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Question #44906

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The Question

Consider two coordination compounds, $\ce{P}$ and $\ce{Q}$ , with empirical formulas $\ce{NiCl2.3en}$ (en = ethylenediamine) and $\ce{PtCl4.2HCl}$ , respectively. When $1.0\text{ mol}$ of $\ce{P}$ is treated with excess $\ce{AgNO3(aq)}$ at $298\text{ K}$ , $2.0\text{ mol}$ of $\ce{AgCl(s)}$ precipitates. When $1.0\text{ mol}$ of $\ce{Q}$ is treated with excess $\ce{AgNO3(aq)}$ under identical conditions, no precipitate is formed. Based on Werner's theory and Crystal Field Theory (CFT), identify the correct set of parameters for the complex ions of $\ce{P}$ and $\ce{Q}$ . The parameters are given in the order: [Hybridization, Spin-only magnetic moment ( $\mu_s$ ), Total number of stereoisomers].

Options

A.
$\ce{P}$ : $sp^3d^2$ , $2.83\text{ BM}$ , $2$ ; $\ce{Q}$ : $d^2sp^3$ , $0\text{ BM}$ , $1$
B.
$\ce{P}$ : $d^2sp^3$ , $0\text{ BM}$ , $2$ ; $\ce{Q}$ : $sp^3d^2$ , $4.90\text{ BM}$ , $1$
C.
$\ce{P}$ : $sp^3d^2$ , $2.83\text{ BM}$ , $1$ ; $\ce{Q}$ : $sp^3d^2$ , $4.90\text{ BM}$ , $2$
D.
$\ce{P}$ : $d^2sp^3$ , $0\text{ BM}$ , $1$ ; $\ce{Q}$ : $d^2sp^3$ , $0\text{ BM}$ , $2$

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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

Werner’s Theory Postulates|Valence Bond Theory Hybridization

Why This Question Is Tricky

Common Mistakes

The question tests the boundaries of Valence Bond Theory (VBT) and Crystal Field Theory (CFT) against common heuristic rules.

Trap 1: Students often memorize 'strong field ligand = inner orbital ( $d^2sp^3$ )'.

For $\ce{Ni^{2+}}$ ( $3d^8$ ), inner orbital hybridization is impossible without unfeasible electron promotion, so it must be $sp^3d^2$ despite the strong field $\ce{en}$ ligand.

Trap 2: Students memorize 'halides = weak field = outer orbital ( $sp^3d^2$ )'.

For $\ce{Pt^{4+}}$ ( $5d^6$ ), the high effective nuclear charge and larger $5d$ orbitals result in a massive $\Delta_o$ , making all ligands act as strong field.

Thus, it forms a low-spin $d^2sp^3$ complex.

Additionally, it tests the stereochemical visualization of tris-chelate complexes ( $\ce{M(AA)3}$ ), which are always chiral.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Coordination Compounds - Class 12
Concept	Werner’s Theory Postulates
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free