Determine the coordination number and geometry

The complex is $[\mathrm{RhCl}(\mathrm{CO})(\mathrm{PPh}_3)(\mathrm{NH}_3)]$

It has 4 different monodentate ligands around Rh:

• $\mathrm{Cl}^-$
• $\mathrm{CO}$
• $\mathrm{PPh}_3$
• $\mathrm{NH}_3$

So the coordination number is $4$.

Possible geometries for coordination number 4

For a 4-coordinate complex, the common geometries are:

• tetrahedral
• square planar

Since Rh is a transition metal that commonly forms square planar complexes (especially for $d^8$ systems), this complex can show square planar arrangement. Also, with four different ligands, tetrahedral arrangement may give optical isomerism, but geometrical isomerism is associated with square planar arrangement here.

Count geometrical isomers in square planar case

For a square planar complex of type $[MABCD]$ where all four ligands are different, geometrical isomerism arises from different choices of which ligand is trans to which.

In a square plane, each ligand has exactly one ligand trans to it. Distinct isomers are obtained by choosing trans pairs.

Let the ligands be: $A=\mathrm{Cl},\quad B=\mathrm{CO},\quad C=\mathrm{PPh}_3,\quad D=\mathrm{NH}_3$

The possible distinct trans pairings are:

• $A$ trans $B$, and $C$ trans $D$
• $A$ trans $C$, and $B$ trans $D$
• $A$ trans $D$, and $B$ trans $C$

Thus, the total number of geometrical isomers is: $3$

Final answer

$\boxed{3}$

Comparison with stored correct answer

Stored correct answer = $3$

My derived answer also = $3$

Hence, they agree.