Why does the #NO_2# molecule not follow the octet rule?

Answer 1

Well, basic arithmetic dictates that it is a 17 valence electron compound. There MUST be a lone electron.

The #NO_2# species is a radical species, which displays a lone electron (somewhere!).

Because one of the oxygens has a negative charge and the nitrogen has a positive charge, the Lewis structure requires charge separation.

#O=N^(+)-O^-# is a neutral species; there is a single lone electron on the nitrogen atom, and, given that there are only 4 valence electrons associated with this centre (6 electrons in total), nitrogen bears a positive charge. On the other hand, the singly bound oxygen owns 9 electrons (rather than the normal 8), and bears a formal negative charge.

#O=N^(+)-O^-# undergoes dimerization, in which the lone electron on each nitrogen couple to form a nitrogen-nitrogen bond:

#2NO_2 rightleftharpoons (""^(-)O)O=N^(+)-^(+)N=O(O^-)#

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Answer 2

Owen Ambrose

Because NO₂'s valence shell contains an odd number of electrons, it does not adhere to the octet rule, which results in an unpaired electron and violates the octet stability principle.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.