Concept:
The third ionization enthalpy ($\Delta_i H_3$) represents the energy required to remove an electron from a doubly charged gaseous cation ($\text{M}^{2+} \rightarrow \text{M}^{3+} + e^-$). Its magnitude depends heavily on the electronic configuration of the $\text{M}^{2+}$ ion. Removing an electron from a completely filled or half-filled subshell requires an exceptionally high amount of energy due to the extra exchange energy and symmetric stability of such configurations.
Step 1: Write out the electronic configurations for each $\text{M}^{2+}$ cation.
Let's look at the outer electronic configurations for the transition metal ions after losing two electrons (from their $\text{4s}$ subshells):
• $\text{V}^{2+} (Z=23) \rightarrow [\text{Ar}] 3\text{d}^3$
• $\text{Cr}^{2+} (Z=24) \rightarrow [\text{Ar}] 3\text{d}^4$
• $\text{Mn}^{2+} (Z=25) \rightarrow [\text{Ar}] 3\text{d}^5$
• $\text{Fe}^{2+} (Z=26) \rightarrow [\text{Ar}] 3\text{d}^6$
Step 2: Analyze the stability to identify the highest energy barrier.
The $\text{Mn}^{2+}$ ion has a $3\text{d}^5$ valence shell configuration. The $\text{d}$ subshell is exactly half-filled, which gives it extra thermodynamic stability and high exchange energy. Removing a third electron from $\text{Mn}^{2+}$ requires disrupting this stable, symmetric half-filled shell, creating an exceptionally large energy barrier.
In contrast, removing an electron from $\text{Fe}^{2+}$ ($3\text{d}^6$) leaves a stable half-filled $3\text{d}^5$ configuration, which occurs much more readily. Therefore, Manganese possesses the highest third ionization enthalpy.