Identify Wilkinson's catalyst

Wilkinson's catalyst is: $[\text{RhCl}(\text{PPh}_3)_3]$ where $\text{PPh}_3$ is triphenylphosphine.

Determine oxidation state of Rh

Let the oxidation state of Rh be $x$.

• $\text{Cl}^-$ contributes $-1$
• Each $\text{PPh}_3$ is a neutral ligand
• Overall complex is neutral

So, $x + (-1) = 0$ $x = +1$

Thus, the metal ion is $\text{Rh}^{+1}$.

Find the electronic configuration of $\text{Rh}^{+1}$

Rhodium $(Z=45)$ has ground-state configuration approximately: $\text{Rh}: [\text{Kr}]\,4d^8 5s^1$

Removing one electron gives: $\text{Rh}^{+1}: [\text{Kr}]\,4d^8$

Determine coordination number

In $[\text{RhCl}(\text{PPh}_3)_3]$:

• $1$ chloride ligand
• $3$ phosphine ligands

Total coordination number $=4$.

Geometry for a $d^8$ metal ion with coordination number 4

For $4d$ and $5d$ transition metals with $d^8$ configuration, the preferred geometry is generally square planar due to larger crystal field splitting.

Hence Wilkinson's catalyst is square planar.

Hybridization

Square planar complexes correspond to: $dsp^2$

Evaluate options

• A: $sp^3d$, trigonal bipyramidal $\rightarrow$ incorrect
• B: $sp^3$, tetrahedral $\rightarrow$ incorrect
• C: $dsp^2$, square planar $\rightarrow$ correct
• D: $d^2sp^3$, octahedral $\rightarrow$ incorrect

Final answer

Wilkinson's catalyst has:

• Hybridization: $dsp^2$
• Shape: square planar