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Byju's Answer

Standard XII

Chemistry

Smelting

Aqueous solut...

Question

Aqueous solution of a metallic nitrate $X$ reacts with $N H_{4} O H$ to form $Y$ which dissolves in excess $N H_{4} O H$ . The resulting complex is reduced by acetaldehyde to deposit the metal. $X$ and $Y$ , respectively, are:

C s (N O_{3})

and

C s O H

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Z n {(N O_{3})}_{2}

and

Z n O

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A g N O_{3}

and

{A g}_{2} O

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M g {(N O_{3})}_{2}

and

M g {(O H)}_{2}

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Solution

The correct option is C $A g N O_{3}$ and ${A g}_{2} O$

$A g N O_{3} + N H_{4} O H \to N H_{4} N O_{3} + A g O H$

$2 A g O H \to A g_{2} O + H_{2} O$

$A g_{2} O + 2 N H_{4} O H \to 2 [A g (N H_{3})_{2}] O H (S o l u b l e) + 3 H_{2} O$

$C H_{2} O H | (C H O H_{4}) | C H O G l u c o s e + A g_{2} O N H_{4} O H - ---- \to$ $C O O H | (C H O H_{4}) | C O O H G l u c o s e + 2 A g) (s i l v e r m i r r o r)$

Hence option $C$ is correct.

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Similar questions

Q. An aqueous solution of metal ion (X) reacts seperately with reagents A and B in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion (Y) always form tetrahedral complexes with these reagents and on reaction with small amount of C reagent gives precipitate which dissolves in excess of C.

X, A, B respectively are:

Q. A metal

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dissolves both in dilute

H C l

and dilute

N a O H

to liberate

H_{2}

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and excess

N H_{4} O H

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N H_{4} C l

to the

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Q. The solution of a chemical compound

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reacts with

A g N O_{3}

solution to form a white precipitate of

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which dissolves in

N H_{4} O H

to give a complex

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Q. The aqueous solution of an inorganic compound (X) gives white precipitate with

N H_{4} O H

which does not dissolve in excess of

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Aqueous solution of a metallic nitrate X reacts with NH4OH to form Y which dissolves in excess NH4OH. The resulting complex is reduced by acetaldehyde to deposit the metal. X and Y, respectively, are:

The correct option is C AgNO3 and Ag2OAgNO3+NH4OH→NH4NO3+AgOH2AgOH→Ag2O+H2OAg2O+2NH4OH→2[Ag(NH3)2]OH(Soluble)+3H2OCH2OH|(CHOH4)|CHOGlucose+Ag2ONH4OH−−−−−→COOH|(CHOH4)|COOHGlucose+2Ag)(silvermirror)Hence option C is correct.

Aqueous solution of a metallic nitrate $X$ reacts with $N H_{4} O H$ to form $Y$ which dissolves in excess $N H_{4} O H$ . The resulting complex is reduced by acetaldehyde to deposit the metal. $X$ and $Y$ , respectively, are: