Understanding Third Ionization Enthalpy

Ionization enthalpy is the energy required to remove an electron from an isolated gaseous atom. The third ionization enthalpy (IE₃) is the energy needed to remove the third electron. For transition metals, this depends on electronic configuration stability.

Step 1: Write Electronic Configurations

First, write the ground state electronic configurations of the given elements:

• Mn (Z=25): [Ar] 4s² 3d⁵
• Fe (Z=26): [Ar] 4s² 3d⁶
• Co (Z=27): [Ar] 4s² 3d⁷
• Ni (Z=28): [Ar] 4s² 3d⁸

Step 2: Determine Configurations After Two Ionizations

After removing two electrons (IE₁ and IE₂), the configurations become:

• Mn²⁺: [Ar] 3d⁵ (half-filled, stable)
• Fe²⁺: [Ar] 3d⁶
• Co²⁺: [Ar] 3d⁷
• Ni²⁺: [Ar] 3d⁸

Step 3: Analyze Stability for Third Ionization

Removing a third electron (IE₃) from these ions:

• Mn²⁺ has a stable half-filled d⁵ configuration. Removing an electron disrupts this stability, so IE₃ is very high.
• Fe²⁺ has d⁶. Removing an electron gives Fe³⁺ which is d⁵ (half-filled, stable). This requires less energy compared to Mn.
• Co²⁺ (d⁷) and Ni²⁺ (d⁸) do not achieve extra stability upon losing another electron, so their IE₃ values are higher than Fe but lower than Mn.

Thus, Fe has the minimum third ionization enthalpy because IE₃ leads to a stable half-filled configuration.

Final Answer

Fe (Iron) has the minimum third ionization enthalpy among the given options.

Related Topics

• Ionization Energy Trends in Periodic Table
• Electronic Configuration and Stability
• Transition Metal Chemistry
• Hund's Rule and Half-filled Stability

Key Formulae and Theory

Ionization enthalpy for the n^th electron: $\mathrm{IE}n=E(M^{(n-1)}+)-E(M^{\left(n\right)}+)$

Half-filled (d⁵) and fully-filled (d¹⁰) subshells have extra stability due to symmetric distribution and exchange energy.