Question
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.
JEE Main
Chemistry
p Block Elements (Group 13 & 14)
2026
JEE Main 2026 (21 January Shift 1)
JEE Main Chemistry Question (2026) — Solution
Options
- A. \( Na _2 [ ~Pb ( OH )_2 ], PbCrO O _4\) and \( ( NH _4 )_2 [ ~Pb ( CH _3 COO )_4 ]\)
- B. Na _ 2 [ ~Pb ( OH )_ 2 ], PbCrO _ 4 and [ Pb ( NH _ 3 )_ 4 ] SO _ 4
- C. PbCrO _ 4 , Na _ 2 [ ~Pb ( OH )_ 4 ] and ( NH _ 4 )_ 2 [ ~Pb ( CH _ 3 COO )_ 4 ]
- D. PbCrO _ 4 , Na _ 2 [ ~Pb ( OH )_ 4 ] and [ Pb ( NH _ 3 )_ 4 ] SO _ 4
Answer
C. PbCrO _ 4 , Na _ 2 [ ~Pb ( OH )_ 4 ] and ( NH _ 4 )_ 2 [ ~Pb ( CH _ 3 COO )_ 4 ]
Step-by-step solution
Reaction 1: PbCl_2 + K_2CrO_4 PbCrO_4 + 2KCl A = PbCrO_4 (lead chromate) Reaction 2: PbCrO_4 + 4NaOH Na_2[Pb(OH)_4] + Na_2CrO_4 B = Na_2[Pb(OH)_4] (sodium tetrahydroxoplumbate) Reaction 3: PbSO_4 + 4CH_3COONH_4 (NH_4)_2SO_4 + (NH_4)_2[Pb(CH_3COO)_4] X = (NH_4)_2[Pb(CH_3COO)_4] (ammonium tetraacetatoplumbate)
Related: Chemistry — p Block Elements (Group 13 & 14) · All PYQ Banks