Question

An inorganic Compound 'X' on treatment with concentrated \(\mathrm{H_2SO_4}\) produces brown fumes and gives dark brown ring with \(\mathrm{FeSO_4}\) in presence of concentrated \(\mathrm{H_2SO_4}\). Also Compound 'X' gives precipitate 'Y', when its solution in dilute HCl is treated with \(\mathrm{H_2S}\) gas. The precipitate 'Y' on treatment with concentrated \(\mathrm{HNO_3}\) followed by excess of \(\mathrm{NH_4OH}\) further gives deep blue coloured solution, Compound 'X' is :

• A. \(\mathrm{Co(NO_3)_2}\)
• B. \(\mathrm{Pb(NO_2)_2}\)
• C. \(\mathrm{Cu(NO_3)_2}\)
• D. \(\mathrm{Pb(NO_3)_2}\)

Hint:

Brown fumes plus the brown-ring test indicate nitrate; a group II sulphide that dissolves in \(\mathrm{HNO_3}\) and gives a deep blue ammoniacal solution is copper.

Answer 1

The Correct Option is C

Brown fumes of \(\mathrm{NO_2}\) with conc. \(\mathrm{H_2SO_4}\) and the dark brown ring with \(\mathrm{FeSO_4}\) + conc. \(\mathrm{H_2SO_4}\) (formation of \(\mathrm{[Fe(H_2O)_5NO]^{2+}}\)) show the anion is \(\mathrm{NO_3^-}\) (the brown ring test is the confirmatory test for nitrate).
With \(\mathrm{H_2S}\) in dilute HCl a precipitate forms only for group II cations, so \(\mathrm{Co^{2+}}\) (group IV, needs alkaline medium) is ruled out. Possible: \(\mathrm{Cu^{2+}}\) giving black CuS, or \(\mathrm{Pb^{2+}}\) giving black PbS.
CuS dissolves in conc. \(\mathrm{HNO_3}\) to give \(\mathrm{Cu(NO_3)_2}\); excess \(\mathrm{NH_4OH}\) forms the deep blue \(\mathrm{[Cu(NH_3)_4]^{2+}}\). \(\mathrm{Pb^{2+}}\) would give a white precipitate of \(\mathrm{Pb(OH)_2}\) that does not dissolve in excess ammonia, and no blue colour.
So X is \(\mathrm{Cu(NO_3)_2}\).
Answer: (3) \(\mathrm{Cu(NO_3)_2}\)