[ultranova3]

i am wondering if My Molecular orbital for BeF2 is correct:
http://bp0.blogger.com/_2ADAC4zLTbc/RzMqsW3umEI/AAAAAAAAABU/_ZpSg96slcs/s1600-h/Mo1.bmp
I have tried constructing it but i found out that the Bond order is 2 instead of 1. Is there any problem with my MOs?
Please help me
I need to submit it by Monday
thank you

PS I don't know how to insert pictures so I use hyperlink. Sorry for the inconvenience 

[TheBigF]

I too don't know how to get a picture in - The MO diagram that you show looks okay - I am assuming that you have already taken combinations of the orbitals to get the combined AO's, and that Be is on the left and 2F on the right. The biggest problem to getting the right bond order is the number of electrons. Assume that Be provides 1s2 2s2 (of which 1s2 are core) and that each fluorine provides 1 electron, then you have 4 in total, all in bonding orbitals, so your total bond order is 2.

[AWK]

BeF2 is a triatomic molecule and zou have two bonds!

[ultranova3]

That is the part that I am not sure about:
1. will core electrons of Be (1s2) participate in bonding. Should we just consider valence electrons for bonding.
2. The bond order from my MO diagram is 4 (Be=F). But lewis structure is only 2 (Be-F single bond). Is there any possibility that F donates and shares its 2 electrons via dative covalent in order for Be to achieve octet configuration?
3. How about the energy level of each MOs? I have not learned the mathematical calculation so I am not sure that my qualitative MO diagram will be accurate.

Here is my linear combination of F orbitals and symmetry of each orbitals:
Be (2s)                      σg
Be (2pz)             σu
Be(2px,y)             πu
F(2sa + 2sb)            σg
F(2sa – 2sb)             σu
F(2pza + 2pzb)            σu
F(2pza - 2pzb)             σg
F(2px,ya+ 2px,yb)   πu
F(2px,ya – 2px,yb)   πg

Is there anything wrong with my orbital symmetry?

Thank you

[stranger]

Perhaps it is too late, though better later than never.
There is one small but important mistake in your drawings, I guess.
IE(1) for Berillium   is 899 kJ/mol and for Fluorine  it is 1681.  Therefore energy level of Berillium 2S orbital must be much higher than Fluorines 2p orbital, which is not the case on the picture.
To take this into account and all the MOs will be shifted. You'll get less bonding orbitals (some of them are actually nonbonding).
Some preliminary ideas can be taken from
Housecroft & Sharpe "Inorganic Chemistry" (Chapter 4 would be very useful)