Question

In $\text{Fe}{\left(\text{CO}\right)}_5$, the $\text{Fe}-\text{C}$ bond possess:
A. $\pi$ character only
B. both $\pi$ and $\sigma$ characters
C. ionic character
D. $\sigma$ character only

Answer 1

Hint: The $\text{Fe}{\left(\text{CO}\right)}_5$ compound forms metal-carbonyl bonds with each other. The structure of this compound is the same as others but the bonding is different. This bonding involves the d-orbitals and antibonding orbitals of the metal atom.

Complete step by step answer:
- In $\text{Fe}{\left(\text{CO}\right)}_5$, there are same ligands which is $\text{CO}$ so, $\text{Fe}{\left(\text{CO}\right)}_5$ is a homoleptic carbonyl compound. Such compounds are formed by most of the transition metals or d-block elements and they have simple and well defined structures. The shape of the compound is trigonal bi-pyramidal. The IUPAC name of the compound is Pentacarbonyliron $\left(0\right)$. The structure is

Let us discuss the special type of bonding in $\text{Fe}{\left(\text{CO}\right)}_5$:

The special type of bonding is named as ‘synergic bonding’. The ligand $\left(\text{CO}\right)$ donates its lone pair of electrons to the vacant orbitals of the central metal atom which is an iron atom and forms the sigma-bond with it. As the iron atom also possesses some electrons in its d-orbitals. The configuration of iron is $1{\text{s}}^2\text{2}{\text{s}}^2\text{2}{\text{p}}^6\text{3}{\text{s}}^2\text{3}{\text{p}}^6\text{4}{\text{s}}^2\text{3}{\text{d}}^6$ or $\begin{array}{ccccc}\uparrow \downarrow & \uparrow & \uparrow & \uparrow & \uparrow \end{array}$.
It back donates these electrons to the molecular orbitals of the ligand forming a $\pi$-bond. The $\text{M}-\text{C}$ $\pi$ bond is formed by the donation of electrons from a filled d-orbital of metal into the vacant antibonding ${\pi }^{\ast }$ orbital of carbon monoxide. The bonding looks like

This creates a synergic effect between the metal to ligand which strengthens the bond between $\text{CO}$ and the metal. The metal-carbon bond in metal carbonyls possesses both $\sigma$ and $\pi$ character.
So, the correct answer is “Option B”.

Note: One important point to keep in mind is that the metal atom donates its electron pairs to the antibonding MO of $\text{CO}$, so the $\text{CO}$ bond is weakened by this synergic bonding which leads to a larger $\text{CO}$ bond length in the complex.