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Topic: Complex Isomerism  (Read 6709 times)

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Offline Big-Daddy

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Re: Complex Isomerism
« Reply #15 on: April 03, 2013, 09:00:48 AM »
What process should I follow to find the isomers for octahedral complexes?

Is it simply the same, but if rotating any opposite bonded pairs on Configuration 2 can take me back to Configuration 1, then Configuration 2 is the same molecule as Configuration 1? i.e. I follow the same process as for square planar, except I can also rotate the 2 new bonds that are present (and opposite each other) in the same way I rotate the opposite bonds for square planar molecules ...

Offline Dan

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Re: Complex Isomerism
« Reply #16 on: April 03, 2013, 10:03:08 AM »
Yes, it's the same process, just more difficult to visualise mentally.
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Offline Big-Daddy

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Re: Complex Isomerism
« Reply #17 on: April 21, 2013, 11:21:19 AM »
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?

Offline Big-Daddy

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Re: Complex Isomerism
« Reply #18 on: April 28, 2013, 04:51:32 PM »
So ... is there a systematic way of doing this or what?

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