ionic and neutral electron-counting conventions
Suppose two accountants tally the same company and both must reach the same bottom line, but one assumes every employee was hired as a full team while the other assumes everyone came as freelancers. As long as each is consistent, they agree on the final figure. Electron counting in organometallic chemistry has exactly two such bookkeeping styles, the ionic convention and the neutral convention, and the trick is to pick one and follow it through without mixing.
In the neutral, or covalent, convention you imagine every bond breaking evenly. The metal is treated as a neutral atom and contributes its group number of valence electrons, while each ligand is taken as a neutral fragment: CO donates 2, a hydrogen radical or methyl radical donates 1, eta-5 cyclopentadienyl donates 5, and the overall charge of the complex is subtracted if it is a cation, added if an anion. In the ionic, or donor-pair, convention you imagine bonds breaking heterolytically so the ligand keeps both electrons. The metal is now an ion contributing its d-electron count for that oxidation state, while anionic ligands donate even pairs: a hydride donates 2, a methyl anion 2, chloride 2, and cyclopentadienyl as Cp-minus donates 6. Both methods, done carefully, give the identical valence electron total.
Why keep two? Because each shines for a different question. The neutral convention is quick when you just want the electron count and care about radical-type reaction steps, while the ionic convention makes the metal's oxidation state fall straight out, which matters for oxidative addition, reductive elimination and redox reasoning. The classic beginner's mistake is to mix them — counting a methyl as an anion donating 2 while also giving the metal its full neutral group number — which double-counts electrons and gives nonsense. Choose a lane and stay in it.
Count ferrocene, Fe(C5H5)2. Neutral method: iron (group 8) gives 8, each neutral Cp radical gives 5, total 8 + 5 + 5 = 18. Ionic method: iron is Fe2+ with six d electrons, each Cp-minus gives 6, total 6 + 6 + 6 = 18. Same answer, different routes.
Two conventions, the same complex, the same final count of eighteen electrons.
The fatal mistake is mixing the two methods in one count; either treat the metal and ligands all as neutral fragments or all as ions, never half and half.