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Offline lilmul123

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simple question on bond order definition
« on: December 17, 2006, 04:36:08 PM »
My chem teacher didn't go into explicit detail about bond orders, and I need some help figuring it out.

I have the NO3- molecule.  What is it's bond order? 

With some research, I've come up with the idea that it's just the number of bonds between a pair of atoms.  So, I'd have bond orders of 1 and 2 for the molecule, but I'd like to be certain of this fact.

Was I correct?

Offline Yggdrasil

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Re: simple question on bond order definition
« Reply #1 on: December 17, 2006, 04:56:48 PM »
There are two ways to define bond order.  The most technical definition of bond order is 1/2 * (# of electrons in bonding orbitals - # of electrons in antibonding orbitals), but this definition requires knowledge of molecular orbital theory to interpret.

The more simple definition is as you said, the number of bonds between a pair of atoms.  For example, since the carbon in CO2 has a double bond to the oxygens, the bond order of the C-O bond in carbon dioxide is 2.  Now, with NO3-, you are forgetting about resonance structures.  Remember that no one resonance structure correctly represents the molecule you are representing (in fact, resonance structures do not exist in real life at all); rather, the molecule is most like an average of all the resonance structures of the molecule (i.e. a resonance hybrid).  So, to find the bond order of NO3-, I would draw out all of the resonance structures (hint: there are three), then come up with the average number of bonds between the N and a specific O.

Offline lilmul123

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Re: simple question on bond order definition
« Reply #2 on: December 17, 2006, 06:06:47 PM »
*scratches head*
So is the bond order....1.3 repeating decimal?

Offline Yggdrasil

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Re: simple question on bond order definition
« Reply #3 on: December 17, 2006, 07:49:47 PM »
Yup.  The bond order is a fraction (1 1/3) because of resonance.  The bonds in NO3- are partially double bonds.  However, because the electrons in the pi bond are delocalized over all three bonds, the pi bond is weaker than a normal pi bond.  Basically, the two electrons in the double bond that you see in one of the resonance structures are actually shared between all three bonds, so each N-O bond consists of one sigma bond and one third of a pi bond.  (This explanation may not make sense if you haven't studied molecular orbital theory and don't know what sigma and pi bonds are)

Offline FeLiXe

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Re: simple question on bond order definition
« Reply #4 on: December 18, 2006, 06:19:08 AM »
in case you are wondering:

this is what the pi-MO-scheme would look like. you have one bonding, two nonbonding, and no antibonding orbital populated. that's the 1/3 bond besides that you have a sigma-bond for each oxygen

                   
                ___             
___                         
                               
  N         _||_  _||_                     
                                 ___ ___ ___
                               
                 _||_                    O

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