JEE MainChemistryCoordination Compounds
Consider the two octahedral cobalt complexes A = [CoF₆]³⁻ and B = [Co(NH₃)₆]³⁺ . The spin-only magnetic moments of A and B, and the magnitude of the difference in their Crystal Field Stabilization Energies (CFSE, ignoring pairing energy) in terms of _o , respectively, are: (Given Atomic Number: Co = 27 )
Options
- A24 BM , 0 BM , 2.0 _o
- B24 BM , 24 BM , 0 _o
- C24 BM , 0 BM , 0.8 _o
- D24 BM , 0 BM , 2.8 _o
Correct answer
A. 24 BM , 0 BM , 2.0 _o
Step-by-step solution
In both complexes, the central metal ion is Co³⁺ , which has a 3d^6 electronic configuration. For complex A ( [CoF₆]³⁻ ), F^- is a weak field ligand. The octahedral splitting is _o Number of unpaired electrons, n = 4 . _A = 4(4+2) = 24 BM . CFSE _A = (-0.4 4 + 0.6 2) _o = -0.4 _o . For complex B ( [Co(NH₃)₆]³⁺ ), NH₃ acts as a strong field ligand with Co³⁺ . The splitting is _o > P . The electronic configuration is t_ 2g ^6 e_g^0 . Number of unpaired electrons, n = 0 . _B = 0(0+2) = 0 BM . CFSE _B = (-0.4 6 + 0.6 0