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:

$\ce{PbCl2}$ is sparingly soluble in water.
$\ce{PbCl2}$ + little amount of dil. $\ce{HCl}$ decreases its solubility due to common ion effect.
Upon further addition of dil. $\ce{HCl}$ or $\ce{PbCl2}$ + conc. $\ce{HCl}$ leads to formation of soluble lead complexes: $\ce{[PbCl3]-}$ and $\ce{[PbCl4]^2-}.$
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 : $$\ce{6 SnCl2 + O2 + H2O  -&gt; SnCl4 + 4 Sn(OH)Cl}$$ and, in ($\ce{B}$),  $\ce{SnO2}$ is soluble in $\ce{KOH}$ solutions according to $$\ce{SnO2 + 2 KOH + 2 H2O -&gt; K2[Sn(OH)6]}$$ (D) is wrong, because the reaction is the following $$\ce{Pb3O4 + 4 HNO3 -&gt; PbO2 + 2 Pb(NO3)2 + 2 H2O}$$ 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 $$\ce{2 PbO + 4 HNO3 -&gt; 2 Pb(NO3)2 + 2 H2O}$$