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

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SUMMARY

The discussion centers on the Lewis structure of Nitric Oxide (NO), specifically the placement of the unpaired electron between nitrogen and oxygen. It is established that the Lewis structure features a double bond between nitrogen and oxygen, with a total of 11 valence electrons. The formal charge rule is crucial in determining the optimal placement of the unpaired electron, which can either reside on nitrogen or oxygen. The preferred configuration maintains charge neutrality, resulting in nitrogen having a formal charge of zero when the unpaired electron is placed on it.

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  • Understanding of Lewis structures
  • Familiarity with formal charge calculations
  • Knowledge of valence electrons for nitrogen and oxygen
  • Concept of resonance in molecular structures
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  • Research the formal charge rule in detail
  • Explore resonance structures in nitric oxide and similar molecules
  • Study the implications of electron delocalization in molecular stability
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Chemistry students, educators, and professionals interested in molecular structure analysis and the principles of chemical bonding.

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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
 
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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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