Four successive members of the first-row transition elements are listed below with their atomic numbers. Which one of them is expected to have the highest third ionization enthalpy?

Manganese (Z = 25)

The correct answer is option 2. Manganese (Z = 25).

To understand which of the given elements has the highest third ionization enthalpy, we need to delve into the concept of ionization energy and the stability of electron configurations, especially after the removal of the second electron. Here is a detailed breakdown:

Ionization Energies and Electron Configurations

The ionization energy is the energy required to remove an electron from an atom or ion. The first ionization energy is for the removal of the first electron, the second for the second electron, and the third for the third electron.

For transition metals, the electron configurations and the stability of these configurations are key to understanding their ionization energies.

Vanadium (Z = 23)

Electron configuration: \([Ar] 3d^3 4s^2\)

After removing two electrons: \([Ar] 3d^3\)

Third ionization removes an electron from the \(3d^3\) configuration, which is relatively stable but not as stable as a half-filled d subshell.

Chromium (Z = 24)

Electron configuration: \([Ar] 3d^5 4s^1\)

After removing two electrons: \([Ar] 3d^4\)

Third ionization removes an electron from the \(3d^4\) configuration, which is less stable than a half-filled or completely filled subshell but more stable than a less-than-half-filled subshell.

Manganese (Z = 25)

Electron configuration: \([Ar] 3d^5 4s^2\)

After removing two electrons: \([Ar] 3d^5\)

Third ionization removes an electron from the half-filled \(3d^5\) configuration, which is highly stable. The half-filled d subshell \((3d^5)\) is particularly stable due to electron exchange energy and symmetrical distribution.

Iron (Z = 26)

Electron configuration: \([Ar] 3d^6 4s^2\)

After removing two electrons: \([Ar] 3d^6\)

Third ionization removes an electron from the \(3d^6\) configuration, which is stable but not as stable as a half-filled \((3d^5)\) configuration.

Stability of Electron Configurations

Half-filled and fully filled subshells: These configurations are exceptionally stable due to exchange energy (parallel spin electrons in degenerate orbitals lower the energy of the system) and symmetrical electron distribution.

Manganese (Mn) has a half-filled \(3d^5\) configuration after losing two electrons. Removing another electron from this stable half-filled configuration requires significantly more energy compared to other configurations.

Comparison

1. Vanadium: After losing two electrons, it has a \(3d^3\) configuration. Removing an electron from here is relatively straightforward compared to breaking a half-filled or fully filled subshell.

2. Manganese: After losing two electrons, it has a \(3d^5\) half-filled configuration, which is highly stable. The third ionization energy is very high because breaking this stable configuration requires a lot of energy.

3. Chromium: After losing two electrons, it has a \(3d^4\) configuration. Removing one more electron to break a stable configuration is harder but not as difficult as breaking a half-filled \(3d^5\) configuration.

4. Iron: After losing two electrons, it has a \(3d^6\) configuration. Removing one more electron makes it less stable than breaking a half-filled configuration, but it is not as challenging as breaking a half-filled \(3d^5\) configuration.

Manganese (Mn), with its half-filled d^5 configuration after the second ionization, will have the highest third ionization enthalpy. The energy required to remove an electron from a half-filled d^5 subshell is significantly higher than from other configurations due to its exceptional stability.

Therefore, the correct answer is Manganese (Z = 25).