JEE Main202621 January 2026Morning ShiftChemistryp Block Elements (Group 13 & 14)Actual
Consider the following reactions. PbCl ₂+ K ₂ CrO ₄ ~A +2 KCl (Hot solution) A + NaOH B + Na ₂ CrO ₄ PbSO ₄+4 CH ₃ COONH ₄ ( NH ₄ )₂ SO ₄+ X In the above reactions, A , B and X are respectively.
Options
- A( Na ₂ [ ~Pb ( OH )₂ ], PbCrO O ₄ ) and ( ( NH ₄ )₂ [ ~Pb ( CH ₃ COO )₄ ] )
- BNa ₂ [ ~Pb ( OH )₂ ], PbCrO ₄ and [ Pb ( NH ₃ )₄ ] SO ₄
- CPbCrO ₄, Na ₂ [ ~Pb ( OH )₄ ] and ( NH ₄ )₂ [ ~Pb ( CH ₃ COO )₄ ]
- DPbCrO ₄, Na ₂ [ ~Pb ( OH )₄ ] and [ Pb ( NH ₃ )₄ ] SO ₄
Correct answer
C. PbCrO ₄, Na ₂ [ ~Pb ( OH )₄ ] and ( NH ₄ )₂ [ ~Pb ( CH ₃ COO )₄ ]
Step-by-step solution
Reaction 1: PbCl₂ + K₂CrO₄ PbCrO₄ + 2KCl A = PbCrO₄ (lead chromate) Reaction 2: PbCrO₄ + 4NaOH Na₂[Pb(OH)₄] + Na₂CrO₄ B = Na₂[Pb(OH)₄] (sodium tetrahydroxoplumbate) Reaction 3: PbSO₄ + 4CH₃COONH₄ (NH₄)₂SO₄ + (NH₄)₂[Pb(CH₃COO)₄] X = (NH₄)₂[Pb(CH₃COO)₄] (ammonium tetraacetatoplumbate)