2. Identify the first-row transition metal for the following 18-electron species: (a) [M(CO) (PPho)l (b) [(n5-CsHs)M(CO)32 (assume a single MM bond) Solution (a) [M(CO)3(PPh3)] - PPh3 = 2 e- 1 PPh3 = 1*2 = 2e- CO = 2 e- 3 CO = 3*2 = 6e- Metal = x e- 1 M = 1*x = xe- (assuming zero oxidation state) Oxidaion State = 1 e- = 1e- ------------------------------------------------- 8 e- + 1 e- + x e- = 18e- x = 9e- = s 2 d 7 metal = Cobalt (Co) metal (b) [(n 5 -C 5 H 5 )M(CO) 3 ] 2 calculation done around 1 Metal (n 5 -C 5 H 5 ) = 5 e- 1 (n 5 -C 5 H 5 ) = 1*5 e- = 5 e- (only one metal is considered, else this would be 10e-) CO = 2 e- 3 CO = 3*2 e- = 6 e- (only one metal is considered, else this would be 12e-) M-M = 1 e- 1 M-M = 1*1 e- = 1 e- (only one metal is considered, else this would be 2e-) M = x e- 1 M = 1*x e- = x e- (only one metal is considered, else this would be 2x e-) ---------------------------------------------------------------- 12 e- + x e- = 18 e-Â Â (only one metal is considered, else this would be 18*2 = 36 e-) x e- = 18-12 = 6 e- thus x = 6 e- = s 2 d 4 metal = Chromium (Cr) metal .