Question

The complex, $[Pt(Py)(NH_3)Br\,Cl]$ will have how many geometrical isomers?

• A. 4
• B. 0
• C. 2
• D. 3

Answer 1

The Correct Option is D

 The given complex is \([Pt(Py)(NH_3)BrCl]\). To determine the number of geometrical isomers, let's analyze the structure of the complex.

This is a square planar complex of platinum (Pt) with four different ligands: pyridine (Py), ammonia (NH₃), bromide (Br), and chloride (Cl). In square planar complexes, geometrical isomerism is common due to the different possible arrangements of ligands around the metal center.

Since the complex has four different ligands, it can exhibit geometrical isomerism. Here is a step-by-step approach to find the number of geometrical isomers:

1. Consider \(Py\) and \(NH_3\) bonded to the platinum as adjacent to each other and opposite to each other. Similar treatment is done with \(Br\) and \(Cl\).
2. The possible arrangements can be:
   • Py at 90° to NH₃, Br at 90° to Cl (py-Cl, NH₃-Br).
   • Py at 180° to NH₃, Br at 180° to Cl (py-NH₃, Br-Cl).
   • Py at 90° to NH₃, Br at 180° between py-Cl (py-Br, Cl at 180°).
3. For square planar complexes with four different ligands, you can have three unique geometrical isomers based on these permutations.

Below are representations of these isomers:

Geometrical Isomer	Description
I	Pyridine adjacent to ammonia, bromide adjacent to chloride.
II	Pyridine opposite to ammonia, bromide opposite to chloride.
III	Pyridine adjacent to bromide, chloride opposite to ammonia.

Therefore, the complex \([Pt(Py)(NH_3)BrCl]\) can have a total of 3 geometrical isomers.

Correct Answer: 3