An inorganic compound ‘X’ on treatment with concentrated H2SO4 produces brown fumes and given 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 given deep blue coloured solution, Compound ‘X’ is:

Cu(NO3)2

Pb(NO3)2

Co(NO3)2

Pb(NO2)2

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Block D and F Elements Ncert Solutions Chemistry Class 12th Chemistry Class 12th

An inorganic compound ‘X’ on treatment with concentrated H₂SO₄ produces brown fumes and given 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 given deep blue coloured solution, Compound ‘X’ is:

Option 1 -

Cu(NO₃)₂

Option 2 -

Pb(NO₃)₂

Option 3 -

Co(NO₃)₂

Option 4 -

Pb(NO₂)₂

2 Views | Posted 3 months ago

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Answered by

alok kumar singh | Contributor-Level 10

3 months ago

Correct Option - 1

Detailed Solution:

$X \to_{H_{2} S O_{4}}^{C o n c}$ Brown fumes + Brown ring test with FeSO₄/H₂SO₄

$x \to_{H C l}^{H_{2} S} Y (p p t) \to_{H N O_{3}}^{C o n c} d i s s o l v e d \to_{N H_{4} O H}^{}$ Deep blue colour solution

In cation analysis of Cu⁺ ions, precipitate formed is CuS on treating with H₂S and HCl which dissolved in HNO₃ and produced blue colour complex solution ${[C u {(N H_{3})}_{4}]}^{+ +}$ in NH₄OH. Brown fumes and brown ring performance given by nitrate sample. Hence salt is Cu (NO₃)₂.

<!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1816169331"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mi>X</mi> <munderover> <mo>→</mo> <mrow> <msub> <mrow> <mi>H</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mi>S</mi> <msub> <mrow> <mi>O</mi> </mrow> <mrow> <mn>4</mn> </mrow> </msub> </mrow> <mrow> <mi>C</mi> <mi>o</mi> <mi>n</mi> <mi>c</mi> </mrow> </munderover> </mrow> </math> Brown fumes + Brown ring test with FeSO4/H2SO4 <math> <mrow> <mi>x</mi> <munderover> <mo>→</mo> <mrow> <mi>H</mi> <mi>C</mi> <mi>l</mi> </mrow> <mrow> <msub> <mrow> <mi>H</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mi>S</mi> </mrow> </munderover> <mi>Y</mi> <mrow> <mo> (</mo> <mrow> <mi>p</mi> <mi>p</mi> <mi>t</mi> </mrow> <mo>)</mo> </mrow> <munderover> <mo>→</mo> <mrow> <mi>H</mi> <mi>N</mi> <msub> <mrow> <mi>O</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msub> </mrow> <mrow> <mi>C</mi> <mi>o</mi> <mi>n</mi> <mi>c</mi> </mrow> </munderover> <mi>d</mi> <mi>i</mi> <mi>s</mi> <mi>s</mi> <mi>o</mi> <mi>l</mi> <mi>v</mi> <mi>e</mi> <mi>d</mi> <munderover> <mo>→</mo> <mrow> <mi>N</mi> <msub> <mrow> <mi>H</mi> </mrow> <mrow> <mn>4</mn> </mrow> </msub> <mi>O</mi> <mi>H</mi> </mrow> <mrow></mrow> </munderover> </mrow> </math> Deep blue colour solutionIn cation analysis of Cu+ ions, precipitate formed is CuS on treating with H2S and HCl which dissolved in HNO3 and produced blue colour complex solution <math> <mrow> <msup> <mrow> <mrow> <mo> [</mo> <mrow> <mi>C</mi> <mi>u</mi> <msub> <mrow> <mrow> <mo> (</mo> <mrow> <mi>N</mi> <msub> <mrow> <mi>H</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>4</mn> </mrow> </msub> </mrow> <mo>]</mo> </mrow> </mrow> <mrow> <mo>+</mo> <mo>+</mo> </mrow> </msup> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1816169333"> </o:OLEObject></xml><! [endif]->in NH4OH. Brown fumes and brown ring performance given by nitrate sample. Hence salt is Cu (NO3)2.

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