Given below are 22 tips to never miss a question on stereoisomerism in complexes. Understand and remember following points. [A and B are monodentate ligands. AA is symmetrical bidentate ligand like en (ethylenediamine or ethane-1,2-diamine), ox (oxalato) etc. AB is unsymmetrical bidentate ligand like gly(glycinato) etc.]
1.Tetrahedral complexes do not exhibit geometrical isomerism as all positions are equidistant or adjacent to each other. however, if all four ligands are different then it exhibits optical isomerism.
2. Tetrahedral Be(acac)2 exhibits optical isomerism.
3.Square planar MA2B2 type has two geometrical isomers (in cis form, similar groups occupy adjacent positions at angle 90 degree while in trans form they occupy opposite positions at 180 degree) for example, Pt(NH3)2Cl2.
4.Square planar MA2BC type has cis-trans isomers.
5.Square planar MABCD type has three geometrical isomers. depending upon the fact whether B, C or D is trans to A.
6.Square planar M(AB)2 type has two geometrical isomers for example, Pt(Gly)2.
7. Tetrahedral M(AB)2 type has a pair of enantiomers.
8.Octahedral MA4B2 or MA4BC type has cis, trans isomers. [Cr(NH3)4Cl2]+, [Co(H2O)4Cl2]+
9. MA3B3 type has fac, mer isomers. Facial isomer has identical ligands and central metal atom are on same face of octahedral. Meridional isomer has identical ligands and central metal atom are on same plane.
10.Octahedral MA2B2C2 type has total five geometrical isomers. All cis-isomer is optically active & exists as a pair of enantiomers.
11.M(AA)3 has cis-d & cis-l isomers. It’s trans form is not possible. It exhibits optical isomerism but does not exhibit geometrical isomerism. (AA is symmetrical bidentate ligand) Cr(en)3, [Co(Ox)3[3-
12. M(AB)3 has cis & trans isomers and both are optically active hence total four isomers. It exhibits both optical isomerism and geometrical isomerism. Cr(gly)3
- M(AA)2B2 or M(AA)2BC has total three stereoisomers, cis-d, cis-l & trans. It exhibits both geometrical & optical isomerism. Cr(en)2Cl2
- M(AA)B2C2 has three geometrical isomers – two trans forms & one (all) cis-form. M(AA)B2C2 has one pair of enantiomers. Cis-form is optically active. So, it has total four stereoisomers.
- M(acac)3 have a pair of enantiomers.
- M(EDTA) have a pair of enantiomers. [Co(EDTA)]-
- M(ABCDEF) have a pair of enantiomers.
- M(AA)B3C type has cis-trans isomers but no optical isomerism.
19. MA2B3C type has three geometrical isomers.
20. Cr(NH3)4ClBr]Br- has four isomers (including ionization isomers).
21. [Cr(NH3)4Br2]NO2- has four isomers (including Linkage + Stereoisomers)
22. Square planar complexes do not exhibit optical isomerism due to presence of plane of symmetry.