[d1wedemeier]

According to my text the electronegativity between B and F is 4.0-2.0=2.0.  This should result in an ionic bond because it is greater than 1.7.  Other sites on the internet say that it is a polar covalent bond.  Is BF3 some sort of exception or is it an ionic bond as the work from the text indicates?  

[Mitch]

Its somewhere in the middle. But definately on the covalent side. Here is a question for you. How can BF3 exist? Isn't it violating the octet rule, try to reason why it is "stable". Others please don't answer for him.

[d1wedemeier]

Could it be a double covalent bond?

[Mitch]

Good job, some of the electrons from the non-bonding electrons on flourine donate into the empty p-orbital of Boron. Its not exactly a double bond, but not exactly a single bond either.

[d1wedemeier]

In my text book it says that any bond that exceeds  1.7 difference in electronegativity is ionic.  I saw that it violated the octet rule leaving me to think that it had to be ionic.  How can I put down polar covalent as the answer when I write out the difference and it falls in the ionic category?

[Mitch]

How can I put down polar covalent as the answer when I write out the difference and it falls in the ionic category?

I don't know.

[Donaldson Tan]

in my opinion, the B-F bond is described as covalent for simplification (and convenience) in high school chemistry syllabus. i don't think a high school student is expected to understand it to such rigourous level.

With reference to the attached MO diagram, all MO energy levels are filled with electron pairs, except the two highest ones. there are 6 pairs of electrons not involved in bonding (see orbitals 1a2', 1e'', 2e' and 3a1'). There are 12pairs of valence electrons in all, meaning 6pairs are involved in making a molecular orbital. This means each F contribute 2 pairs of electron. However, to not actually violate octet rule, it means each F atom contribute one electron in making a B-F bond, and contribute an electron pair partially into boron orbitals.