JEE Advanced 2022 · Paper 2 · Chemistry Q17

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ChemistrySingle correct+3 / -1Question 53 of 54

The reaction of $\ce{Pb(NO3)2}$ 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 the formation of

\ce{PbCl2}

\ce{PbCl4}

\ce{[PbCl4]^{2-}}

\ce{[PbCl6]^{2-}}

Answer and worked solution

Official answer: C

Precipitate. $\ce{Pb^{2+} + 2Cl- -> PbCl2 v}$ (white).

Dissolution in HCl. With excess chloride, it forms the soluble tetrachloroplumbate(II) complex:
$\ce{PbCl2 + 2Cl- -> [PbCl4]^{2-}}.$
Lead stays in the +2 state, so $\ce{PbCl4}$ and $\ce{[PbCl6]^{2-}}$ (Pb(IV)) are ruled out. (C).

Topics:Group 14 Stability and applications

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JEE Advanced 2022 · Paper 2, Chemistry question 17