JEE MainChemistryCoordination Compounds
Match List I with List II. List I (Complex) List II (Geometry and Magnetic Moment) (A) [ CoF ₆ ]³⁻ (I) Tetrahedral, =0 (B) [ Ni ( CO )₄ ] (II) Octahedral, =0 (C) [ Cu ( NH ₃ )₄ ]²⁺ (III) Square planar, =1.73 ~BM (D) [ Fe ( CN )₆ ]⁴⁻ (IV) Octahedral, =4.9 ~BM Choose the correct answer from the options given below:
Options
- AA-II, B-I, C-III, D-IV
- BA-IV, B-III, C-I, D-II
- CA-IV, B-I, C-III, D-II
- DA-IV, B-II, C-III, D-I
Correct answer
C. A-IV, B-I, C-III, D-II
Step-by-step solution
(A) For [ CoF ₆ ]³⁻ : Co is in +3 oxidation state ( 3d ⁶ ). F ⁻ is a weak field ligand, so no pairing occurs. Number of unpaired electrons = 4 . = 4(4+2) = 4.9 ~BM . Hybridisation is sp ³ d ² (Octahedral). (IV) (B) For [ Ni ( CO )₄ ] : Ni is in 0 oxidation state ( 3d ⁸ 4s ² ). CO is a strong field ligand, causing the 4 s electrons to pair in the 3 d orbitals, resulting in a 3d ¹⁰ configuration. Number of unpaired electrons = 0 . = 0 . Hybridisation is sp ³ (Tetrahedral). (I) (C) For [ Cu ( NH ₃ )₄ ]²⁺ : Cu is in +2