The third ionization enthalpy of manganese ( Z=25 ) is...

The atomic number of manganese ($\mathrm{Mn}$) is $25$. Its ground state electronic configuration is $[\mathrm{Ar}] 3\mathrm{d}^5 4\mathrm{s}^2$.

During the removal of the first two electrons to form $\mathrm{Mn}^{2+}$, the $4\mathrm{s}$ electrons are removed, resulting in the configuration $[\mathrm{Ar}] 3\mathrm{d}^5$. The third ionization enthalpy involves the removal of an electron from $\mathrm{Mn}^{2+}$ to form $\mathrm{Mn}^{3+}$. This corresponds to the transition $3\mathrm{d}^5 \rightarrow 3\mathrm{d}^4$. Since the $3\mathrm{d}^5$ configuration is exactly half-filled and exceptionally stable, removing an electron from it requires a very high amount of energy.

The atomic number of iron ($\mathrm{Fe}$) is $26$. Its ground state electronic configuration is $[\mathrm{Ar}] 3\mathrm{d}^6 4\mathrm{s}^2$.

Removing two electrons forms $\mathrm{Fe}^{2+}$ with the configuration $[\mathrm{Ar}] 3\mathrm{d}^6$. The third ionization enthalpy involves the removal of an electron from $\mathrm{Fe}^{2+}$ to form $\mathrm{Fe}^{3+}$. This corresponds to the transition $3\mathrm{d}^6 \rightarrow 3\mathrm{d}^5$. Because the resulting $3\mathrm{d}^5$ configuration is half-filled and highly stable, the removal of this third electron is energetically favorable, making the third ionization enthalpy of iron comparatively lower.

Thus, the contrast is best explained by the transitions $\mathrm{d}^5 \rightarrow \mathrm{d}^4$ in $\mathrm{Mn}^{2+}$ and $\mathrm{d}^6 \rightarrow \mathrm{d}^5$ in $\mathrm{Fe}^{2+}$.

Answer: $\mathrm{d}^5 \rightarrow \mathrm{d}^4$ in $\mathrm{Mn}^{2+}$ and $\mathrm{d}^6 \rightarrow \mathrm{d}^5$ in $\mathrm{Fe}^{2+}$