Part 2
Geometrical Isomers of Square Planar and Tetrahedral Complexes
In the square planar stereochemistry all 4 donor atoms of the ligand are located at the corners of the square having the metal ion at the centre. When two kinds of ligand are present in the same square planar complex, two different arrangements are possible. If two like ligands are at the corners of the same side of a square, the structure is described as a cis-structure (from the Latin cis, on the same side). If the two like ligands are at opposite corners of a square, the structure is called a trans-structure (from the Latin trans, across). These two different spatial arrangements give rise to two isomeric isomers known as cis-trans isomers The two forms of cis-trans isomers usually differ in certain physical and chemical properties, in particular their solubilities, colours and dipole moments.
The two general spatial arrangements of a square planar complex with the formula [Ma2a2], where M is the central metal atom of the complex, and "a" and "a" are the coordinating groups, are shown below.
Geometrical isomerism is possible in square planar complexes because, although the 4 coordinating groups are equidistant from the central atom, they are not equidistant from one another. Thus, in the complex [Ma2b2], the b-to-b distance "x" in the cis isomer is different from the b-to-b distance (2½x) in the trans isomer. In a tetrahedral arrangement, however, all 4 coordinating groups are equidistant from one another. Thus, cis-trans isomerism is not possible when a complex of the type [Ma2b2] has a tetrahedral structure.
An example of cis-trans isomerism is provided by the two isomeric square planar compounds of Pt(II), having the formula PtCl2(NH3)2 (shown below).
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| cis and trans-PtCl2(NH3)2 |
(The cis compound is marketed as "Cisplatin", one of the more effective drugs for treating certain kinds of cancer.)
Question 6.
How many geometrical isomers can exist for PtBrCl(NH3)2? Are the terms cis and trans still applicable to them?Geometrical Isomers of Octahedral Complexes
For octahedral complexes which contain two or more different ligands geometric isomerism is possible. For example in complexes of the type [MA4B2] and [MA3B3] where M is the central metal ion, and A and B are the ligands, we also find cis-trans and fac-mer geometric isomers:
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| cis and trans-[Co(NH3)4Cl2]+ | fac and mer-Co(NH3)3Cl3 | ||
(a) |
(b) |
(c) |
(d) |
Question 7.
Look at the four diagrams labelled (a) to (d) above. Give their formulae including the appropriate cis, trans, fac or mer designation. (Assume the ligands are Cl and NH3 and that the metal ion Co3+.)So far we have only touched on ligands which are monodentate, ie. form only one sigma bond with the central metal ion. There are however many ligands capable of forming two or more sigma bonds with the central metal ion. Such ligands are called polydentate; eg., ethylenediamine (NH2CH2CH2NH2) is a bidentate ligand since each N atom of the molecule can form one sigma bond to a metal. Tridentate, tetradentate, and so on, ligands will not be discussed here; although they are of great importance in coordination chemistry (Have you ever seen a pictures of metalloproteins such as vitamin B12 or hemoglobin? They are also coordination compounds with elaborate polydentate ligands).
OPTICAL ISOMERISM OF COMPLEXES
One of the more important properties of some coordination compounds their ability to rotate plane-polarized light either clockwise or ounterclockwise. Compounds which have the same molecular formula and the same geometrical stereochemistry but which differ in their rotation of plane-polarized light are known as optical isomers and are said to be optically active. While there are many optically active compounds having a tetrahedral geometry in organic chemistry, e.g. any carbon compound of the type CABDE where A, B, D, and E are any different groups one can think of, and C is a carbon atom. D-+-Glyceraldehyde, the left isomer shown below, can be considered as the building block for the optically active sugars.
Usually, optical isomers of tetrahedral inorganic complexes cannot be isolated because of rapid site exchange. There are some exceptions involve strongly bound unsymmetrical bidentate ligands. Such a compound might be (M(AB)2 where M might be beryllium and AB is an unsymmetrical (but not necessarily optically active) bidentate ligand eg. NH2CH2CH2COOH, an amino acid.
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| Optical Isomers of Be(NH2CH2CH2COOH)2 |
The molecules depicted are optical isomers, that is they are mirror images of each other, and they are separable because they are not easily interconverted.
Question 8.
Which of the two isomers, the one on the left, or the one in the centre, is depicted on the right?Optical isomers occur only rarely among compounds having a square planar geometry. The few known cases involve unusual polydentate ligands. These are not discussed further here.
The most common cases of optical isomerism occur among compounds having the octahedral coordination geometry. Optical isomers can occur in complexes with three or more different kinds of monodentate ligand and no more than two of any one kind, that is: [Ma2b2c2], [Ma2b2cd], [Ma2bcde], and [Mabcdef]. For example in the case of [Ma2b2c2] we have one geometric isomer, the cis, cis, cis form that has non-superimposable mirror images. Optical isomers of [CoBr2Cl2(NH3)2]2- are shown below:
The requirement for optical activity is the absence of a plane of symmetry or a centre of symmetry. By far the most work on optical activity has been carried out on octahedral compounds of Co(III) or Cr(III) containing bidentate ligands; eg. the complex ions shown below:
D-[Co(NH2CH2CH2NH2)3]3+ |
L-[Co(NH2CH2CH2NH2)3]3+ |
[Co(NH2CH2CH2NH2)2Br2]+
The molecule above has only two bidentate ligands, but it can still be classified as either L or D by analogy with the pair with three bidentate ligands. Which is it?Question 10.
There is one other isomer of the molecule with the two bidentate ligands above which is not optically active. Sketch it, and suggest how its formula might be written to specify its geometry.