Why second ionization enthalpies of chromium and copper are exceptionally higher than those of their neighbouring elements?
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Solution and Explanation
Concept: Ionization enthalpy is the energy required to remove an electron from an atom or ion. It depends strongly on the stability of the electronic configuration, especially the presence of half-filled or fully-filled subshells.
Step 1: Write electronic configurations. \[ \text{Cr} (Z=24): [Ar]\,3d^5\,4s^1 \] \[ \text{Cu} (Z=29): [Ar]\,3d^{10}\,4s^1 \]
Step 2: Remove the first electron (formation of $M^+$). The first ionization removes the $4s$ electron: \[ \text{Cr}^+: [Ar]\,3d^5 \quad (\text{half-filled}) \] \[ \text{Cu}^+: [Ar]\,3d^{10} \quad (\text{fully-filled}) \]
Step 3: Analyze second ionization enthalpy ($IE_2$). The second electron must now be removed from the $d$-subshell:
- In Cr$^+$: removal from stable half-filled $3d^5$
- In Cu$^+$: removal from highly stable fully-filled $3d^{10}$
Both configurations are exceptionally stable due to symmetry and high exchange energy.
Step 4: Conclusion. Breaking these stable half-filled and fully-filled configurations requires extra energy. Hence, the second ionization enthalpy is unusually high for both Cr and Cu.
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