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Question #44591

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The Question

An aqueous solution of $\ce{CuSO4}$ is treated separately with excess $\ce{KCN}$ (Reaction 1) and concentrated $\ce{HCl}$ (Reaction 2) to form the major complex anions $\mathbf{P}$ and $\mathbf{Q}$ , respectively. A student incorrectly assumes both $\mathbf{P}$ and $\mathbf{Q}$ retain the $+2$ oxidation state of copper and applies standard Valence Bond Theory to predict their structures. Based on Crystal Field Theory and the actual oxidation states, which of the following correctly identifies the geometry and the $d$ -orbital splitting configuration of the central metal in $\mathbf{P}$ and $\mathbf{Q}$ ?

Options

A.
$\mathbf{P}$ : Perfect Tetrahedral, $e^4 t_2^6$ ; $\mathbf{Q}$ : Flattened Tetrahedral, $e^4 t_2^5$
B.
$\mathbf{P}$ : Square Planar, $e_g^4 a_{1g}^2 b_{2g}^2 b_{1g}^1$ ; $\mathbf{Q}$ : Perfect Tetrahedral, $e^4 t_2^5$
C.
$\mathbf{P}$ : Perfect Tetrahedral, $e^4 t_2^6$ ; $\mathbf{Q}$ : Square Planar, $e_g^4 a_{1g}^2 b_{2g}^2 b_{1g}^1$
D.
$\mathbf{P}$ : Flattened Tetrahedral, $e^4 t_2^5$ ; $\mathbf{Q}$ : Flattened Tetrahedral, $e^4 t_2^5$

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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

Tetrahedral Splitting Patterns|Valence Bond Theory Hybridization|Oxidation State Traps

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Oxidation State Trap' combined with 'Jahn-Teller Distortion' in coordination chemistry.

Students relying blindly on Valence Bond Theory (VBT) heuristics will assume $\ce{CN^-}$ (strong field) forces pairing in a $d^9$ system to give a $dsp^2$ Square Planar complex, and $\ce{Cl^-}$ (weak field) gives an $sp^3$ Perfect Tetrahedral complex.

This fails on two fronts: (1) $\ce{CN^-}$ reduces $\ce{Cu^{2+}}$ to $\ce{Cu^+}$ , changing the system to $d^{10}$ (Perfect Tetrahedral, no distortion). (2) The $d^9$ $\ce{[CuCl4]^{2-}}$ complex is orbitally degenerate ( $e^4 t_2^5$ ) and undergoes a Jahn-Teller distortion to a flattened tetrahedron, rendering the 'perfect' $sp^3$ assumption obsolete.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Coordination Compounds - Class 12
Concept	Tetrahedral Splitting Patterns
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free