[SOLVED] Problem about Lewis structure

[nihong]

HI~all
Who can explain what is unusual about the bonding in compound in borane and boron trifluoride?:?::-?
I know boron have 3 valence electrons and two in 2s orbital and one in 2p orbital. H has only one valence eletrons in 2p orbital.F has 7 valence eletrons,2s orbital ,2px orbitai and 2py orbital is full.Only has one unshared pair electrons in 2pz orbital.
But how to explain the unusual about this?I have been confused with this problem for at least one night. Can you help me to solve this problem:-?:-?

 

[jpanhalt]

This is listed as a "solved" question. Is it solved or still open? If solved, can you share the answer you got?

I find boron chemistry one of the more interesting of the second-row elements.

John

 

[nihong]

I had find a answer in YAHOO answer.
This is the answer online:
The boron atoms in both of these molecules are special because neither obey the octet rule. With the borane, there are six valence electrons, which form bonds between the B and each of the H's, which only accounts for six electrons for the B. In the boron trifluoride, there are 24 valence electrons, six of which go to each F (18 of them), and two of which go to a bond between the B and F (the other six). Again, this means there are only three pairs around the B. But, these are the correct structures for both molecules.
And i think this would be a suitable answer.
But i want to know not only what their unusual but also why they can be so unusual.

 

[jpanhalt]

What that author probably considers unusual is that the so-called octet rule is violated. There are only 6 electrons involved in the bonds. The compounds are non-ionic. Refer to the Wikipedia discussions of each:

https://en.wikipedia.org/wiki/Trifluoroborane
https://en.wikipedia.org/wiki/Borane

In terms of boron, many of its interesting reactions are related to its properties as a Lewis acid. That is, it accepts a pair of bonding electrons giving it an octet, 4 bonds, and a negative charge. A simple reaction to illustrate that is the reaction of boric acid, which can be written as H₃BO₃ or B(OH)₃, with NaOH. By analogy to the reaction of formic acid (H₂CO₂ or HCO₂H), one might think that boric acid loses a proton (i.e., acts as Bronsted acid) to give B(OH)₂O^- + Na⁺ + H₂O .

There is strong evidence, however, that a hydroxyl adds to the boron (i.e., boron acts as a Lewis acid) to give B(OH)₄^- + Na⁺. B(OH)₄^- follows the octet rule and is tetrahedral. You will probably see that structure used more commonly (See "properties" and discussion of conflicting theories at: https://en.wikipedia.org/wiki/Boric_acid ). The latter reaction mechanism is useful for explaining the propensity of boric acid and organic derivatives of boric acid (e.g., boronic acids) to form both charged and uncharged adducts with adjacent hydroxyl groups in compounds such as carbohydrates.

Another useful compound of boron in which it acts as a Lewis acid is sodium borohydride (NaBH₄)(https://en.wikipedia.org/wiki/Sodium_borohydride). Because that compound can donate a hydride (H^-), it is a powerful reducing agent.

John

Edit: Fixed superscripts and sunbscipts