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Topic: Nitro Group  (Read 5580 times)

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Offline isenrich

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Nitro Group
« on: May 02, 2012, 09:17:02 PM »
I was wondering why the nitro group Lewis structure is drawn as it is. The link provided shows the nitro group: link.

My point is that basic rules of resonance say that the separation of unlike charges is an undesired effect and any contributing structures that do not have this (keeping all else the same) are more contributing to the resonance hybrid. The separation of unlike charges costs energy. If one of the lone pairs of the oxygen with a negative formal charge were used to make a double bond to nitrogen all atoms of the nitro group would have 0 formal charges. Is the reason the nitro group is represented like this because having filled octets is more important to resonance than having all 0 formal charges and exceeding the octet?

Edit by Borek: link hidden in url tags.
« Last Edit: May 03, 2012, 03:34:55 AM by Borek »

Offline AWK

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Re: Nitro Group
« Reply #1 on: May 02, 2012, 09:52:40 PM »
How do you explain that both N-O bonds  show the same lenghth (intermediate between single and double bond)?
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Offline isenrich

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Re: Nitro Group
« Reply #2 on: May 02, 2012, 11:28:25 PM »
The Lewis structure I posted a link to is just a representation of the nitro group. In reality we are expected to assume that the nitro group is a resonance hybrid of the Lewis structure and the other contributing structure (where the double and single bonds switch between the two oxygens). So the resonance hybrid will be a weighted average of both contributing structures (don't misunderstand that to mean that the nitro group interconverts between the two resonance structures) and the N-O bond lengths will be equivalent. One could also draw a resonance hybrid Lewis structure with a single bond from the nitrogen to each oxygen and a dotted line curving from the first oxygen to the nitrogen to the second oxygen to represent that the pi-bonding character is delocalized over the set of three atoms. I attached a link to such a representation: link.

My question is more geared toward knowing whether a full octet (not exceeding it, especially since nitrogen isn't a third period or higher element) takes precedence over all 0 formal charges in the valuation of a resonance structure.

Edit by Borek: link hidden in url tags.
« Last Edit: May 03, 2012, 03:34:17 AM by Borek »

Offline fledarmus

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Re: Nitro Group
« Reply #3 on: May 03, 2012, 07:45:32 AM »
Exceeding an octet on anything up through Ne is extremely high energy - atoms really like empty shells or full shells, and those extra electrons have to open up a new shell. You can completely disregard any resonance structures that exceed octets on fluorine, oxygen, nitrogen, carbon, and boron; they will have a high enough energy to make no contribution to the actual structure.

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