# Does lead(II) chloride dissociate into Pb²⁺ and Cl⁻ ions in HCl?

While solving JEE Advanced 2020 (Paper 2) questions, I came across this particular question:

Q.10 Choose the correct statement(s) among the following.

(A) $$\ce{SnCl2 · 2 H2O}$$ is a reducing agent.
(B) $$\ce{SnO2}$$ reacts with $$\ce{KOH}$$ to form $$\ce{K2[Sn(OH)6]}.$$
(C) A solution of $$\ce{PbCl2}$$ in $$\ce{HCl}$$ contains $$\ce{Pb^2+}$$ and $$\ce{Cl-}$$ ions.
(D) The reaction of $$\ce{Pb3O4}$$ with hot dilute nitric acid to give $$\ce{PbO2}$$ is a redox reaction.

According to the final answer key, its answer is options (A), (B).

Coming to the option (C), after doing some research, I have been able to conclude the following:

1. $$\ce{PbCl2}$$ is sparingly soluble in water.
2. $$\ce{PbCl2}$$ + little amount of dil. $$\ce{HCl}$$ decreases its solubility due to common ion effect.
3. Upon further addition of dil. $$\ce{HCl}$$ or $$\ce{PbCl2}$$ + conc. $$\ce{HCl}$$ leads to formation of soluble lead complexes: $$\ce{[PbCl3]-}$$ and $$\ce{[PbCl4]^2-}.$$
4. Hence, option (C) is incorrect as in one case, $$\ce{PbCl2}$$ doesn't dissociate at all and even in the other, $$\ce{Pb^2+}$$ and $$\ce{Cl-}$$ ions are not formed (so the dilemma of the $$\ce{HCl}$$ doesn't exist anymore as this option is incorrect regardless of the $$\ce{HCl}$$ being dilute or concentrated).

Are the above four points accurate? It'd be appreciated if someone could add more details and correct the above points if needed.

Your analysis about $$\ce{PbCl2}$$ in four points is correct.
Then ($$\ce{A}$$ and ($$\ce{B}$$) are correct, if they are in aqueous solution. The following equation shows that $$\ce{SnCl2}$$ is a reducing agent that is easily oxidized by air : $$\ce{6 SnCl2 + O2 + H2O -> SnCl4 + 4 Sn(OH)Cl}$$ and, in ($$\ce{B}$$), $$\ce{SnO2}$$ is soluble in $$\ce{KOH}$$ solutions according to $$\ce{SnO2 + 2 KOH + 2 H2O -> K2[Sn(OH)6]}$$ (D) is wrong, because the reaction is the following $$\ce{Pb3O4 + 4 HNO3 -> PbO2 + 2 Pb(NO3)2 + 2 H2O}$$ This reaction can be deduced by dealing with $$\ce{Pb3O4}$$ as if it was a mixture of $$\ce{PbO2 + 2 PbO}$$ so that only the two $$\ce{PbO}$$ react with $$\ce{HNO3}$$, without changing its oxidation number $$\ce{2 PbO + 4 HNO3 -> 2 Pb(NO3)2 + 2 H2O}$$