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Question #44164

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The Question

Consider the two linkage isomers of a neutral square planar $\ce{Pt(II)}$ complex dissolved in a dilute, non-polar solvent at 298 K: Complex A is $trans-\ce{[Pt(NH_3)_2(NCS)_2]}$ and Complex B is $trans-\ce{[Pt(NH_3)_2(SCN)_2]}$ . Let $\mu_A$ and $\mu_B$ be their respective observed time-averaged dipole moments in solution. Which of the following correctly describes the relationship between their dipole moments?

Options

A.
$\mu_A = 0, \mu_B > 0$
B.
$\mu_A = 0, \mu_B = 0$
C.
$\mu_A > 0, \mu_B = 0$
D.
$\mu_A > 0, \mu_B > 0$

Check Your Answer

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

Dipole Moment Vector Addition|Ambidentate Linkage Isomerism

Why This Question Is Tricky

Common Mistakes

Students typically memorize the heuristic that ' $trans$ isomers of square planar complexes always have a zero dipole moment' (e.g., $trans-\ce{[PtCl_2(NH_3)_2]}$ ).

This question introduces a 'Hidden Variable': the internal geometry of the ambidentate ligand.

Students fail to account for the fact that linkage isomerism fundamentally redefines the bond angle at the metal-ligand interface.

The N-bound $\ce{NCS^-}$ is linear, preserving the center of inversion, whereas the S-bound $\ce{SCN^-}$ is bent.

This bent geometry, combined with free rotation in solution, generates non-centrosymmetric conformers, resulting in a net non-zero dipole moment.

This invalidates the 'fixed scalar ligand' assumption.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Chemical Bonding and Molecular Structure - Class 11 Part 1
Concept	Dipole Moment Vector Addition
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free