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Question #44288

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The Question

The Lanthanide Contraction, driven by the poor shielding of 4f4f electrons, profoundly alters the expected periodic trends in the 5d5d transition series. However, the magnitude of this anomaly is not static across the dd-block. Consider the highest oxidation state ions of Group 4 (Zr4+,Hf4+Zr^{4+}, Hf^{4+}), Group 5 (Nb5+,Ta5+Nb^{5+}, Ta^{5+}), and Group 6 (Mo6+,W6+Mo^{6+}, W^{6+}). Assuming a coordination number of 6 for all species, which of the following correctly represents the relationship between the ionic radii (rr) of these homologous pairs?

Options

  • A.

    r(Zr4+)>r(Hf4+)r(Zr^{4+}) > r(Hf^{4+}) ; r(Nb5+)≈r(Ta5+)r(Nb^{5+}) \approx r(Ta^{5+}) ; r(Mo6+)<r(W6+)r(Mo^{6+}) < r(W^{6+})

  • B.

    r(Zr4+)<r(Hf4+)r(Zr^{4+}) < r(Hf^{4+}) ; r(Nb5+)≈r(Ta5+)r(Nb^{5+}) \approx r(Ta^{5+}) ; r(Mo6+)>r(W6+)r(Mo^{6+}) > r(W^{6+})

  • C.

    r(Zr4+)≈r(Hf4+)r(Zr^{4+}) \approx r(Hf^{4+}) ; r(Nb5+)<r(Ta5+)r(Nb^{5+}) < r(Ta^{5+}) ; r(Mo6+)<r(W6+)r(Mo^{6+}) < r(W^{6+})

  • D.

    r(Zr4+)>r(Hf4+)r(Zr^{4+}) > r(Hf^{4+}) ; r(Nb5+)>r(Ta5+)r(Nb^{5+}) > r(Ta^{5+}) ; r(Mo6+)>r(W6+)r(Mo^{6+}) > r(W^{6+})

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Atomic vs Ionic Radii|Isoelectronic Energy Ordering|Quantum Number Constraints

Why This Question Is Tricky

Common Mistakes

Students commonly memorize the Lanthanide Contraction as a static rule, assuming that 4d and 5d homologous pairs are always identical in size (e.g., Zr≈HfZr \approx Hf, Mo≈WMo \approx W).

This problem tests the 'Hidden Variable' of differential contraction rates.

Because the 4d series has a lower baseline atomic number, the addition of protons across the period represents a larger fractional increase in ZeffZ_{eff} compared to the 5d series.

Consequently, the 4d series shrinks faster, creating a 'crossover' effect where the 5d ion starts smaller (Group 4), becomes equal (Group 5), and ends up larger (Group 6) than its 4d counterpart.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterClassification of Elements and Periodicity in Properties - Class 11 Part 1
ConceptAtomic vs Ionic Radii
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree