I recently stumbled upon a question asking how iron destabilizes the bonds in a nitrogen molecule. I have searched the web and my textbook for quite a while now and I cannot find a satisfactory answer. The only explanation I could muster was that the d-orbitals of the Iron atom interact with the HOMO and LUMO of the nitrogen molecule, due to the overlap of the molecular orbitals with d-orbitals. I am not sure if this makes sense. I'd really appreciate any answer or advice!

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    – airhuff
    Mar 15, 2017 at 16:07

1 Answer 1


You are on the right track with your answer.MO Diagram for N2

Notice that the LUMO is a $\pi^*$ orbital, meaning that filling it with electrons would weaken the bonding. When nitrogen and iron interact, electrons from the d-orbitals on iron are well aligned, in terms of symmetry and energy, to fill into $\ce{N_2}$'s $\pi^*$ orbital. This forms a bond between $\ce{Fe}$ and $\ce{N_2}$, but at the cost of weakening the bond between the nitrogens. This is a very important concept in inorganic chemistry referred to as Pi Backbonding.

The classic example given of a ligand that exhibits $\pi$ backbonding in metal complexes is $\ce{CO}$, which is isoelectronic to $\ce{N_2}$ and has the same molecular orbital diagram (though the orbitals themselves look different due to the asymmetry of $\ce{CO}$).

  • $\begingroup$ If $d$ orbitals interact with $\pi^*$ then new molecular orbitals are created and electrons go to the bonding MO. I don't think it makes much sense to talk about bond weakening just because the electrons went into a $\pi^*$. In fact they don't even fill the $\pi^*$ of CO. They fill the MO that result from the combination of $\pi^*$ (CO) and $d$ (metal). $\endgroup$
    – ado sar
    Nov 12, 2022 at 23:10

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