Meaning of 3rd ionisation enthalpy

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

So, we must compare the stability of the $M^{2+}$ ions of the given transition metals. The more stable $M^{2+}$ is, the more difficult it is to remove the third electron, and hence the higher the 3rd ionisation enthalpy.

Write electronic configurations

Atomic configurations:

• $V : [Ar]3d^3 4s^2$
• $Cr : [Ar]3d^5 4s^1$
• $Mn : [Ar]3d^5 4s^2$
• $Fe : [Ar]3d^6 4s^2$

First, remove $4s$ electrons to form $M^{2+}$:

• $V^{2+} : [Ar]3d^3$
• $Cr^{2+} : [Ar]3d^4$
• $Mn^{2+} : [Ar]3d^5$
• $Fe^{2+} : [Ar]3d^6$

Compare stability of $M^{2+}$ ions

Among these, $Mn^{2+}$ has configuration: $[Ar]3d^5$ which is a half-filled d-subshell, and hence exceptionally stable.

Therefore, removing one more electron from $Mn^{2+}$ to form $Mn^{3+}$ is especially difficult.

Hence, 3rd ionisation enthalpy

Since: $Mn^{2+}([Ar]3d^5) \rightarrow Mn^{3+}([Ar]3d^4) + e^-$ destroys the extra stability of the half-filled $d^5$ configuration, this requires the maximum energy among the given options.

Check options

• A: $Mn$ — highest 3rd ionisation enthalpy due to very stable $Mn^{2+}(3d^5)$ ✔️
• B: $Fe$ — $Fe^{2+}$ is $3d^6$, not as stable as half-filled $3d^5$ ✘
• C: $Cr$ — $Cr^{2+}$ is $3d^4$, less stable than $Mn^{2+}$ ✘
• D: $V$ — $V^{2+}$ is $3d^3$, less stable than $Mn^{2+}$ ✘

Final answer

The transition metal with the highest 3rd ionisation enthalpy is: $\boxed{Mn}$