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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?

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  • $\begingroup$ @LightYagami The first part of your question was easy to understand, I have commented on it there; the second is where my question lies. $\endgroup$ – Meta xylene Apr 3 at 15:27
  • $\begingroup$ @Rishi what does that have to do with the question? Is the given explanation wrong? $\endgroup$ – Meta xylene Apr 3 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... $\endgroup$ – Harry Potter Apr 3 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. $\endgroup$ – Meta xylene Apr 3 at 15:43
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    $\begingroup$ 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 at 10:30

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