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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?

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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.

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

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