JEE Advanced 2020 · Paper 1 · Chemistry Q3

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ChemistrySingle correct+3 / -1Question 21 of 54

A colorless aqueous solution contains nitrates of two metals, X and Y. When it was added to an aqueous solution of NaCl, a white precipitate was formed. This precipitate was found to be partly soluble in hot water to give a residue P and a solution Q. The residue P was soluble in aq. $\ce{NH3}$ and also in excess sodium thiosulfate. The hot solution Q gave a yellow precipitate with KI. The metals X and Y, respectively, are

Ag and Pb

Ag and Cd

Cd and Pb

Cd and Zn

Answer and worked solution

Official answer: A

White chlorides. NaCl gives white precipitates only with $\ce{Ag+}$ , $\ce{Pb^{2+}}$ (and $\ce{Hg2^{2+}}$ ). The mixture contains $\ce{AgCl}$ and $\ce{PbCl2}$ .

Hot water. $\ce{PbCl2}$ dissolves, so the solution Q gives yellow $\ce{PbI2}$ with KI.

Residue P. $\ce{AgCl}$ dissolves in aqueous ammonia, as $\ce{[Ag(NH3)2]+}$ , and in excess thiosulfate, as $\ce{[Ag(S2O3)2]^{3-}}$ .

So X = Ag and Y = Pb. (A)

Topics:Cation analysis

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