Bonding in Diborane: Unique 3c-2e Bonds & Hybridization

[aim1732]

The bonding in diborane is unique as it involves two 3c-2e bonds. Most books show the overlapping as between two sp3 orbitals of B atoms and s orbital of H atom.But the bond angle between the terminal H atoms and central B atom is 120, as if the B atom were sp2 hybrid. Any ideas?

 

[DrDu]

That's only a difference of 11 deg. Not much given that the model is incredibly oversimplified.

 

[aim1732]

Hmmm.. But it's a difference still.

 

[DrDu]

Of course there is a difference as in any stained system like e.g. cyclobutane or -propane.
For a precise valence bond calculation have a look e.g. here:

http://pubs.acs.org/doi/abs/10.1021/j100179a024
"Electronic structure of diborane and octahydrotriborate(1-): boron-hydrogen-boron bridges and closed boron-boron-boron bonds"
Maurizio Sironi, Mario Raimondi, David L. Cooper, Joseph Gerratt
J. Phys. Chem., 1991, 95 (26), pp 10617–10623
DOI: 10.1021/j100179a024

 

[Gokul43201]

The bonding in diborane is unique as it involves two 3c-2e bonds. Most books show the overlapping as between two sp3 orbitals of B atoms and s orbital of H atom.But the bond angle between the terminal H atoms and central B atom is 120, as if the B atom were sp2 hybrid. Any ideas?

The bond angle being 120 deg between the outer H atoms is not such a big deal, since diborane does not have a (BH₃)₂ structure, like ethane.

The four bonds between the two B atoms and the two "internal" H atoms want to get close to a square, making the bond angle between the internal H atoms settle on some value between 90 deg (to reduce steric strains among the internal bonds) and 109.5 deg (from tetrahedral symmetry of the sp3 orbital). As it turns out, this internal H-B-H bonds angle is a reasonable 97 degrees. This reduction from the tetrahedral angle allows the exterior H-B-H bond to spread out farther than 109.5 deg. So, a number like 120 deg doesn't sound unreasonable. In short, the reason for the non-tetrahedral angle is that there are steric effects from the inside atoms forming a cyclic structure, just as Dr Du mentioned above.

That the angle is almost exactly the sp2 bond angle is just an unrelated coincidence.

 

[DrDu]

Instead of an equal sp3 hybridization for all 4 bonds you could alternatively consider two sp2 hybrids for the terminal bonds and a rehybridization of the third sp2 with the p to form two sp5 hybrids which an opening angle of 101.5 degree. Something exactly similar is possible in the case of ethane where the sigma and pi bonds are rehybridized to form two banana bonds.
Would be nice to compare the two alternatives with an ab initio valence bond program like VB2000.

 

[chemisttree]

When I saw this question my thoughts immediately went to "bananna bonds". Really weird...