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?

  • $\begingroup$ @LightYagami The first part of your question was easy to understand, I have commented on it there; the second is where my question lies. $\endgroup$
    – m-Xylene
    Apr 3, 2021 at 15:27
  • $\begingroup$ @Rishi what does that have to do with the question? Is the given explanation wrong? $\endgroup$
    – m-Xylene
    Apr 3, 2021 at 15:28
  • $\begingroup$ @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... $\endgroup$
    – V.G
    Apr 3, 2021 at 15:40
  • $\begingroup$ @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. $\endgroup$
    – m-Xylene
    Apr 3, 2021 at 15:43
  • 4
    $\begingroup$ 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. $\endgroup$
    – H.Linkhorn
    Apr 4, 2021 at 10:30

3 Answers 3


A common measure of $\ce{C-O}$ bond strength is $\ce{C-O}$ distance. For $\ce{[Fe(CO)5]}$ the data can be found, for example, in paper 1, where values of 1.12-1.14 Å are reported. For $\ce{[V(CO)6]-}$ in paper 2 values around 1.17-1.28 Å are reported.

As for rationalization, there are several factors in play here with opposite effect. It is not possible to guess which one wins beforehand. Here are two factors

  1. The iron complex shares 8 electrons between 5 carbonyl ligands while the vanadium complex shares 6 electrons between 6 ligands, so iron provides more electrons per ligand
  2. The vanadium complex has the central atom withlower electronegativity and higher negative charge, so donation from vanadium is easier.

First of all your answer, :

The order of CO bond order is : $\ce{[Mn(CO)6]+}>\ce{[Fe(CO)5]}>\ce{[Cr(CO)6]}>\ce{[V(CO)6]-}$

Surprising position of $\ce{[Fe(CO)5]}$ !!


CO bond length is compared by studying it's stretching frequency . I won't go into the details, but to give an idea, CO bond order is directly proportional to it's stretching frequency.

[Note: Although it's frequency, but typical units of wave number are used in IR Spectra]

For details on stretching frequency:

Stretching frequencies of some complexes are:

  • $\ce{[V(CO)6]-} = 1859 cm^{-1}$
  • $\ce{[Cr(CO)6]} = 2000 cm^{-1}$
  • $\ce{[Mn(CO)6]+}= 2100 cm^{-1}$

$\ce{[Fe(CO)5]}$ complex is characterized by two stretching frequencies of 2034 and 2014 $cm^{-1}$ due to axial and equatorial CO ligands. But both the stretching frequencies are greater than that of $\ce{[Cr(CO)6]}$.

Hence the order of CO bond strength is $\ce{[Mn(CO)6]+}>\ce{[Fe(CO)5]}>\ce{[Cr(CO)6]}>\ce{[V(CO)6]-}$.

So weakest bond of CO is in $\ce{[V(CO)6]-}.$

Theoretically, the order $\ce{[Fe(CO)5]}>\ce{[Cr(CO)6]}>\ce{[V(CO)6]-}$ can be understood as:

  • $\ce{[V(CO)6]-}$ has negative charge as compared to other to so it donates most electron density in $\pi^{*}$ orbital of CO ligand and hence CO has least bond order in $\ce{[V(CO)6]-}$
  • Other two complexes are neutral. The order of effective nuclear charge is $\ce{Fe} > \ce{Cr} $ due to more $d e^{-}s $ in in $\ce{Fe}$. Hence $\ce{Fe}$ has less donation tendency then $\ce{Cr}$ and so it donates less electron density in $\pi^{*}$ orbital of CO and therefore CO has higher bond order in $\ce{[Fe(CO)5]}$.

Hope helpful :)

Source for $\ce{[Fe(CO)5]}$ stretching frequency: Adams, R. D.; Barnard, T. S.; Cortopassi, J. E.; Wu, W.; Li, Z. "Platinum-ruthenium carbonyl cluster complexes" Inorganic Syntheses 1998, volume 32, pp. 280-284.


Credits: Dr. Michael Evans

The given answer is incorrect. From the article:

As we move left to right, the d orbital energies decrease and the energies of the $\textrm{d}\pi$ and $\pi^{*}$ orbitals separate. As a result, the backbonding orbital interaction becomes worse (remember that strong orbital interactions require well-matched orbital energies) as we move toward the more electronegative late transition metals!

  • $\begingroup$ You should explain it fully ,,and not just give a vague reference $\endgroup$
    – Amit
    Mar 4 at 16:29

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