Which of the following statements is correct?

Options

A. `[CoF_(6)]^(3-) and [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 the correct statement regarding the two coordination complexes, we will analyze each complex step by step.

### Step 1: Analyze the first complex \([CoF_6]^{3-}\)

1. **Determine the oxidation state of cobalt (Co)**:
- Let the oxidation state of Co be \(x\).
- Fluoride (F) has a charge of \(-1\).
- The overall charge of the complex is \(-3\).
- The equation is:
\[ x + 6(-1) = -3 \]
\[ x - 6 = -3 \implies x = +3 \]

2. **Electron configuration of Co in +3 oxidation state**:
- The electron configuration of neutral Co (atomic number 27) is \([Ar] 4s^2 3d^7\).
- For Co in +3 state:
\[ [Ar] 4s^0 3d^6 \]

3. **Hybridization and geometry**:
- Fluoride is a weak field ligand, so it does not cause pairing of electrons.
- The hybridization involves the outer orbitals:
\[ \text{Hybridization: } sp^3d^2 \]
- The geometry is octahedral.

4. **Magnetic properties**:
- Since there are 6 unpaired electrons, the complex is **paramagnetic**.

5. **Classification**:
- This is an **outer orbital complex** because the 4d orbitals are involved in hybridization.

### Step 2: Analyze the second complex \([Co(NH_3)_6]^{3+}\)

1. **Determine the oxidation state of cobalt (Co)**:
- Let the oxidation state of Co be \(x\).
- Ammonia (NH3) is a neutral ligand.
- The overall charge of the complex is \(+3\).
- The equation is:
\[ x + 0 = +3 \implies x = +3 \]

2. **Electron configuration of Co in +3 oxidation state**:
- The electron configuration remains the same as before:
\[ [Ar] 4s^0 3d^6 \]

3. **Hybridization and geometry**:
- Ammonia is a strong field ligand, which causes pairing of electrons.
- The hybridization involves the inner orbitals:
\[ \text{Hybridization: } d^2sp^3 \]
- The geometry is also octahedral.

4. **Magnetic properties**:
- Since there are no unpaired electrons, the complex is **diamagnetic**.

5. **Classification**:
- This is an **inner orbital complex** because the 3d orbitals are involved in hybridization.

### Step 3: Evaluate the statements

Now, we can evaluate the statements provided in the question:

1. Both are paramagnetic: **False** (first is paramagnetic, second is diamagnetic).
2. Both are high spin complexes: **False** (the second is low spin due to strong field ligand).
3. Both are octahedral and pentagonal geometry: **False** (both are octahedral).
4. One is an outer orbital complex and the other is an inner orbital complex: **True** (first is outer, second is inner).

### Conclusion

The correct statement is: **One is an outer orbital complex and the other is an inner orbital complex.**

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