Shape of BF3

(Polar molecules, Non-polar molecules, etc.)

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Shape of BF3

Postby JulieAljamal1E » Sat Nov 24, 2018 3:14 pm

When drawing the Lewis structure for Boron Trifluoride, why do we allow it to have an incomplete octet? I get that adding a lone pair to the Boron would mess up its formal charge, but why can it then have an incomplete octet? This impacts determining the molecular shape because it involves regions of electron density that's why I'm asking here.

Kathryn Wilhem 1I
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Joined: Fri Sep 28, 2018 12:27 am

Re: Shape of BF3

Postby Kathryn Wilhem 1I » Sat Nov 24, 2018 3:23 pm

If you look at the valence electrons of B and F, Boron has 3 valence electrons and Fluorine has 3(7) valence electrons, totaling 24 total electrons to use in your lewis structure. If you use 6 electrons for B-F bonds, then you have 18 left over. F won't be stable unless it has a complete octet, so you must use all of the 18 electrons for fluorine atoms. This leaves you with none left for boron. If you look at formal charges, this is a stable lewis structure because both F and B have a formal charge of 0.

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