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Question #44273

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The Question

In the context of Valence Bond Theory (VBT) and orbital energetics, consider the formation of the diamagnetic liquid $\ce{[Fe(CO)_5]}$ at 298 K. For the isolated $\ce{Fe}$ atom, the $4s$ subshell is filled before the $3d$ subshell according to the $(n+l)$ rule. However, upon complexation with $\ce{CO}$ , the $4s$ electrons are relocated.

Which of the following correctly identifies the hybridization of $\ce{Fe}$ in $\ce{[Fe(CO)_5]}$ and the primary thermodynamic/electronic driving forces for this specific electron relocation?

Options

A.
$\ce{dsp^3}$ ; In the complexed state, the $3d$ orbitals are lower in energy than the $4s$ orbital. The strong ligand field of $\ce{CO}$ forces the $4s$ electrons into the $3d$ subshell ( $\ce{3d^8}$ ). This vacates the $4s$ and one $3d$ orbital for $\sigma$ -bonding, while the filled $3d$ orbitals maximize $M \rightarrow C$ $\pi$ -backbonding.
B.
$\ce{sp^3d}$ ; The $4s$ electrons are excited to the $4d$ subshell to avoid the high pairing energy of the $3d$ subshell. The $\ce{CO}$ ligands donate into the vacated $4s$ , $4p$ , and one $4d$ orbital, while the $\ce{3d^6}$ core remains paramagnetic, which is quenched by the synergic effect.
C.
$\ce{d^2sp^3}$ ; The $4s$ electrons are completely transferred to the $\pi^*$ antibonding orbitals of the $\ce{CO}$ ligands, oxidizing the metal to $\ce{Fe^{2+}}$ . The remaining $\ce{3d^6}$ electrons pair up, vacating two $3d$ orbitals to accommodate six $\ce{CO}$ ligands in a polymeric structure.
D.
$\ce{sp^3d}$ ; The $(n+l)$ rule is strictly maintained. The $4s$ electrons cannot transfer to the higher energy $3d$ subshell. Instead, the $\ce{CO}$ ligands donate electron pairs into the empty $4p$ and $4d$ orbitals, maintaining the $\ce{3d^6 4s^2}$ configuration and minimizing interelectronic repulsion.

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Explanation

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Mechanism_Trap

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Concepts Behind This Question

Energy of Subshells (n+l rule)|Valence Bond Theory Hybridization|Synergic Bonding in Carbonyls

Why This Question Is Tricky

Common Mistakes

Students often rigidly apply the $(n+l)$ rule to determine the valence configuration of transition metals, assuming the $4s$ orbital remains lower in energy than $3d$ and retains its electrons during complexation.

They fail to account for two hidden variables: 1) In the complexed state, $3d$ drops below $4s$ in energy. 2) The strong synergic back-bonding ( $M \rightarrow C$ $\pi^*$ ) and strong ligand field of CO force the $4s$ electrons to pair up inside the $3d$ subshell.

This alters the expected hybridization from a hypothetical $sp^3d$ (using outer $d$ orbitals) to $dsp^3$ (using inner $d$ orbitals), fundamentally changing the geometric and magnetic interpretation of the complex.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Structure of Atom - Class 11 Part 1
Concept	Energy of Subshells (n+l rule)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free