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A molecule can have resonance structures when it has a lone pair or a double bond on the atom next to a double bond.
Whenever you can draw two or more Lewis structures for a molecule, differing only in the locations of the electrons, the actual structure is none of the structures but is a hybrid of them all.
For example, what is the Lewis structure of the nitrite ion, NO₂⁻?
1. Decide which is the central atom in the structure. That will normally be the least electronegative atom (N).
2. Draw a skeleton structure in which the other atoms are single-bonded to the central atom: O-N-O.
3. Draw a trial structure by putting electron pairs around every atom until each gets an octet. In this editor, I will have to write it as ::Ö-N(::)-Ö::
4. Count the in your trial structure (20).
5. Now count the you actually have available.
1 N + 2 O + 1 = 1×7 + 2×6 +1 = 18.
The trial structure has two more electrons than are available.
6. Draw a new trial structure, this time inserting one double bond for each extra pair of electrons: O-N=O and O=N-O (two possibilities) — Right here, the two possibilities tell us that that we will have two resonance structures.
7. As before, add valence electrons to give each atom an octet:
8. Assign formal charges to each atom.
Each atom “owns” its lone pair electrons and half of the shared pairs.
The right hand O atom has 6 lone pair electrons plus 1 from the single bond. This is one more electron than in an isolated O atom. So the formal charge on O is -1.
The two resonance structures are:
The actual structure is none of these. It is a resonance hybrid of them both.
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