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Question #44284
The Question
Using Slater's rules, the effective nuclear charge () for the valence electrons of \ce{Ca^{2+}} () and \ce{Cu+} () are calculated as and , respectively. Based on this, one might incorrectly predict that \ce{Cu+} is larger than \ce{Ca^{2+}}. However, accounting for the actual radial distribution functions (where the orbital lacks radial nodes) and the poor shielding of electrons by and electrons, what is the true decreasing order of the ionic radii for \ce{K+}, \ce{Ca^{2+}}, \ce{Cu+}, and \ce{Zn^{2+}}?
Options
- A.
- B.
- C.
- D.
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Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Numerical_Precision
Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.
Concepts Behind This Question
Effective Nuclear Charge (Zeff)|Atomic vs Ionic Radii|Isoelectronic Energy Ordering
Why This Question Is Tricky
Common Mistakes
The question tests the limitations of Slater's rules when comparing s/p-block ions with d-block ions.
Students often blindly apply Slater's rules, which assign a shielding constant of 1.0 to all inner electrons (including 3s and 3p) for a 3d electron.
This incorrectly predicts a lower Z_eff for Cu+ compared to Ca2+, leading to the false conclusion that Cu+ is larger (Option C).
The 'Hidden Variable' is the actual radial distribution function: the 3d orbital has zero radial nodes and its maximum probability density is closer to the nucleus than the 3p orbital.
Consequently, 3s and 3p electrons shield 3d electrons very poorly, resulting in a massively higher actual Z_eff for Cu+ (~13.5) than Ca2+ (~8.0), making Cu+ significantly smaller.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Classification of Elements and Periodicity in Properties - Class 11 Part 1 |
| Concept | Effective Nuclear Charge (Zeff) |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |