[Big-Daddy]

How many isomers (restricted to optical and geometric isomerism) exist for the coordination compound (PPh3)₂PtCl₂ (Ph = phenyl), known to be square planar?

If you imagine the first bond drawn going into the page on the top left of the central atom, label this substituent a; second bond going into the page on the top right, label this substituent b; third bond going out of the page on the bottom right, label this substituent c; fourth bond going out of the page on the bottom left, label this substituent d.

I thought optical isomers were possible for both tetrahedral and octahedral complexes, but not square planar ones. Thus the two solutions should simply be: {a=PPh₃, b=PPh₃, c=Cl, d=Cl}, {a=PPh₃, b=Cl, c=PPh₃, d=Cl}. And it should not matter whether the bonds are going in or out of the page, just the geometric positions. But the mark scheme has different solutions: {a=PPh₃, b=PPh₃, c=Cl, d=Cl}, {a=PPh₃, b=Cl, c=Cl, d=PPh₃}. What is going on?

[Borek]

Looks to me like the mark scheme is wrong. Unless I am missing something big.

[Big-Daddy]

Looks to me like the mark scheme is wrong. Unless I am missing something big.

Thanks, that is reassuring.

Am I right to say that whatever the composition of the square planar ([Mabcd], [Ma₃b], [Ma₄], [Ma₂bc], [Ma₂b₂], even with polydentate ligands) it will not exhibit optical isomerism? For any composition of tetrahedral, it will not exhibit geometric isomerism? And what types do the trigonal-bypramidal or square-pyramidal (each has 5 ligands) exhibit?

[Sossy]

Since the square planar complex has a plane of symmetry, the plane containing Pt, the chloride ions and the P atoms, it cannot have optical isomers. The two geometric isomers of the complex are cis and trans.