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Cindy Nguyen 1L
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Postby Cindy Nguyen 1L » Wed May 23, 2018 1:02 am

In the textbook, it says, "As well as a bonding pair of electrons, a fluorine molecule also possesses lone pairs of electrons; that is, pairs of valence electrons that do not take part in bonding. The lone pairs on one F atom repel the lone pairs on the other F atom, and this repulsion is almost enough to overcome the favorable attractions of the bonding pair that holds the atoms together. This repulsion is one of the reasons why fluorine gas is so reactive: the atoms are bound together as F2 molecules only very weakly."

I understand the repulsion strength of lone pairs is high, but just to confirm, when it says fluorine gas is so reactive, does it mean this covalent bond would break apart in favor of bonding to something else other than another fluorine atom? Also, I would like some clarification on "almost enough to overcome." Does this mean the fluorine atoms' covalent bond is greater than the strength of the repulsion? So that's why two fluorine atoms would bond? That having its octet filled is basically "worth" the lone pair repulsion?

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Re: Fluorine

Postby Chem_Mod » Wed May 23, 2018 9:32 am

You are correct, when it says that fluorine gas is so reactive it is saying that the covalent bond that holds the two fluorine molecules together is so weak that it will easily break and react with other molecules or atoms. Similarly, when it says that the lone pair repulsion is almost enough to overcome the covalent bond it is saying that the lone pair repulsion is almost too strong, to the point that the two fluorine atoms would not bind at all. It is basically another way of saying that the binding between the two fluorine atoms is incredibly weak, so weak that it almost doesn't happen at all but the covalent bond is slightly favorable over not binding so the two atoms will bind. Your explanation as to why the covalent bond still happens is accurate, having a filled octet is worth the lone pair repulsion.

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