JEE Advanced 2022 Paper 2 Q17 Chemistry Inorganic Chemistry Qualitative Analysis Medium

JEE Advanced 2022 Paper 2 · Q17 · Qualitative Analysis

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

  1. A. PbCl$_2$
  2. B. PbCl$_4$
  3. C. [PbCl$_4$]$^{2-}$
  4. D. [PbCl$_6$]$^{2-}$
Reveal answer + step-by-step solution

Correct answer:C

Solution

Step 1 — Pb(NO$_3$)$_2$ + 2 NaCl $\rightarrow$ PbCl$_2$ (white ppt, sparingly soluble) + 2 NaNO$_3$.

Step 2 — PbCl$_2$ + 2 HCl $\rightarrow$ H$_2$[PbCl$_4$], i.e. the soluble tetrachloroplumbate(II) complex anion [PbCl$_4$]$^{2-}$. The dissolution is due to complex-ion formation.

Pb(II) (5d$^{10}$6s$^2$, lone pair on Pb in +2 oxidation state) forms 4-coordinate [PbCl$_4$]$^{2-}$ in concentrated HCl. The +4 oxidation state [PbCl$_6$]$^{2-}$ would require an oxidising agent (HCl is not a strong enough oxidiser to take Pb(II) to Pb(IV) under these conditions).

Answer: (C) [PbCl$_4$]$^{2-}$.

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