Which Atom Holds the Unpaired Electron in Nitric Oxide's Lewis Structure?

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Discussion Overview

The discussion centers on the Lewis structure of nitric oxide (NO), specifically regarding the placement of the unpaired electron and the implications for formal charge on nitrogen and oxygen. Participants explore the acceptability of different Lewis structure variants and the concept of resonance in this context.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions whether the unpaired electron in the Lewis structure of nitric oxide belongs to nitrogen or oxygen, suggesting both variants may be acceptable.
  • Another participant explains the total number of valence electrons in NO and discusses the use of formal charge to determine the placement of the unpaired electron, indicating a preference for formal charge neutrality.
  • A later reply introduces a rule of thumb regarding the stability of octets in Lewis structures, noting that in the case of NO, achieving two octets is not possible, and emphasizes the importance of charge neutrality.

Areas of Agreement / Disagreement

Participants express differing views on the placement of the unpaired electron and the implications for formal charge, indicating that multiple competing perspectives remain without a clear consensus.

Contextual Notes

The discussion involves assumptions about formal charge calculations and the interpretation of resonance, which may not be fully resolved within the context provided.

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Take Nitric Oxide for example. When I've drawn out the Lewis structure, how do I know whether the unpaired electron belongs in the nitrogen or the oxygen? Are both variants of the NO Lewis structure acceptable? Furthermore, if both of these variants are accepted, is there a "resonance" of NO in which the extra unpaired electron is delocalized between the two atoms?
 
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An Oxygen atom has 6 valance shell electrons and a Nitrogen atom has 5 valance shell electrons for a total of 11 electrons in the NO molecule.
The Lewis structure of the molecule has a double bond between the Nitrogen and the Oxygen atom, but in order to figure out how many lone pairs and unbonded single electrons I believe this is a situation where one would use the "formal charge" rule (is that the name of it?).

It is "preferable" for the atoms to "see" the same number of electrons around them after the bonds are made as they originally had (for a formal charge of zero), if they "see" too many, they have a negative formal charge, if not enough, then a positive formal charge. By "see" I mean to say the total of their unbonded electrons + 1 electron for every bond that was made.

So if you get to the point of the Lewis structure where you have a double bond between N and O and then have placed two lone pairs around the Oxygen and 1 long pair around the Nitrogen...where to place that last electron?
As it is, Oxygen would have a formal charge of zero since it has 2 lone pairs and a 2 bonds. Nitrogen gas a formal charge of +1 since it has two bonds plus 1 long pair. By placing the unbonded electron around Nitrogen you change its formal charge back to zero and complete the Lewis structure.

NO Lewis structure:
http://upload.wikimedia.org/wikipedia/en/9/9a/Nitricoxide.png
 
Last edited by a moderator:
Oh yeah, forgot about that. Thanks.
 
There's this rule of thumb that you can keep handy for drawing up Lewis structures.

From greatest stability to least: neutral octets > charged octets > neutral non-octets > charge non-octets

In the case of NO, you can not have two octets. The best you can do is have one octet and ensure charge neutrality on both atoms.
 

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