JEE MAIN - Chemistry (2021 - 20th July Morning Shift - No. 5)
An inorganic Compound 'X' on treatment with concentrated H2SO4 produces brown fumes and gives 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 excess of NH4OH further gives deep blue coloured solution, Compound 'X' is :
Co(NO3)2
Pb(NO2)2
Cu(NO3)2
Pb(NO3)2
Explanation
Compound ‘X’ is copper nitrate, i.e. Cu(NO3)2 is an inorganic
compound. On treatment with concentrated H2SO4 produces brown
fumes and gives dark brown ring with FeSO4
in presence of
concentrated H2SO4. Chemical reaction is as follows
_20th_July_Morning_Shift_en_5_1.png)
Cu2+ is a group II cation with group II reagents (HCl/H2S) , it gives black coloured precipitates. These precipitates gives blue colour solution on treatment with HNO3 followed by excess of NH4OH.
_20th_July_Morning_Shift_en_5_2.png)
Cu2+ is a group II cation with group II reagents (HCl/H2S) , it gives black coloured precipitates. These precipitates gives blue colour solution on treatment with HNO3 followed by excess of NH4OH.
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