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Question #44284

Difficulty: 8/10 | 🟢 Free

The Question

Using Slater's rules, the effective nuclear charge (ZeffZ_{eff}) for the valence electrons of \ce{Ca^{2+}} (3p3p) and \ce{Cu+} (3d3d) are calculated as 8.758.75 and 7.857.85, 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 3d3d orbital lacks radial nodes) and the poor shielding of 3d3d electrons by 3s3s and 3p3p 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.

    \ceK+>\ceCa2+>\ceCu+>\ceZn2+\ce{K+} > \ce{Ca^{2+}} > \ce{Cu+} > \ce{Zn^{2+}}

  • B.

    \ceCu+>\ceZn2+>\ceK+>\ceCa2+\ce{Cu+} > \ce{Zn^{2+}} > \ce{K+} > \ce{Ca^{2+}}

  • C.

    \ceK+>\ceCu+>\ceCa2+>\ceZn2+\ce{K+} > \ce{Cu+} > \ce{Ca^{2+}} > \ce{Zn^{2+}}

  • D.

    \ceK+>\ceCa2+>\ceZn2+>\ceCu+\ce{K+} > \ce{Ca^{2+}} > \ce{Zn^{2+}} > \ce{Cu+}

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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

SubjectChemistry
ChapterClassification of Elements and Periodicity in Properties - Class 11 Part 1
ConceptEffective Nuclear Charge (Zeff)
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree