[BeepoGirl]

A question in a tutorial sheet I can't seem to find the answer for. Maybe I'm being really dim but I am assuming they mean the boron is withdrawing electron density from the other atoms, but I didn't think it was that electronegative as an element? Does anyone have any ideas?

It also asks how B2H6 and BF3 alleviate the electron deficiency. I'm guessing once I know the first part however this will be obvious. 

Thanks a lot.

[BeepoGirl]

Ah. Just thought, maybe just to do with the face that boron compounds have 6 electrons in, not fulfilling the octet rule. Deary me.

[aldoxime_amine]

Ah. Just thought, maybe just to do with the face that boron compounds have 6 electrons in, not fulfilling the octet rule. Deary me.

Yes boron compounds behave like Lewis acids for that reason.

What about the second part of the question?

[BeepoGirl]

I suppose the fluorine donates some electron density to the empty orbital to fill it, but I'm not sure about B2H6. I looked up the structure, it appears that the 2 hydrogens form 2 bonds each to each of the boron atom in a bridge. Which fulfills the octet rule for boron but where do the extra electrons come from and can hydrogen make 2 bonds like that?

[craiglen]

You're right, the Hydrogens do bond to both borons, it's unusual. The two borons both have three valence electrons each. They have normal bonds with the two outer Hydrogens, as you would expect. This leaves the borons with one remaining electron each. The bridging hydrogen atoms share electrons with the two borons, so you have B-H-B 3-electron-2-center bonding. These bonds are weaker than standard boron hydrogen bons but they are bonds nonetheless.

Wikipedia covers it well: http://en.wikipedia.org/wiki/Diborane#Structure_and_bonding