# Comparing bond lengths in metal carbonyls

Question:

It's solution is given as:

Now I don't understand which d$$\pi$$ electrons have been considered here. For example, in $$\ce{Fe(CO)_5}$$, according to this, number of d$$\pi$$ electrons would be 4. I cannot see how it is 4. Secondly, is it the right method to compare the bond lengths?

I thought of this in terms of electron density, more is the electron density in the complex, stronger would be the metal ligand bond, hence weaker the carbon-oxygen bond, hence more the C-O bond length. But, since the atoms are different, I don't know if this idea would work in this case.

• Related: chemistry.stackexchange.com/questions/64999/…. There is a good answer to find bond order. – Mathew Mahindaratne Mar 28 at 17:07
• @MathewMahindaratne Thanks for the link, but unfortunately I don't understand MOT in much detail, just the basics. For example, I am only familiar with $\ce{t_{2g}}$ and $\ce{e_g}$ orbitals. I have now got the part how the electron pairs are 4. It is because since CO is a strong field ligand, the configuration of $\ce{Fe}$ would be $\ce{[Ar] 3d^8}$. But still, my question remains, is this a correct method? – Light Yagami Mar 28 at 17:51
• Each $\mathrm{d} \pi$ pair contributed to back bonding with $\ce{C#O}$ carbon so triple bond character get weaken (e.g., $\ce{Fe=C=O}$). That may be the theory behind this calculation. It is a crude method but works. – Mathew Mahindaratne Mar 28 at 20:23
• For the $Fe(CO)_{5}$ part, note that the formula is number of dπ electron $pairs$, hence $0.8$ indicates $4$ pairs of electrons or 8 electrons in total, as seen in the $3d^{8}$ configuration in $Fe$ in the presence of the strong field ligand $CO$. I have the same question in the second part however. – Meta xylene Apr 3 at 15:22