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Question #44901

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The Question

When metallic Platinum and Palladium are treated with excess $F_2(g)$ under rigorous conditions, Pt yields a volatile, dark red hexafluoride ( $PtF_6$ ), whereas Pd yields only a tetrafluoride ( $PdF_4$ ). Similarly, Tungsten hexafluoride ( $WF_6$ ) exhibits greater thermodynamic stability than Molybdenum hexafluoride ( $MoF_6$ ) at 298 K. Given that the atomic radii of 4d and 5d congeners are nearly identical due to the lanthanoid contraction, which of the following best explains the anomalous stabilization of the +6 oxidation state in 5d transition metals?

Options

A.
The lanthanoid contraction increases $Z_{eff}$ in 5d metals, inducing relativistic expansion of the 5d orbitals which facilitates superior spatial overlap and stronger covalent bonding with fluorine's 2p orbitals.
B.
The post-lanthanoid radii equality significantly lowers the successive ionization enthalpies of 5d metals compared to 4d metals, thermodynamically favoring the formation of the $M^{6+}$ cation.
C.
The 4d orbitals possess more radial nodes than 5d orbitals, which sterically hinders the approach of six fluorine atoms, restricting 4d metals to lower coordination numbers and oxidation states.
D.
Fluorine acts as a strong-field ligand for 5d metals but a weak-field ligand for 4d metals, maximizing the Crystal Field Stabilization Energy (CFSE) for $PtF_6$ and $WF_6$ in octahedral geometries.

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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

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

Why This Question Is Tricky

Common Mistakes

The question tests the student's ability to reconcile the Lanthanoid Contraction with the stability of high oxidation states in 5d metals.

A common trap is assuming that similar radii (Mo vs W) imply similar chemical reactivity or that 5d metals have lower ionization energies.

The 'hidden variable' is the Relativistic Effect: while the overall atomic radius is constrained, the 5d orbitals specifically expand.

This leads to exceptionally strong covalent bonds with electronegative ligands (F, O) that thermodynamically drive the formation of high oxidation states despite higher ionization energies.

Distractor D also traps students who blindly apply Coordination Chemistry concepts without checking the d-electron count ( $d^0$ has no CFSE).

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	D and F Block Elements - Class 12
Concept	Stability of High O.S. Fluorides/Oxides
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free