JEE Advanced 2022 — Paper 2 — CHEM-III solution

JEE Advanced2022Paper 2CHEM-III

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

A
PbCl $_{2}$
B
PbCl $_{4}$
C
[PbCl $_{4}$ ] $^{2 -}$
D
[PbCl $_{6}$ ] $^{2 -}$

Solution

Pb $^{2 +}$ reacts with Cl $^{-}$ to give a white precipitate of PbCl $_{2}$ : Pb $^{2 +}$ + 2 Cl $^{-} \to$ PbCl $_{2} (s)$ .

On adding concentrated HCl, the excess chloride forms the soluble tetrachloroplumbate(II) complex:

PbCl $_{2}$ + 2 Cl $^{-} \to$ [PbCl $_{4}$ ] $^{2 -}$ .

Hence the dissolution arises from complex formation: option (C).

← Back to full paper

Practice more CHEM-III

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →