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Question #44867

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The Question

Zirconium ( $Zr$ ) and Hafnium ( $Hf$ ) are often considered 'chemical twins' due to the lanthanide contraction, resulting in nearly identical ionic radii for $Zr^{4+}$ ( $72$ pm) and $Hf^{4+}$ ( $71$ pm). However, their highest oxidation state oxides, $ZrO_2$ and $HfO_2$ , exhibit distinct thermodynamic properties at $298$ K and standard pressure. Which of the following statements correctly identifies and explains this difference?

Options

A.
$HfO_2$ has a higher melting point and greater lattice stability than $ZrO_2$ because the poor shielding of $4f$ electrons and relativistic effects increase the ionic character of the $Hf-O$ bond.
B.
$ZrO_2$ has a higher melting point than $HfO_2$ because the absence of $4f$ electrons in $Zr$ allows for stronger covalent $d\pi-p\pi$ orbital overlap with oxygen.
C.
$HfO_2$ is more easily reduced to $Hf_2O_3$ than $ZrO_2$ to $Zr_2O_3$ because relativistic stabilization of the $6s$ orbital induces a strong inert pair effect in Hafnium.
D.
Both oxides possess identical melting points and lattice energies, as the lanthanide contraction perfectly equalizes their effective nuclear charges and bonding characteristics.

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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

Post-Lanthanoid Radii Equality|Variable Oxidation State Stability|Stability of High O.S. Fluorides/Oxides

Why This Question Is Tricky

Common Mistakes

This problem exploits the 'Lanthanide Contraction' trap.

Students are taught that $Zr$ and $Hf$ are 'chemical twins' with identical radii, leading them to assume identical chemical and physical properties (Option D).

The hidden variable is the relativistic effect and poor shielding of $4f$ electrons in $Hf$ , which alters bonding energetics and increases the ionic character and lattice stability of $HfO_2$ .

It also tests the misconception that the inert pair effect applies to $d$ -block elements (Option C), whereas heavier transition metals actually have more stable higher oxidation states.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	D and F Block Elements - Class 12
Concept	Post-Lanthanoid Radii Equality
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free