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Question: Answered & Verified by Expert
Aqueous solution of metallic nitrate $X$ reacts with $\mathrm{NH}_{4} \mathrm{OH}$ to form $\mathrm{Y}$ which dissolves in excess $\mathrm{NH}_{4} \mathrm{OH}$. The resulting complex is reduced by acetaldehyde to deposit the metal. $X$ and $Y$, respectively, are
ChemistryAldehydes and KetonesKVPYKVPY 2013 (SB/SX)
Options:
  • A $\mathrm{Cs}\left(\mathrm{NO}_{3}\right)$ and $\mathrm{CsOH}$
  • B $\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}$ and $\mathrm{ZnO}$
  • C $\mathrm{AgNO}_{3}$ and $\mathrm{Ag}_{2} \mathrm{O}$
  • D $\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}$ and $\mathrm{Mg}(\mathrm{OH})_{2}$
Solution:
1983 Upvotes Verified Answer
The correct answer is: $\mathrm{AgNO}_{3}$ and $\mathrm{Ag}_{2} \mathrm{O}$
$\mathrm{AgNO}_{3}+\mathrm{NH}_{4} \mathrm{OH} \longrightarrow \mathrm{AgOH} \frac{\mathrm{NH}_{4} \mathrm{OH}}{(\text { excess })}\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]+$
Tollen's reagent $+\mathrm{R}-\mathrm{CHO} \longrightarrow \mathrm{R}-\mathrm{COOH}$

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