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Why can't fluorine be the central atom in inter-halogen compounds?

A $\ce{F-F}$ bond is weaker than a $\ce{F-X}$ bond and thus, fluorine should be happy to form inter-halogen compounds. But, why doesn't it act as the central atom?

In my textbook, the reason given is because of the high electronegativity of fluorine atoms. But, how does that affect fluorine being the central atom or not?

If fluorine is the central atom, it can draw electrons from other less electronegative halogens and be happy. Does this lead to instability of the compound?

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    $\begingroup$ Fluorine being smallest than all other halogens can't act as the center. Majority of Inter halogen compounds are formed by $Br$ and $I$, which combine with small size atoms such as fluorine, since it is possible to associate more small size atoms around a large central atom. Your text book is right, more electronegative atoms are smaller in size, as like fluorine. $\endgroup$
    – Sensebe
    Jan 13, 2014 at 15:17

5 Answers 5

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Why can't fluorine be the central atom in inter-halogen compounds?

First off, fluorine can be at the "center of things." Examples would include the strongly hydrogen bonded hydrofluoric acid

enter image description here

and the very relevant example of the trifluoride anion

$$\ce{[F-F-F]^-}$$

The trifluoride anion example is critical as it demonstrates that fluorine can be hypercoordinated (hypervalent). Of course, in this last example we've "tricked" the fluorine into occupying the middle position, so your question remains (slightly altered), why doesn't fluorine occupy a central position more often?

In order to answer this question, two general cases need to be considered,

  • The anionic situation $\ce{[X-F-X]^-}$ vs. $\ce{[X-X-F]^-}$
  • The cationic situation $\ce{[X-F-X]^+}$ vs. $\ce{[X-X-F]^+}$

where X is Cl, Br or I.

In the case of the anion, Lewis dot structures suggest a negative charge on the central atom.

enter image description here

However, computational analysis of the electron distribution in the triiodide anion (see p. 156) suggests that the central atom is positively charged (+0.27 electron), while the terminal atoms are negatively charged (-0.63 electrons on each terminal atom). Assuming that the same general trend would be applicable to mixed trihalide anions, then in the case of $\ce{[X-F-X]^-}$ we would be placing positive charge on highly electronegative fluorine atom instead of the larger, more polarizable (e.g. more capable of bearing a positive charge) bromine, chlorine or iodine atom.

In the case of the cation, similar arguments come into play. Again, calculations for a variety of trihalide cations (mixed and homogenous) suggest substantial positive charge on the central atom (this time in agreement with Lewis Dot structures). The same arguments made above would operate here to destabilize structures with a central fluorine atom. This analysis also found extremely long (less stabilizing) bonds resulted when fluorine (or the most electronegative of the 3 atoms) occupied the central position.

Arguments that don't explain the infrequent occurrence of fluorine at a central position in inter-halogen compounds include:

  • the lack of "d-orbitals"; as other's have commented, this is an old, invalid argument; the existence of $\ce{F3^-}$ demonstrates that fluorine is capable of hypercoordinate bonding without the need for d-orbital involvement
  • inefficient 2p-3p overlap; $\ce{FCl}$ has a higher bond strength and lower heat of formation than $\ce{Cl2}$
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  • $\begingroup$ Wait, your analysis of the triodide ion's charges doesn't add up to -1. $\endgroup$
    – Dissenter
    Oct 4, 2014 at 23:39
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    $\begingroup$ In the anion there are 6 + 6 + 6 = 18 lone pair electrons, plus 4 bonding pair electrons for a total of 22. In neutral fluorine there are 7 electrons; 3 x 7 = 21. 21 - 22 = -1. Am I missing your point? $\endgroup$
    – ron
    Oct 4, 2014 at 23:43
  • $\begingroup$ I mean when you say that the central atom in the trihalide ion triiodide is +0.27 and the peripheral atoms is -0.63 - that adds up to -0.99. $\endgroup$
    – Dissenter
    Oct 4, 2014 at 23:45
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    $\begingroup$ Yes, just a rounding error. $\endgroup$
    – ron
    Oct 5, 2014 at 0:16
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    $\begingroup$ Explanations that you mentioned in the end are still taught in many high schools :( $\endgroup$
    – jonsno
    Sep 9, 2017 at 4:40
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As a practical example there is (1) a fluorine-bridged iodine structure of $\ce{[R_\text{f} - I - F - I - R_\text{f}]-}$ composition with 5-centered 6-electron bond stabilized by fluorine in the middle:

enter image description here

Here is an isolated anionic part of tris(diethylamino)sulfonium bis((perfluorophenyl)iodo)fluoride, $\ce{F - I}$ distance is approx. 2.5 Å:

enter image description here

This is not exactly an interhalogen compound, but the closest known analogue I found with fluorine in the middle.

(1) Farnham, W. B.; Dixon, D. A.; Calabrese, J. C. J. Am. Chem. Soc. 1988, 110 (25), 8453–8461.DOI: 10.1021/ja00233a022

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Several examples of species with fluorine in the middle of a chain are reported here. What they have in common: the fluorine is embedded in an environment of high or enhanced electronegativity, which makes fluorine in the center more energetically favorable.

  • The fluorine-iodine bridged structure in andselisk's answer. Iodine itself is not very electronegative, but here it is incorporated into a functional group that is rendered electron-withdrawing via both the ring of $sp^2$ carbon atoms and the attached terminal fluorines.

  • Gold heptafluoride, discussed here. This low-temperature compound is a difluorine complex, having a structure that may be rendered as $\ce{F5Au - F - F}$. The geometry at the bridging fluorine atom is bent. The $\ce{AuF5}$ fragment is so electronegative it imparts partial positive charges to the fluorine atoms in the $\ce{F2}$ ligand.

  • Caesium trifluoride, $\ce{Cs(F3)}$ (not yet experimentally known). This proposed high-pressure compound would contain a true polyhalide species, specifically $\ce{F3^-}$. It is calculated to be an intermediate stage towards other salts and complexes where the caesium becomes a $5p$ electron donor and assumes oxidation states greater than $+1$. See Ref. [1].

Reference

1. Miao, M.S. "Caesium in high oxidation states and as a p-block element." Nat Chem. 2013 Oct; 5(10):846-52. https://doi.org/10.1038/nchem.1754. PMID: 24056341.

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Fluorine can't be the central atom in inter-halogen compounds because it is an element from the period 2 in the periodic table. It can not have more than 8 valence electrons. And as it has 7 valence electrons, it can only form one bond. This is not the case of the other halogens $\ce{Cl, Br, I,...}$ where these elements are below period 2 in the periodic table. That means they can have an expanded octet. They can have more than 8 valence electrons. So they can be central atoms in inter-halogen compounds.

From an orbital viewpoint, we can attribute this to the fact that in the case of $\ce{Cl, Br, I,...}$ the orbitals $\ce{3d}$ , $\ce{4d}$ and $\ce{5d}$ respectively are available for hybridization and can interact with the valence orbitals $\ce{ns, np}$ of these atoms. So these halogen atoms have the possibility to form more than one bond with other atoms. In the case of fluorine, the valence orbitals $\ce{2s, 2p}$ can not interact with $\ce{3d}$ due to the big difference of energy.

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    $\begingroup$ The use of d orbitals for hypercoordination is, as ron mentioned this in his answer, already successfully disproved. As this answer continues, the existence of $\ce{F3-}$ also disproves your one bond theory. $\endgroup$
    – Martin - マーチン ♦
    Oct 7, 2014 at 16:41
  • $\begingroup$ @martin my interpretation would still be fluorine forms just one net bond. However, this can be delocalized over multiple linkages. $\endgroup$
    – Oscar Lanzi
    Dec 13, 2021 at 17:09
  • $\begingroup$ @oscar yes, maybe, probably. Over the years and the further and farther I looked I lost sight of "the bond" and realised there are multiple interpretations, view points, opinions, etc. and no good way to reconcile any of that. However, on a purely mathematical level, the argument with d orbitals falls through - there simply is a simpler way to do it (pun intended). Therefore I remain with my vote on this answer: it's wrong. $\endgroup$
    – Martin - マーチン ♦
    Dec 15, 2021 at 0:47
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fluorine is high electronegativity so when it combine with other halogen which is less electronegativity then fluorine can take electron from other atom due to this compound become unstable that's why fluorine can't be central atom in interhalogen compound.

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  • $\begingroup$ This makes little sense. If I understood your answer, you're implying that when fluorine reacts, an unstable compound is formed than the reactants, which is usually not the case $\endgroup$
    – M.A.R.
    Jul 4, 2017 at 16:22

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