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Question #44902

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The Question

Consider the reaction of Potassium perrhenate ( $\ce{KReO4}$ ) with an excess of Lithium aluminium hydride ( $\ce{LiAlH4}$ ) in dry diethyl ether at $298\text{ K}$ , followed by an aqueous $\ce{KOH}$ workup. A diamagnetic, homoleptic transition metal complex $\mathbf{Z}$ is isolated as the major rhenium-containing product. Which of the following correctly identifies the oxidation state of Rhenium in $\mathbf{Z}$ , its coordination geometry, and the primary reason the metal center resists reduction to a lower oxidation state?

Options

A.
O.S. = +3; Octahedral; The high crystal field splitting energy ( $\Delta_o$ ) of the $5d$ series stabilizes the $t_{2g}^4$ configuration.
B.
O.S. = +7; Tricapped trigonal prismatic; The formation of a highly covalent, 18-electron closed-shell complex kinetically and thermodynamically stabilizes the high oxidation state.
C.
O.S. = +4; Tetrahedral; The strong covalent character of the $\ce{Re-O}$ bonds prevents complete cleavage, resulting in a mixed oxo-hydrido complex.
D.
O.S. = -1; Square antiprismatic; $\ce{LiAlH4}$ acts as a powerful reducing agent, driving the metal to a negative oxidation state due to the strong $\pi$ -acceptor properties of the hydride ligands.

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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

Stability of High O.S. Fluorides/Oxides|Oxidation State Traps|Reduction Selectivity (Hydrides)

Why This Question Is Tricky

Common Mistakes

This problem tests a profound exception to standard redox intuition.

Students reflexively associate strong reducing agents like LiAlH4 with the reduction of high oxidation state metals (e.g., Cr(VI) to Cr(III)).

The trap assumes the metal must lower its oxidation state.

However, with certain 4d/5d metals, hydrides can act purely as coordinating ligands.

The hidden variable is the 18-electron rule: the formation of the d0 [ReH9]2- complex perfectly satisfies the 18-electron rule, conferring immense stability that halts any actual electron transfer (reduction) to the metal center.

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