Predicting bond-strength of metal carbonyls

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.


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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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