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Question #44859

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following reaction sequence involving Cobalt complexes under specific thermal and solvent constraints:

I. Anhydrous $\ce{CoCl2}$ reacts with excess $\ce{KSCN}$ in acetone to form a deep blue complex $\mathbf{P}$ . II. An aqueous solution of $\ce{[Co(NH3)5(H2O)]^3+}$ reacts with $\ce{KSCN}$ at $0^\circ\text{C}$ to yield complex $\mathbf{Q}$ as the major kinetic product. III. Upon gentle heating, complex $\mathbf{Q}$ completely isomerizes to the thermodynamically stable complex $\mathbf{R}$ .

Identify the correct set of parameters for complexes $\mathbf{P}$ , $\mathbf{Q}$ , and $\mathbf{R}$ respectively, in the order: [Hybridization of central metal, Donor atom of the ambidentate ligand].

Options

A.
$\mathbf{P}$ : ( $sp^3$ , N); $\mathbf{Q}$ : ( $d^2sp^3$ , S); $\mathbf{R}$ : ( $d^2sp^3$ , N)
B.
$\mathbf{P}$ : ( $dsp^2$ , S); $\mathbf{Q}$ : ( $sp^3d^2$ , N); $\mathbf{R}$ : ( $d^2sp^3$ , S)
C.
$\mathbf{P}$ : ( $sp^3$ , S); $\mathbf{Q}$ : ( $d^2sp^3$ , N); $\mathbf{R}$ : ( $sp^3d^2$ , S)
D.
$\mathbf{P}$ : ( $sp^3$ , N); $\mathbf{Q}$ : ( $sp^3d^2$ , S); $\mathbf{R}$ : ( $d^2sp^3$ , N)

Check Your Answer

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

Variable Oxidation State Stability|Coordination Isomerism|Ambidentate Linkage Isomerism|Valence Bond Theory Hybridization

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of Linkage Isomerism (Kinetic vs.

Thermodynamic control) and Valence Bond Theory/Crystal Field Theory.

The primary trap lies in Complex $\mathbf{Q}$ .

Students often memorize that S-donor ligands (like $\ce{SCN^-}$ ) are weak-field and assume they force outer-orbital ( $sp^3d^2$ ) hybridization in octahedral complexes (Option D).

However, they overlook the 'Hidden Variable': the average ligand field strength.

With five strong-field $\ce{NH3}$ ligands, the overall $\Delta_o$ remains large enough to pair the $d^6$ electrons of $\ce{Co^3+}$ , keeping the complex inner-orbital ( $d^2sp^3$ ) regardless of the linkage isomer.

Additionally, it rigorously tests the HSAB principle where the softer S-atom acts as a better nucleophile (kinetic control), but the harder N-atom forms a stronger bond with the hard $\ce{Co^3+}$ acid (thermodynamic control).

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	D and F Block Elements - Class 12
Concept	Variable Oxidation State Stability
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free