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P Block Elements question

2022 · Shift 2 · Q17
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P Block Elements question

2022 · Shift 2 · Q17

JEE AdvancedChemistryP Block ElementsMCQ+3 / −1
The reaction of Pb(NO3)2\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}Pb(NO3​)2​ and NaCl\mathrm{NaCl}NaCl in water produces a precipitate that dissolves upon the addition of HCl\mathrm{HCl}HCl of appropriate concentration. The dissolution of the precipitate is due to the formation of
  1. A
    PbCl2\mathrm{PbCl}_{2}PbCl2​
  2. B
    PbCl4\mathrm{PbCl}_{4}PbCl4​
  3. C
    [PbCl4]2−\left[\mathrm{PbCl}_{4}\right]^{2-}[PbCl4​]2−
  4. D
    [PbCl6]2−\left[\mathrm{PbCl}_{6}\right]^{2-}[PbCl6​]2−
View written solutionFree

Correct answer: C

  1. Initial precipitation reaction

When aqueous lead nitrate reacts with aqueous sodium chloride, lead(II) chloride precipitates:

Pb(NO3)2(aq)+2NaCl(aq)→PbCl2(s)+2NaNO3(aq)\mathrm{Pb(NO_3)_2(aq) + 2NaCl(aq) \rightarrow PbCl_2(s) + 2NaNO_3(aq)}Pb(NO3​)2​(aq)+2NaCl(aq)→PbCl2​(s)+2NaNO3​(aq)

So the precipitate formed is PbCl2\mathrm{PbCl_2}PbCl2​.

  1. Why does the precipitate dissolve in excess chloride/HCl?

PbCl2\mathrm{PbCl_2}PbCl2​ is sparingly soluble in water, but in the presence of a sufficiently high concentration of chloride ions (provided by HCl\mathrm{HCl}HCl), it forms a soluble chloro-complex.

This is a common behavior of metal halides:

PbCl2(s)+2Cl−⇌[PbCl4]2−(aq)\mathrm{PbCl_2(s) + 2Cl^- \rightleftharpoons [PbCl_4]^{2-}(aq)}PbCl2​(s)+2Cl−⇌[PbCl4​]2−(aq)

Thus, the precipitate dissolves because the equilibrium shifts toward formation of the soluble complex ion.

  1. Check the options

  • A: PbCl2\mathrm{PbCl_2}PbCl2​
    This is the precipitate itself, not the species responsible for dissolution.

  • B: PbCl4\mathrm{PbCl_4}PbCl4​
    Neutral lead(IV) chloride is not formed here in aqueous hydrochloric acid.

  • C: [PbCl4]2−\mathrm{[PbCl_4]^{2-}}[PbCl4​]2−
    This is the soluble tetrachloroplumbate(II) complex formed in excess chloride. This explains dissolution.

  • D: [PbCl6]2−\mathrm{[PbCl_6]^{2-}}[PbCl6​]2−
    This is not the usual species considered for dissolution of PbCl2\mathrm{PbCl_2}PbCl2​ in concentrated chloride medium at this level.

  1. Conclusion

The precipitate dissolves due to formation of:

[PbCl4]2−\boxed{\mathrm{[PbCl_4]^{2-}}}[PbCl4​]2−​

So the correct option is C.

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