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Why is the $\ce{F-F}$ bond relatively weak? And why does bond strength decrease from $\ce{Cl-Cl}$ to $\ce{I-I}$?

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Does this have anything to do with "lone-pair" repulsion on the rather small $\ce{F-F}$ molecule? I haven't asked my professor but I'm suspecting this is going to the first thing on his mind. I'm sure that many of you here will have rebuttals to this argument.

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On the other hand does $\ce{F2}$ even have lone pairs? There are no non-bonding orbitals populated on the molecular orbital diagram of fluorine. I wouldn't expect there to be any non-bonding orbitals populated either, as this is a simple diatomic we're dealing with and the atoms involved are identical ... so each atom's orbital should be able to interact with the other atom's orbital effectively.

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    $\begingroup$ My initial guess would be that given the high electronegativity of fluorine, the electron density will have an appreciable restriction of motion that will hinder the overlap of electron density in the singly-occupied orbitals used in MO theory. This subdued overlap leads to a weaker bond with respect to the other halogens. $\endgroup$
    – LordStryker
    Commented Oct 9, 2014 at 15:48
  • $\begingroup$ What about the lone pair argument? Yay or nay? @Lordstryker $\endgroup$
    – Dissenter
    Commented Oct 9, 2014 at 15:51
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    $\begingroup$ As for fluorine: antibonding unbond a bit more strong than bonding bond and $\pi$-interaction are the strongest for the second row. As for bromine and iodine: orbitals become increasingly diffuse down a column, so any covalent bonding becomes weaker down a column. $\endgroup$
    – permeakra
    Commented Oct 9, 2014 at 15:51
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    $\begingroup$ See also chemistry.stackexchange.com/questions/16276/… for a discussion of "charge shift bonding" in $\ce{F2}$ which seems to be a more accurate representation than either VB (lone pairs) or MO in this case. $\endgroup$
    – Geoff Hutchison
    Commented Oct 9, 2014 at 15:59
  • $\begingroup$ @Dissenter I would imagine that given the decreased 'diffuseness' of electrons around Fluorine (again F is highly electronegative so it holds on 'tightly' to these electrons more than other elements), and at lower energy orbitals, this would lead to larger electron-electron repulsions (since you are effectively restricting the space electrons can be found in). $\endgroup$
    – LordStryker
    Commented Oct 10, 2014 at 14:49

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I see as I write the answer, that many comments are spot-on.

Whether or not $\ce{F2}$ has lone pairs depends on your view of molecular orbitals and valence bond (or Lewis) descriptions of bonding.

Covalent radii for fluorine compounds are usually around 0.64 Å for F, but yet the F-F bond distance is 1.43 Å.

If we go for the Lewis or valence bond view, we'd argue that the large number of lone pairs on a small atom will be highly repulsive.

If I go for the molecular orbital view, I'd point out that in reality, the anti bonding orbitals $\pi^*$ and $\sigma^*$ are higher in energy than their bonding counterparts, particularly in the 2nd row, so the F-F molecule will be weaker than a simple single bond.

There's a really nice article by Peter Politzer in JACS that I like to use when discussing this "anomaly": "Anomalous properties of fluorine" J. Am. Chem. Soc., 1969, 91 (23), pp 6235–6237

It appears rather that the anomaly involves the fluorine atom, in whatever circumstances, and its interaction with an external electron which enters its outer shell. [...] In any case, it is suggested that because of the exceptionally small size of the fluorine atom (see Table I), the entering electron feels unusually large forces of attraction from the nucleus and repulsion from the electrons already associated with the atom. It appears that the magnitude of the anomaly in the repulsive forces is greater than in the attractive forces.

In other words, regardless of the MO or VB view, the issue is one of electrostatic repulsion.

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  • $\begingroup$ There's also a nice article I just found on The Antibonding Effect. $\endgroup$
    – Geoff Hutchison
    Commented Oct 9, 2014 at 15:58
  • $\begingroup$ Aren't anti bonding orbitals always higher in energy than their corresponding bonding orbitals ? $\endgroup$
    – Dissenter
    Commented Oct 9, 2014 at 18:22
  • $\begingroup$ Yes. But in your figure, the bonding orbitals have the same amount of stabilization as the amount of destabilization in the anti-bonding orbitals. In fact, the $\Delta E$ between the atomic orbital levels and the anti bonding MOs is much greater. I guess I'd say the anti-bonding orbitals are less stable than you'd expect simply from the bonding stabilization. $\endgroup$
    – Geoff Hutchison
    Commented Oct 9, 2014 at 18:33
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    $\begingroup$ Yes. But this depends on the magnitude of the overlap of the orbitals (large in the 2nd row). Since $\ce{F2}$ has a lot of occupied anti bonding orbitals, there's a large amount of destabilization. That is, from an MO picture, the "bond order" might be 1, but it's a weaker single bond than say $\ce{B2}$ because of this destabilization. $\endgroup$
    – Geoff Hutchison
    Commented Oct 9, 2014 at 18:48
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Geoff Hutchison
    Commented Oct 10, 2014 at 2:35

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