Question

The type of hybridisation of Co in \([CoF_6]^{3-}\) complex ion is

• A. \(dsp^2\)
• B. \(sp^3d^2\)
• C. \(d^2sp^3\)
• D. \(sp^2d\)
• E. \(sp^3\)

Hint:

Weak field ligands (\(F^-, Cl^-, H_2O\)) usually lead to \(sp^3d^2\) (outer orbital).
Strong field ligands (\(CN^-, NH_3, CO\)) usually lead to \(d^2sp^3\) (inner orbital) for octahedral complexes.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The hybridization and geometry of a complex depend on the oxidation state of the metal and the nature of the ligand (strong field vs. weak field).

Step 3: Detailed Explanation:
1. Oxidation State: In \([CoF_6]^{3-}\), Cobalt is in the +3 state (\(Co^{3+}\)).
2. Electronic Configuration: \(Co^{3+}\) is \(3d^6\).
3. Ligand Nature: Fluoride (\(F^-\)) is a weak field ligand. It cannot cause the pairing of \(3d\) electrons.
4. Orbital Availability: Since \(3d\) electrons are not paired, the metal uses outer orbitals (\(4s, 4p, 4d\)) for bonding.
5. Hybridization: It uses one \(4s\), three \(4p\), and two \(4d\) orbitals to form six \(sp^3d^2\) hybrid orbitals.
This forms an outer orbital, high-spin complex.

Step 4: Final Answer:
The hybridization is \(sp^3d^2\).