NTA Abhyas JEE Main2020ChemistryCoordination CompoundsPractice
The CFSE for [ C o C l 6 ] 4 - complex is 18000 c m - 1 . The Δ for [ C o C l 4 ] 2 - will be
Options
- A9000 c m - 1
- B4000 c m - 1
- C8000 c m - 1
- D2000 c m - 1
Correct answer
C. 8000 c m - 1
Step-by-step solution
The relation of CFSE for tetrahedral and octahedral complex is given as Δ t = 4 9 Δ 0 So Δ t for [ C o C l 4 ] 2 - = 4 9 × 18000 c m - 1 = 8000 c m - 1 .