The third ionization enthalpy is minium for:

Options

A. Co
B. Ni
C. Fe
D. Mn

Answer

To determine which element has the minimum third ionization enthalpy among cobalt (Co), nickel (Ni), iron (Fe), and manganese (Mn), we will analyze the electronic configurations of these elements and their ions after the removal of electrons.

### Step-by-Step Solution:

1. **Understanding Ionization Enthalpy**:
- Ionization enthalpy is the energy required to remove an electron from an atom or ion. The third ionization enthalpy refers to the energy needed to remove the third electron.

2. **Write the Electronic Configurations**:
- **Cobalt (Co)**: Atomic number = 27
- Ground state configuration: \( \text{[Ar]} 4s^2 3d^7 \)
- After removing 2 electrons (to form \( \text{Co}^{2+} \)): \( \text{[Ar]} 3d^7 \)
- For the third ionization (\( \text{Co}^{3+} \)): Remove from \( 3d^7 \).

- **Nickel (Ni)**: Atomic number = 28
- Ground state configuration: \( \text{[Ar]} 4s^2 3d^8 \)
- After removing 2 electrons (to form \( \text{Ni}^{2+} \)): \( \text{[Ar]} 3d^8 \)
- For the third ionization (\( \text{Ni}^{3+} \)): Remove from \( 3d^8 \).

- **Iron (Fe)**: Atomic number = 26
- Ground state configuration: \( \text{[Ar]} 4s^2 3d^6 \)
- After removing 2 electrons (to form \( \text{Fe}^{2+} \)): \( \text{[Ar]} 3d^6 \)
- For the third ionization (\( \text{Fe}^{3+} \)): Remove from \( 3d^6 \).

- **Manganese (Mn)**: Atomic number = 25
- Ground state configuration: \( \text{[Ar]} 4s^2 3d^5 \)
- After removing 2 electrons (to form \( \text{Mn}^{2+} \)): \( \text{[Ar]} 3d^5 \)
- For the third ionization (\( \text{Mn}^{3+} \)): Remove from \( 3d^5 \).

3. **Analyzing Stability of Configurations**:
- Half-filled and fully filled configurations are more stable.
- **Manganese (\( \text{Mn}^{2+} \))** has a half-filled \( 3d^5 \) configuration, making it stable. Removing an electron from this configuration will require more energy.
- **Iron (\( \text{Fe}^{2+} \))** has a \( 3d^6 \) configuration. Removing an electron will lead to a half-filled \( 3d^5 \) configuration, which is stable and thus easier to remove.
- **Cobalt (\( \text{Co}^{2+} \))** has a \( 3d^7 \) configuration, and removing an electron will lead to \( 3d^6 \), which is less stable than \( 3d^5 \).
- **Nickel (\( \text{Ni}^{2+} \))** has a \( 3d^8 \) configuration, and removing an electron will lead to \( 3d^7 \), which is also less stable than \( 3d^5 \).

4. **Conclusion**:
- The third ionization enthalpy will be minimum for **Iron (Fe)** because after removing the third electron, it attains a half-filled \( 3d^5 \) configuration, which is energetically favorable.

### Final Answer:
The third ionization enthalpy is minimum for **Iron (Fe)**.