p Block Elements Part 3

Concept:Double displacement and complex formation reactions involving lead compounds.Explanation:Reaction 1: $\mathrm{PbCl_2}$ reacts with $\mathrm{K_2CrO_4}$ to form a yellow precipitate of $\mathrm{PbCrO_4}$.This is a double displacement reaction where $\mathrm{Pb^{2+}}$ and $\mathrm{CrO_4^{2-}}$ combine.Thus, $\mathrm{A = PbCrO_4}$.Reaction 2: $\mathrm{PbCrO_4}$ reacts with excess $\mathrm{NaOH}$ in a reversible manner.Lead is amphoteric and forms a complex with hydroxide ions.The balanced equation is: $\mathrm{PbCrO_4 + 4NaOH \rightleftharpoons Na_2[Pb(OH)_4] + Na_2CrO_4}$.Hence, $\mathrm{B = Na_2[Pb(OH)_4]}$ (sodium tetrahydroxoplumbate).Reaction 3: $\mathrm{PbSO_4}$ reacts with ammonium acetate, which acts as a complexing agent.Four acetate ions coordinate with $\mathrm{Pb^{2+}}$ to form a soluble tetraacetatoplumbate(II) complex.The equation is: $\mathrm{PbSO_4 + 4CH_3COONH_4 \rightarrow (NH_4)_2SO_4 + (NH_4)_2[Pb(CH_3COO)_4]}$.Thus, $\mathrm{X = (NH_4)_2[Pb(CH_3COO)_4]}$.Answer:$\mathrm{A = PbCrO_4}$, $\mathrm{B = Na_2[Pb(OH)_4]}$, $\mathrm{X = (NH_4)_2[Pb(CH_3COO)_4]}$; Option C is correct.