}B=PbC{l}_{2}C=PbSD=Pb{l}_2$

Since the crystalline compound (B) dissolves in hot water and gives a yellow precipitate withNaI, it should be lead chloride, $PbC{l}_2$ and the solution (A) consists a lead salt.

$PbC{l}_2+2Nal(\to )Pb{l}_2+2NaCl$

(B) (D)

The compound (A) does not give any gas with dilute HCl but liberates a reddish brown gas on heating, it should be lead nitrate, $Pb(N{O}_3{)}_2$

$2Pb\left(N{O}_3{)}_2\right(\to )2PbO+2N{O}_2+O_2$

(A) Reddish brown gas

Lead chloride is sparingly soluble in water.When H₂S is passed, it gives a black precipitate of lead sulphide, PbS.

$PbC{l}_2+H_{2}S(\to )PbS+2HCl$

Black

Thus

A) is lead nitrate, $Pb\left(N{O}_3{)}_{2}B\right)$is lead chloride, $PbC{l}_2$

C) is lead Sulphide, PbSD) is lead Iodide, $Pb{l}_2$