Before You Begin

Question #44306

Difficulty: 8/10 | 🟢 Free

The Question

Transition metals exhibit their highest oxidation states in compounds with highly electronegative elements. For instance, Tungsten forms $\ce{WF6}$ , and Osmium forms $\ce{OsF6}$ . However, while Manganese forms the oxide $\ce{Mn2O7}$ (+7 oxidation state) and Osmium forms $\ce{OsO4}$ (+8 oxidation state), the corresponding fluorides $\ce{MnF7}$ and $\ce{OsF8}$ are unknown. Based on Pauling's electronegativity scale, bond energy polarity, and structural thermodynamics at 298 K, which of the following statements correctly explains this apparent contradiction?

Options

A.
Fluorine's high electronegativity induces a large ionic resonance energy that maximizes the $M-F$ bond enthalpy, compensating for high ionization energies. However, its inability to form multiple bonds necessitates high coordination numbers, leading to severe F-F steric repulsions that prevent the formation of $\ce{MnF7}$ or $\ce{OsF8}$ .
B.
The high electronegativity of fluorine ensures purely covalent $M-F$ bonds with maximum orbital overlap. The absence of $\ce{MnF7}$ is due to the thermodynamic instability of the $\ce{Mn^{7+}}$ cation, which cannot be stabilized by any ligand, including oxygen, without undergoing spontaneous reduction.
C.
Fluorine stabilizes high oxidation states primarily because the $F-F$ bond dissociation energy is exceptionally high, making the formation of metal fluorides thermodynamically favorable. Oxygen surpasses fluorine only because of its higher electron affinity.
D.
The stabilization is driven by the high lattice enthalpy of molecular fluorides like $\ce{WF6}$ and $\ce{OsF6}$ . Oxygen forms $\ce{Mn2O7}$ because the $\ce{O^{2-}}$ ion has a smaller ionic radius than $\ce{F^-}$ , allowing higher coordination without steric hindrance.

Check Your Answer

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

Electronegativity Scales|Bond Energy vs Bond Length|Stability of High O.S. Fluorides/Oxides

Why This Question Is Tricky

Common Mistakes

The question tests the 'Chemical Exception' where students over-rely on the heuristic that Fluorine, being the most electronegative element, must always form the highest oxidation state compounds.

The hidden variable is the steric constraint (F-F repulsion) and coordination number limits.

While Fluorine's electronegativity provides the necessary bond enthalpy (via ionic resonance energy) to compensate for high ionization energies, its inability to form multiple bonds means high oxidation states require impossibly high coordination numbers, leading to destabilizing steric crowding.

This explains why Oxygen, despite being less electronegative, stabilizes the absolute highest oxidation states (+7, +8) via multiple bonding.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Classification of Elements and Periodicity in Properties - Class 11 Part 1
Concept	Electronegativity Scales
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free