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Topic: Regarding stable molecules having an incomplete octet.  (Read 5361 times)

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Offline Alcoinito

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Regarding stable molecules having an incomplete octet.
« on: August 10, 2011, 02:16:39 AM »
BeF2 and BF3 both are stable yet each has incomplete octet. Explain. Also ascertain why BeF2 is more stable of the two.

I am aware that these molecules are exceptions to the octet rule. But "explaining why" has particularly stumped me. I have googled my problem, and have also checked  reference book "Inorganic chem" by J.D.Lee, but to no avail.
I would be glad if someone could help me  :)

Offline gertrudetrumpet

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Re: Regarding stable molecules having an incomplete octet.
« Reply #1 on: August 12, 2011, 04:07:41 PM »
i think it is because the fluorine does share electrons with the boron and beryllium. it undergoes resonance. The beryllium is more stable because both fluorine share their electrons, creating a balance in charge. Also, since the molecular is linear, there is less electron repulsion between the fluorines.

Offline DevaDevil

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Re: Regarding stable molecules having an incomplete octet.
« Reply #2 on: August 12, 2011, 04:17:55 PM »
Beryllium is a metal, this makes Beryllium Fluoride a salt, aka an ionic compound. (Beryllium fluoride  has a crystal lattice in which each beryllium is surrounded by 4 fluorides)



Offline 408

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Re: Regarding stable molecules having an incomplete octet.
« Reply #3 on: August 12, 2011, 07:56:38 PM »
BF3 has short B F bonds.  electrons in p orbitals on F are perpendicular to the planar BF3 molecule.  These two factors combined means the p electron density has significant overlap with the empty p orbital on B, partially filling it.

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