The third ionization enthalpy is minimum for:

The third ionization enthalpy is the energy needed to pull out the third electron from a gaseous neutral atom. Symbolically we write the three successive steps as

$$\text{M}(g) \xrightarrow{\Delta_iH_1} \text{M}^+(g) + e^-,$$

$$\text{M}^+(g) \xrightarrow{\Delta_iH_2} \text{M}^{2+}(g) + e^-,$$

$$\text{M}^{2+}(g) \xrightarrow{\Delta_iH_3} \text{M}^{3+}(g) + e^-.$$

Here $$\Delta_iH_3$$ is the quantity we are comparing for Co, Fe, Ni and Mn.

Electrons are always removed first from the higher-energy $$4s$$ subshell and only afterwards from the $$3d$$ subshell. So, for any first-row transition element, the first two ionizations take away the two $$4s$$ electrons, and the third ionization removes the first $$3d$$ electron.

Let us write the ground-state electronic configurations (in the compact form using the argon core $$[\,\text{Ar}\,]$$):

Co : $$[\,\text{Ar}\,]\,3d^7\,4s^2$$

Fe : $$[\,\text{Ar}\,]\,3d^6\,4s^2$$

Ni : $$[\,\text{Ar}\,]\,3d^8\,4s^2$$

Mn : $$[\,\text{Ar}\,]\,3d^5\,4s^2$$

Now we remove electrons step by step.

Cobalt

First electron: $$[\,\text{Ar}\,]\,3d^7\,4s^2 \longrightarrow [\,\text{Ar}\,]\,3d^7\,4s^1$$

Second electron: $$[\,\text{Ar}\,]\,3d^7\,4s^1 \longrightarrow [\,\text{Ar}\,]\,3d^7$$

Third electron (from $$3d$$): $$[\,\text{Ar}\,]\,3d^7 \longrightarrow [\,\text{Ar}\,]\,3d^6$$

The resulting $$\text{Co}^{3+}$$ ion has configuration $$3d^6$$, which is neither half-filled ($$3d^5$$) nor fully filled ($$3d^{10}$$). Therefore no extra stability is gained, so $$\Delta_iH_3$$ is not especially low.

Iron

First electron: $$[\,\text{Ar}\,]\,3d^6\,4s^2 \longrightarrow [\,\text{Ar}\,]\,3d^6\,4s^1$$

Second electron: $$[\,\text{Ar}\,]\,3d^6\,4s^1 \longrightarrow [\,\text{Ar}\,]\,3d^6$$

Third electron (from $$3d$$): $$[\,\text{Ar}\,]\,3d^6 \longrightarrow [\,\text{Ar}\,]\,3d^5$$

Now the $$\text{Fe}^{3+}$$ ion possesses the exactly half-filled configuration $$3d^5$$. A half-filled subshell enjoys extra exchange-energy stabilization; hence the energy needed to reach this state (the third ionization enthalpy) is markedly lower than for neighbouring elements.

Nickel

First electron: $$[\,\text{Ar}\,]\,3d^8\,4s^2 \longrightarrow [\,\text{Ar}\,]\,3d^8\,4s^1$$

Second electron: $$[\,\text{Ar}\,]\,3d^8\,4s^1 \longrightarrow [\,\text{Ar}\,]\,3d^8$$

Third electron: $$[\,\text{Ar}\,]\,3d^8 \longrightarrow [\,\text{Ar}\,]\,3d^7$$

The $$3d^7$$ configuration is not especially stable, so $$\Delta_iH_3$$ remains comparatively high.

Manganese

First electron: $$[\,\text{Ar}\,]\,3d^5\,4s^2 \longrightarrow [\,\text{Ar}\,]\,3d^5\,4s^1$$

Second electron: $$[\,\text{Ar}\,]\,3d^5\,4s^1 \longrightarrow [\,\text{Ar}\,]\,3d^5$$

Third electron: $$[\,\text{Ar}\,]\,3d^5 \longrightarrow [\,\text{Ar}\,]\,3d^4$$

Here we destroy a half-filled subshell, so the third ionization enthalpy is actually very high, the opposite of what we want.

Among the four elements considered, only iron attains the extra-stable $$3d^5$$ arrangement after losing its third electron. This additional stability lowers the required energy, making $$\Delta_iH_3$$ minimal for iron.

Hence, the correct answer is Option B.