Identify the CORRECT set of details from the following: A. [ Co ( NH _ 3 )_ 6 ]^ 3+ : Inner orbital complex; d ^ 2 sp ^ 3 hybridized B. [ MnCl _ 6 ]^ 3- : Outer orbital complex; sp ^ 3 ~d ^ 2 hybridized C. [ CoF _ 6 ]^ 3- : Outer orbital complex; d ^ 2 sp ^ 3 hybridized D. [ FeF _ 6 ]^ 3- : Outer orbital complex; sp ^ 3 ~d ^ 2 hybridized E. [ Ni ( CN )_ 4 ]^ 2- : Inner orbital complex; sp ^ 3 hybridized Choose the correct answer from the options given below:
B. A, B & D Only
Analyzing each complex based on Valence Bond Theory (VBT): A. [Co(NH_3)_6]^ 3+ : Co^ 3+ is 3d^6. NH_3 is a strong field ligand (SFL) for Co^ 3+ , causing pairing. This leaves two 3d orbitals vacant. Hybridization is d^2sp^3 (Inner orbital complex). Statement A is correct. B. [MnCl_6]^ 3- : Mn^ 3+ is 3d^4. Cl^- is a weak field ligand (WFL). No pairing occurs. To accommodate 6 ligands, it uses 4s, 4p, and 4d orbitals. Hybridization is sp^3d^2 (Outer orbital complex). Statement B is correct. C. [CoF_6]^ 3- : Co^ 3+ is 3d^6. F^- is a WFL. No pairing occurs. Hybridization is sp^3d^2 (Outer orbital complex). Statement C is incorrect because it mentions d^2sp^3. D. [FeF_6]^ 3- : Fe^ 3+ is 3d^5. F^- is a WFL. No pairing occurs. Hybridization is sp^3d^2 (Outer orbital complex). Statement D is correct. E. [Ni(CN)_4]^ 2- : Ni^ 2+ is 3d^8. CN^- is a SFL, causing pairing of the two unpaired electrons in 3d. This leaves one 3d orbital vacant. Hybridization is dsp^2 (Square planar, Inner orbital complex). Statement E is incorrect because it mentions sp^3. Correct statements are A, B, and D.
Related: Chemistry — Coordination Compounds · All PYQ Banks