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Step-by-Step Explanation
Step 1: Define Second Ionization Enthalpy
The second ionization enthalpy is the amount of energy required to remove one electron from a unipositive ion (M+) of an element, forming a dipositive ion (M2+). Symbolically:
$ \text{M}^+ \rightarrow \text{M}^{2+} + e^- $
Step 2: Write Electronic Configurations of the Elements
Consider the electronic configurations of the ground state atoms of V, Cr, Mn, and Fe (atomic numbers 23, 24, 25, and 26 respectively):
V (Z = 23): [Ar] 3d3 4s2
Cr (Z = 24): [Ar] 3d5 4s1
Mn (Z = 25): [Ar] 3d5 4s2
Fe (Z = 26): [Ar] 3d6 4s2
Step 3: Consider the First Ionization
When these atoms lose one electron, the electron is generally removed from the 4s orbital first (which is at a relatively higher energy level in the ionized state):
V+: [Ar] 3d3 4s1 (or effectively [Ar] 3d4)
Cr+: [Ar] 3d5 (highly stable, half-filled d5)
Mn+: [Ar] 3d5 4s1 (or effectively [Ar] 3d6)
Fe+: [Ar] 3d6 4s1 (or effectively [Ar] 3d7)
Notably, Cr+ adopts the particularly stable half-filled configuration 3d5.
Step 4: Rationalize the Second Ionization Enthalpy
Removing the second electron from a unipositive ion is influenced by the electronic stability of that ion. Since Cr+ has a perfectly half-filled d5 configuration, removing an electron from this stable arrangement requires significantly higher energy. Thus:
$ \text{Cr}^+ (3d^5) \rightarrow \text{Cr}^{2+} (3d^4) + e^- $
This step necessitates a large amount of energy because the half-filled d-orbital is unusually stable. Other ions do not possess this exact level of stability in their unipositive states.
Step 5: Conclusion
Hence, among V, Cr, Mn, and Fe, the highest second ionization enthalpy is expected for chromium (Cr) due to the extra stability of its half-filled 3d5 configuration in Cr+.
