Before You Begin

Question #44130

Difficulty: 8/10 | 🟢 Free

The Question

An actinoid oxide, \ce{M_xO_y} (Compound \textbf{P}), exhibits an average oxidation state of $+5.33$ for the metal \ce{M}. Upon fluorination with excess \ce{F2(g)} at $700\text{ K}$ , \textbf{P} is completely converted into a volatile, diamagnetic, octahedral fluoride, \ce{MF6} (Compound \textbf{Q}). Compound \textbf{Q} is known to be a covalent molecule rather than an ionic solid.

Which of the following correctly identifies the actual oxidation states of \ce{M} in \textbf{P}, and the primary quantum mechanical reason for the stability of the $+6$ oxidation state in \textbf{Q}?

Options

A.
$+4$ and $+6$ ; Indirect relativistic expansion of $5f$ orbitals, allowing their participation in strong covalent \ce{M-F} bonds.
B.
$+4$ and $+6$ ; Relativistic contraction of $5f$ orbitals, which shields the nuclear charge and stabilizes the \ce{M^{6+}} ionic lattice.
C.
$+5$ and $+6$ ; High electronegativity of fluorine overcoming the inert pair effect of the $7s$ electrons.
D.
$+5.33$ (delocalized); The small size of the \ce{F^-} ion minimizes steric hindrance around the central \ce{M} atom.

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

Average vs. Actual Oxidation State|Actinoid Oxidation State Variety|Stability of High O.S. Fluorides/Oxides

Why This Question Is Tricky

Common Mistakes

This question tests the distinction between mathematical average oxidation states and actual discrete oxidation states in mixed-valence compounds (e.g., \ce{U3O8}).

The primary trap lies in the misapplication of 'Relativistic Contraction'.

Students often memorize that relativistic effects cause contraction (like the inert pair effect in post-transition metals).

However, they fail to realize that while $s$ and $p$ orbitals contract, $d$ and $f$ orbitals undergo indirect relativistic expansion.

This expansion is the exact reason why early actinoids (like U, Np, Pu) can utilize their $5f$ orbitals for covalent bonding to achieve unusually high oxidation states, unlike their lanthanoid counterparts where the $4f$ orbitals are deeply buried and chemically inert.

Furthermore, assuming high oxidation state fluorides are stabilized by 'ionic lattice energy' is a classic Fajans' rule violation.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Redox Reactions - Class 11 Part 2
Concept	Average vs. Actual Oxidation State
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free