Which of the following statements is correct?

Options

A. `[CoF_(6)]^(3-) annd [Co(NH_(3))_(6)]^(3+)` both are paramagnetic complexes
B. `[CoF_(6)]^(3-) and [Co(NH_(3))_(6)]^(3+)` both are high spin complexes
C. `[CoF_(6))]^(3-) ` is octahedral while `[Co(NH_(3))_(6)]^(3+)` has a pentagonal pyramid shape
D. `[CoF_(6)]^(3-)` is outer orbital complex while `[Co(NH_(3))_(6)]^(3+)` is inner orbital complex

Answer

To determine which statement regarding the coordination complexes is correct, we will analyze the information provided about the ligands and the cobalt(III) cation.

### Step-by-Step Solution:

1. **Identify the Ligands and Their Strength**:
- The ligands mentioned are fluorine (F) and ammonia (NH3).
- Fluorine is classified as a weak field ligand, while ammonia is a strong field ligand.

**Hint**: Remember that weak field ligands tend to produce high spin complexes, while strong field ligands lead to low spin complexes.

2. **Determine the Electronic Configuration of Cobalt(III)**:
- Cobalt in its +3 oxidation state has the electronic configuration of [Ar] 3d^6.

**Hint**: The oxidation state of the metal ion is crucial for determining its electronic configuration.

3. **Analyze the Spin States**:
- For the weak field ligand (fluorine), the d electrons will occupy the available orbitals singly before pairing occurs, resulting in a high spin complex.
- For the strong field ligand (ammonia), the d electrons will pair up in the lower energy t2g orbitals before occupying the eg orbitals, resulting in a low spin complex.

**Hint**: High spin complexes have unpaired electrons, while low spin complexes have all electrons paired.

4. **Hybridization**:
- In the case of the ammonia complex (strong field), the hybridization will be sp^3d^2, indicating it is an outer orbital complex.
- In the case of the fluorine complex (weak field), the hybridization will be d^2sp^3, indicating it is an inner orbital complex.

**Hint**: The type of hybridization can help determine the geometry of the complex.

5. **Determine the Magnetic Properties**:
- The complex with fluorine (high spin) will have unpaired electrons, making it paramagnetic.
- The complex with ammonia (low spin) will have all paired electrons, making it diamagnetic.

**Hint**: The presence of unpaired electrons is what defines whether a complex is paramagnetic or diamagnetic.

6. **Evaluate the Statements**:
- **Statement 1**: Both complexes are paramagnetic. (Incorrect, as one is paramagnetic and the other is diamagnetic.)
- **Statement 2**: Both are high spin complexes. (Incorrect, as one is high spin and the other is low spin.)
- **Statement 3**: One is octahedral and the other is pentagonal. (Incorrect, as both are octahedral complexes.)
- **Statement 4**: The correct statement is that one complex is high spin (fluorine) and the other is low spin (ammonia).

**Hint**: Carefully analyze each statement against the properties discussed.

### Conclusion:
The correct statement is that one complex is high spin (with fluorine) and the other is low spin (with ammonia).