JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Practical and Analytical Chemistry

An inorganic compound 'X' on treatment with concentrated

H_{2} SO_{4}

produces brown fumes and gives dark brown ring with

FeSO_{4}

in presence of concentrated

H_{2} SO_{4}

. Also compound 'X' gives precipitate 'Y', when its solution in dilute HCl is treated with

H_{2} S

gas. The precipitate 'Y' on treatment with concentrated

HNO_{3}

followed by excess of

NH_{4} OH

further gives deep blue coloured solution, Compound 'X' is:

(A) $Cu (NO_{3})_{2}$ ✓ Correct
(B) $Pb (NO_{3})_{2}$
(C) $Co (NO_{3})_{2}$
(D) $Pb (NO_{2})_{2}$

Answer: (A)

Solution

Brown fumes with concentrated

H_{2} S O_{4}

and the brown ring

[Fe (H_{2} O)_{5} NO]^{2 +}

identify the anion as nitrate, which rules out the nitrite salt. Precipitation by

H_{2} S

from a solution acidified with dilute HCl places the cation in the second analytical group, so cobalt, whose sulphide appears only from ammoniacal solution, is excluded. The sulphide Y dissolves in hot concentrated

HN O_{3}

, and adding excess ammonia to that solution gives a deep blue colour, which belongs to the tetraamminecopper(II) ion

[Cu (N H_{3})_{4}]^{2 +}

; lead, the other group II cation offered, would have given a white precipitate with ammonia instead. Compound X is

Cu (N O_{3})_{2}

, the precipitate Y being black CuS.

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Practical and Analytical Chemistry

Solution

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