BH3 <-- Boron trihydride
If you are to draw a Lewis structure, you would see that just a single bond is needed b/ween each B-H bond. Is Boron coordinate covalent bonding in this situation?
Also, its electron configuration before bonding is 1s^2 2s^2 2p^1
After bonding with those 3 Hydrogens, is its valence configuration going to be:
1s^2 2s^0 2p^0 OR
Does it hybridize and we get:
[e-, ][e-, ][e-, ] <--- sp^2 [e-, ] <--- p
What I am not understanding is how does Boron's valence configuration change before bonding versus after bonding?
Indeed B-H bond is covalent (the difference in electronegativity is small), the boron is sp^2-hybridised (resulting in BH3 having a trigonal planar structure).
In practice, BH3 readily dimerises to form diborane B2H6. The dimer make up of the difficient electrons by forcing an H from each BH3 to act as a "bridge" between the two borons so the boron become effectively like sp^3 hybridised (technically we have a three-centre two-electron bond involving these H).
Ok so if it is sp^2 hybridizing, it will get this (Assume that "e-" is an electorn, and their seperated by commas around a square bracket that designates that that is an orbital)
sp^2 orbitals [e-,e-
unhybridized p-orbital [ ] Bold designates shared electrons from hydrogen
So Boron is stable like this, despite having its unhybridized p-orbital completely empty?