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Question #44277

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following transition metal ions in an octahedral coordination environment (CN = 6) at 298 K: $Y^{3+}$ , $Zr^{4+}$ , $Lu^{3+}$ , and $Hf^{4+}$ .

These ions form two distinct isoelectronic series. Based on the interplay of effective nuclear charge ( $Z_{eff}$ ) and the Lanthanoid Contraction effect, which of the following represents the correct strictly decreasing order of their ionic radii?

Options

A.
$Y^{3+} > Lu^{3+} > Zr^{4+} > Hf^{4+}$
B.
$Lu^{3+} > Y^{3+} > Hf^{4+} > Zr^{4+}$
C.
$Lu^{3+} > Hf^{4+} > Y^{3+} > Zr^{4+}$
D.
$Y^{3+} > Zr^{4+} > Lu^{3+} > Hf^{4+}$

Check Your Answer

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

Isoelectronic Energy Ordering|Periodic Properties of Lanthanides|Post-Lanthanoid Radii Equality

Why This Question Is Tricky

Common Mistakes

Students typically rely on the heuristic that ionic size strictly increases down a group due to the addition of a principal quantum shell.

While they may know that the Lanthanoid Contraction causes 4d and 5d radii to 'converge' (e.g., Zr and Hf being roughly equal), this problem tests the hidden exception: at the end of the f-block, the contraction actually overcompensates.

This makes the 6th period ions (Lu3+, Hf4+) strictly smaller than their 5th period counterparts (Y3+, Zr4+).

Furthermore, it tests the ability to superimpose this vertical periodic exception over the horizontal isoelectronic trend.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Structure of Atom - Class 11 Part 1
Concept	Isoelectronic Energy Ordering
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free