Geometrical Isomers of Square Planar Rhodium Complex

To determine the total number of geometrical isomers for the complex $[\text{RhCl}(\text{CO})(\text{PPh}_3)(\text{NH}_3)]$:

1. Identify the Geometry of the Complex:

   • Rhodium ($\text{Rh}$) in this complex has an oxidation state of $+1$, giving it a $d^8$ electronic configuration.
   • $d^8$ transition metal complexes with a coordination number of 4 prefer a square planar geometry due to significant crystal field stabilization.

2. Formulation as a General Complex:

   • The given complex is of the type $[Mabcd]$, where $M = \text{Rh(I)}$ and $a, b, c, d$ represent four different monodentate ligands:
     $a = \text{Cl}^-$
     $b = \text{CO}$
     $c = \text{PPh}_3$
     $d = \text{NH}_3$

3. Counting Geometrical Isomers:

   • For a square planar complex $[Mabcd]$, geometrical isomers are formed by placing different ligands directly opposite (trans) to a reference ligand.
   • Fixing ligand $a = \text{Cl}^-$, the ligand occupying the position trans to it can be chosen in 3 distinct ways:
     • Isomer 1: $\text{Cl}^-$ is trans to $\text{CO}$ (consequently, $\text{PPh}_3$ is trans to $\text{NH}_3$).
     • Isomer 2: $\text{Cl}^-$ is trans to $\text{PPh}_3$ (consequently, $\text{CO}$ is trans to $\text{NH}_3$).
     • Isomer 3: $\text{Cl}^-$ is trans to $\text{NH}_3$ (consequently, $\text{CO}$ is trans to $\text{PPh}_3$).

Therefore, the total number of geometrical isomers for $[\text{RhCl}(\text{CO})(\text{PPh}_3)(\text{NH}_3)]$ is $3$.