[Dolphinsiu]

This question appears in my past exam paper. I feel very difficult.

1. Although they have the similar formulas, the compund BF3 exists but the compound BH3 does not. Offer a plausible explaination of this fact. Describe two possible methods to prepare stable BH3.

How to explain?Using Molecular Orbital Theory seems not work.Also hydrogen can exist as H+ and H-, I feel doubtful that BH3 does not exist.

In addition, I have no idea in prepare stable BH3 as I haven't been taught in my class. If this type of question appears in my exam. How can I do?

Such as HN3 and NH3, why NH3 is more stable than HN3? How can I prepare stable HN3?(I think I will abandon this question in my exam!!)

[Dolphinsiu]

No one can give some hints?

[Dan]

BH₃ does not exist in monomeric form. In what form does it exist?

In order to exist in this form, an available empty p orbital on boron is required.

Why might the empty p orbital on B in BF₃ not be available?

[Dolphinsiu]

http://www.madsci.org/posts/archives/dec2001/1009553236.Ch.r.html

From this website, I only know BH3 exist as dimer forms.

But I still don't know why BH3 does not exist as monmeric form!!!

Also electron configuration of boron is (2,3), that is, 1s22s22p1
excited state is 1s22s12p2.

Then your question seems to increase my misunderstanding!

[Dan]

http://www.madsci.org/posts/archives/dec2001/1009553236.Ch.r.html

From this website, I only know BH3 exist as dimer forms.

Indeed.

But I still don't know why BH3 does not exist as monmeric form!!!

Also electron configuration of boron is (2,3), that is, 1s22s22p1
excited state is 1s22s12p2.

Then your question seems to increase my misunderstanding!

Ok, if we consider the hypothetical structure of BH₃, we expect sp2 hybridisation. So the 6 bonding electrons (3 from B and 1 from each H) would be in 3 sp2 orbitals and there would be an empty p orbital (just like a carbocation, but it wouldn't be charged). The Hs would be in a trigonal planar geometry around the B.

Right, problem is, this molecule is very electron deficient and really wants some more electrons to put in that empty p orbital. Next problem, where the hell is it going to get them from? another molecule of BH₃! so the two molecules bond together and form the dimer because they are both so greedy for electrons!

Now, to the question at hand. Why doesn't BF₃ do the same thing? Well, the central B still has an empty p orbital as before, but there is a closer source of electrons to put in it than another molecule.... where do you think that source is? Hint: See if you can think of any resonance structures for BF₃

It is easier to approach this question from the "why doesn't BF₃ dimerise" direction than the "Why does BH₃ dimerise" direction. If you see what I mean...

[Dolphinsiu]

I see your point!

But as you say, in case of BH3, B is electron deficient as empty 2p orbital exists.
BH3 cannot undergo resonance due to empty 1s orbital of H. Is that right?
If so, all hydrogens in another BH3 has also empty 1s orbital.
Why BH3 still can form dimer but not undergo resonance?
What is the bonding in between in dimer?Hydrogen bond?Dative covalent bond?

[Dan]

There isn't really an empty 1s on H, the bonding is covalent, so electrons are shared between B and H in a sigma bonding orbital.
The lack of resonance is simply due to the fact that you can't move any electrons without disconnecting atoms and producing very very high energy species

The way I look at it is that BH₃ forms the dimer because  it is not resonance stabilised. How about BF₃?

The bonding in diborane (B₂H₆) is rather unusual. There are "3 centre, 2 electron" bonds. The MOs for these bonds are made up of combinations of 3 atomic orbitals. Look up "diborane" in a good inorganic chemistry textbook for full details.

[Dolphinsiu]

Thank you very much, Dan!
I feel that I have learnt a lot from you!
I will check this special atom for detail!