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Question #44315

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The Question

Consider the complex \ce{[Bi(EDTA)]^-} formed in an aqueous medium at 298 K. Assuming EDTA acts strictly as a hexadentate ligand (coordination number = 6), the central \ce{Bi^{3+}} ion possesses 6 bonding pairs and 1 valence lone pair. Unlike \ce{XeF_6}, which exhibits a distorted octahedral geometry due to a stereochemically active lone pair, the primary coordination sphere of \ce{[Bi(EDTA)]^-} (ignoring ligand bite-angle constraints) is effectively octahedral. Which of the following correctly identifies the orbital housing the lone pair in \ce{[Bi(EDTA)]^-} and the primary physical phenomenon responsible for its stereochemical inactivity?

Options

A.
Pure $6s$ orbital ; Relativistic contraction of the $6s$ orbital (Inert Pair Effect) prevents hybridization.
B.
$sp^3d^3$ hybrid orbital ; Steric crowding of the hexadentate ligand forces the lone pair into a symmetric position.
C.
Pure $6p$ orbital ; Strong ligand field of EDTA causes spin-pairing in a higher energy unhybridized orbital.
D.
$sp^3d^2$ hybrid orbital ; Lone pair delocalization into the $\pi^*$ anti-bonding orbitals of EDTA via back-bonding.

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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

Inert Pair Effect|Ligand Denticity Variations|Tetrahedral Splitting Patterns

Why This Question Is Tricky

Common Mistakes

This question tests the 'Inert Pair Effect' and its impact on hybridization and VSEPR theory in heavy p-block elements.

Students often blindly apply the VSEPR rule (7 domains = $sp^3d^3$ = distorted octahedral) without accounting for the relativistic contraction of the $6s$ orbital in Period 6 elements like Bi(III).

The trap lies in assuming that steric crowding (Option B) or complex coordination effects (Options C/D) are responsible for the regular geometry, whereas the true hidden variable is the energetic unsuitability of the $6s$ orbital for hybridization, rendering the lone pair stereochemically inactive in a pure spherical $s$ -orbital.

Quick Facts About This Question

Question Information

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