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Step-by-Step Explanation
Step 1: Identify the Species for Third Ionization
The third ionization enthalpy refers to the energy required to remove the third electron from a neutral atom, effectively resulting in an M3+ ion from the neutral element M. In this case, to determine which metal has the highest third ionization enthalpy among V, Cr, Mn, and Fe, we focus on the electronic configurations of their M2+ ions (since removing the third electron goes from M2+ to M3+).
Step 2: Write Configurations of M2+ Ions
The electron configurations of the M2+ ions for the given elements are (removing two electrons from the neutral atom):
V (Z=23): neutral V = [Ar] 3d3 4s2
V2+ = [Ar] 3d3
Cr (Z=24): neutral Cr = [Ar] 3d5 4s1
Cr2+ = [Ar] 3d4
Mn (Z=25): neutral Mn = [Ar] 3d5 4s2
Mn2+ = [Ar] 3d5
Fe (Z=26): neutral Fe = [Ar] 3d6 4s2
Fe2+ = [Ar] 3d6
Step 3: Stability of the Half-Filled Configuration
The next electron to be removed (the third electron from the neutral atom) will come from the 3d orbitals in these M2+ ions. The stability of the 3d orbitals greatly influences the ionization enthalpy. A half-filled d5 configuration is particularly stable because it maximizes exchange energy and maintains symmetrical distribution of electrons.
Among these ions, Mn2+ = [Ar] 3d5 is the only one with a half-filled d5 configuration. Removing an electron from this especially stable arrangement requires significantly more energy, resulting in a higher third ionization enthalpy.
Step 4: Conclusion
Because Mn2+ has the half-filled d5 electronic configuration, manganese (Z = 25) exhibits the highest third ionization enthalpy among the given elements.
Final Answer
Manganese (Z = 25) has the highest third ionization enthalpy.
