Which of the following complex species involves `d^(2)sp^(3)` hybridisation?

Options

A. `[CoF_(6)]^(3-)`
B. `[Co(NH_(3))_(6)]^(3+)`
C. `[Fe(CN)_(6)]^(3-)`
D. `[Cr(NH_(3))_(6)]^(3+)`

Explanation

To determine which complex species involves `d^(2)sp^(3)` hybridization, we will analyze each of the given complexes step by step.

### Step 1: Analyze the first complex, [CoF6]²⁻
1. **Determine the oxidation state of cobalt (Co)**:
- In [CoF6]²⁻, the overall charge is -2. Each fluoride ion (F) has a charge of -1. Therefore, the oxidation state of Co can be calculated as:
\[ x + 6(-1) = -2 \implies x - 6 = -2 \implies x = +4 \]
- Thus, Co is in the +4 oxidation state.

2. **Find the electronic configuration of Co**:
- Cobalt (Co) has an atomic number of 27. Its ground state electronic configuration is [Ar] 4s² 3d⁷.
- For Co in +4 oxidation state, we remove 4 electrons (2 from 4s and 2 from 3d):
\[ 4s^2 3d^7 \rightarrow 3d^5 \]

3. **Identify the ligand field strength**:
- Fluoride (F⁻) is a weak field ligand, which means it does not cause pairing of electrons in the d-orbitals.

4. **Determine the hybridization**:
- With 5 d-electrons and no pairing, the hybridization will be:
\[ sp^3d^2 \]
- Therefore, [CoF6]²⁻ does not involve `d^(2)sp^(3)` hybridization.

### Step 2: Analyze the second complex, [Co(NH3)6]³⁺
1. **Determine the oxidation state of cobalt (Co)**:
- In [Co(NH3)6]³⁺, the overall charge is +3. NH3 is a neutral ligand, so:
\[ x + 6(0) = +3 \implies x = +3 \]
- Thus, Co is in the +3 oxidation state.

2. **Find the electronic configuration of Co**:
- For Co in +3 oxidation state:
\[ 4s^2 3d^7 \rightarrow 3d^6 \]

3. **Identify the ligand field strength**:
- Ammonia (NH3) is a moderate field ligand, which can cause some pairing of electrons.

4. **Determine the hybridization**:
- In the presence of NH3, the 3d electrons will pair up, leading to:
\[ sp^3d^2 \]
- Thus, [Co(NH3)6]³⁺ does not involve `d^(2)sp^(3)` hybridization.

### Step 3: Analyze the third complex, [Fe(CN)6]³⁻
1. **Determine the oxidation state of iron (Fe)**:
- In [Fe(CN)6]³⁻, the overall charge is -3. CN⁻ is a -1 ligand:
\[ x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3 \]
- Thus, Fe is in the +3 oxidation state.

2. **Find the electronic configuration of Fe**:
- For Fe in +3 oxidation state:
\[ 4s^2 3d^6 \rightarrow 3d^5 \]

3. **Identify the ligand field strength**:
- Cyanide (CN⁻) is a strong field ligand, which causes pairing of electrons.

4. **Determine the hybridization**:
- With pairing, the configuration will be:
\[ d^2sp^3 \]
- Therefore, [Fe(CN)6]³⁻ involves `d^(2)sp^(3)` hybridization.

### Conclusion
The complex that involves `d^(2)sp^(3)` hybridization is **[Fe(CN)6]³⁻**.