Cobalt octacarbonyl reacts with hydrogen: Co2(CO)8 + H2 -> 2HCo(CO)4. Treating H as hydride (2-electron donor), the oxidation state of Co and its electron count in HCo(CO)4 are:

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Cobalt octacarbonyl reacts with hydrogen: $Co X_{2} (CO) X_{8} + H X_{2} 2 HCo (CO) X_{4}$ . Treating H as hydride (2-electron donor), the oxidation state of Co and its electron count in $HCo (CO) X_{4}$ are:

With

H X^{-}

and four neutral CO, the oxidation state of Co is

+ 1

, so it is

d^{8}

with 8 electrons. Count =

8 + 2 (H X^{-}) + 4 \times 2 = 18

. Hence Co(+1), 18 electrons.