I'm finding this didn't work for square planar complexes with 2 ligands of one type and 2 of another, or 3 of one type and 1 of another, or 2 of one type and 1 of each of two others. Maybe there is some modification needed to the process I'm using so far, which is:
Imagine you have the square planar complex written in 2D with ligands at position a (top left), b (top right), c (bottom right) and d (bottom left). If you have a ligand at a and a ligand at b, you cannot switch them and be left with the same isomer; however, ligands at a are freely rotatable with those at c, and ligands at b are freely rotatable with those at d. So, if by performing these free rotations you can go from one proposed isomer to another, then they are really the same isomer and you should only write down one of them (you could write down either).
Before we move onto octahedrals (where there are also problems) can you explain why this method won't work with square planar complexes that don't specifically have 4 different ligands?
An example is: (PPh3)2PtCl2, draw all isomers. (Pt is central of course.)
The mark scheme suggests we've got a=PPh3, b=Cl, c=Cl, d=PPh3 (ligands ordered clockwise from top left) as 1 isomer, but then the other is a=PPh3, b=PPh3, c=Cl, d=Cl (ligands ordered clockwise from top left). But wouldn't a simple free rotation of b and d in either one of the isomers bring us back identically to the other?
ChemBuddy | 
Chem Reddit | 
Topic: Complex Isomerism (Read 8829 times)