What are the rules to identify the major resonance structure?

Apr 9, 2015

The best resonance structure (whether there's one or two) minimizes formal charges and gives every atom their proper valence electron distributions. Note that some atoms past Argon can hold more than 8 electrons because they have access to their d-shells.

Examples of tough ones:
$N O$
$N {O}_{2}^{-}$
$N {O}_{3}^{-}$
${N}_{3}^{-}$

1. Count electrons.
2. Include ionic charge. i.e. if 2+, then subtract 2 electrons.
3. Start with no bonds, think about how you want to distribute the atoms and electrons. Generally the atom with the highest bonding capacity should be in the center, like $C l$ in $C l {F}_{5}$, or $N$ in $N {O}_{2}^{-}$.
4. Formal charge is (bonds owned - bonds expected) . So if your nitrogen has 6 electrons surrounding it, it has a formal charge of -1. Note that bonding electrons are assumed to be half assigned per bonded atom; EX: one triple bond = six electrons, three electrons per atom.
5. Shoot for formal charges of 0 if possible, but +/- 1 is okay. If necessary, distribute negative formal charge on the most electronegative (EN) element(s) . EX: EN for carbon, nitrogen, oxygen, and fluorine: 2.5, 3.0, 3.5, and 4.0, respectively.

Here is $N O$'s resonance structure break-down. I can't draw it here but here is a description of what I would do:

1. Count up the electrons to find 5 + 6 = 11.
2. Set up a basic N-O single-bond connection. By "bonding capacity", it just means, what is the highest order bond it can afford? Nitrogen can triple bond at most (always), whereas oxygen can double bond at most (usually but not always).
3. Surround each atom with electrons to form full valence shell depictions. So far, 2 have been used, leaving 9. The more electronegative atom gets the most electrons, so they are distributed more on oxygen (EN = 3.5, vs. EN = 3.0 for Nitrogen). Nitrogen has a radical electron and one lone pair so far.
4. Then, since nitrogen has a formal charge of +1 while oxygen has a formal charge of -1, we can minimize this more by erasing a lone pair on oxygen and forming a double bond with nitrogen. It gives nitrogen 5 owned electrons and oxygen 6 owned electrons, giving formal charges of 0 for both!

So, the final result for the NO example is N=O with 3 electrons on N and 4 electrons on O. Sometimes though, you have to settle on nonzero formal charges if there is nothing better! Furthermore, you may also have to settle with the ugly radical on there. :)