# What is bridge bonding? Does it need backbonding?

Jul 20, 2017

Bridge bonding is basically advanced $\pi$ bonding. A common scenario is in a complex with two identical metals.

For example, consider $\left({\eta}_{5} - {\stackrel{\textcolor{b l u e}{- 1}}{\text{C"_5"H"_5))_2stackrel(color(blue)(+1))("Fe"_2)stackrel(color(blue)(0))(("CO}}}_{4}\right)$: Take the righthand isomer as an example.

The middle $\text{CO}$ ligands are bridge bonding with the iron atoms. When one depicts just the bridge bonding via the bonding molecular orbital, it looks approximately like this: The $\pi$-bonding molecular orbitals of the $\text{CO}$ ligands $\boldsymbol{\pi}$ bond with two $\pi$-type $d$ orbitals at the same time. So, if the $\text{CO}$ and the $\text{Fe}$ were on the $x y$ plane, then:

• The $\text{Fe}$ use their $3 {d}_{x y}$ atomic orbitals to bond with the $\text{CO}$ ${\pi}_{2 p x}$ molecular orbitals if $x$ is horizontal.
• The $\text{Fe}$ use their $3 {d}_{y z}$ atomic orbitals to bond with the $\text{CO}$ ${\pi}_{2 p z}$ molecular orbitals (where $z$ is into the screen and $y$ is vertical).

The metal $d$ orbitals in the above molecular orbital contribution are then of the bonding type with respect to the ligand $\pi$ bonding orbitals, since their signs match up.

In this case, we cannot see how the backbonding occurs yet; we have to draw the antibonding molecular orbital depiction to see this: This is possible, since molecular orbitals really have contributions from multiple phase combinations to certain extents. In other words, both of the above depictions occur at the same time, similar to resonance structures.

This weakens the $\text{C"-"O}$ triple bond (within the ligand itself), and strengthens the $\text{M"-"CO}$ interaction (between the metal and the ligand directly).

You can see that both metals could do that at the same time, so a relatively low oxidation state on them favors this (less than $+ 2$), since they will not get overly electropositive (which is destabilizing).

In this case, $\text{Fe}$ is in a $+ 1$ oxidation state, so this is favorable.