1.2 · Bonding & Lewis

Formal charge and resonance

How to pick the best structure and when one drawing is not enough.

By the end you should be able to:

Assign formal charges and choose the best Lewis structure
Draw resonance structures and find average bond orders and lengths

Formula

Formal charge

$\text{FC} = V - N - \tfrac{1}{2}B$

$V$ = valence electrons of the free atom
$N$ = nonbonding (lone-pair) electrons on the atom
$B$ = bonding electrons around the atom

Quick form: FC = V − (dots + lines), counting each lone-pair electron as a dot and each bond as a line. The formal charges always add up to the overall charge of the species.

Atom	FC = −1	FC = 0	FC = +1
C	3 bonds, 1 lone pair	4 bonds	3 bonds, no lone pair
N	2 bonds, 2 lone pairs	3 bonds, 1 lone pair	4 bonds
O	1 bond, 3 lone pairs	2 bonds, 2 lone pairs	3 bonds, 1 lone pair

Key idea

Choosing the best Lewis structure

When more than one structure obeys the octet rule, prefer, in this order:

formal charges as close to zero as possible (fewest and smallest);
any negative formal charge on the more electronegative atom;
no like charges on adjacent atoms.

For $\ce{CO2}$ , O=C=O (all formal charges 0) beats O≡C–O (+1 on one O, −1 on the other). For the cyanate ion, $\ce{[O-C#N]^-}$ and $\ce{[O=C=N]^-}$ each carry a single −1; rule 2 favours the form with −1 on O. Real species are a blend of such forms, so rule 2 is a tie-breaker rather than a law.

Key idea

Resonance structures

When a multiple bond can be placed in two or more equivalent positions, a single Lewis structure cannot describe the species. Draw every valid form and join them with double-headed arrows ( $\leftrightarrow$ ).

Only electrons (lone pairs and multiple-bond pairs) move; the atoms stay where they are.
The real species is the resonance hybrid, one structure that is the average of the forms. It does not flip back and forth between them.
Equivalent forms contribute equally; forms with worse formal charges contribute less.

Classic cases: $\ce{NO3-}$ and $\ce{CO3^2-}$ (3 forms), $\ce{O3}$ , $\ce{NO2-}$ and $\ce{HCOO-}$ (2 forms), benzene (2 forms).

Formula

Average bond order and bond length

$\text{bond order} = \frac{\text{total bonds over the equivalent positions}}{\text{number of equivalent positions}}$

Count a double bond as 2 and a triple bond as 3.

Species	Bond	Bond order
$\ce{NO3-}$	N–O	$\frac{2+1+1}{3} = \frac{4}{3} \approx 1.33$
$\ce{CO3^2-}$	C–O	$\frac{4}{3} \approx 1.33$
$\ce{O3}$	O–O	$\frac{2+1}{2} = 1.5$
benzene	C–C	1.5

A higher bond order means a shorter, stronger bond. Both O–O bonds in ozone are 128 pm, between the O–O single bond in $\ce{H2O2}$ (148 pm) and the O=O double bond in $\ce{O2}$ (121 pm).

Method

Finding resonance forms and bond orders

Draw one valid Lewis structure and assign formal charges.
Look for a multiple bond that could equally well sit on a neighbouring equivalent atom.
Redraw by converting a lone pair on that atom into a bond and the old π bond into a lone pair; the atoms and the electron total stay fixed.
Repeat until the multiple bond has visited every equivalent position once.
For each set of equivalent bonds, bond order = total bonds ÷ number of positions.

Common mistake

Resonance and formal-charge traps

Wrong: " $\ce{NO3-}$ has one short N=O bond and two long N–O bonds." Right: all three N–O bonds are identical, with bond order 4/3 and a length between single and double.

Wrong: moving an atom, such as shifting an H, to make a new "resonance form". Right: that gives a different compound (an isomer); resonance forms differ only in where electrons are.

Wrong: treating a formal charge as the real charge on an atom. Right: formal charge is bookkeeping that assumes perfectly equal sharing; real partial charges follow electronegativity.

Worked example

Worked example: nitrate, NO₃⁻

Valence electrons: $5 + 3(6) + 1 = 24$ .
N is central with three N–O single bonds (6 e⁻). Completing the O octets uses 18 e⁻, leaving 0, and N has only 6 electrons.
Convert one O lone pair into an N=O double bond: N now has an octet.
Formal charges: N: $5 - 0 - 4 = +1$ ; double-bonded O: $6 - 4 - 2 = 0$ ; each single-bonded O: $6 - 6 - 1 = -1$ . Sum: $+1 + 0 - 2 = -1$ , the ion's charge.
The double bond can sit on any of the three O atoms, so there are 3 equivalent resonance forms.

Answer: N–O bond order $= \frac{4}{3} \approx 1.33$ ; all three N–O bonds have the same length.

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