My textbook(NCERT) says:

With exception of $\ce{Fe2(CO)9}$, all other metal carbonyls are soluble in hydrocarbon solvents.

Weller, M.; Overton, T.; Rourke, J.; Armstrong, F. Inorganic Chemistry, 6th ed. states:

The most striking exception among the common metal carbonyls is nonacarbonyl diiron(0), which has a very low vapour pressure and is insoluble in solvents with which it does not react.

Other metal carbonyls are well soluble in hydrocarbon solvents but not Diiron nonacarbonyl. Why? What makes Diiron nonacarbonyl so exceptional? Is it because it has low vapor pressure?

There are no explanations regarding the insolubility of diiron nonacarbonyl. Is there any experiment conducted to determine the reason for its insolubility or is it just mere observation?


@Orthocresol has told me to draw comparison of diiron nonacarbonyl with $\ce{[Mn2(CO)10]}$ and $\ce{[Co2(CO)8]}$ because of their similar molecular weight. Based on this fact, I went on to research further on this topic and from various handbooks and research notes I delved upon, I have drawn the following solubility comparison table. [Note: Only binuclear metal carbonyls were considered of form $\ce{[M2(CO)_x]}$ because they have same structure and thus should have same physical properties].

\begin{array}{c|c} \mathbf{Metal~carbonyl} & \mathbf{Solubility} \\\hline \ce{Mn2(CO)10} & \mathrm{ether, other~organic~solvents}\\ \ce{Tc2(CO)10} & \mathrm{ether,acetone} \\ \ce{Fe2(CO)9} & \mathrm{Insoluble~in~benzene,ether,petrol.Only~soluble~in~THF}\\ \ce{Rh2(CO)8} & \mathrm{organic~solvents}\\ \ce{Ir2(CO)8} & \mathrm{Ether,CCl_4}\\ \ce{Co2(CO)8} & \mathrm{petrol,benzene,alcohol}\end{array}

Also, a table of comparison for the other two iron carbonyls for a reference.

\begin{array}{c|c} \mathbf{Iron~carbonyl} & \mathbf{Solubility} \\\hline \ce{Fe(CO)5} & \mathrm{soluble~in~all~organic~solvents~like~ether,petroleum.Insoluble~in~water}\\ \ce{Fe3(CO)12} & \mathrm{Insoluble~in~water. Soluble~in~non~polar~organic~solvents. } \end{array}

We can observe that there is a drastic anomaly in case of solubility of diiron nonacarbonyl in both the tables. Any explanation to this?

  • 2
    $\begingroup$ It's solid not liquid like Fe(CO)5 therefore it has low vapor pressure. $\endgroup$
    – Mithoron
    Commented Nov 26, 2016 at 17:34
  • $\begingroup$ @Mithoron Ok, so why it exist as solid whereas other iron carbonyls are liquids? $\endgroup$ Commented Nov 27, 2016 at 4:29
  • $\begingroup$ Well, almost twice as big molecule make wonders ;) $\endgroup$
    – Mithoron
    Commented Nov 27, 2016 at 15:48
  • $\begingroup$ If it's solely based on molecular weight, then I suppose a good comparison would be $\ce{[Mn2(CO)10]}$ or $\ce{[Co2(CO)8]}$. $\endgroup$ Commented Aug 27, 2017 at 16:33
  • $\begingroup$ @orthocresol. I added a comparison table. $\endgroup$ Commented Nov 20, 2017 at 7:26

1 Answer 1


Wikipedia (primary reference) suggests a possible reason for $\ce{Fe2(CO)9}$ dissolving preferentially in THF versus nonpolar solvents: it reacts according to the scheme

$\ce{Fe2(CO)9 + THF <=> Fe(CO)5 + Fe(CO)4 \cdot THF}$

Such a reaction is invoked to account for the dinuclear complex giving mononuclear products with various ligands in THF. A similar reaction with the corresponding cobalt and manganese species would have to produce radical or ionic products, due to the odd atomic number of the metal, and so would be less favored.

This would not, of course, explain the lack of solubility in nonpolar solvents or nonvolatility of $\ce{Fe2(CO)9}$. It is possible, however, that the same $\ce{Fe2(CO)9}$ molecules that bind with the oxygen in THF could also bind intermolecularly through the oxygen atoms, in effect delocalizing the covalent bonding and creating more intermolecular cohesion.

We can represent this type of interaction using the SMILES structural notation. Fr a single $\ce{Fe2(CO)9}$ molecule it would look like this, taken from Wikipedia:


For two molecules we would have


Then we have a contributing structure where an oxygen atom from the left molecule could combine with the right one, displacing an iron pentacarbonyl molecule:


where the blue atoms form an intermolecular bond in this contributing structure. The displaced iron pentacarbonyl molecule is in the last set of brackets.

The Case of Hydrogen Bonding

The delocalization of covalent bonds, hypothesized above, may be seen in a more familiar context: hydrogen bonding. Usually thus is rendered as an electrostatic attraction between oppositely charged atoms. But it may also be considered as a molecular-orbital interaction: a nonbonding electron pair overlaps an adjacent molecule's antibonding orbital in exchange for creating an "intermolecular" covalent bond. With water it would look like this


with the hydrogen-bonded, "intermolecular bonded" contribution


where the intermolecular bond is between the blue atoms. This is also the autoionized structure for water.

  • $\begingroup$ Seems like a possibility however can you share any place else where such a phenomenon is displayed? (The one mentioned in 2nd paragraph) $\endgroup$
    – user78585
    Commented Nov 20, 2019 at 17:05
  • 1
    $\begingroup$ It's actually analogous to hydrogen bonding where molecules of a substance such as water interact through their hydrogen atoms. $\endgroup$ Commented Nov 20, 2019 at 17:30
  • 2
    $\begingroup$ This is all my hypothesis, and we do not know enough about diiron nonacarbonyl to truly say. But xenon difluoride has some interesting characteristics pointing in the same direction. When I get more time I will discuss this case, hopefully this weekend. $\endgroup$ Commented Nov 20, 2019 at 21:22
  • 1
    $\begingroup$ Yes that is a good start. Looking forward for a complete answer. Also, thanks to @StackUpPhysics for holding a bounty on this question. I had been meaning to hold a bounty on this question about an year ago but just forgot about it :(. $\endgroup$ Commented Nov 21, 2019 at 5:45
  • 2
    $\begingroup$ @NilayGhosh No worries, I found a few questions very interesting hence I added a bounty. I am also the one gaining here :) $\endgroup$
    – user78585
    Commented Nov 21, 2019 at 6:09

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