Understanding Octet Rule for NO2 and ClO-

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NO2 does not obey the octet rule, as nitrogen typically forms three bonds, resulting in an incomplete octet. It exists as a resonance hybrid, where the double bond is shared between the two nitrogen-oxygen bonds. ClO- does obey the octet rule, similar to hydroxide, with chlorine achieving a complete octet through lone pairs. The discussion also highlights that NO2 can exist as N2O4 in the liquid phase, which does obey the octet rule, although the N-N bond in N2O4 is weak due to electron repulsion. Overall, the complexities of bonding in these molecules illustrate exceptions to the octet rule.
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Does NO2 obey the octet rule? I know NO2 + does so I figured that NO2 wouldn't I can't figure out how to draw NO2 using formal charges.

Does ClO- obey the rule? I tried drawing it and I think it does but I am not 100% sure
 
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NO2 obeys, but I'm not sure about ClO-, so I won't add to the confusion on that one.

NO2 exists as a resonance hybrid; that is, one double bond is shared between the two N-O bonds. Because of this, the oxygens get 2 bonds most of the time (making them neutral) while nitrogen gets 3 bonds most of the time (making it neutral). This picture shows the hybrid (the dotted lines imply sharing of the double bond):
http://en.wikipedia.org/wiki/Image:Nitrogen-dioxide-2D-dimensions.png"
 
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Cl-O- obeys the octet rule. You can think of it like HO-, aka hydroxide, except hydrogen has been replaced with chlorine. Then add lone pairs of course to make chlorine neutral and have a complete octet.
 
Nitrogen-dioxide-2D-dimensions.png


if the dotted lines represent 'half bonds' then the nitrogen would indeed have 3 bonds but each oxygen atom only has 1 1/2 bonds. shouldn't there be another dotted line between the 2 oxygen atoms?
 
No, the N in NO2 does not obey the octet rule (rules are there to be broken). Neither does N in NO. In the liquid phase (that is somewhere below 4 deg Celsius if I remember correctly). NO2 actually prevails as N2O4 which obeys the octet rule. The bond between the two N's is very weak, however. This is a general feature of the homopolar bonds at the end of the period. E.g., the bond in F2 is also very weak. The problem is that the elements at the end of the period are so crammed up with electrons, that electron repulsion couteracts bond formation.
 
280px-Stickstoffdioxid.svg.png


but the nitrogen always has 3 bonds. How could there be a N-N bond?
 
He's talking about the N-N bond in N2O4
 
so am I

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The upper of the two formulas is the one I am talking about.
Nitrogen does not form compounds with 3 bonds, only. Think of
The Ammonium ion NH_4+, Nitrylchloride NO_2Cl or Nitrylfluoride NO_2F.
 
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