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When a salt is treated with sodium hydroxide solution it gives gas X. On passing gas X through reagent Y a brown…

p-Block Elements (Class 12) · Class 12 · JEE Main Previous Year Question

Question

When a salt is treated with sodium hydroxide solution it gives gas X. On passing gas X through reagent Y a brown coloured precipitate is formed.

X and Y respectively, are

Options
  1. a

    \ceX=NH3\ce{X = NH3} and \ceY=HgO\ce{Y = HgO}

  2. b

    \ceX=NH3\ce{X = NH3} and \ceY=K2HgI4+KOH\ce{Y = K2HgI4 + KOH}

  3. c

    \ceX=NH4Cl\ce{X = NH4Cl} and \ceY=KOH\ce{Y = KOH}

  4. d

    \ceX=HCl\ce{X = HCl} and \ceY=NH4Cl\ce{Y = NH4Cl}

Correct Answerb

\ceX=NH3\ce{X = NH3} and \ceY=K2HgI4+KOH\ce{Y = K2HgI4 + KOH}

Detailed Solution

Step 1: Identify gas X

A salt treated with NaOH solution gives gas X. If the salt is an ammonium salt:

\ceNH4++OH>NH3(g)+H2O\ce{NH4+ + OH- -> NH3(g) + H2O}

Gas X = \ceNH3\ce{NH3} (ammonia — pungent smell, turns moist red litmus blue)

Step 2: Identify reagent Y (Nessler's reagent)

Nessler's reagent = K2[HgI4]{K2[HgI4]} in KOH (alkaline solution)

When \ceNH3\ce{NH3} is passed through Nessler's reagent:

\ceNH3+2K2[HgI4]+3KOH>Hg2O.NH2Iv+7KI+2H2O\ce{NH3 + 2K2[HgI4] + 3KOH -> Hg2O.NH2I v + 7KI + 2H2O}

The product \ceHg2O.NH2I\ce{Hg2O.NH2I} (iodide of Millon's base) is a brown precipitate

Step 3: Conclusion

X = \ceNH3\ce{NH3}, Y = K2[HgI4]{K2[HgI4]} in KOH (Nessler's reagent)

Answer: (b)

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