# How can the 18 electron rule be used to explain that Fe(η5-C5H5)2 doesn't react with hydrogen but Ni(η5-C5H5)2 does?

Fe(η5-C5H5)2 does not react with hydrogen, but its nickel analogue, Ni(η5-C5H5)2 is readily hydrogenated to give Ni(η5-C5H5)(η3-C5H7). Use the 18 electron rule to explain this behaviour.

Fe(η5-C5H5)2 18 electron rule = valence electrons from Fe is 8, electrons from C5H5 is 5 therefore 8 + 5 + 5 = 18 so it is stable.

Ni(η5-C5H5)2 18 electron rule = valence electrons from Ni is 10, electrons from C5H5 is 5 therefore 10 + 5 + 5 = 20 so Ni is not in most stable form

Ni(η5-C5H5)(η3-C5H7) 18 electron rule = valence electrons from Ni is 10, electrons from C5H5 is 5.....

how many electrons from (η3-C5H7) and how does this all explain why the iron complex doesn't react with hydrogen but the nickel one does? It is to do with the fact it has two too many electrons (20e) so it bonds with two protons from hydrogen? Also what does this symbol η mean?

Hydrogenation of NiCp2 takes place on the ligand, not the metal, to reduce donor electron count, giving a low-energy closed-shell 18 electron count.

http://en.wikipedia.org/wiki/Nickelocene
http://en.wikipedia.org/wiki/Ferrocene
http://en.wikipedia.org/wiki/Hapticity
http://www.chem.mun.ca/homes/cmkhome/Organometallics_The%20Basics.pdf
p. 9, 10. "η" is "hapticity," the number of ligand atoms contributing pi-electron orbitals to the metal center.

$\ce{Cp^-}$ is aromatic (draw the Frost diagram) when coordinated $\ce{η^5}$ on-face, or $\ce{η^1}$ non-aromatic as a simple anion coordinated on-vertex. Saturating one localized double bond leaves cyclic delocalized allyl anion coordinated $\ce{η^3}$.

In the case of $\ce{TiCp4}$, two Cp anions are face pentahapto, two are vertex monohapto. If we take a proton or C-13 NMR, will we see discrete protons or carbons for the $\ce{η^1}$-Cp, or will allowed 1,5-shifts scramble all five protons? All five carbons? Will it be temperature dependent if it happens? If it happens, will H and C have different kinetics? 