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Question #44705

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The Question

Consider the square planar complexes $\ce{[Cu(NH3)4]^2+}$ (Complex P) and $\ce{[Ni(CN)4]^2-}$ (Complex Q) at 298 K. In classical Valence Bond Theory (VBT), the square planar geometry of Complex P is rationalized by the promotion of its unpaired electron to a higher energy orbital, $O_{VBT}$ , to facilitate $dsp^2$ hybridization. However, modern Crystal Field Theory (CFT) and EPR spectroscopy reveal that the unpaired electron in Complex P actually resides in the orbital $O_{CFT}$ , with the geometry arising from extreme Jahn-Teller distortion. For Complex Q, let $O_{HOMO}$ be the highest occupied d-orbital according to the standard square planar CFT splitting model. Which of the following options correctly identifies the set of values (Radial Nodes, Angular Nodes, $|m_l|$ ) for the orbitals $O_{VBT}$ , $O_{CFT}$ , and $O_{HOMO}$ , respectively? (Assume the z-axis is the principal axis perpendicular to the square planar field).

Options

A.
$O_{VBT}: (2, 1, 0)$ ; $O_{CFT}: (0, 2, 2)$ ; $O_{HOMO}: (0, 2, 2)$
B.
$O_{VBT}: (1, 1, 1)$ ; $O_{CFT}: (0, 2, 0)$ ; $O_{HOMO}: (0, 2, 2)$
C.
$O_{VBT}: (2, 1, 0)$ ; $O_{CFT}: (0, 2, 2)$ ; $O_{HOMO}: (0, 2, 0)$
D.
$O_{VBT}: (3, 0, 0)$ ; $O_{CFT}: (1, 2, 2)$ ; $O_{HOMO}: (1, 2, 2)$

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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

Quantum Number Constraints|Valence Bond Theory Hybridization|Geometric Isomerism Requirements

Why This Question Is Tricky

Common Mistakes

The question tests the 'magnetic moment paradox' and the failure of VBT for $d^9$ square planar complexes.

Students often rote-memorize $dsp^2$ hybridization without realizing it implies an easily oxidizable promoted electron ( $4p_z$ ), which contradicts the stability of Cu(II).

CFT resolves this via Jahn-Teller distortion, keeping the electron in the $3d_{x^2-y^2}$ orbital.

Furthermore, it tests the exact CFT splitting energy order for square planar complexes, specifically identifying $d_{xy}$ as the HOMO for $d^8$ systems, trapping students who might incorrectly assume $d_{z^2}$ is higher than $d_{xy}$ .

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