Naked Science Forum
Non Life Sciences => Chemistry => Topic started by: Gregorian on 19/03/2009 02:53:29
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Hello,
Why doesn't BF3 follow the octet rule? It is not in the third period!!!!
As far as I know, elements in the third period can disobey the octet rule. Is Boron a moron?
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Rules are made to be broken. [:)]
A facile explanation invokes the symmetry-allowed overlap of a p orbital on the boron atom with the in-phase combination of the three similarly oriented p orbitals on fluorine atoms.
http://en.wikipedia.org/wiki/Boron_trifluoride#Structure_and_bonding
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Violation of the Octet Rule for Period 3 and up (or is it down based on the Periodic Table) is because of the presence of d-Orbitals, and usually the valency is in excess of 8 if I remember correctly.
Boron Trifluoride and Beryllium Chloride are both violations of this octet rule. Exactly why, I do not know but I assume it has something to do with stability. >.<
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Thank you guys
Now how come NO3- violates the octet rule in one of its resonance forms!!?? (as in the attached picture).. and yes this form exists. It was in my Inorganic Chem book [::)]
Thanks
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Do you know about orbitals/subshells? s, p, d, f etc....?
Perhaps you can have a read of http://www.thenakedscientists.com/forum/index.php?topic=19467.msg217305#msg217305 [:)]
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NO3- doesn't violate the octet rule, all atoms have 8 electrons.
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Ya Chemistry4me is correct that you have to read more about Atomic struture, periodic tables.. and more Hybridization and also VESPER theory
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That's VSEPR (Valence shell electron pair repulsion).
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oh correct !!!!
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Nitrogen does not have 8 electrons.. it only has 6!! why is that?
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Oh sorry, I didn't look at that diagram very closely. [:I]
But after looking at Wikipedia (http://en.wikipedia.org/wiki/Nitrate#Chemical_properties) and other websites I'm not sure that, that the resonance structure of your's exists.
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The Nitrate Ion is arranged in a plane at all times as it is sp2 hybridized. In that state, it is just like Boron Trifluoride. In which case Chem4Me's first answer and mine should work as well.
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Also, I just realized that Aluminium Chloride has a similar 'problem' of trivalency. However, we could argue that because it is in a plane and the p-Orbitals perpendicular to this plane can all overlap, there is a degree of 'delocalization'. This may also be applicable for Boron Trifluoride but less likely because Fluorine is highly electronegative.
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boron and aluminum in BF3 and AlCl3 for example. Both are electron deficient (only 6 valence shell electrons)
Al in AlCl3 has a +3 oxidation state, but that doesn't make it isoelectronic with neon. The Al has no valence electrons, each Cl has 2 valence electrons for a total of six which is electronic deficient.
Many elements exceed the octet rule, especially those in the third row of the periodic table and below. S in H2SO4 and P in H3PO4 are examples like the other guy said, but those certainly aren't the only ones.
H and He are the obvious/simple answers because their valence shell can only hold two electrons. If you are in Grade 10 or 11, this may be the answer your teacher is looking for.
Basically, I don't think this question has only 2 answers. There are many elements that do not follow the octet rule. In fact, the octet rule is generally not correct, except for simple (1st and 2nd row) compounds
regards
Raghavendra