What was the JEE Advanced 2022 p-block question on Pb(NO3)2 and NaCl precipitate dissolving in HCl?
Lead nitrate and sodium chloride mixed in water give an immediate white precipitate. That precipitate redissolves when HCl of suitable concentration is added. The dissolution occurs because one of four possible lead-chloride species is formed. It appeared in Paper 2 of JEE Advanced 2022.
Students recall that PbCl2 is sparingly soluble and expect the common-ion effect to suppress solubility further. They get anxious when excess Cl– dissolves the precipitate instead, unsure which charged complex is correct. The official key selects the species that satisfies both charge balance and the +2 oxidation state of lead.
What is the complete official solution for the 2022 JEE Advanced lead complex question?
The official solution identifies [PbCl4]2− as the species formed through this exact sequence.
Complex formation lowers free Pb²⁺ concentration. The solubility equilibrium therefore shifts right and more solid dissolves.
Oxidation-state verification confirms the assignment. Let the oxidation state of Pb be x. Then
Lead remains in the +2 state, exactly as in the starting Pb²⁺ ion. Hence the correct species responsible is [PbCl4]²⁻.
What method mistake selects the wrong lead-chloride species in that question?
One method mistake is stopping at the neutral formula PbCl4 without balancing charge after attaching four Cl⁻ to Pb²⁺. Four chloride ions contribute −4 while lead contributes +2, so the complex must carry 2− charge.
Candidates who treat the reaction as simple double displacement forget to check for soluble anionic complex formation. Skipping the oxidation-state verification compounds the error. The calculation forces x = +2 and rules out any species implying Pb(IV). Writing the full ionic equation first, then verifying charge and oxidation state, eliminates the wrong option.
Why does excess Cl– dissolve PbCl2 despite the common-ion effect?
Excess Cl– dissolves PbCl2 because formation of [PbCl4]²⁻ removes free Pb²⁺ and pulls the solubility equilibrium forward. Common-ion effect normally decreases solubility of PbCl2 because extra Cl⁻ pushes the dissolution equilibrium back toward the solid.
When Kf for [PbCl4]²⁻ is sufficiently large, complexation overrides that suppression. Pb(II) prefers coordination number 4 in this chloro complex, consistent with inert-pair effect. The official argument rests only on the direction of the equilibrium shift once the complex forms.
What practice questions test the same solubility and complexation reasoning?
Practice Question 1
Tin(II) chloride dissolves in concentrated HCl to give a clear solution. The species formed is most likely
A) SnCl₂
B) SnCl₄
C) [SnCl₆]²⁻
D) [SnCl₃]⁻
Write the ionic equation, check charge on the complex, and confirm the oxidation state of tin. The equilibrium shift mirrors the PbCl2 case exactly.
Practice Question 2
PbSO4 is insoluble in dilute H₂SO4 but dissolves readily in ammonium acetate solution. Explain this observation using only complex or acetate formation.
These two questions let you test the identical charge-balance habit required for the 2022 question. You can expect to solve them without hesitation once that habit is automatic. For more solved examples from recent Advanced papers, search the past-paper archive by chapter or year.
The 2021 metallurgy PYQ already solved on this site follows the same logic of selective dissolution through complexation: [/blog/p-block-elements-jee-2021-metallurgy-question-solution/].
What habits prevent losing marks on similar p-block solubility questions in JEE Advanced?
Always write the full ionic equation and check charge balance before choosing an option. Verify oxidation state of the central atom in every complex ion listed. Remember that Pb²⁺ forms [PbCl4]²⁻ while higher oxidation states or other metals may adopt coordination number 6.
When a precipitate dissolves in excess reagent, first ask whether a soluble anionic complex is possible rather than assuming further precipitation. Practicing this sequence on every group-14 halide question will make the charge and oxidation-state checks automatic.
Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).
Read next: P-Block Elements JEE 2021: Metallurgy Question Solution.
Frequently asked questions
Why does PbCl2 precipitate dissolve in excess HCl in JEE Advanced 2022?
The precipitate dissolves because PbCl2(s) reacts with excess Cl– to form the soluble complex [PbCl4]2–. This lowers free Pb²⁺ concentration and shifts the solubility equilibrium forward. Oxidation state calculation confirms Pb remains +2: x + 4(−1) = −2 gives x = +2.
What is the species formed when PbCl2 dissolves in HCl for JEE 2022?
The correct species is [PbCl4]2−. Four Cl– ions contribute −4 charge while Pb²⁺ contributes +2, requiring an overall 2− on the complex. This satisfies both charge balance and the +2 oxidation state of lead in the p-block reaction.
Why does excess Cl– dissolve PbCl2 despite common ion effect?
Complexation of Pb²⁺ with Cl– to form [PbCl4]2– removes free Pb²⁺ ions from solution. This pulls the PbCl2 dissolution equilibrium to the right, overriding the suppressing effect of extra Cl–. The formation constant of the complex is large enough for net dissolution to occur.
How to verify oxidation state of Pb in [PbCl4]2- for JEE p-block questions?
Let oxidation state of Pb be x. For [PbCl4]2− the equation is x + 4(−1) = −2, so x = +2. This matches the starting Pb²⁺ from Pb(NO3)2 and rules out Pb(IV) species. Always write the full ionic equation and check charge balance first.