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Topic: Bonding Action  (Read 3402 times)

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James123

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Bonding Action
« on: March 19, 2006, 01:25:40 PM »
Assume we have a molecule AB where B has a higher electronegativity value than A.  In this case, the bonding pair electrons will move closer to B than to A.  This will cause a partial change in the electron distribution of both atoms.  With a shift of negative charge to B, this will cause B to have partial negative diapole.  On the other hand, the lost of negative charge from A will cause it to have partial postive diapole.  

Furthermore, with more negative charge in B, this will cause B to have a larger radii.  Conversely, with more positive charge in A, this will cause A to have a smaller radii.  When this happen, will the larger atom (B) now push A out?  If this is the case, would this not weaken the bond between A and B?  Or would the bond strength remain the same and why?

When 2 atoms bond together, the one with the higher electronegativity value will pull the other one to it which cause the outer orbitals to overlap.  When there is a double bond, does this mean that one of the atom has a very high electronegativity value which pulls the other atom in further than a single bond?  Is this true for triple bonds?

In theory when bonds form, it releases energy.  In any bonding, one atom will shrink while the other one increases in size.  I always assume the energy is being released by the shrinking atom.  Does energy get release from both atoms?  If so, how does it do it?  If energy gets released from the shrinking atom, does this energy get picked up by the expanding atom?



Thanks.



 


 


Offline Mitch

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Re:Bonding Action
« Reply #1 on: March 19, 2006, 01:49:00 PM »
Quote
Furthermore, with more negative charge in B, this will cause B to have a larger radii.  Conversely, with more positive charge in A, this will cause A to have a smaller radii.  When this happen, will the larger atom (B) now push A out?  If this is the case, would this not weaken the bond between A and B?  Or would the bond strength remain the same and why?

No, because the bond length is still the same. The only thing that is different is we are not saying, the A and B radius is exactly half of the A-B bond, but that B's radius is a greater component than 50% in the A-B bond.
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