Question

Show how electron counting may be used to determine the total valence electron count for each of the following compounds, draw a structure for each compound AND determine the oxidation state of the metal in each compound. Do not use abbreviations please. (a) [Ir(CO)H(PPh3)3]

Answer 1

- IH Answer) The effective number of electrons in the valence Shell of the metal complexes is counted as a) (Ix (CO)HCP Ph3.)3] 15 Count the number of electron in the Valence shell of the metal atom Electron configuration of Iridium & Structure Ir = [Xe] 484 5dr 682 Valence electron = 9 ii) oxidation state of metal in compound [Ir(CO) HCPP H3)3] co and Tri Phenyl Phosphine are neutral 2 + 100)+1(-1) + 360) - O x-1=0 x= +1 so Iridium have ti oxidation state c) the number of electrons contributed by the ligend Effective number of Elections in the valency shell of [IM (CO) HCP Ph 3)3) Metal complexe Namber of electrons in Number of electrons Number of Valency shell of Free metal shell of Metal in oxidation atom CO= 2 [Ir(CO) HCPPH3)3] IRI H = 2 19-1=8 3 PPh z=G in valency Electrons contribute by ligands State Total number of Electrons in the Valency shell of contral Metal in [Ircco) HCPPh ₂ - IXI = 8 CO = 2 H = 2 3 PPh3 - 6 Total = 18 e