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Question #44704

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The Question

A heavy transition metal $M$ in its ground state possesses exactly 20 electrons in orbitals characterized by the quantum number constraint $(n+l) = 7$ . Upon reaction with excess \ce{O2}, $M$ forms a highly volatile, diamagnetic, tetrahedral oxide $X$ . The lighter 3d congener of $M$ cannot achieve this maximum oxidation state. Which of the following correctly identifies $X$ , the number of radial nodes in the valence d-orbitals of $M$ , and the primary thermodynamic justification for the anomalous stability of the maximum oxidation state in $M$ compared to its 3d congener?

Options

A.
\ce{OsO4} ; 2 ; Poor shielding by 4f electrons increases $Z_{eff}$ and ionization energies, but the spatially diffuse 5d orbitals form exceptionally strong covalent bonds with oxygen, overcoming the ionization penalty.
B.
\ce{OsO4} ; 2 ; Poor shielding by 4f electrons decreases $Z_{eff}$ , which anomalously lowers the successive ionization energies required to reach the +8 oxidation state.
C.
\ce{RuO4} ; 3 ; The 4d orbitals experience minimal shielding from core electrons, allowing for complete ionization of the valence shell without a significant energy barrier.
D.
\ce{WO3} ; 2 ; Lanthanide contraction reduces the atomic radius, significantly increasing the lattice enthalpy of the oxide and stabilizing the +6 state.

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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

Quantum Number Constraints|d-block Variable Oxidation States|d-orbital contraction

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of quantum number constraints and the thermodynamic anomalies of heavy transition metals.

The 'Hidden Variable' is the misconception that higher oxidation state stability in 5d elements is due to lower ionization energies.

Students often falsely assume that poor shielding makes electrons easier to remove.

In reality, poor 4f shielding increases Z_eff and ionization energies.

The true driver for the stability of high oxidation states like Os(VIII) is the immense bond enthalpy generated by the highly effective overlap of diffuse 5d orbitals with ligand orbitals, which overcomes the ionization penalty.

Quick Facts About This Question

Question Information

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