The reaction of Pb(NO₃)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 formation of

The correct option is [PbCl₄]^2-.

Step-by-Step Explanation:

1. Formation of the initial precipitate:
When aqueous solutions of lead(II) nitrate, $\text{Pb}{\left({\text{NO}}_3\right)}_2$, and sodium chloride, $\text{NaCl}$, are mixed, a double displacement reaction takes place. This results in the formation of a white precipitate of lead(II) chloride, ${\text{PbCl}}_2$:

${\text{Pb}}^{2+}\left(\text{aq}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)\to {\text{PbCl}}_2\left(\text{s}\right)$

2. Dissolution upon addition of concentrated HCl:
When hydrochloric acid ($\text{HCl}$) of appropriate concentration is added to the white precipitate of ${\text{PbCl}}_2$, the excess chloride ions (${\text{Cl}}^{-}$) react further with ${\text{PbCl}}_2$. This leads to the formation of a soluble coordination complex ion, tetrachloroplumbate(II):

${\text{PbCl}}_2\left(\text{s}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)\to {\left[{\text{PbCl}}_4\right]}^{2-}\left(\text{aq}\right)$

Because the complex ion ${\left[{\text{PbCl}}_4\right]}^{2-}$ is soluble in water, the solid precipitate dissolves, resulting in a clear solution.