[souro10]

I'm trying to explain the 3c-2e bridge-bonds in Diborane using L.C.A.O. method, to construct M.O. and show that the situation is bonding.

Now,for simplicity let us try to explain only one of the bridge-bonds in diborane.

I am considering 2 sp3 hybrid orbitals from the two boron atoms, and the s-orbital of hydrogen- involving a total of 3 atomic orbitals. We are to fill these with two electrons.

I understand that number of atomic orbitals combining should give the same number of molecular orbitals, right? That means I should get three molecular orbitals.

Now- I can-not get three MOs by LCAO. I'm getting 4.

One of them is bonding - which is formed by in-phase overlap of all three orbitals.
Two of them are non-bonding- each one being formed by in-phase overlap of the s-orbital and one sp3 orbital but out-of-phase overlap of the other sp3 orbital.
One of them is anti-bonding - which is formed by out of phase overlap of both the sp3 hybrid orbitals with the s-orbital of hydrogen.

Where am I going wrong? What should be the proper combination?

[Corribus]

Each boron hybrizes to sp3 and there are three electrons (so one of them will be empty).  Two of these bond with the 1s orbitals of the peripheral hydrogens.  That leaves two sp3 orbitals on each boron (with one electron total) plus the single 1s oribtal on each central hydrogen (each having one electron).  This is six total orbitals.  Divide by two = 3 total orbitals to participate in each bond.

See. e.g. http://voh.chem.ucla.edu/vohtar/fall03/classes/30B/pdf/diborane.pdf

[souro10]

Each boron hybrizes to sp3 and there are three electrons (so one of them will be empty).  Two of these bond with the 1s orbitals of the peripheral hydrogens.  That leaves two sp3 orbitals on each boron (with one electron total) plus the single 1s oribtal on each central hydrogen (each having one electron).  This is six total orbitals.  Divide by two = 3 total orbitals to participate in each bond.

See. e.g. http://voh.chem.ucla.edu/vohtar/fall03/classes/30B/pdf/diborane.pdf

I guess you did not understand my question..

My question is simple, I am getting 4 MOs from 3 AOs because I am can think of two possible non-bonding interactions. Clearly, I'm wrong somewhere, but I can't figure out why they are showing only one possible non-bonding interaction. I'm getting two non-bonding orbitals, one bonding and one-antibonding.

[Dan]

Two of them are non-bonding- each one being formed by in-phase overlap of the s-orbital and one sp3 orbital but out-of-phase overlap of the other sp3 orbital.

These two are the same. You can not create more LCAOs than the AOs you start with.

This is the way I think about it (an organic chemist's naive interpretation):

B H B
+ + +     Same as - - -
+ + -     Same as - - + or + - - or - + +
- + -      Same as + - +

Similarly for dihydrogen there will be only two LCAOs:

H H
+ +    Same as - -
+ -    Same as - +

[souro10]

Two of them are non-bonding- each one being formed by in-phase overlap of the s-orbital and one sp3 orbital but out-of-phase overlap of the other sp3 orbital.

These two are the same. You can not create more LCAOs than the AOs you start with.

This is the way I think about it (an organic chemist's naive interpretation):

B H B
+ + +     Same as - - -
+ + -     Same as - - + or + - - or - + +
- + -      Same as + - +

Similarly for dihydrogen there will be only two LCAOs:

H H
+ +    Same as - -
+ -    Same as - +

Thanks, it was really helpful! Clear now

[souro10]

But it would be helpful if the foundation behind regarding the "Same as" as only one interaction be explained.

[Enthalpy]

Boranes are known to make unnormal bonds. Some sources tell about divalent hydrogen in boranes.

Sorry if I put nonsense, I didn't even try to understand what I read. But for sure, valences don't work with boranes.