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Old Nov 12th 2011, 08:38 PM   #1
Skyrim
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Hybrdization of atoms that do not follow octet rule

So far I am clear on how hybridization works for "most" atoms. But some atoms do not obey the octet rule...

Some atoms have more than an octet of electrons, in which case it's either spppd or spppdd hybridization. That I understand.

Some atoms though have fewer than an octet of electrons. Boron in boron trifluoride for example (BF3). The boron only has 3 valence electrons, and bonded with each flourine there are 6 total electrons available to boron rather than 8.

So what will Boron's hybridization be in this case? I'm guessing sp2 with an unfillled unhybridized p-orbital? This is in agreement with the fact that BF3 has a trigonal planar molecular geometry (and no nonbonding electron pairs).

And also, what about atoms that have an odd number of valence electrons? Nitrogen monoxide for example? The nitrogen atom has 7 total electrons available, of which 4 are shared with oxygen via a double bond. Out of the remaining 3 electrons, two are involved in a nonbonding electron pair. But there is one final electron. Will that electron occupy a hybridized or an unhybridized orbital??

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Old Nov 13th 2011, 06:12 AM   #2
Unknown008
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1. Right.

2. It'll be an SP2 hybridisation with the lone pair in the hybridised orbital. Though NO tends also to get the lone electron (delocalisation) in the bond, which then is an SP hybridised structure.
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