[Mel22]

Why is the Lumo of BBr3 lower in energy than the one of BF3?
Would anyone mind giving me a hint?
Thanks a lot!

[GCT]

Well, what are the characteristics of the LUMO, e.g. antibonding, pi, sigma...and you might also want to consider whether the atomic orbitals of F are higher or lower in energy then that of Br. 

[Mel22]

Yep, I know that the F AOs are lower in energy than the Br AOs. The antibonding LUMO is polarised towards the B atom, since it is higher in energy than both the Br and F AOs. This effect is stronger for the BF3 than for BBr3 (due to stronger bond polarity), which implies that the antibonding LUMO of the latter is lower in energy.

Can one argue like that? Or is it better to simply say: the more polarised the bond the more is the antibonding MO raised in energy?
And what does the energy have to do with whether it is a sigma or a pi bond?

Thanks a lot!

[GCT]

Yep, I know that the F AOs are lower in energy than the Br AOs. The antibonding LUMO is polarised towards the B atom, since it is higher in energy than both the Br and F AOs. This effect is stronger for the BF3 than for BBr3 (due to stronger bond polarity), which implies that the antibonding LUMO of the latter is lower in energy.

Can one argue like that? Or is it better to simply say: the more polarised the bond the more is the antibonding MO raised in energy?
And what does the energy have to do with whether it is a sigma or a pi bond?

Thanks a lot!

I haven't referenced this specifically, however, it may be that there is more stabilization with BBr3 then with the flourine analog.  The resulting molecular orbital with BBr3 may be more covalent in character, while that of BF3 is more ionic, you may have studied the differences between molecular orbital diagrams that are ionic in nature as opposed to more covalent, that is when the two atomic orbitals are more similar in energy; you may want to refer to the acid base chapter in your inorganic text.  That if the relevant atomic orbitals that form the LUMO between B and Br are more similar in energy, the resulting LUMO may have more stabilization to it as a result of the interactions, rather than that with B and F which, according to your statements, seems to be more separated in energy.  There are alternate explanations but this is what I'm proposing for now, until, it's verifiably wrong.

[Winga]

I think the problem is to know from which B's AOs(HOMO & LUMO) interact with F's/Br's AOs(HOMO & LUMO).
That means B's HOMO interacts with F's LUMO and B's LUMO interacts with F's HOMO (same case for Br's).
Then, to see which one has a greater interaction, greater stability for bonding orbital, greater destability for antibonding orbital.

[letk0]

Winga sounds correct to me.  The p-orbital orthogonal to the B-X bonds is able to accept electron density from either halogen in order to get closer to having an octet.  F's atomic orbitals are closer in energy to that of B compared with Br, so the donating effect is greater and lowers the HOMO for the BF₃ complex.  The lowering of the HOMO causes the LUMO's energy to increase.

[Dan]

I would argue it like this:

I would have expected BBr₃ to have the higher energy LUMO, because Brs p orbitals are higher in energy than Fs.

However, the opposite order is observed, so there must be a stronger factor at work here.

Here's an MO diagram for BF₃ http://www.brynmawr.edu/Acads/Chem/sburgmay/chem231/BF3mos.jpg

The lumo is 2a₁", which arises from p-p interactions. In BF₃ these interations are strong, due to the small size of Fs p orbitals (good overlap with Bs p orbitals), resulting in a very large splitting - and a high energy 2a₁".

in comparison, Brs p orbitals are large, so the net overlap between B and Brs p orbitals is relatively poor, p-p interactions are relatively weak resulting in a much smaller splitting - and therefore a lower energy 2a₁".

Presumably, this effect outweighs the the fact that Brs p orbitals are higher in energy than Fs