Consider the following reactions. PbCl_2 + K_2CrO_4 rightarrow A + 2KCl$PbCl_{2} + K_{2}CrO_{4} \rightarrow A + 2KCl$ (Hot solution) A + NaOH rightleftharpoons B + Na_2CrO_4$A + NaOH \rightleftharpoons B + Na_{2}CrO_{4}$ PbSO_4 + 4CH_3COONH_4 rightarrow (NH_4)_2SO_4 + X$PbSO_{4} + 4CH_{3}COONH_{4} \rightarrow (NH_{4})_{2}SO_{4} + X$ In the above reactions, A, B and X are respectively

### Core Logic The precipitation and complex formation reactions of Lead are: mathrmPbCl_2 + mathrmK_2mathrmCrO_4 rightarrow mathrmPbCrO_4 + 2mathrmKCl quad (textHot solution) quad textso, A is mathrmPbCrO_4$$\mathrm{PbCl}_2 + \mathrm{K}_2\mathrm{CrO}_4 \rightarrow \mathrm{PbCrO}_4 + 2\mathrm{KCl} \quad (\text{Hot solution}) \quad \text{so, A is } \mathrm{PbCrO}_4$$ mathrmPbCrO_4 + 4mathrmNaOH \ (excess) rightarrow mathrmNa_2[mathrmPb(OH)_4] + mathrmNa_2mathrmCrO_4 quad textso, B is mathrmNa_2[mathrmPb(OH)_4]$$\mathrm{PbCrO}_4 + 4\mathrm{NaOH \ (excess)} \rightarrow \mathrm{Na}_2[\mathrm{Pb(OH)}_4] + \mathrm{Na}_2\mathrm{CrO}_4 \quad \text{so, B is } \mathrm{Na}_2[\mathrm{Pb(OH)}_4]$$ mathrmPbSO_4 + 4mathrmCH_3mathrmCOONH_4 rightarrow (mathrmNH_4)_2 [mathrmPb(CH_3COO)_4] + (mathrmNH_4)_2mathrmSO_4 quad textso, X is (mathrmNH_4)_2[mathrmPb(CH_3COO)_4]$$\mathrm{PbSO}_4 + 4\mathrm{CH}_3\mathrm{COONH}_4 \rightarrow (\mathrm{NH}_4)_2 [\mathrm{Pb(CH_3COO)}_4] + (\mathrm{NH}_4)_2\mathrm{SO}_4 \quad \text{so, X is } (\mathrm{NH}_4)_2[\mathrm{Pb(CH_3COO)}_4]$$ ### Pattern Recognition Lead forms a yellow precipitate of lead chromate (A), which is amphoteric and dissolves in excess NaOH to form soluble plumbate(II) complex (B). It also forms a stable soluble complex with ammonium acetate (X). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: p-Block Elements Class 12 Chemistry: d and f Block Elements