`sp^(3)d^(2)` hybridisation is present in

Options

A. `[CoF_(6)]^(-3)`
B. `[Ni(CO)_(4)]`
C. `[Co(NH_(3))_(6)]^(+2)`
D. `[Ni(CN)_(4)]^(-2)`

Explanation

To determine which coordination complex exhibits `sp^(3)d^(2)` hybridization, we can follow these steps:

### Step 1: Understand Hybridization
Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals. The notation `sp^(3)d^(2)` indicates that one s orbital, three p orbitals, and two d orbitals are involved in hybridization.

### Step 2: Identify the Coordination Number
The coordination number is the number of ligand atoms that are bonded to the central metal atom. For `sp^(3)d^(2)` hybridization, the coordination number is typically 6, which suggests an octahedral geometry.

### Step 3: Determine the Nature of Ligands
The type of ligands (weak or strong) affects whether the d orbitals involved in hybridization are from the outer or inner shells. Strong field ligands cause pairing of electrons in the lower energy d orbitals (inner orbitals), while weak field ligands do not cause pairing.

### Step 4: Analyze the Given Options
1. If the complex has a coordination number of 4, it can either be tetrahedral (sp³) or square planar (dsp²). In both cases, `sp^(3)d^(2)` hybridization is not applicable.
2. If the complex has a coordination number of 6, we need to check the ligands:
- If the ligands are strong (like NH₃), they will cause pairing and lead to inner orbital complexes (d²sp³).
- If the ligands are weak (like F⁻), they will not cause pairing, allowing the use of outer d orbitals, leading to `sp^(3)d^(2)` hybridization.

### Step 5: Conclusion
Based on the analysis, the complex that has `sp^(3)d^(2)` hybridization must have a coordination number of 6 with weak field ligands. A suitable example is a complex with fluorine ligands.

### Final Answer
The coordination complex that exhibits `sp^(3)d^(2)` hybridization is one that has a central metal atom with weak field ligands, such as fluorine, resulting in an outer orbital complex.

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