= Solution
Oxidizing <cobaltocene> removes its antibonding nineteenth <electron> and gives the particularly favourable 18-electron <cobaltocenium> <cation>. <iodine> can accept the <electrons>, giving \b[<cobaltocenium> iodide] at the simple <stoichiometry>
$$
\boxed{2\mathrm{CoCp_2}+\mathrm I_2\longrightarrow2[\mathrm{CoCp_2}]^+\mathrm I^- .}
$$
Each <cobalt> changes from Co(II), $d^7$, to Co(III), $d^6$; each <iodine> atom changes from 0 to $-1$. Excess <iodine> can change the isolated counterion to a polyiodide such as $\mathrm I_3^-$. In contrast, oxidation of 18-electron <ferrocene> would remove an <electron> from its occupied closed-shell manifold to form 17-electron ferrocenium. <iodine> is insufficient to drive that reaction under the conditions implicit here. <ferrocene> can still be oxidized by suitable stronger oxidants; an <electron> count is a useful rationale for the comparative redox behaviour, not a universal proof of a numerical electrode potential.
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