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What is the valence electron count for [FeCp(CO)2]2?(show me how to calculate it) Cp is -1,...

What is the valence electron count for [FeCp(CO)2]2?(show me how to calculate it)

Cp is -1, CO is 0, so Fe is +1, d7,7+2+2*2=13, but how do i consider the 2 outside the bracket

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Compound can be written as Cp2Fe2(CO)4OR (η5-C5H5)2Fe2(CO)4 & the structure is

O=0 11 CEO Q -0=0

Finding oxidation state of ‘Fe’:

Each ‘Cp’ contributes a charge of ‘-1’ & each ‘CO’ contributes a charge of ‘0’.

Let ‘x’ be the oxidation state of ‘Fe’ in the complex. So,

2Fe + 2Cp + 4CO   = 2x + 2(-1) + 4(0) = 0

x = +1 = oxidation state of each ‘Fe’

Electron count:

Each ‘Fe’ = 7 e-; Each ‘CO’ = 2 e-; Each Cp = 6 e-; Fe-Fe bond = 2e-

2Fe + 4CO + 2Cp + 1Fe-Fe = (2 x 7) + (4 x 2) + (2 x 6) + 2 = 36 e-

So, Each Fe will get 18 e-

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