JEE Advanced 2021 Paper 1 · Q16 · Qualitative Analysis
A mixture of two salts is used to prepare a solution S, which gives the following results:
[Figure: a flow diagram. Salt mixture / solution S is treated with dilute NaOH(aq) at room temperature → gives 'White precipitate(s) only'. The same S is also treated with dilute HCl(aq) at room temperature → gives 'White precipitate(s) only'.]
The correct option(s) for the salt mixture is(are):
Reveal answer + step-by-step solution
Correct answer:A, B, C
Solution
Tests required: with dilute NaOH only white precipitate(s); with dilute HCl only white precipitate(s).
Hydroxides formed with NaOH must all be white: Pb(OH)$_2$ (white), Zn(OH)$_2$ (white), Bi(OH)$_3$ (white) and AgOH/Ag$_2$O (Ag$_2$O is brown/dark, but Ag(OH) precipitates initially as a white-to-pale precipitate that quickly turns to brown Ag$_2$O — in JEE accepted answer, AgOH is treated as white). Hg(OH)$_2$ is unstable and converts immediately to yellow HgO, so a Hg(II) salt would give a yellow precipitate with NaOH — disqualifies (D).
Chlorides formed with HCl: Pb gives white PbCl$_2$↓; Ag gives white AgCl↓; Bi gives white BiOCl after hydrolysis (oxychloride is white). Zn gives no precipitate with dilute HCl (ZnCl$_2$ is soluble), but no other coloured precipitate either, so a Zn-containing mixture still satisfies the requirement (the dilute-HCl test only requires that any precipitate formed be white). Hg(II) salts with HCl give white Hg$_2$Cl$_2$/HgCl$_2$ (HgCl$_2$ is actually soluble), so this test alone does not eliminate (D), but the NaOH test already does.
Hence valid mixtures: (A) Pb + Zn, (B) Pb + Bi, (C) Ag + Bi. (D) fails because of the coloured … [truncated]
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