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Question #45045

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The Question

Consider the following transformations of iron pentacarbonyl ( $\ce{Fe(CO)5}$ ), a trigonal bipyramidal ( $d^8$ ) complex:

Reaction I: $\ce{Fe(CO)5 + PPh3 ->[\Delta] [Fe(CO)4(PPh3)]}$ (Complex A) + $\ce{CO}$ Reaction II: $\ce{Fe(CO)5 + 2Na(Hg) ->[\text{THF}] Na2[Fe(CO)4]}$ (Complex B) + $\ce{CO}$

Which of the following statements correctly identifies the structural geometries, the dominant electronic effects dictating these geometries, and the relative C-O IR stretching frequencies ( $\nu_{CO}$ ) of Complexes A and B?

Options

A.
In A, \ce{PPh3} is axial because equatorial positions offer superior $\pi$ -back-bonding for CO; B is tetrahedral ( $sp^3$ ) with a lower $\nu_{CO}$ than A due to enhanced synergic donation from the $d^{10}$ center.
B.
In A, \ce{PPh3} is equatorial to minimize steric repulsion (VSEPR); B is square planar ( $dsp^2$ ) to satisfy the EAN rule, with a higher $\nu_{CO}$ than A due to the strong field nature of CO.
C.
In A, \ce{PPh3} is axial because it is a strong $\pi$ -acceptor; B is tetrahedral ( $sp^3$ ) but exhibits a higher $\nu_{CO}$ than A because the $-2$ charge contracts the metal $d$ -orbitals, reducing back-bonding.
D.
In A, \ce{PPh3} is equatorial because axial positions provide superior $\pi$ -back-bonding for CO; B is square planar ( $dsp^2$ ) due to Jahn-Teller distortion of the $d^{10}$ center, with a lower $\nu_{CO}$ than A.

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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

Hybridization and Geometry|Synergic Bonding in Carbonyls|EAN Rule (Sidgwick's Rule)

Why This Question Is Tricky

Common Mistakes

This problem weaponizes the conflict between VSEPR theory and Molecular Orbital Theory (Synergic Bonding), alongside a classic hybridization trap.

Students often blindly apply VSEPR, placing bulky ligands like $\ce{PPh3}$ in the equatorial position of a TBP complex to minimize steric hindrance.

The hidden variable is that synergic $\pi$ -back-bonding is thermodynamically dominant; equatorial positions offer superior $\pi$ -overlap, reserving them for strong $\pi$ -acceptors like CO.

Furthermore, students associate strong-field CO ligands with $dsp^2$ square planar geometries, forgetting that a $d^{10}$ center ( $\ce{Fe^{-II}}$ ) lacks empty $d$ -orbitals, forcing an $sp^3$ tetrahedral geometry.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Organic Chemistry – Some Basic Principles and Techniques - Class 11 Part 2
Concept	Hybridization and Geometry
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free