Multi-electron Orbital Energy And Penetration
In a many-electron atom, the energy of the 4s orbital is lower than that of 3d in the ground state, yet electrons are removed from 4s before 3d during ionization. Which combination of factors best explains both observations?
Select the correct option:
Solution
4s has greater penetration to the nucleus lowering its energy, but 3d contraction upon ionization makes 3d lower energy in the cation
This question probes a nuanced and often misunderstood aspect of atomic orbital energies in multi-electron systems. In the neutral atom, 4s electrons have a higher probability density near the nucleus (greater penetration) compared to 3d electrons, which experience stronger shielding. This higher penetration lowers the effective energy of 4s below 3d, explaining why 4s fills before 3d per the Aufbau principle. However, upon ionization to form a cation (e.g., Cu^2+, Fe^2+), the removal of one or more electrons reduces electron-electron repulsion and allows the nuclear charge to contract the 3d orbitals more effectively. In transition metal cations, 3d orbitals become lower in energy than 4s, so the remaining ionizations preferentially remove 4s electrons. This is confirmed experimentally by the electron configurations of transition metal ions (e.g., Fe^2+ is [Ar] 3d^6, not [Ar] 3d^4 4s^2). Option (A) incorrectly attributes removal order to principal quantum number alone, ignoring the dynamic change in orbital energies upon ionization. Option (C) confuses Hund's rule (which governs spin multiplicity within a subshell) with the filling order between subshells. Option (D) is factually wrong; 4s and 3d are not degenerate in neutral atoms. This represents a JEE Advanced-level question on the interplay between penetration, shielding, and orbital energy in multi-electron systems, directly tested in several past Advanced papers. Plausibility check: Fe^2+ has configuration [Ar] 3d^6 (not [Ar] 3d^4 4s^2), confirming that 4s is lost first in ionization.
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About This Question
- Subject
- chemistry
- Chapter
- atomic structure
- Topic
- multi-electron orbital energy and penetration
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
4s has greater penetration to the nucleus lowering its energy, but 3d contraction upon ionization makes 3d lower energy in the cation
This question probes a nuanced and often misunderstood aspect of atomic orbital energies in multi-electron systems. In the neutral atom, 4s electrons have a higher probability density near the nucleus (greater penetration) compared to 3d electrons, which experience stronger shielding. This higher penetration lowers the effective energy of 4s below 3d, explaining why 4s fills before 3d per the Aufbau principle. However, upon ionization to form a cation (e.g., Cu^2+, Fe^2+), the removal of one or more electrons reduces electron-electron repulsion and allows the nuclear charge to contract the 3d orbitals more effectively. In transition metal cations, 3d orbitals become lower in energy than 4s, so the remaining ionizations preferentially remove 4s electrons. This is confirmed experimentally by the electron configurations of transition metal ions (e.g., Fe^2+ is [Ar] 3d^6, not [Ar] 3d^4 4s^2). Option (A) incorrectly attributes removal order to principal quantum number alone, ignoring the dynamic change in orbital energies upon ionization. Option (C) confuses Hund's rule (which governs spin multiplicity within a subshell) with the filling order between subshells. Option (D) is factually wrong; 4s and 3d are not degenerate in neutral atoms. This represents a JEE Advanced-level question on the interplay between penetration, shielding, and orbital energy in multi-electron systems, directly tested in several past Advanced papers. Plausibility check: Fe^2+ has configuration [Ar] 3d^6 (not [Ar] 3d^4 4s^2), confirming that 4s is lost first in ionization.
This hard difficulty chemistry question is from the chapter atomic structure, covering the topic of multi-electron orbital energy and penetration. It appeared in the 2025 exam.
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