Talk:VSEPR theory
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[edit] Removed text
I removed the following text because I didn't know where it would fit better in the article. Please add it back if you figure out.
- If the connectivity, bond lengths, bond angles and torsional angles are found, a molecules exact geometry is known. For any non-linear molecule with N atoms, 3N - 6 internal coordinates need to be specified in order to know the exact geometry of the molecule.
-- Rune Welsh ταλκ 21:40, July 28, 2005 (UTC)
Is this correct? I always thought that VSEPR was part of Valence Bond Theory. I thought sigma and pi bonding were part of Molecular Orbital Theory.
"VSEPR theory is usually compared and contrasted with valence bond theory, which addresses molecular shape through orbitals that are energetically accessible for bonding. Valence bond theory concerns itself with the formation of sigma and pi bonds."
Captain Video 20:52, 9 August 2006 (UTC)Captain Video
VSEPR theory is not part of Valence Bond theory. See what the original authors od VSEPR say. It does however suggest similarities to some people. Sigma and pi bonding are part of both valence bond and molecular orbital approaches. I would keep the first sentence and delete the second. --Bduke 23:39, 9 August 2006 (UTC)
[edit] More removed text
I removed the following as unsourced, and lacking in good grammar. I'm no expert in VSEPR so I can't make a decision as to whether it is acceptable/correct or not. Anyone with more experience is invited to correct and reinsert material where appropriate.Hyenaste (tell) 22:07, 16 August 2006 (UTC)
- It must be emphasised that the VESPR is not really a theory, in spite of its often being referred to in this manner. It is a convenient method to get a rough idea about what the probable shape of a molecule might be. The actual shape that real molecules achieve is the result of quantum mechanics and the geometry of orbitals. One aspect of it that is particularly bad from a theoretical standpoint is the idea that "non bonding electrons repel more strongly than bonded ones." The real reason that stibine (SbH3) exhibits 90 degree bond angles is because the antimony (Sb) atom uses its p orbitals for forming the Sb-H bonds, and these are mutually perpendicular. Ammonia exhibits a 107 degree bond angle because the energy difference between 2s and 2p orbitals are small so hybridised orbitals can form. For s and p hybridised bonds the bond angle is given by acos(-s/p) where s is the "s" fraction of the orbital and p is the "p" component.
[edit] Rating
I've given the article a quick assessment after adding the chemistry template. I only gave it a start class rating since it's quite short and lacks structure (sections). A little on the history of the theory would help with both these issues, for example. It's quite a good start class article all the same, maybe even a low B-class. Richard001 04:21, 16 January 2007 (UTC)
How is this possible?
