Complex isomers


by Big-Daddy
Tags: complex, isomers
Big-Daddy
Big-Daddy is offline
#1
Aug26-13, 01:50 PM
P: 334
How do I find all the different geometric and optical isomers of a general octahedral complex ion with only monodentate ligands? I know well the method for finding geometric isomers in a square planar complex: any pair of ligands positioned diagonally to each other are opposite and their positions freely rotatable. However, I need a similar method to apply for geometric isomerism in octahedral complexes (and preferably, geometric isomerism in general) and also optical isomerism which I don't know how to include when in conjunction with geometric.
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eigenperson
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#2
Aug26-13, 08:55 PM
P: 131
I'm not a chemist, so I don't know what aspects of this situation are interesting to chemists. But I know the mathematics.

First of all, the real answer to the question is that you should use the Pólya enumeration theorem to count the number of geometric or geometric and optical isomers. This works for any geometry.

But if you're not comfortable with that, I'll try to do it another way.

The number of isomers (counting optical isomers as different) of an octahedron with 6 different ligands is 30. To construct them, start by putting ligand 1 at the north pole (without loss of generality). Then put one of the remaining ligands at the south pole (5 choices). Of the remaining 4 ligands, put one at the vertex nearest you (without loss of generality). Put one of the remaining 3 ligands opposite that one (3 choices). Finally, the last two ligands can be put in one of two configurations (2 choices).

If you count optical isomers as equivalent (i.e. you only want to count geometric isomers), then there are 15. The process for constructing them is the same as before, except now when you get to the last step of placing the final two ligands, it doesn't matter which way you do it since the two choices produce optical isomers.

Now, what if you don't have 6 different ligands? You could go through a similar logic to the above, adjusted as needed -- basically, just counting very carefully. Alternatively, you could temporarily pretend that the identical ligands are different by giving them different lables. Then you could draw all 30 or 15 potential isomers on paper. Finally, erase the labels, and group the structures into equivalence classes.

And yes, that's a bit of work, and there is some potential for error, but I don't know how to do it more easily without Pólya's theorem.
Big-Daddy
Big-Daddy is offline
#3
Aug28-13, 05:38 AM
P: 334
Quote Quote by eigenperson View Post
Now, what if you don't have 6 different ligands? You could go through a similar logic to the above, adjusted as needed -- basically, just counting very carefully. Alternatively, you could temporarily pretend that the identical ligands are different by giving them different lables.
Ok, so could we just go from the set of all 30 isomers for the complex with 6 different ligands and then, for instance, for the case Ma4bc, set 3 others of the ligands all equal to a (so that 4 ligands are a, one is b and one is c in the each isomer) and then collect together complexes which are now identical?

And the set of 30 isomers for the complex with 6 different ligands is found by holding one particular ligand in the north pole position. But then shouldn't there be 5! different isomers for this complex? (Rather than 30)

eigenperson
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#4
Aug28-13, 11:09 AM
P: 131

Complex isomers


Ok, so could we just go from the set of all 30 isomers for the complex with 6 different ligands and then, for instance, for the case Ma4bc, set 3 others of the ligands all equal to a (so that 4 ligands are a, one is b and one is c in the each isomer) and then collect together complexes which are now identical?
Yes. That's the only way I know of to do it without group theory. In that particular case there are a lot of identical complexes -- in the end there are only 2 different geometric isomers (the two unique ligands could be at opposite poles, or they could be adjacent) and no optical isomers.

(This would become apparent fairly quickly if you had a model of the octahedron to play with. If you don't have an actual octahedron, a cube has exactly the same symmetries as long as you label the faces of the cube instead of the vertices. So, for example, take a cube, 4 red dots, 1 yellow dot, and 1 blue dot. How many different "isomers" can you make by sticking one dot on each face? If you experiment, you'll immediately see that there's really only one choice to make, which is where to put the blue dot relative to the yellow dot, and there are only two possible choices.)

And the set of 30 isomers for the complex with 6 different ligands is found by holding one particular ligand in the north pole position. But then shouldn't there be 5! different isomers for this complex? (Rather than 30)
No, because there are still ways to rotate the octahedron even while the north pole is kept fixed, so some of these 5! different arrangements will be equivalent under rotation. (Rotate about an axis through the north and south poles.) In fact, there are 4 different positions the octahedron can be rotated to like this, even after fixing the north pole, so the 5! = 120 different arrangements will be grouped into equivalence classes of size 4. That leaves 120/4 = 30 equivalence classes (isomers).
Big-Daddy
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#5
Aug28-13, 03:19 PM
P: 334
Quote Quote by eigenperson View Post
No, because there are still ways to rotate the octahedron even while the north pole is kept fixed, so some of these 5! different arrangements will be equivalent under rotation. (Rotate about an axis through the north and south poles.) In fact, there are 4 different positions the octahedron can be rotated to like this, even after fixing the north pole, so the 5! = 120 different arrangements will be grouped into equivalence classes of size 4. That leaves 120/4 = 30 equivalence classes (isomers).
How should I tell which of the 5! arrangements will be equivalent? Perhaps some of the bonds can be rotated or switched and that is the key to this problem. What are the steps to take then? 1) Fix the ligand at the north pole position; for 2) I was thinking about fixing the ligand at the vertex closest to me, but that leaves just 4!=24 isomers.

If I were to label the positions around the metal atom 1 (north pole), 2, 3, 4, 5 and 6 in a clockwise direction on paper, which positions would I find ligands in that I could switch with one another to give me the same isomer? Let's say that (1) is a fixed ligand. Then (4), the south-pole, can be any of the 5 others. So the question comes down to the other 4 positions.
eigenperson
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#6
Aug28-13, 08:38 PM
P: 131
Denote the ligands as follows: A, B, C, D, E, F
Denote the positions as follows:

1 = the vertex at the North Pole
2 = the vertex furthest from you
3 = the vertex to your right
4 = the vertex at the South Pole
5 = the vertex closest to you
6 = the vertex to your left

(I believe this is the same notation you were using, but please check.)

Here is how to write down the 30 configurations.

Step 1: Put A at vertex 1.

Step 2: Any one of the remaining ligands could be at vertex 4. There are 5 different choices to make here, and each will generate 6 isomers. I'll demonstrate for ligand B at vertex 4, but you'll have to do the same thing with for ligand C, D, E, and F at vertex 4.

Step 3: You have already attached two ligands, and have 4 left to attach. In this case they're C, D, E, and F. Attach them in the following ways:

2C 3E 5D 6F
2C 3F 5D 6E
2C 3D 5E 6F
2C 3F 5E 6D
2C 3D 5F 6E
2C 3E 5F 6D

If stereoisomers are to be treated the same, write down only the bold ones. If stereoisomers are to be treated as different, write them all down.

----------

You got as far as step 2, but you got stuck at that point. The point is that the other 4 vertices form a square, which you are allowed to rotate, but not flip over (unless you treat stereoisomers as the same). Of the 4! = 24 different ways to place CDEF around the square, only 6 of them are actually different. For example, CDEF and DEFC are really the same, because they differ only by a rotation.
Big-Daddy
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#7
Aug29-13, 07:55 AM
P: 334
That is hugely clarifying. So, of the 4 non-polar ligand positions, the formed square is freely rotating, so, for one isomer, if we can rotate each of the ligands at these 4 positions by 90 degrees (so that we get the ligand at 2 switching to position 3; the ligand at 3 switching to 5; and the ligand at 5 switching to 6) and reach the configuration of "another" isomer, the two isomers are really the same (given the same ligand at the south pole and by extension, having used all 5 already, at the north pole). So, since we choose one of the 4 vertices of the square as a constant ligand when finding all the isomers (it does not matter which, for each choice of south-pole ligand, we can use any of the others as the fixed one, excepting the north pole ligand of course), we get 3! combinations for each of the 5 ligands we can choose to be the south-pole. This covers both geometric and optical isomerism.

Am I right to think optical isomerism around the metal centre cannot exist for square planar complexes? If so, then I should have now covered the geometric and optical isomerism of square planar and octahedral complexes, for all monodentate ligands. Is there a similar method for tetrahedral complexes?
eigenperson
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#8
Aug29-13, 09:27 PM
P: 131
Quote Quote by Big-Daddy View Post
Am I right to think optical isomerism around the metal centre cannot exist for square planar complexes?
That's right (as long as each ligand is itself achiral, of course).

Is there a similar method for tetrahedral complexes?
Yes, but it's much simpler. If you have 4 different ligands, then there are two optical isomers. Otherwise, there is only one isomer.
Big-Daddy
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#9
Aug30-13, 05:11 AM
P: 334
Ok, I assume this is in the same way as for organic compounds around a chiral centre (if tetrahedral with 4 different ligands).

If we introduce linkage isomerism, can we simply say that, within each geometric and optical isomer we found before, we may have versions with different atoms from the ligand making the link, and that's it? e.g. if we had 30 isomers before (6 different ligands), and one of the ligands was NO2, then we now have 60 isomers (each of the isomers before, but in two different versions: one where the N links the NO2 ligand, and one where an O links the NO2 ligand). And a similar method works if a ligand has chirality itself (we would just need to consider each isomer with each of the enantiomers of the ligand, one by one). So the existence of chirality in the ligand does not affect the geometric or optical isomerism around the metal centre, and nor does the linkage. Is this all ok?
Big-Daddy
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#10
Sep2-13, 03:01 PM
P: 334
Hi, can you help me with this?
ChiralWaltz
ChiralWaltz is offline
#11
Sep2-13, 06:21 PM
P: 114
Quote Quote by Big-Daddy View Post
So the existence of chirality in the ligand does not affect the geometric or optical isomerism around the metal centre, and nor does the linkage. Is this all ok?
I think you have the right idea. However, you contradicted yourself with the linkage statement. The linkage isomerism will affect geometric and optical isomerism depending on whether the N or O is connected in the nitro group.

Once you start picking up multiple chiral centers in a molecule, there is a potential for meso compunds.
Big-Daddy
Big-Daddy is offline
#12
Sep3-13, 12:52 PM
P: 334
Ok but we just need to consider NO2 where N is the bonding atom and NO2 where O is the bonding atom as separate ligands and then consider the possibilities present for how many arrangements there could be?
ChiralWaltz
ChiralWaltz is offline
#13
Sep4-13, 01:26 PM
P: 114
For linkage isomerization of NO2, we can use A-F or A-G without F.

F= bonding point N
G= bonding point O

All you have to do is multiply by 2 to generate the second set. I'll see what more I can find on the mathematics behind it and get back to you.


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