Verified Solution • 2025–26 Session moderate

QJEEAdv22Ch12cf: JEE Advanced 2022 Chemistry - Class 12 JEEAdv22Ch12cf Solution: The reaction of and in water produces a precipitate that…

The reaction of Pb(NO3)2\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} and NaCl\mathrm{NaCl} in water produces a precipitate that dissolves upon the addition of HCl\mathrm{HCl} of appropriate concentration. The dissolution of the precipitate is due to the formation of

Multiple Choice Options

  • A PbCl2\mathrm{PbCl}_{2}
  • B PbCl4\mathrm{PbCl}_{4}
  • C [PbCl4]2−\left[\mathrm{PbCl}_{4}\right]^{2-}
  • D [PbCl6]2−\left[\mathrm{PbCl}_{6}\right]^{2-}

Step-by-Step Verified Solution

Pb2+\mathrm{Pb}^{2+} on reaction with Cl−\mathrm{Cl}^{-} , produces white precipitate of PbCl2\mathrm{PbCl}_2

Pb2++2Cl−⟶PbCl2↓\mathrm{Pb}^{2+}+2 \mathrm{Cl}^{-} \longrightarrow \mathrm{PbCl}_2 \downarrow (White)

This precipitate is soluble in concentrated hydrochloric acid due to formation of tetrachloroplumbate (II) ion

PbCl2↓+2Cl−⟶[PbCl4]2−\mathrm{PbCl}_2 \downarrow+2 \mathrm{Cl}^{-} \longrightarrow\left[\mathrm{PbCl}_4\right]^{2-}

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