Question

Why second ionization enthalpies of chromium and copper are exceptionally higher than those of their neighbouring elements?

Hint:

Breaking a $d^5$ or $d^{10}$ stable core always results in a massive spike in ionization energy.

Answer 1

Step 1: Concept
Ionization enthalpy depends heavily on the thermodynamic stability of the specific electronic configuration from which the electron is being removed.

Step 2: Meaning
Exactly half-filled ($d^5$) and fully filled ($d^{10}$) subshells possess immense exchange energy and spherical symmetry, making them exceptionally stable.

Step 3: Analysis
The ground state configurations are $Cr$ ($3d^5 4s^1$) and $Cu$ ($3d^{10} 4s^1$).
When they lose their first electron (the 4s electron),
they form the univalent ions $Cr^+$ ($3d^5$) and $Cu^+$ ($3d^{10}$).
The second ionization enthalpy ($IE_2$) involves removing an electron from these resulting $Cr^+$ and $Cu^+$ ions.

Step 4: Conclusion
Because the second electron must be extracted from a highly stable, symmetrical half-filled ($3d^5$) or completely filled ($3d^{10}$) core, an exceptionally large amount of energy is required compared to neighboring elements lacking this extra stability.

Final Answer: After losing the first electron, Cr and Cu form $Cr^+$ ($3d^5$) and $Cu^+$ ($3d^{10}$), which have highly stable half-filled and fully filled d-subshells, respectively. Removing a second electron disrupts this immense stability, requiring exceptionally high energy.