Question

53. Compound I + H2C0. -PI(NH3)2C04] Compound II - H.C:0:-P (NH3)2(HC,04)2] P!(NH3)C1,) is found to exist as two isomers designated as I and II. which are found to react with oxalic acid as indicated above. From this information, one is able to conciude that (A) I is tetrahedral and Il is square planar (B) I and II are both tetrahedral and I has the trans configuration (C) I and II are both tetrahedral and I has the cis configuration (D) I and II are both square planar and I has the trans configuration (E) I and II are both square planar and I has the cis configuration A 8

Answer 1

537 Compound I + H₂C2O4 → [PL-CN MadG04] cis form Compound 1 + H2604 → [P(94) (196202] Thans Form When Both the form is guane planan geometry Cis on Thans (Pt (Aritz) (NO3),] isomes is beaching with another ligand the leaving group is not going to be NH3 because the Pt- Ntz bond is too strong and the thans effect of Noo is not high enog enough to break the bond. In lis and Thans form the noge ligand which is a weak base is likely to dissociate. There reaching of lis form with oxalic ā should give (PL- (NH3)₂ CC204)] in which oxalate is acting as bidentale ligand through oxygen atom both carboxylic group. For the trans form howeren, if not possible for the oralale lo be bidentate ligand and healing of the I trung form with oxalic ā i's expeeled to produce toams - [Prenm;)&ca) In which oxalate in act as an monodentate ligand. so option ® will be connect. I and it are both squane planah and I has the Cis-Configurand.