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An inorganic Compound 'X' on treatment with concentrated $\text {H}_{2}\text{SO}_{4}$ produces brown fumes and gives dark brown ring with $\text {FeSO}_{4}$ in presence of concentrated $\text {H}_{2}\text{SO}_{4}$ . Also Compound 'X' gives precipitate 'Y', when its solution in dilute HCl is treated with $\text {H}_{2}\text{S}$ gas. The precipitate 'Y' on treatment with concentrated $\text {HNO}_{3}$ followed by excess of $\text {NH}_{4}\text{OH}$ further gives deep blue coloured solution, Compound 'X' is:
Option: 1 $\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}$
Option: 2 $\mathrm{Pb}\left(\mathrm{NO}_{2}\right)_{2}$
Option: 3 $\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}$
Option: 4 $\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$

Answers (1)

best_answer

Acconding to the information given in the question

$X\xrightarrow[\text{Conc.}]{\text{H}_{2}\text{SO}_{4}}\text{Brown fumes}$

$X\xrightarrow[\text{Ring}]{\text{Brown}}\text{(Positive)}$

This confirms the presence of Nitrate ions $(\text{NO}_{3}^{-})$ as the acidic Radical.

$X\xrightarrow[\text{dil.HCl}]{\text{H}_{2}\text{S}}\text{Y(ppt)}$

$X\xrightarrow[\text{Conc.}]{\text{HNO}_{3}}\xrightarrow[\text{excess}]{\text{NH}_{4}\text{OH}}\text{Blue Solution}$

The precipitation with $\text{H}_{2}\text{S}$ in presence of dil. HCl tells us that the Basic radical is of Group (2).

Futher reaction of the precipitate with $\text{HNO}_{3}$ and excess $\text{NH}_{4}\text{OH}$ giving a blue solution confirms the presence of $\text{Cu}^{2+}$ ions

$\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}: \text { Deep blue color }$

Hence, the correct answer is option (3)

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