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  1. 1. Module – 24.3 Isomerism Two or more substances having the same molecular formula but different structural or spatial arrangement are called isomers. They are of two types. a. Structural isomerism b. Stereo isomerism a. Structural isomerism: Two or more complexes have same molecular composition, but different properties are known as isomerism. Those are called as isomers. The isomers differ in the arrangement of ligands within the complest are called structural isomers and the phenomenon is called structural isomerism. 1. Ionization isomerism: This is due to difference in ionisable groups. Example: i) [Co(NH3)5Cl]SO4 and [Co(NH3)5SO4]Cl ii) [Pt(NH3)4Cl2]Br2 and [Pt(NH3)4Br2]Cl2 2. Polymerization isomerism: Compounds having the same emperial formula, but different molecular weights. Example: [Pt(NH3)2Cl] and [Pt(NH3)4][PtCl4] 3. Hydrate isomerism: Isomerism differing in the number of water molecules attached the metal ion as ligands in coordination sphere Example: [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2H2O Violet Green
  2. 2. 4. Linkage isomerism: Isomers of this type have different kinds of linkages of a ligand to the central mental ion. Example: [Co(NH3)5(NO2)]Cl2 and [Co(NH3)5CoNO]Cl2 5. Coordination isomerism: This isomerism caudes by the interchange of ligands between the two complex ions. Example: [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6] 6. Ligand isomerism: Some ligands capable of existing isomerism can give different isomeric complexes. Stereo isomerism in complex compounds: Two or more complexes having same molecular composition but different in arrangement of atoms (or) groups in space is called as stereo isomerism. It is of two types. a. Geometrical isomerism b. Optical isomerism Stereo isomerism in coordina coordination number 6 compounds:
  3. 3. The arrangement of six ligands in a complex around central metal ion in two different ways. Those are regular hexagon and regular octahedron. If the complex Ma4b2 has hexagonal structure it can give three isomers corresponding to (1, 2), (1, 3) and (1, 4) positions of b, while octahedral arrangement can give only two isomers. X – ray analysis, confirmed that coordination number 6 complexes exhibit octahedral arrangement by giving two isomers of Ma4b2 type. Hence coordination number 6 complexes have octahedral structure. 1. No isomerism is possible in [Ma6] or Ma5b of complexes Example: [Co(NH3)6]3+; [Co(NH3)5Cl]2+ 2. In [Ma4b2] and [Ma4bc] type complexes can give two isomers. Those are Cis and trans isomers. Example: [Co(NH3)4Cl2]+ gives Cis and trans isomers. 3. In [Ma3b3] type of complexes can exhibit geometrical isomerism.
  4. 4. Example: [Co(NH3)3Cl3] gives Cis – trans isomers. 4. In [M(aa)2b2] type of complexes having two bidentate ligands, can exhibit geometrical isomerism. Example: [Co(en)2Cl2]+ gives Cis and trans isomers. Optical isomerism: 1. The Cis isomer of [M(aa)2b2] type complexes has no plane of symmetry. So it can give two isomers (d and l)
  5. 5. 2. Complexes of [M(aa)3] type having three bidetate ligands are also unsymmetric and gives optical isomerism Example: [Pt(en)3]4+ gives optical isomers. Stereo isomerism in coordination numbers “4” compounds: Complex compounds of coordination number “4” type gives either tetrahedral or square planar structure. 1. No isomers are possible [Ma4], [Ma3b] types complexes Example: Ni(Co)4, [Cu(NH3)4]2+ 2. [Ma2b2] and [Ma2bc] complexes of some metals like Pt (II), Pd (II), Ni (II), Cu (II) can give square plannnar structure. These gives dsp2 hybridisation Example: [Pt(NH3)2(CN)2] gives Cis trans isomers The geometrical isomers of [Pt(NH3)2Cl2] are
  6. 6. Some complex compounds of Cu, Zn, Ni can give tetrahedral structure. These gives sp3 hybridisation [Mabcd] type of complex gives optical isomerism, if thay have tetrahedral. Three isomers are possible, if they have square planner 5 structure. Example: [Pt(NH3)(Py)ClBr] gives three isomers. Applications of complex compounds: I. Applications in equilibrium analysis: a. Separation of AgCl from Hg2Cl2: In the Ist group silver ions precipitated as white ppt of AgCl. It is soluble in ammonia due to the formation of [Ag(NH3)2]Cl On the other hand, white p of Hg2Cl2 turns black on treatment with ammonia. ppt
  7. 7. b. Separation of Cu and Cd: Candmium is separated from a mixture of copper and cadmium ion by adding KCN. Copper ion forms a stable complex with KCN in the solution. Where as cadmium complex being unstable, decomposes to give Cd+2 ions. These ions gives yellow precipitate with H2S w c. Test for Fe+2 (ferrous ions) and Fe+3 (ferric ion): Ferrous salt solution gives deep blue colour precipitate with potassium ferry cyanide. These tests indicate the presence of ferrous ion. Ferric salt precipitate with potassium ferro cyanide. This indicates the presence of ferric ion. d. Test for K+ and ion: Potassium salt solution gives yellow precipitate with sodium cobalt nitrate. Thie test indicates the presence of potassium ion. Nessler’s reagent is used for identification of ammonium ion. It gives reddish identification brown precipitate. e. Test for nickel ions: Nickel salts react with dimethyl glyoxime in presence of ammonium hydroxide to give red precipitate of nickel dimethyl glyoxime
  8. 8. f. The complex “Ferroin” is used as indicator in some redox titrations. ” [Fe(phenanthroline)3]2+ II. Applications in quantitative analysis: a. Estimation of several cations like Mg+2, Zn+2, Cu+2, Ni2+ using EDTA as titrant in presence of suitable indicator Example: Erio chrome black – T b. Dimethyl glyoxime is used for estimation of nickel. Example: 8 – hydroxyl quinolene is used for estimation of zinc c. Hardness of water can be calculated by EDTA Assignment questions: 1. Give any three applications of complex compounds in quatitative analysis 2. List various types of structural isomerism possible for coordination compounds, giving one example each 3. Draw the structures of optical isomers of a. [Pt(en)3]4+ b. [Co(en)2Cl2]+ Example set: 1. [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br exhibit a. Hydrate isomerism
  9. 9. b. Ionization isomerism c. Ligand isomerism d. Co – ordination isomerism Solution: b) 2. Blue colour/precipitate will be obtained when K4[Fe(CN)6] reacts with a. Fe (II) ions b. Cu (II) ions c. Fe (III) ions d. Cu (I) ions Solution: c) 3. Which one of the following square planner complexes will show Cis – trans isomerism a. Ma4 b. Ma3b c. Ma2b2 d. Mabcd Solution: c) 4. The number of isomers possible for square plannar complex. K2[PdClBr2(SN)] a. 2 b. 4 c. 5 d. 6 Solution: a) Problem set: 1. Silver chloride is soluble in ammonia due to the formation of
  10. 10. a. [Ag(NH3)2]2+ b. [Ag(NH3)2]+ c. [Ag(NH3)]+ d. [Ag(NH3)]+2 Solution: d) 2. Geometrical isomerism would be expected form? a. [Zn(NH3)4]2+ b. [Cu(CN)4]3- c. [Pt(NH3)2Cl2] d. [Au(NH3)2]+ Solution: c) 3. [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2H2O exhibit a. Hydrate isomerism b. Ionization isomerism c. Ligand isomerism d. Coordination isomerism Solution: a) 4. Which of the following octahedral complexes does not show geometrical isomerism (A and B are monodentate ligands)? a. Ma3b3 b. Ma4b2 c. Ma5b d. Ma2b4 Solution: c) 5. Name the type of isomerism exhibited by the following isomers a. [Cr(NH3)6] [Co(CN)6] and [Co(CN)6][Cr(NH3)6] b. [Co(Py)2(H2O)2Cl2]Cl and [Co(Py)2(H2O)Cl3]H2O c. [Pt(NH3)4Br2]Cl2 and [Pt(NH3)4Br2]Cl2 d. [Co(NH3)5NO2]Cl2 and [Co(NH3)5Ono)Cl2
  11. 11. Solution: a. Coordination isomerism b. Hydrate isomerism c. Ionization isomerism d. Linkage isomerism 6. Draw structures of geometrical isomers [Fe(NH3)2(CN)4] Solution: Exercise questions: 1. Why is geometrical isomerism not possible in tetrahedral complexes having two different types of unidentate ligands coordinated with the central metal ion? 2. Give evidence the [Co(NH3)5Cl)SO4 and [Co(NH3)5SO4]Cl are ionization isomers? 3. Indicate the types of isomerism exhibited by the following complexes and draw the structure for these isomers a. [Co(NH3)5NO2](NO3)2 b. Pt[(NH3)(H2O)Cl2] 4. Platinum (II) forms square planar complexes and platinum (IV) gives octahedral complexes. How many geometrical isomers are possible for each of the possible following complexes? Describe their structures? a. [Pt(NH3)3Cl]+ b. [Pt(NH3)Cl5]- c. [Pt(NH3)2ClNO2] d. [Pt(NH3)4ClBr]+2
  12. 12. Solutions to exercise questions: 1. Tetrahedral complexes do not show geometrical isomerism because the relative positions of the unidentate ligands attached to the central metal atom are the same w.r.t each other. 2. When ionization isomers are dissolved in water, they ionize to give different ions. These ions then react differently with different reagents to give different products. 3. a. A pair of optical isomers It can also show linkage isomerism [Co(NH3)5(NO2)](NO3)2 and [Co(NH3)5(ONO)](NO3)2 It can also show ionization isomerism [Co(NH3)5(NO2)](NO3)2 and [Co(NH3)5](NO3)(NO2) b. Geometrical (Cis-, trans isomers of [Pt(NH3)(H2O)Cl2] can exist , trans-)
  13. 13. 4. a. No isomers are possible for a square planar complex of th type Ma3 the b. No isomers are possible for an octahedral complex of the type Mab5 c. Cis and trans isomers are possible for a square planar complex of the type Ma2bc d. Cis and trans isomers are possible for an octahedral complex of the type Ma4bc