JEE MainChemistryCoordination Compounds
Match List I with List II. List I (Complex) List II (Spin-only magnetic moment) A. [FeF₆]³⁻ I. 0 BM B. [Co(NH₃)₆]³⁺ II. 1.73 BM C. [NiCl₄]²⁻ III. 2.83 BM D. [Cu(NH₃)₄]²⁺ IV. 5.92 BM Choose the correct answer from the options given below:
Options
- AA-II, B-I, C-IV, D-III
- BA-IV, B-I, C-III, D-II
- CA-IV, B-III, C-I, D-II
- DA-I, B-IV, C-II, D-III
Correct answer
B. A-IV, B-I, C-III, D-II
Step-by-step solution
For A: In [FeF₆]³⁻ , Fe is in +3 oxidation state. Fe³⁺ is 3d^5 . Since F^- is a weak field ligand, it forms a high spin complex with 5 unpaired electrons. = 5(5+2) = 5.92 BM. (A IV) For B: In [Co(NH₃)₆]³⁺ , Co is in +3 oxidation state. Co³⁺ is 3d^6 . NH₃ acts as a strong field ligand for Co³⁺ , causing pairing of electrons. Thus, 0 unpaired electrons. = 0 BM. (B I) For C: In [NiCl₄]²⁻ , Ni is in +2 oxidation state. Ni²⁺ is 3d^8 . Cl^- is a weak field ligand, forming a tetrahedral complex with 2 unpaired electrons.