AP Chem 15-16 Unit 3 Structural formulas, Formal charge, and resonance
An outline of how to determine the "best" Lewis structure for an example, NO3- is given
below:
1. Determine the total number of valence electrons in a structure. If an ion, remember to
add or remove electrons based on the charge.
2. Draw a skeleton for the molecule which connects all atoms using only single bonds. In
simple molecules, the atom with the most available sites for bonding is usually the central
atom, however the OFFICIAL rule is to place the least electronegative atom in the center
(unless it’s Hydrogen—Hydrogen is never a central atom). The number of bonding sites is
determined by considering the number of valence electrons and the ability of an atom to
expand it's octet. As you become better, you will be able to recognize that certain groups
of atoms prefer to bond together in a certain way. See exceptions below.
3. Of the 24 valence electrons in NO3-, 6 were required to make the skeleton. Consider
the remaining 18 electrons and place them so as to fill the octets of as many atoms as
possible (start with the most electronegative atoms first then proceed to the more
electropositive atoms).
4. Are the octets of all the atoms filled? If not then fill the remaining octets by making
multiple bonds (make a lone pair of electrons, located on a more electronegative atom,
into a bonding pair of electrons that is shared with the atom that is electron deficient).
5. Check that you have the lowest FORMAL CHARGES possible for all the atoms,
without violating the octet rule;
(valence e-) - (1/2 bonding e-) - (lone electrons).
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AP Chem 15-16 Unit 3 Structural formulas, Formal charge, and resonance
IMPORTANT : no Lewis diagram is complete without formal charges. Lewis diagrams are
drawn to examine mechanisms so knowing which parts of a molecule are electron
deficient (+) and which are electron rich (-) is vital.
It is best to have a formal charge of 0 for as many of the atoms in a structure as possible,
and a structure with the lowest formal charges possible is most likely as long as:
 If a formal charge is negative, it must be on the most electronegative atoms in the
structure (conversely, a positive formal charge would be on more electropositive
atoms)
 The sum of the formal charges of each individual atom must equal the net charge of
the structure (i.e. Since NO3 -1 has a net charge of -1, the formal changes of all 4
atoms must sum to -1).
CAUTION : octets can be expanded to minimize formal charges but only for where the
highest occupied energy level is n=3 or greater. For instance in our example, N cannot
expand its octet so keeps a formal charge of 1+ and both singly bonded oxygens a formal
charge of 1-. If our molecule were SO3 , however, it would be possible to minimize all
formal charges by having the sulfur expand its octet.
6. You may find that the best Lewis diagram (the one with the lowest formal charges and
all octets satisfied) is given in a number of different ways. For NO3-, three different
diagrams are given below. From left to right they start with the most complete Lewis
diagram to the most simplified.
Why so many different ways? Depends on the need of the chemist. For instance,
complete structures are more useful for the novice organic chemist learning to appreciate
the mechanism of a reaction while simplified versions may be preferred by inorganic
chemists.
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AP Chem 15-16 Unit 3 Structural formulas, Formal charge, and resonance
Resonance
A reasonable Lewis structure for the nitrate polyatomic ion, NO3−, is below.
This Lewis structure shows two different types of bonds, single and double.
Because it takes more energy to break a double bond than a single bond, we
say that a double bond is stronger than a single bond. Double bonds also
have a shorter bond length, the distance between the nuclei of the two atoms
in the bond, than single bonds do. Thus, if the above Lewis structure for
nitrate were correct, the nitrate polyatomic ion would have one bond that is
shorter and stronger than the other two.
This is not the case. Laboratory analyses show all three of the bonds in the
nitrate ion to be the same strength and the same length. Interestingly, the
behavior of the bonds suggests they are longer than double bonds and
shorter than single bonds. They are also stronger than single bonds but not
as strong as double bonds. In order to explain how this is possible for the
nitrate ion and for molecules and polyatomic ions like it, the valence-bond
model had to be expanded.
The model now allows us to view certain molecules and polyatomic ions as if
they were able to resonate between two or more different structures. For
example, the nitrate ion can be viewed as if it resonates between the three
different structures below. Each of these structures is called a resonance
structure. The hypothetical switching from one resonance structure to another
is called resonance, and the convention is to separate the resonance
structures with double headed arrows.
It is important to stress that the nitrate ion is not really changing from one
resonance structure to another, but chemists find it useful, in an intermediate
stage in the process of developing a better description of the nitrate ion, to
think of it as if it were doing so. In actuality, the ion behaves as if it were a
blend of the three resonance structures.
We can draw a Lewis-like structure that provides a better description of the
actual character of the nitrate ion by blending the resonance structures into a
single resonance hybrid:
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AP Chem 15-16 Unit 3 Structural formulas, Formal charge, and resonance

Draw the skeletal structure, using solid lines for the bonds that are found in all of the
resonance structures.

Where there is sometimes a bond and sometimes not, draw a dotted line.

Draw only those lone pairs that are found on every one of the resonance structures.
(Leave off the lone pairs that are on one or more resonance structure but not on all
of them.)
The resonance hybrid for the nitrate polyatomic ion is
The actual geometry of the polyatomic ion is trigonal planar with bond angles
of 120°.
Resonance Structures and the Resonance Hybrid
Resonance is possible whenever a Lewis structure has a multiple bond and
an adjacent atom with at least one lone pair. The following is the general form
for resonance in a structure of this type. The arrows show how you can think
of the electrons shifting as one resonance structure changes to another.
You can follow these steps to write resonance structures.

Shift one of the lone pairs on an adjacent atom down to form another bond.

Shift one of the bonds in a double or triple bond up to form a lone pair. (You might
find it useful to draw arrows indicating the hypothetical shift of electrons.)

Draw additional resonance structures by repeating this process for each adjacent
atom with a lone pair.

Separate the resonance structures with double-headed arrows.
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AP Chem 15-16 Unit 3 Structural formulas, Formal charge, and resonance
For example, the two resonance structures for the formate ion, HCO2− are
To generate the second resonance structure from the first, we imagine one
lone pair dropping down to form another bond, and pushing an adjacent bond
off to form a lone pair. The arrows show this hypothetical shift of
electrons. These resonance structures lead to the resonance hybrid below.
Exceptions to the Octet Rule
There are three situations where the octet rule.
1. Molecules with an odd number of electrons (i.e. NO2).
If you have an odd number of electrons there is no way to satisfy the octet rule. Try
it if you don’t believe me.
2. Molecules where an atom has less than an octet (i.e. H2, BF3, BeH2,).
This only happens to atoms near the boundary between metals and non-metals, such
as H, B, and Be. H is stable with 2 e-, B with 6e-, and Be with 4e-. The
electronegativity of these atoms is not high enough to force more electronegative
non-metals into forming double and triple bonds. When working with such atoms
never draw multiple bonds and you will be able to get the correct Lewis dot
structures.
3. Molecules where an atom has more than an octet of electrons (i.e. ClF3, PCl5, XeF2).
Atoms with an occupied 3rd energy level or higher can expand to accept bonding
electrons into their empty d orbitals if this provides a more stable structure.
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AP Chem 15-16 Unit 3 Structural formulas, Formal charge, and resonance
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AP Chem 15-16 Unit 3 Structural formulas, Formal charge, and resonance
For each formula below:
a) Calculate the total quantity of electrons available for bonding
b) Draw the most likely (most stable) structural formula
c) Calculate the formal charge for every atom in the formula and write this formal charge near the atom
in the structure.
1)
PBr3
5)
2)
BFBr2
6)
3)
NH4 +1
4)
CH3OH
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CO3-2
CO2
7)
C2H4
8)
BFS
AP Chem 15-16 Unit 3 Structural formulas, Formal charge, and resonance
9)
HBr
10)
C2H5OH (ethanol) (carbon is the center atom)
11)
13)
SF6
14)
PCl4-1
N2F4 (put Nitrogens in the center – linear
molecule)
15) PCl4+1
12)
H2S
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AP Chem 15-16 Unit 3 Structural Formulas, Formal Charge, Resonance, and Bond Order
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AP Chem 15-16 Unit 3 Structural Formulas, Formal Charge, Resonance, and Bond Order
Lewis Structures Practice Worksheet - Solutions
Please forgive my very poor drawing skills!
1)
PBr3
2)
N2H2
3)
CH3OH
4)
NO2-1
5)
C2H4
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AP Chem 15-16 Unit 3 Structural Formulas, Formal Charge, Resonance, and Bond Order
6)
BSF
7)
HBr
8)
C2H5OH (ethanol)
9)
N2F4
10)
SF6
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AP Chem 15-16 Unit 3 Structural Formulas, Formal Charge, Resonance, and Bond Order
11)
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