Question

Explain formation of [CoF\(_6\)]\(^{3-}\) complex with respect to:
i. Hybridisation
ii. Magnetic properties
iii. Inner/outer complex
iv. Geometry

Hint:

Weak-field ligands like F\(^-\) cause high-spin complexes; check ligand field strength for hybridisation.

Answer 1

For [CoF\(_6\)]\(^{3-}\), Co is in +3 oxidation state (\( \text{Co}^{3+} \), d\(^6\)).
i. Hybridisation: F\(^-\) is a weak-field ligand, so high-spin complex. Co\(^{3+}\) uses 3d, 4s, and 4p orbitals for d\(^2\)sp\(^3\) hybridisation (outer orbital complex).
ii. Magnetic Properties: d\(^6\) high-spin: 4 unpaired electrons (\( t_{2g}^4 e_g^2 \)). Magnetic moment: \[ \mu = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9 \, \text{BM}. \] iii. Inner/Outer Complex: Uses 4s and 4p orbitals, so it’s an outer orbital complex (d\(^2\)sp\(^3\)).
iv. Geometry: Octahedral, as six F\(^-\) ligands coordinate around Co\(^{3+}\).