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Q5.12:
Write all the geometrical isomers of ([Pt(NH_{3})(Br)(Cl)(py)]) and how many of these will exhibit optical isomers?

Solution :

Step 1: Analyzing the Coordination Entity

The complex is $[Pt(NH_{3})(Br)(Cl)(py)]$. The central metal ion is Platinum ($Pt^{2+}$), which possesses a coordination number of 4. Complexes with coordination number 4 and $d^8$ configuration (like $Pt^{2+}$) typically adopt a Square Planar Geometry. The general formula is $[Mabcd]$, where $a=NH_{3}$, $b=Br$, $c=Cl$, and $d=py$ (pyridine).

Step 2: Determining Geometrical Isomers

For a square planar complex of the type $[Mabcd]$, the number of geometrical isomers is 3. We fix one ligand (e.g., $NH_{3}$) at one position and vary the positions of the other three ligands ($Br, Cl, py$) relative to it.

  • Isomer 1: $NH_{3}$ and $Br$ are trans to each other. ($Cl$ and $py$ are trans).
  • Isomer 2: $NH_{3}$ and $Cl$ are trans to each other. ($Br$ and $py$ are trans).
  • Isomer 3: $NH_{3}$ and $py$ are trans to each other. ($Br$ and $Cl$ are trans).

Step 3: Visual Representation of Geometrical Isomers

NH₃ Br Cl py Isomer 1 (trans-Br) NH₃ Cl Br py Isomer 2 (trans-Cl) NH₃ py Br Cl Isomer 3 (trans-py)

Step 4: Analysis of Optical Isomerism

Optical isomerism requires the absence of a plane of symmetry (chirality). In square planar complexes ($[Mabcd]$):

  • The coordination sphere is strictly planar.
  • A square planar molecule possesses a molecular plane that acts as a plane of symmetry ($\sigma$).
  • Since a plane of symmetry exists for every configuration, the complex cannot be superimposable on its mirror image in a chiral sense.
  • [Per the principles of molecular symmetry], square planar complexes of the type $[Mabcd]$ are achiral.

Final Solution: The complex $[Pt(NH_{3})(Br)(Cl)(py)]$ exhibits 3 geometrical isomers. None of these isomers will exhibit optical isomerism because the square planar geometry inherently possesses a plane of symmetry.


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