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Question #44329

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The Question

The phenomenon of lanthanoid contraction is classically illustrated by the nearly identical atomic radii of Zirconium ( $\ce{Zr}$ , $Z=40$ , $r_{\ce{Zr}} \approx 160\text{ pm}$ ) and Hafnium ( $\ce{Hf}$ , $Z=72$ , $r_{\ce{Hf}} \approx 159\text{ pm}$ ). However, when analyzing their chemistry in the maximum $+4$ oxidation state, a deviation from this 'identical size' heuristic is observed.

Which of the following correctly identifies the relationship between their ionic radii ( $r_{\ce{M^{4+}}}$ ) for a given coordination number, and the primary quantum-mechanical rationale?

Options

A.
$r_{\ce{Zr^{4+}}} < r_{\ce{Hf^{4+}}}$ ; The removal of valence electrons eliminates the screening effect of the $6s$ shell, allowing the $n=5$ principal quantum shell of $\ce{Hf^{4+}}$ to dictate a larger ionic size.
B.
$r_{\ce{Zr^{4+}}} \approx r_{\ce{Hf^{4+}}}$ ; The lanthanoid contraction perfectly offsets the addition of the $n=5$ shell, maintaining identical spatial extensions for both the atomic and $+4$ ionic states.
C.
$r_{\ce{Hf^{4+}}} < r_{\ce{Zr^{4+}}}$ ; Poor shielding by the $4f^{14}$ electrons drastically increases the effective nuclear charge ( $Z_{eff}$ ) experienced by the $5s^2 5p^6$ core of $\ce{Hf^{4+}}$ , contracting it to a size slightly smaller than the $4s^2 4p^6$ core of $\ce{Zr^{4+}}$ .
D.
$r_{\ce{Hf^{4+}}} \ll r_{\ce{Zr^{4+}}}$ ; The $4f^{14}$ subshell in $\ce{Hf^{4+}}$ undergoes severe relativistic expansion, forcing the outermost $5d$ electrons to collapse into the nucleus.

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Screening Effect in Inner Shells|Lanthanoid Contraction cause|d-block Variable Oxidation States

Why This Question Is Tricky

Common Mistakes

The trap exploits the student's reliance on the 'identical size' heuristic for 4d and 5d elements caused by the lanthanoid contraction.

While $r_{\ce{Zr}} \approx r_{\ce{Hf}}$ is true for neutral atoms, students incorrectly assume this equality holds (or that the $n=5$ shell makes Hf larger) for their ions.

The hidden variable is the dominant role of $Z_{eff}$ in high oxidation states.

Stripping the valence electrons exposes the core to the poorly shielded $+32$ nuclear charge increment, contracting the $5s^2 5p^6$ shell of $\ce{Hf^{4+}}$ to a radius strictly smaller than the $4s^2 4p^6$ shell of $\ce{Zr^{4+}}$ .

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Classification of Elements and Periodicity in Properties - Class 11 Part 1
Concept	Screening Effect in Inner Shells
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free