- The Figure Below Shows The Distributions Of Equatorial Ligands The Distances Of Three Ru Ru Bond And Bond Angles In Ru3 1 (46.38 KiB) Viewed 126 times
- The Figure Below Shows The Distributions Of Equatorial Ligands The Distances Of Three Ru Ru Bond And Bond Angles In Ru3 2 (20.92 KiB) Viewed 126 times
- The Figure Below Shows The Distributions Of Equatorial Ligands The Distances Of Three Ru Ru Bond And Bond Angles In Ru3 3 (16.44 KiB) Viewed 126 times
The figure below shows the distributions of equatorial ligands, the distances of three Ru-Ru bond and bond angles in Ru3(CO)12, obtained from its single- crystal X-ray diffraction 033 034 103 100296 024 Rua olis 96.32 971 1050 Ru2 Rut 2.859 98.8 98.99 1040 023 Figure 1 are substituted with a 1,1- If two co ligands in Figure 1 bis(diphenylphosphino)methane, (a) Predict the bond distances of three Ru-Ru bonds. (6 marks) (b) Predict the bond angles. (6 marks)
(c) Explain the two main differences, in term of geometry parameters, between Figure 1 and complex stated in (a)(0) (6 marks) (d) Substitution reactions of Figure 1 with monodentate and bidentate ligands yield new trisubstituted complexes, where the trinuclear core of Ruz structure remains unchanged. LI Ru 11 where, L1-L2: dppm L: PPH3, PEt3, PCy3, AsPhz PPrz CO L Ru 3 Ru 2 L 2 8 Co CO
i. Discuss why monodentate group 15 ligands are able to coordinate with bidentate ligand effectively in the metal cluster. ii. Differentiate the Ru-Ru and Ru-P bond lengths for (L = PPh3) and (L = PCy3) if Ru-Ru and Ru-P bond lengths for (L PEt3) are 2.8733 A and 2.3480 A, respectively. iii. Explain your answer in (ii). iv. Compare and contrast the 'H, 130 and 31P{1H} nuclear magnetic resonance (NMR) for complexes (L = PPh3), (L = AsPh3) and (L = PEt3). (25.5 marks)