# Why does cyclobutadieneiron tricarbonyl behave aromatically?

It is said that $$\ce{(\eta^4-C4H4)Fe(CO)3}$$ can undergo electrophilic substitution reactions. Therefore, it displays aromaticity.
For the iron atom, it has $$8$$ electrons in its outer shell initially and it receives $$6$$ electrons from three carbonyls. It achieves the stable state of $$18e$$ after bonding with cyclobutadiene. Therefore, the oxidation state of iron should be zero, and the $$\pi$$ electrons of cyclobutadiene in the complex is still $$4$$.
But this result contradicts with Huckel's rule, which states that the $$\pi$$ electrons should be $$4n+2$$. Where did I go wrong?

• You imply that during bonding the C4H4 gives its 4 electrons to iron for nothing in return. Now there is a catch: bonding is a game for two. – Ivan Neretin Oct 1 '19 at 5:00
• @IvanNeretin One of my professors once said ‘you notice that that, what doesn’t work with humans, namely socialism, works beautifully with electrons.’ – Jan Oct 1 '19 at 14:24
• @Jan True, or so it seems. Then again, I never had a chance to discuss that with an electron. – Ivan Neretin Oct 1 '19 at 14:29
• @IvanNeretin Exactly, this was what I thought. I have never thought about the delta bond. Now I understand, with the help of the paper, the beautiful game for these two fragments and the exactly same symmetry of MOs. – Kemono Chen Oct 2 '19 at 3:07