2.6 - Resonance and Formal Charge

Overview

Sometimes the steps from Topic 2.5 give you more than one valid Lewis diagram. If the diagrams are equivalent, the real molecule is a blend of all of them, called a resonance hybrid. If they aren't equivalent, the octet rule and formal charge tell you which one is the best model. This topic also covers where the Lewis model breaks down, especially for species with an odd number of electrons. On the AP exam these ideas show up as "draw the resonance structures," "which diagram is best? justify using formal charge," and bond length comparisons.

Resonance

When one diagram isn't enough

Take ozone,

\mathrm{O_3}

, with

3(6) = 18

valence electrons. Following the steps, the central O needs a double bond to one of the end O atoms. But which one? Both choices are equally valid.

These are resonance structures: valid Lewis diagrams with the same arrangement of atoms that differ only in where the electrons are placed. The double-headed arrow between them means "these are resonance structures." Each one has 3 bonding pairs + 6 lone pairs = 9 pairs = 18 electrons.

The resonance hybrid

Real ozone isn't either structure, and it doesn't flip back and forth between them. It's a resonance hybrid, an average of the structures, all the time. The electrons in the "extra" bond are delocalized: spread over all three atoms instead of sitting in one bond.

This is why the CED calls resonance a refinement that's needed for accurate predictions. A single structure predicts one O-O single bond (long) and one O=O double bond (short). Experiments show both bonds in ozone are the same length, 128 pm, which falls between a typical O-O single bond (148 pm) and an O=O double bond (121 pm). Only the hybrid predicts that.

Bond order in a resonance hybrid

Bond order is the number of electron pairs shared between two atoms. With resonance, the shared pairs are spread evenly over the equivalent bonds:

\text{bond order} = \frac{\text{bonding pairs shared among the equivalent bonds}}{\text{number of equivalent bonds}}

Ozone has 3 bonding pairs spread over 2 O-O bonds, so each bond has a bond order of

\frac{3}{2} = 1.5

. A higher bond order means a shorter, stronger bond (Topic 2.7), so 1.5 fits a length between a single and a double bond.

Rules for drawing resonance structures

Only electrons move: lone pairs and the extra pairs of multiple bonds. Atoms never move. If atoms are connected differently, you've drawn a different compound, not a resonance structure.

Every structure must have the same total number of electrons and the same overall charge.

Draw all of them, with a double-headed arrow between each pair, and brackets around each structure for an ion.

Example 1: Resonance in the nitrate ion

Draw all the resonance structures of

\mathrm{NO_3^-}

and find the N-O bond order.

Step 1: Count valence electrons, adding one for the negative charge.

5 + 3(6) + 1 = 24 \text{ valence electrons}

Step 2: N goes in the center with three single bonds, which use 6 electrons:

24 - 6 = 18

left.

Step 3: Each O gets three lone pairs:

18 - 3(6) = 0

left. N has only 6 electrons.

Step 4: One O lone pair moves in to make an N=O double bond. Any of the three O atoms can do it, so there are three equivalent structures.

Step 5: Find the bond order. There are 4 bonding pairs spread over 3 N-O bonds.

\text{bond order} = \frac{4}{3} \approx 1.33

Answer: three resonance structures, each with 4 bonding pairs + 8 lone pairs = 24 electrons. All three N-O bonds in the real ion are identical, with a bond order of

\frac{4}{3}

: longer than an N=O double bond and shorter than an N-O single bond.

Formal charge

What formal charge means

Formal charge is the charge an atom would have if the electrons in every bond were split evenly between the two bonded atoms. It's bookkeeping that lets you compare Lewis diagrams. It isn't the real charge on the atom; real charge depends on electronegativity, as in Topic 2.1.

\text{formal charge} = V - \left(L + \frac{B}{2}\right)

$V$ : valence electrons in the free atom (from the group number)

$L$ : electrons in lone pairs on the atom (count every dot)

$B$ : electrons in bonds to the atom, so $\frac{B}{2}$ is just the number of bond lines touching it

A quick way to say it: formal charge = valence electrons minus (dots + lines). The formal charges in a structure always add up to the overall charge, 0 for a neutral molecule and the ion's charge for an ion. Use that as a check every time.

Example 2: Formal charges in ozone

Find the formal charge on each atom in the first ozone structure above (double bond on the left).

Atom	V	Dots	Lines	Formal charge
Central O	6	2	3	$6 - (2 + 3) = +1$
Left O (double bond)	6	4	2	$6 - (4 + 2) = 0$
Right O (single bond)	6	6	1	$6 - (6 + 1) = -1$

Answer:

+1

0

, and

-1

, which add up to 0 for the neutral molecule. In the hybrid, the

-1

is shared by the two end atoms (about

-\frac{1}{2}

each), which fits the delocalized picture.

Choosing the best Lewis diagram

When the possible structures aren't equivalent, the best model for predicting structure and properties is the one that:

Gives every atom an octet (2 for H). Period 2 atoms (C, N, O, F) never go past eight.

Has formal charges as close to zero as possible: the fewest and the smallest.

Puts any negative formal charge on the most electronegative atom (and any positive charge on the least).

Avoids formal charges of the same sign on neighboring atoms.

Example 3: Which skeleton is right for dinitrogen monoxide?

Dinitrogen monoxide,

\mathrm{N_2O}

(laughing gas), could be connected N-N-O or N-O-N. Use formal charge to decide.

Step 1:

2(5) + 6 = 16

valence electrons.

Step 2: Draw each skeleton with two double bonds. Both give every atom an octet, with 4 bonding pairs + 4 lone pairs = 16 electrons. The formal charges are in red.

Step 3: Work out the formal charges.

Skeleton	End atom	Middle atom	Other end	Sum
N-N-O	N: $5 - (4 + 2) = -1$	N: $5 - (0 + 4) = +1$	O: $6 - (4 + 2) = 0$	0
N-O-N	N: $5 - (4 + 2) = -1$	O: $6 - (0 + 4) = +2$	N: $5 - (4 + 2) = -1$	0

Answer: N-N-O is the better model. Its formal charges are smaller, while N-O-N puts a

+2

charge on oxygen, the most electronegative atom in the molecule. Every possible N-O-N diagram has this problem, since the central O has to form four bonds. Experiments confirm the atoms are connected N-N-O.

Example 4: The best structure for the cyanate ion (AP-style)

Three Lewis diagrams can be drawn for the cyanate ion,

\mathrm{OCN^-}

. (a) Find the formal charge on each atom in each diagram. (b) Which diagram is the best model? Justify your answer.

All three have

6 + 4 + 5 + 1 = 16

valence electrons (4 bonding pairs + 4 lone pairs), and every atom has an octet. The formal charges are in red.

(a) For example, in (b) the N has 6 dots and 1 line, so its formal charge is

5 - (6 + 1) = -2

, and the O has 2 dots and 3 lines, so its formal charge is

6 - (2 + 3) = +1

Diagram	O	C	N	Sum
(a) O=C=N	$0$	$0$	$-1$	$-1$
(b) O≡C-N	$+1$	$0$	$-2$	$-1$
(c) O-C≡N	$-1$	$0$	$0$	$-1$

(b) Diagram (c) is the best model. Diagram (b) has the largest formal charges and puts a positive charge on oxygen, so it's out. Diagrams (a) and (c) each have a single

-1

, but (c) puts it on oxygen, which is more electronegative than nitrogen (3.44 vs. 3.04). Since (a) and (c) are both valid but not equivalent, the real ion is a hybrid that looks more like (c).

When the octet rule and formal charge pull in different directions

\mathrm{BF_3}

(Topic 2.5), boron has only 6 electrons. A B=F double bond would give boron an octet, but it would also put a

+1

formal charge on fluorine, the most electronegative element, and

-1

on boron. The structure with single bonds has every formal charge at zero, so it's the accepted one.

Carbon monoxide goes the other way. The only structure that gives both atoms an octet is the triple-bonded one, which has formal charges of

-1

on C and

+1

on O. A double-bonded version has zero formal charges but leaves carbon with only 6 electrons, so the octet rule wins and CO is drawn with a triple bond. These criteria are guidelines; the exam sticks to cases where they point clearly to one answer.

Limitations of the Lewis model

Every model simplifies, and the CED points to species with an odd number of valence electrons as the clearest place the Lewis model falls short.

Odd-electron species

\mathrm{NO_2}

has

5 + 2(6) = 17

valence electrons. With an odd total, one electron can't be paired, so at least one atom ends up with fewer than eight. The usual diagram puts the unpaired electron on N, which then has only 7 electrons around it. There are two resonance structures:

Check: 3 bonding pairs (6 electrons) + 5 lone pairs (10 electrons) + 1 unpaired electron = 17 electrons. The formal charges are

+1

on N,

0

on the double-bonded O, and

-1

on the single-bonded O.

A species with an unpaired electron is called a free radical, and radicals are usually very reactive. Two

\mathrm{NO_2}

molecules readily join into

\mathrm{N_2O_4}

, which pairs up the odd electrons:

\mathrm{2NO_2(g) \rightleftharpoons N_2O_4(g)}

Other odd-electron examples are

\mathrm{NO}

(11 valence electrons) and

\mathrm{ClO_2}

(19).

Other limits of the model

A Lewis diagram is a flat map of which atoms are connected. It doesn't show the 3-D shape; VSEPR theory (Topic 2.7) does.

One diagram can't show delocalized electrons. Resonance is the patch.

Formal charges aren't real charges, so a Lewis diagram doesn't show bond polarity.

Some properties come out wrong. The Lewis diagram of $\mathrm{O_2}$ shows every electron paired, but liquid oxygen is attracted to a magnet, which means $\mathrm{O_2}$ has unpaired electrons. Explaining this takes molecular orbital theory, which the AP exam doesn't test.

Common mistakes

Thinking the molecule switches back and forth between resonance structures. It's always the hybrid.

Moving atoms when drawing resonance structures. Only electrons move.

Treating every set of possible diagrams as equally good. Equivalent structures are resonance; nonequivalent ones get ranked with the octet rule and formal charge.

Using the number of bonds in the whole molecule for $B$ . $B$ is the bonding electrons on that one atom.

Forgetting to check that the formal charges add up to the overall charge.

Calling formal charge the real charge on an atom.

Giving C, N, O, or F more than eight electrons to make the formal charges smaller.

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