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

Options

A. Vanadium `(Z = 23)`
B. Manganese `(Z = 25)`
C. Chromium `(Z = 24)`
D. Iron `(Z = 26)`

Explanation

To determine which of the four successive members of the first row transition elements has the highest third ionization enthalpy, we will analyze the electronic configurations of the elements and their stability after ionization.

### Step-by-Step Solution:

1. **Identify the Elements and Their Atomic Numbers**:
The four successive members given are:
- Vanadium (V) - Atomic Number 23
- Chromium (Cr) - Atomic Number 24
- Manganese (Mn) - Atomic Number 25
- Iron (Fe) - Atomic Number 26

2. **Write the Electronic Configurations**:
- Vanadium (V): \( [Ar] 3d^3 4s^2 \)
- Chromium (Cr): \( [Ar] 3d^5 4s^1 \) (due to half-filled stability)
- Manganese (Mn): \( [Ar] 3d^5 4s^2 \)
- Iron (Fe): \( [Ar] 3d^6 4s^2 \)

3. **Understand Ionization Enthalpy**:
- The first ionization involves removing one electron from the outermost shell.
- The second ionization involves removing a second electron, which may come from the \(4s\) or \(3d\) subshell.
- The third ionization involves removing a third electron, which will typically come from the \(3d\) subshell after the \(4s\) electrons have been removed.

4. **Analyze the Third Ionization**:
- For Vanadium (V): After removing two electrons (from \(4s^2\)), the configuration becomes \( [Ar] 3d^3 \). The third ionization will remove an electron from \(3d^3\).
- For Chromium (Cr): After removing two electrons (from \(4s^1\)), the configuration becomes \( [Ar] 3d^5 \). The third ionization will remove an electron from \(3d^5\).
- For Manganese (Mn): After removing two electrons (from \(4s^2\)), the configuration becomes \( [Ar] 3d^5 \). The third ionization will also remove an electron from \(3d^5\).
- For Iron (Fe): After removing two electrons (from \(4s^2\)), the configuration becomes \( [Ar] 3d^6 \). The third ionization will remove an electron from \(3d^6\).

5. **Determine Stability and Ionization Enthalpy**:
- Manganese (Mn) has a half-filled \(3d^5\) configuration, which is particularly stable. Removing an electron from a half-filled subshell requires more energy (higher ionization enthalpy).
- Chromium (Cr) also has a half-filled \(3d^5\) configuration, but it has already lost one \(4s\) electron, making it less stable than Mn in terms of the third ionization.
- Iron (Fe) has a \(3d^6\) configuration, which is less stable than the half-filled configurations of Mn and Cr.
- Therefore, the third ionization enthalpy will be highest for Manganese (Mn) due to its stable half-filled \(3d^5\) configuration.

### Conclusion:
The element expected to have the highest third ionization enthalpy among the given options is **Manganese (Mn)**.