1. What is third ionization enthalpy?

   The third ionization enthalpy is the energy required for: $M^{2+} \rightarrow M^{3+} + e^-$

   So we must compare how easily the species $\text{Mn}^{2+}, \text{Fe}^{2+}, \text{Co}^{2+}, \text{Ni}^{2+}$ lose one more electron.

2. Write the electronic configurations

   Neutral atoms:

   • $\text{Mn} : [Ar]3d^5 4s^2$
   • $\text{Fe} : [Ar]3d^6 4s^2$
   • $\text{Co} : [Ar]3d^7 4s^2$
   • $\text{Ni} : [Ar]3d^8 4s^2$

   After losing two electrons (the two $4s$ electrons), we get:

   • $\text{Mn}^{2+} : [Ar]3d^5$
   • $\text{Fe}^{2+} : [Ar]3d^6$
   • $\text{Co}^{2+} : [Ar]3d^7$
   • $\text{Ni}^{2+} : [Ar]3d^8$

3. Now examine removal of the third electron

   Third ionization means removing one electron from these $3d$ configurations.

   • For $\text{Mn}^{2+} = 3d^5$: $\text{Mn}^{2+} \rightarrow \text{Mn}^{3+} : 3d^5 \rightarrow 3d^4$ Since $3d^5$ is a half-filled stable configuration, removing one electron is difficult. So third ionization enthalpy is high.

   • For $\text{Fe}^{2+} = 3d^6$: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} : 3d^6 \rightarrow 3d^5$ This produces the stable half-filled configuration $3d^5$. Therefore removal is comparatively easier, so the third ionization enthalpy is low.

   • For $\text{Co}^{2+} = 3d^7$: $3d^7 \rightarrow 3d^6$ No special extra stability is gained, so it is not minimum.

   • For $\text{Ni}^{2+} = 3d^8$: $3d^8 \rightarrow 3d^7$ Again, no special half-filled or fully-filled stability is obtained.

4. Conclusion from stability

   The easiest third electron removal occurs for Fe because: $\text{Fe}^{2+}([Ar]3d^6) \rightarrow \text{Fe}^{3+}([Ar]3d^5)$ and $3d^5$ is especially stable.

5. Option check

   • A: Ni — No
   • B: Co — No
   • C: Mn — No, actually high because $\text{Mn}^{2+}$ is already half-filled
   • D: Fe — Yes

Therefore, the third ionization enthalpy is minimum for Fe.