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

Work backwards from the deep blue solution: among the cations listed only \(\mathrm{Cu^{2+}}\) (as \(\mathrm{[Cu(NH_3)_4]^{2+}}\)) gives a deep blue colour with excess ammonia. So Y = CuS, and X contains \(\mathrm{Cu^{2+}}\).
Of the options only (3) contains copper. Its anion \(\mathrm{NO_3^-}\) gives brown \(\mathrm{NO_2}\) fumes with conc. \(\mathrm{H_2SO_4}\) and the brown ring test, consistent with the data.
Hence X = \(\mathrm{Cu(NO_3)_2}\).
Answer: (3) \(\mathrm{Cu(NO_3)_2}\)