# Comparison of C=O bond strength in metal carbonyls

Find the weakest $$\ce{C=O}$$ bond among $$\ce{[Mn(CO)6]+},$$ $$\ce{Fe[(CO)5]},$$ $$\ce{[Cr(CO)6]}$$ and $$\ce{[V(CO)6]-}.$$

I thought the $$\ce{C=O}$$ bond strength would be lowest in $$\ce{[V(CO)6]-}$$ since the negative charge meant that vanadium had an excess of electrons, hence the metal-carbon back-bonding would be the strongest. As the result, the $$\ce{C=O}$$ bond order would be the least.

However, the answer given states that iron in $$\ce{Fe[(CO)5]}$$ has a $$\mathrm{3d^8}$$ configuration, which means four electron pairs for back-bonding as compared to $$\mathrm{3d^6}$$ in $$\ce{V-},$$ hence it would have a greater degree of back-bonding and thus its carbonyl bonding would be the weakest.

Is this correct or does $$\ce{[V(CO)6]-}$$ have the strongest metal-carbon synergistic bond? If the former is correct, then does a synergistic bond depend on the number of lone pairs first and a negative charge is only considered if the number of lone pairs is the same?

• @LightYagami The first part of your question was easy to understand, I have commented on it there; the second is where my question lies. Apr 3 at 15:27
• @Rishi what does that have to do with the question? Is the given explanation wrong? Apr 3 at 15:28
• @Metaxylene Also see this question, I wonder, why nobody provides a definite answer to this question, maybe due to lack of data of bond lengths? So, it may just become handwaving... Apr 3 at 15:40
• @LightYagami In the isoelectronic case, I see that it depends on the electron density. So maybe in other non-isoelectronic cases, the number of d$\pi$ orbital electrons is deciding. I agree though- it seems like no one knows for sure what the trend/property is. Apr 3 at 15:43
• That comparing $\ce{[Fe(CO)5]}$ and the octahedral complexes isn't a fair comparison. So your explanation that you deduced in your question only applies to the octahedral complexes and you should then treat $\ce{[Fe(CO)5]}$ separately. The drop is coordination number also means the metals $d$ electrons are "shared" among less CO ligands and so the degree of backbonding can be expected to be greater. Apr 4 at 10:30