In: Chemistry
For each of the following (real or imaginary) complexes determine a0 the number of electrons in the coordination sphere and b0 the oxidation state of the metal in each of the complexes.
a. (CO)5Cr=C(OEt)Me
b. NbBr3(PPh3)2(=CHCMe3)
c. (Me3CO)3W(triple bond CCMe3)
d. [(n5-C5H5)RuCl(Me)(PMe3)]-
e. [(n5-C5H5)Ir(PMe3)(=CH2)]+
(a) (CO)5Cr=C(OEt)Me
Electron inside the coordination sphere count
5 CO = 10 e, Cr = 6e, =C = 2e, Total electron 18, and oxidation state +2
(b) NbBr3(PPh3)2(=CHCMe3)
Electron count inside coordination sphere: Nb = 5e, 3Br = 3, 2PPh3 = 4e, =C = 2 ; Total electron count = 14 e, Oxidation state of Nb is +5.
(c) (Me3CO)3W(triple bondCCMe3)
Electron count inside coordination sphere: 3e from 3(Me3CO), 6e from W, and 3e from triple bondCCMe3 . Total electron count =12 e; Oxidation state of metal ion is +6.
(d) [(n5-C5H5)RuCl(Me)(PMe3)]-
Electron count inside coordination sphere: 5e from (n5-C5H5), 8e from Ru, 1e from Cl, 1e from Me, 2e from PMe3 ; Total electron count = 17+1(from negative charge) =18 e; Oxidation state of metal ion is +2.
e. [(n5-C5H5)Ir(PMe3)(=CH2)]+
Electron count inside coordination sphere: 5 e from (n5-C5H5), 9e from Ir, 2e from PMe3, 2e from (=CH2 ). Total electron count = 18-1(positive charge) =17 , Oxidation state of metal ion is +4.