Atomic number of `Mn, Fe, Co` and `Ni` are 25, 26, 27 and 28 respectively. Which of the following outer orbital octahedral complexes have same number of unpaired electrons ?

Options

A. `[MnCl_(6)]^(3-)`
B. `[FeF_(6)]^(3-)`
C. `[CoF_(6)]^(3-)`
D. `[Ni(NH_(3))_(6)]^(2+)`

Explanation

To solve the problem, we need to determine the number of unpaired electrons in the outer orbital octahedral complexes of manganese (Mn), iron (Fe), cobalt (Co), and nickel (Ni) based on their oxidation states. We will analyze each metal complex step by step.

### Step 1: Determine the oxidation state of each metal in the complex.
- **Manganese (Mn)**: In the complex with chlorine (Cl) as a ligand, the oxidation state can be calculated as follows:
\[ X - 6(-1) = -3 \implies X + 6 = -3 \implies X = +3 \]
- **Iron (Fe)**: In the complex with fluorine (F) as a ligand:
\[ X - 6(-1) = -3 \implies X + 6 = -3 \implies X = +3 \]
- **Cobalt (Co)**: In the complex with fluorine (F) as a ligand:
\[ X - 6(-1) = -3 \implies X + 6 = -3 \implies X = +3 \]
- **Nickel (Ni)**: In the complex with ammonia (NH₃) as a ligand:
\[ X + 0 = +2 \implies X = +2 \]

### Step 2: Determine the electronic configuration of each metal after losing electrons.
- **Manganese (Mn)**: Atomic number = 25
- Electronic configuration: \([Ar] 4s^2 3d^5\)
- After losing 3 electrons: \(4s^0 3d^4\) (d4 configuration)

- **Iron (Fe)**: Atomic number = 26
- Electronic configuration: \([Ar] 4s^2 3d^6\)
- After losing 3 electrons: \(4s^0 3d^5\) (d5 configuration)

- **Cobalt (Co)**: Atomic number = 27
- Electronic configuration: \([Ar] 4s^2 3d^7\)
- After losing 3 electrons: \(4s^0 3d^6\) (d6 configuration)

- **Nickel (Ni)**: Atomic number = 28
- Electronic configuration: \([Ar] 4s^2 3d^8\)
- After losing 2 electrons: \(4s^0 3d^8\) (d8 configuration)

### Step 3: Determine the number of unpaired electrons in each configuration.
- **Manganese (Mn)**: \(d^4\) configuration in a high spin state:
- Filling order: ↑↓ ↑↓ ↑ ↑
- Number of unpaired electrons = 4

- **Iron (Fe)**: \(d^5\) configuration in a high spin state:
- Filling order: ↑↓ ↑↓ ↑ ↑ ↑
- Number of unpaired electrons = 5

- **Cobalt (Co)**: \(d^6\) configuration in a high spin state:
- Filling order: ↑↓ ↑↓ ↑ ↑ ↑
- Number of unpaired electrons = 4

- **Nickel (Ni)**: \(d^8\) configuration in a high spin state:
- Filling order: ↑↓ ↑↓ ↑↓ ↑ ↑
- Number of unpaired electrons = 2

### Step 4: Compare the number of unpaired electrons.
- **Manganese (Mn)**: 4 unpaired electrons
- **Iron (Fe)**: 5 unpaired electrons
- **Cobalt (Co)**: 4 unpaired electrons
- **Nickel (Ni)**: 2 unpaired electrons

### Conclusion:
The outer orbital octahedral complexes of manganese and cobalt both have the same number of unpaired electrons (4).

### Final Answer:
Manganese (Mn) and Cobalt (Co) have the same number of unpaired electrons.