For platinum (II) and (IV) a large number of complexes is known. For most of them the isomers were i — Inorganic Chemistry Chemistry Question
Substitution in square planar complexes
For platinum (II) and (IV) a large number of complexes is known. For most of them the isomers were isolated.
Explain what is the reason for the existence of different isomers and draw the structures for all Pt(NH3)2Cl2Br2 species.
Model Answer
12.1 The isomers can be easily isolated for inert complexes only. An octahedral composition MA2B2C2 has five geometric isomers.
cis-diammine-cis-dichloro-trans-dibromo-platinum(IV)
trans- diammine-cis-dichloro-cis-dibromo-platinum(IV)
cis-diammine-trans-dichloro-cis- dibromo-platinum(IV)
trans-diammine-trans-dichloro-trans-dibromo-platinum(IV)
cis-diammine-cis-dichloro-cis-dibromo-platinum(IV)
The reaction of thiourea with [Pt(amine)2Cl2] isomers results in different products. Explain this fact and give the reaction scheme.
Model Answer
12.2 In the cis-isomer, all the ligands are substituted by thiourea due to a high trans-activity of the entering ligand. In the trans-isomer the amine ligands remain intact.
Give an example of PtX(amine)Cl2 isomeric complexes reacting with thiourea with the formation of one and the same product.
Model Answer
12.3 The complexes containing groups with high trans-effect (such as alkenes) react with thiourea giving tetrathioureates:
Explain why do the reactions of [PtCl4] 2– and [AuCl4] – with iodide result in different products.
Model Answer
12.4 Platinum(+2) is a weak oxidizer, hence in the case of platinum only the substitution of chloride by iodide ligands occurs. In the case of tetrachloroaurate(+3), the rate of electron transfer exceeds the rate of substitution, so the redox process occurs:
2 [AuCl4] – + 6 I– → 2 AuI + 2 I2 + 8 Cl–.