JEE Main20236 Apr 2023Morning ShiftChemistryGeneral Organic ChemistryActual
Match List-I with List-II. List-I Element detected List-II Reagent used/Product formed A Nitrogen I Na 2 Fe CN 5 NO B Sulphur II AgNO 3 C Phosphorus III Fe 4 Fe CN 6 3 D Halogen IV NH 4 2 MoO 4 Choose the correct answer from the options given below:
Options
- AA → III ; B → I ; C → IV ; D → II
- BA → II ; B → IV ; C → I ; D →   III  
- CA → IV ; B → II ; C → I ; D → III
- DA → II ; B → I ; C → IV ; D → III
Correct answer
A. A → III ; B → I ; C → IV ; D → II
Step-by-step solution
The nitrogen present in the organic compound on fusion with sodium metal gives sodium cyanide ( NaCN ) soluble in water. This is converted into sodium ferrocyanide by the addition of sufficient quantity of ferrous sulphate. 6 NaCN   +   FeSO 4   →   Na 4 [ Fe ( CN ) 6 ] Sodium ferrocyanide   +   Na 2 SO 4 Ferric ions generated during the process react with ferrocyanide to form Prussian blue precipitate of ferric ferrocyanide. Na 4 [ Fe ( CN ) 6 ]   +   Fe 3 +