Iridium Complex Reaction with O₂

The reaction of trans-[IrX(CO)(PPh₃)₂] with O₂ is a classic example of oxidative addition. In this reaction, the square planar $\text{Ir(I)}$ complex, which has a $d^8$ electron configuration, reacts with $\text{O}_2$ to form an octahedral $\text{Ir(III)}$ complex, which has a $d^6$ electron configuration.

The general reaction can be represented as:

$\text{trans-[IrX(CO)(PPh}_3\text{)}_2\text{]} + \text{O}_2 \rightarrow \text{[IrX(CO)(O}_2\text{)(PPh}_3\text{)}_2\text{]}$

The rate of oxidative addition reactions is significantly influenced by the electron density at the metal center. Electron-rich metal centers tend to undergo oxidative addition more readily.

Halide Ligand Effect on Rate

The halide ligand ($\text{X}$) attached to the iridium center influences the electron density of the iridium atom through its inductive effect. Halogens are electronegative elements and withdraw electron density from the metal center. The extent of electron withdrawal depends on the electronegativity of the halogen:

• Fluorine ($\text{F}$) is the most electronegative halogen.
• Chlorine ($\text{Cl}$) is less electronegative than Fluorine.
• Bromine ($\text{Br}$) is less electronegative than Chlorine.
• Iodine ($\text{I}$) is the least electronegative halogen among the series $\text{F}$, $\text{Cl}$, $\text{Br}$, $\text{I}$.

Therefore, the order of electron-withdrawing strength of the halide ligands is $\text{F} > \text{Cl} > \text{Br} > \text{I}$.

A more electron-withdrawing ligand reduces the electron density on the iridium center, making it less favorable for oxidative addition. Conversely, a less electron-withdrawing ligand keeps the iridium center more electron-rich, facilitating oxidative addition.

Determining the Rate Order

Based on the electron-donating ability (relative to electron withdrawal) or the reverse order of electronegativity, the ligands can be arranged:

• Iodine ($\text{I}$) withdraws the least electron density, leaving Iridium most electron-rich.
• Bromine ($\text{Br}$) withdraws more than I, but less than Cl and F.
• Chlorine ($\text{Cl}$) withdraws more than Br and I, but less than F.
• Fluorine ($\text{F}$) withdraws the most electron density, leaving Iridium least electron-rich.

Since a more electron-rich metal center favors oxidative addition, the reaction rate will be highest when $\text{X}$ is $\text{I}$ and lowest when $\text{X}$ is $\text{F}$.

Thus, the correct order of reaction rate with respect to the halide $\text{X}$ is in the order of decreasing electron withdrawal (or increasing electron density on Ir):

$\text{Rate}(\text{I}) > \text{Rate}(\text{Br}) > \text{Rate}(\text{Cl}) > \text{Rate}(\text{F})$

This corresponds to the order:

$\text{I} > \text{Br} > \text{Cl} > \text{F}$