An inorganic compound 'X' on treatment with concentrated H2SO4 produces brown fumes and gives dark b

Q: Why do we use dilute HCl before passing $H_2S$ gas?

Dilute HCl provides an acidic medium, which suppresses the ionization of $H_2S$ due to the common ion effect. This ensures that only the highly insoluble sulfides of Group II cations precipitate, preventing the precipitation of Group IV cations.

Q: What is the chemical formula of the brown ring?

The brown ring is a coordination complex named pentaaquanitrosyliron(II) sulfate. Its formula is $[Fe(H_2O)_5(NO)]SO_4$.

Q: Could the black precipitate be Lead Sulfide ($PbS$)?

Yes, $PbS$ is also a black precipitate formed in Group II. However, $Pb^{2+}$ does not form a deep blue solution with excess $NH_4OH$; instead, it forms a white precipitate of $Pb(OH)_2$.

Q: Why does $CuS$ dissolve in concentrated $HNO_3$?

Concentrated $HNO_3$ is a strong oxidizing agent. It oxidizes the sulfide ion ($S^{2-}$) in $CuS$ to elemental sulfur ($S$) or sulfate ($SO_4^{2-}$), thereby breaking down the precipitate and bringing $Cu^{2+}$ into the solution.

Q: What is the structure of the deep blue complex?

The deep blue complex is $[Cu(NH_3)_4]^{2+}$, which has a square planar geometry. The ammonia molecules act as ligands, donating electron pairs to the central copper ion.

Question

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

Options:

(A) $Co(NO_3)_2$

(B) $Pb(NO_2)_2$

(C) $Cu(NO_3)_2$

(D) $Pb(NO_3)_2$

Sigma Percentile · Accepted Answer

Imagine you are a chemical detective, and you are handed a mysterious vial containing an unknown inorganic compound, 'X'. Your mission is to uncover its true identity using a series of chemical interrogations. This problem takes us through a classic sequence of qualitative analysis tests, where every color change and precipitate is a clue.

### Analyzing the Anion: The Brown Fumes and the Ring

Our first clue comes from treating compound 'X' with concentrated sulfuric acid ($H_2SO_4$). The reaction produces brown fumes. In the realm of inorganic chemistry, brown fumes upon the addition of concentrated $H_2SO_4$ strongly point towards the presence of a nitrate ($NO_3^-$) ion, as it decomposes to release nitrogen dioxide ($NO_2$) gas.

To be absolutely certain, we perform the legendary **Brown Ring Test**. By adding freshly prepared ferrous sulfate ($FeSO_4$) and carefully pouring concentrated $H_2SO_4$ down the side of the test tube, a dark brown ring forms at the junction of the two liquids.

$$NO_3^- + 3Fe^{2+} + 4H^+ \rightarrow NO + 3Fe^{3+} + 2H_2O$$

$$[Fe(H_2O)_6]^{2+} + NO \rightarrow [Fe(H_2O)_5(NO)]^{2+} + H_2O$$

This ring is the complex ion $[Fe(H_2O)_5(NO)]^{2+}$, confirming beyond a shadow of a doubt that our anion is indeed nitrate.

### Hunting the Cation: The Black Precipitate

Next, we dissolve 'X' in dilute hydrochloric acid (HCl) and pass hydrogen sulfide ($H_2S$) gas through it. A black precipitate, 'Y', emerges. This specific reagent combination (dilute HCl + $H_2S$) is the group reagent for Group II cations in qualitative analysis.

The formation of a black precipitate narrows our suspects down to a few candidates, most notably copper ($Cu^{2+}$) or lead ($Pb^{2+}$), which form $CuS$ and $PbS$, respectively.

### The Final Confirmation: The Deep Blue Solution

To distinguish between the remaining suspects, we subject the black precipitate 'Y' to further testing. We dissolve it in concentrated nitric acid ($HNO_3$), which oxidizes the sulfide to sulfur and brings the cation back into solution.

The decisive moment arrives when we add an excess of ammonium hydroxide ($NH_4OH$). The solution turns a striking, deep blue color.

$$Cu^{2+} + 4NH_4OH \rightarrow [Cu(NH_3)_4]^{2+} + 4H_2O$$

This is the signature of the tetraamminecopper(II) complex ion, $[Cu(NH_3)_4]^{2+}$. If the cation had been lead, we would have seen a white precipitate of lead hydroxide that does not dissolve in excess ammonia. The deep blue color is the undeniable fingerprint of copper.

### The Verdict

Having identified the anion as nitrate ($NO_3^-$) and the cation as copper ($Cu^{2+}$), we can confidently unmask our unknown compound 'X' as copper(II) nitrate, $Cu(NO_3)_2$.

This problem beautifully illustrates the logical deduction required in qualitative analysis, where each test acts as a filter, systematically eliminating possibilities until only the truth remains.

Select Answer:

Visualized Solution

$X \to Unknown Compound$

$Anion Identification$

$Nitrate Reactions$

$Cation Identification$

$Confirmatory Test for C u^{2 +}$

$Final Compound$

$Key Takeaways$

The Sigma Insight: d-block Elements

Analyzing the Anion

Hunting the Cation

The Final Confirmation

The Verdict

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X→Unknown Compound

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Anion Identification

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Nitrate Reactions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Cation Identification

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Confirmatory Test for Cu2+

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Compound

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Key Takeaways