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Question #44247

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The Question

A $2.0\text{ mol}$ sample of the coordination compound \ce{K3[Fe(SCN)6]} is cooled to absolute zero ( $0\text{ K}$ ). Magnetic susceptibility measurements indicate that the central metal ion retains its high-spin state. Structural analysis reveals a 'frozen disorder' in the crystal lattice: each of the six ambidentate thiocyanate ligands can independently coordinate via either the sulfur (\ce{-SCN}) or nitrogen (\ce{-NCS}) atom. Assuming both linkage modes are energetically degenerate, and that the spin microstates of the metal ion remain completely degenerate at $0\text{ K}$ (negligible zero-field splitting), what is the total residual entropy of the $2.0\text{ mol}$ sample at $0\text{ K}$ , and what is the spin-only magnetic moment ( $\mu_{eff}$ ) of the complex?

Options

A.
$14 R \ln 2 + 2 R \ln 3$ and $\sqrt{35}\text{ BM}$
B.
$12 R \ln 2$ and $\sqrt{35}\text{ BM}$
C.
$14 R \ln 2$ and $\sqrt{3}\text{ BM}$
D.
$0$ and $\sqrt{35}\text{ BM}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Third Law Application|Ambidentate Linkage Isomerism

Why This Question Is Tricky

Common Mistakes

The question tests the statistical thermodynamic definition of entropy ( $S = k_B \ln W$ ) applied to a non-ideal crystal at $0\text{ K}$ .

Students often blindly apply the Third Law of Thermodynamics ( $S=0$ at $0\text{ K}$ ) or only account for the configurational 'frozen disorder' of the ambidentate linkage isomers ( $12 R \ln 2$ ).

The 'Hidden Variable' is the magnetic residual entropy arising from the degenerate spin microstates ( $2S+1$ ) of the high-spin \ce{Fe^{3+}} ion, which must be multiplied with the configurational microstates to find the true ground-state degeneracy.

Distractor C perfectly traps students who incorrectly assume a low-spin $d^5$ configuration ( $\mu = \sqrt{3}\text{ BM}$ , $W_{mag} = 2$ , yielding $14 R \ln 2$ ).

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Thermodynamics - Class 11 Part 1
Concept	Third Law Application
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free