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Question
An inorganic compound 'X' on treatment with concentrated H2SO4 produces brown fumes and gives a dark brown ring with FeSO4 in presence of concentrated H2SO4. Also, compound 'X' gives precipitate 'Y', when its solution in dilute HCl is treated with H2S gas. The precipitate 'Y' on treatment with concentrated HNO3 followed by an excess of NH4OH further gives a deep blue coloured solution, compound 'X' is:
Options
Cu(NO3)2
Pb(NO3)2
Pb(NO2)2
Co(NO3)2
Solution
Cu(NO3)2
Explanation:
When compound X and compound H2SO4 combine, dark odours are produced. Using FeSO4 and conc. H2SO4, compound X generates a brown ring.
Therefore, compound X must contain nitrate ion (NO3) as its anion. The brown fumes evolved are of NO2 gas. The brown ring occurs due to formation of [Fe(H2O)5 (NO)]SO4.
\[\ce{\underset{(Anion)}{\underset{X}{NO}^-_3} + \underset{(Conc.)}{H2SO4} -> \underset{Brown fumes}{NO2 \uparrow + H2O}}\]
\[\ce{FeSO4 + \underset{conc.}{H2SO4} + \underset{X}{NO}^-_3 -> \underset{(dark brown ring)}{[Fe(H2O)5 (NO)]SO4}}\]
Compound X creates a precipitate of compound Y when it combines with reagent II, a mixture of diluted HCl and H2S. The chemical Y produces a vivid blue-coloured solution when it is treated with concentrated HNO3 and excess NH4OH.
The above reactions suggest that compound X must contain Cu2+ as a cation. The compound Y is CuS which has black coloured precipitate. CuS on reaction with cone. HNO3 forms Cu(NO3)2 which is a soluble compound. Cu(NO3)2 on reaction with an excess of NH4OH, produces deep blue-coloured solution of [Cu(NH3)4]2+.
\[\ce{\underset{(cation)}{\underset{X}{Cu^{2+}}} + \underset{(Group - II reagent)}{(dil. HCl + H2S)} -> \underset{(Y)}{\underset{(Black ppt)}{CuS}}}\]
\[\ce{\underset{(Y)}{CuS} ->[Conc.][HNO3] \underset{Soluble}{Cu(NO3)2} + NO2 + S + H2O ->[Excess][NH4(OH)] \underset{Deep blue colour solution}{[Cu(NH3)4]^{2+}}}\]
Therefore, the cation and anion in compound X are Cu2+ and \[\ce{NO^-_3}\]. Therefore compound X is \[\ce{Cu(NO_3)_2}\].
