The metal carbonyls (and similar organometallic compounds) involve a combination of sigma bond, a pi bond and backbonding. The bond strengths under consideration are the metal-carbon bond and the carbon oxygen bond. Here is the original question:-

Predict the order of $\ce{C-O}$ bond strength in the following:-
I) $[\ce{Mn(CO)_6}]^+$
II) $[\ce{Cr(CO)_6}]$
III) $[\ce{V(CO)_6}]^-$

All of these are isoelectronic, and hence whatever the difference of bond strengths, must arise from the difference in the resultant nuclear force of attraction. Since $\ce{Mn+}$ has the highest charge density, I would assume that the $\ce{M-C}$ bond strength of the manganese compound is the strongest (vanadium being the weakest) and therefore, the $\ce{C-O}$ bond strength must follow the opposite order, as the strengthening of the metal carbonyl bond should weaken the carbon oxygen bond.

But the answer given is III<II<I, exactly the opposite of my prediction. Why is it so? A relevant factor which I didn't consider is the back bonding, but I am unsure of how to integrate that into my prediction. Also, how would we make a prediction if the given compounds were not isoelectronic?


As the π-antibonding orbital of CO is filled by electrons from the metal, they weaken the C-O bond compared with free CO. The metal-carbon bond is strengthened in kind. That's what the books says.

  • 1
    $\begingroup$ Considering that, how can we predict the order of bond-strength? $\endgroup$ – stochastic13 Feb 19 '14 at 8:01

It is due to negative charge on vanadium. Vanadium in this comples ion is negatively charged and thus has some extra electrons readily available for donation to the $\pi$-anti bonding molecular orbitals of carbon in $\ce{CO}$ molecule. Thus $\ce{V-C}$ bond in $\ce{[V(CO)5]-}$ is much stronger as compared to other two and hence $\ce{C-O}$ bond order is least in this among the three. If this might help. This what my teacher taught me.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.