[Nescafe]

The B-F bond, I am guessing the nature of this bond is the "sigma" type? I have been looking on the web to find the orbital energy diagram for BF3 and I find nothing. The reason i am looking for this, first of all, I find drawing molecular orbitals for N2, F2, NO etc to be easy while I find them to be very difficult for whole molecules like CH3NH2. Any suggestions on how to improve is welcomed.

Anyways back to my question, I read that BF3 has an empty P orbital thus why it can accept a pair of electrons from lets say trimethyl amine (another example where I failed to draw out the MO energy diagram of). So I wanted to see it for myself so I tried drawing the MO energy diagram and failed.

Boron has three valance electrons: 1s2 2s2 2p1.
Fluorine has 7: 1s2 2s2 2p5

How are these sigma bonds when most of the electrons are in the p orbital. I am probably misunderstanding something :S How would you go about drawing the energy diagram for BF3?

Help

Nescafe.

[Dan]

I put "BF3 MO" in google images and this was the first hit:

[cheese (MSW)]

But where are the e⁻s?  What is the HOMO?  The LUMO?
How do you explain that BF3 is a weaker Lewis acid than BCl3? 

[Dan]

But where are the e⁻s?

Molecular orbitals, combinations of atomic orbitals. They are pictured on the right panel.

  What is the HOMO?  The LUMO?

How many bonding electrons? Fill in the orbital in order of energy and see what you get.

How do you explain that BF3 is a weaker Lewis acid than BCl3? 

How would you expect the the MO diagram of boron trichloride to differ from the one I posted?

[cheese (MSW)]

I now the answer Dan: I am challenging you to do it. 

[cheese (MSW)]

BF3 is an AX3 system which from VSEPR and as found is trigonal planar.  There are only six valence e⁻s about the central B atom and so it does not obey the Octet Rule.   The B is said to be sp^2 hybridized.  However hybridization is an artifact and has no place in modern bonding theory:
The z axis is taken as the axis of highest symmetry: the C3 axis that is perpendicular to the BF3 plane and passes through the B atom.   The bonding AOs on B are the 2s and 2p AOs and likewise for the F subsituents.  Each F atom has a σ AO with 1 e⁻.  The lowest energy MO is formed by the overlap of the B 2s AO (+) with a linear combination of the three F AOs (+++) to give totally symmetric  4a’ and the corresponding antibonding MO (5a1’).  The B 2px(+-) and B2py (+-) AOs are overlapped with suitable linear combinations of the F σ AOs (++-;-++) to give the degenerate pair of MOs (3e’) and the corresponding 5e’ antibonding MO (not shown on MO diagram given; at still higher energy than 5a’).   The 4a’ (2e⁻) and 3e’(4e⁻) represent the three sp^2 σ B:F bonds and as you can see give quite a different picture to the convention description.
But it gets more interesting because the empty B 2pz AO can interact with the lone pairs in the 2pz AOs of the F atoms to give a π bonding  MO (1a2’’; 2e⁻) that extends over the complete BF3 framework.  This gives the B-F linkage partial double bond character which along with the good σ overlap accounts for why it has the strongest “single” bond known (~650 kJ mol^-1).  The antibonding MO corresponding to 1a2’’ is the 2a2’’ because it is a weak interaction this antibonding MO is still low in energy (I would lower it to what is given in the MO diagram).  The 1a’ (2e⁻), 1e’’(4e⁻) and 4e’(4e⁻) are essentially nonbonding MOs that have mainly F 2p character, and can be equated with are lone pairs on the F atoms.   BF3 has 18 valence e⁻s (ignore the F 2s e⁻).  The HOMO is the 1a2’ MO and the LUMO the 2a’’ MO. 
As can be seen the latter has mainly B 2pz character and can accept a pair of e⁻s to give a Lewis acid-Lewis base adduct.   L: →BF3.  Boron trifluoride is a strong Lewis acid.  Nevertheless it is a weaker Lewis acid than BCl3 which is completely unexpected if arguments based on the electronegativity of X in BX3 are considered.   But it is readily rationalized from our MO scheme.  The B2pz-Cl3pz overlap is much less that the B2pz-F2pz overlap so the antibonding MO 2a’’ is not raised as much in energy and is better able to accept an  e⁻ pair from a Lewis base (perturbation MO theory).

I believe the chemical bond is not so simple as some people seem to think.  Robert S. Mulliken

[discodermolide]

Well Mr cheese (MSW) "not a full member", Thanks for clarifying the two posts from poor Mr Nescafe. I don't know how he feels but I did not understand one word of what you posted.
As a simple practical organic chemist could you please re-post your two replies in simpler terms? Words of one syllable would be fine for me.
Thank you.

[cheese (MSW)]

discodermolide:  Like Borek implied: three years of MO theory and you might comprehend it.
It wasn't me that posted the MO scheme.   To get you started:
Is It Time To Retire the Hybrid Atomic Orbital?
A. Grushow J. Chem. Educ., 2011, 88 (7), pp 860–862
MO Treatment of CH4: Advanced Organic Chemistry F. A. Carey and R. J. Sundberg. 5th ed. New York: Springer, 2007 p 37-39.
Professor Cheese (Ministry of Silly Walks) NFM.

[discodermolide]

discodermolide:  Like Borek implied: three years of MO theory and you might comprehend it.
It wasn't me that posted the MO scheme.   To get you started:
Is It Time To Retire the Hybrid Atomic Orbital?
A. Grushow J. Chem. Educ., 2011, 88 (7), pp 860–862
MO Treatment of CH4: Advanced Organic Chemistry F. A. Carey and R. J. Sundberg. 5th ed. New York: Springer, 2007 p 37-39.
Professor Cheese (Ministry of Silly Walks) NFM.

Thank you for the references.

[Dan]

three years of MO theory and you might comprehend it.

Come on, it's not that complicated, BX₃ is a standard case study in undergrad MO theory lectures, I'm pretty sure it was in my 2nd year inorganic exams, so it certainly doesn't require 3 years to possibly comprehend.

The Chem Ed article was an interesting rant, thanks for that. I think there will be a shift to MO theory eventually. I guess the problem is the organic chemists, but as the Author points out, hybridisation only really provides nomenclature for efficient communication. Changing the language will take time though. I always took the advice  of Ian Fleming: "It [hybridisation] should only be used when it offers a considerable simplification. It is not difficult or inconvenient simply to remember that there is no single orbital representing the C-H bond, but rather several orbitals contributing to the bonding between these two atoms" - from Frontier orbitals and organic chemical reactions.

[cheese (MSW)]

I didn't cover the MO diagram of BF3 in my second-year course on MO theory.
This may be because there is a difference in the quality of student at a UK
and a Canadian University? I did briefly mention that CH4 is not sp^3 hybridized.
The MO of CH4 etc is covered in a 4th y course (organic prof).
BF3 may have been mentioned in a 3rd y inorg course (different instructor) we had
but we did away with that course and put in a course on materials.
In order to understand the MO diagram you have to be familiar with quantum theory,
and the Bohr model, wave functions and the SWE, the LCAO method, and symmetry.
Not three years of constant study but usually met in your third year of a BSc if at all.
The link to the "rant" is
http://pubs.acs.org/doi/full/10.1021/ed100155c

[cheese (MSW)]

An elementary MO diagram for BF3 (suitable for 2nd yr undergraduates) would hybridized the B 2s, 2px and 2py AOs into three equivalent AOs each with one e⁻. These AOs are overlapped in typical LCAO fashion with three F 2p AOs (at lower energy) each with one e⁻ to give three B:F σ bonding MOs at lower energy than the constituent  AOs, and three σ* MOs at higher energy.  The six valence e⁻s fill the three σ MOs.  The B2pz is strictly nonbonding and has the same energy as the starting AO (ignore lone pairs of the F atoms).  This is the LUMO that accepts an electron pair in a L:→BF3 adduct.

[Dan]

Maybe that's how we did it, I don't remember it was several years ago, but I'm pretty sure we were taught the MO diagram as I posted (my inorganic notes are 500 miles away in storage). We did diatomic Mos 1st year and polyatomics 2nd for sure though. But that's irrelevant anyway, we can discuss it in chem ed if you have any particular interest.

I think if you and understand an MO diagram you don't need to hybridise. It is not conceptually more difficult without hybridisation, which is one of the arguments in Grushows paper you posted.

But yes, I agree that the take-home message is that it is the interaction between the empty P_z orbital on B and the P_z of the halogen that explains the relative Lewis acidity of the series.