Lewis Structures (Including Formal Charges)
Guidelines for Drawing Lewis Structures Given the Molecular Formula
1. Determine whether or not the compound of interest is ionic. If it is not ionic, proceed as indicated below. If it is
ionic, show the molecular formulas of the ions separately, with the appropriate ionic charges. Ionic compounds can be
recognized by the presence of a metal ion such as Na+, K+, Ca+2, etc., or of the ammonium ion, NH4+. Treat each ion
as a separate entity and do not connect then with a bond. Examples:
Li2CO3 contains lithium metal, so write it as Li+ CO3-2
Li+
NH4NO2 contains the ammonium ion, so write it as NH4+
NO2–
Note: If the above molecular formula were given to you as H4N2O2, you could not be sure that it contained the
ammonium ion, and there could be an acceptable Lewis structure with only covalent bonds.
2. Calculate the total number of valence electrons in the structure or in each ion by adding the valence electrons
of all of the atoms. For positive ions, subtract an electron for each unit of positive charge. For negative ions, add an
electron for each unit of negative charge. Double-check your calculations. A mistake here will cause problems.
3. Arrange the atoms and connect them by single bonds.
This is sometimes the trickiest part and there can often be more than one way to arrange them to give an acceptable
Lewis structure. (These different ways are called isomers, as you will learn later.) Some pointers:
a) Oxy-acids and their anions usually have the oxygen atoms arranged around a central atom. Any hydrogens
are usually attached to oxygens. Examples:
O
O
HClO4 would start out as H O Cl O
H 2SO3 would start out as H O S O H
O
b) Avoid structures with long strings of oxygen or nitrogen atoms. These tend to be unstable. Recall that
hydrogen peroxide, H-O-O-H, is already rather unstable and reactive. Hydrazine, H2N-NH2, is a rocket fuel.
Example:
A Lewis structure for H 2SO4 that meets all of the rules for bonding is H-S-O-O-O-O-H. This
doesn't exist, and if it did you couldn't get near it.
4. Distribute remaining electrons in pairs on the outer atoms so that these have an octet (except hydrogen). Any
electrons left after this operation are placed on the central atom.
5. If the central atom has fewer than 8 electrons, move non-bonded electrons from the outer atoms to form
double or triple bonds so that the central atom has an octet.
6. Calculate the formal charge for each atom (below), and show any that are non-zero as a charge enclosed in a
circle, positioned close to the atom. (The reason for using the circle is that it clearly distinguished formal charges from
overall ionic charges.)
7. Atoms in the 2nd row of the periodic table and beyond (commonly sulfur or phosphorous) may have
“expanded valence shells”. (This means more than 8 electrons around it in its compounds. The reason for this will
we discussed later.). If such and atom ends up with a formal positive charge, eliminate it as much as possible by
moving unshared electrons from adjacent atoms with a formal negative charge.
For example, a phosphorous atom has 5 valence electrons. It can use all of them for bonding so that it can have 10
electrons around it in a Lewis structure. Sulfur, with 6 valence electrons, can have up to 12 electrons around it.
Example:
P O
P O
goes to
8 e's aeound P
+1 formal charge
10 e's around P
0 formal charge
8. Double-check your finished structure.
a) Count up the bonded and non-bonded electrons and compare the total with your calculation in Step 2. If
they don’t agree, either your structure is wrong or your calculation in Step 2 was wrong.
b) Confirm that the atoms in your structure conform to the octet rule. As mentioned previously, atoms in the
second row and beyond may have expanded valence shells. Also, an atom may be “electron deficient” (fewer than 8
electrons) because there are simply not enough electrons available. Boron and aluminum often show this behavior
because these atoms have only 3 electrons in the valence shell.
HH H
H
C
H
H
H C Al C H
Example:
H B H
H
H
6 e's around B
6 e's around Al
Guidelines for Calculating Formal Charges
1. For the atom of interest, count up the number of electrons that the atom “owns” by assuming that it owns all of the
unshared electrons around it plus one-half of the electrons in its bonds.
2. Compare what the electron owns with what it “want” in order to be neutral. What it wants to be neutral is the
number of valence electrons it would have as an isolated neutral atom.
3. Indicate any excess of electrons as negative formal charge on the atom, and any deficiency as positive formal
charge. Don’t label atoms with zero formal charge.
The complete procedure for drawing a Lewis structure, including formal charges, is illustrated below.
Step 1
Example: NH 4HSO4
+
For NH 4
Step 3
+
NH4
Π
HSO4
It's ionic.
H
H N H
H
H owns 1
H wants 1
 0 formal charge
Step 2
+
NH4
4H + N Π 1e =
8 electrons
Step 6
H
H N H
H
N owns 4
N wants 5
 +1 formal charge
Step 8
Π
HSO4
H + S + 4O + 1e =
32 electrons
1. Has 8 valence electrons
2. Octet rule obeyed
3. All OK !
For
Π
HSO4
Step 3
O
H O S O
O
O
H O S O
O
Step 4
S owns 4
S wa nts 6
 +2 formal charge
O owns 6
O wants 6
 0 formal charge
O owns 7
O wants 6
 Π1 formal charge
1. Has 32 valence electrons
2. Octet rule obeyed, S can have 12
3. All OK !
Your completed structure should look like this.
Step 6
H
H N H
H
O
H O S O
O
Step 7
Step 8
O
H O S O
O
S now owns 6, wants 6,
0 formal charge
O
H O S O
O
Note that you could also have shown HSO4– as either of the two Lewis structures below. The three different
arrangements of electrons on the same atomic framework represent what are called resonance structures.
O
H O S O
O
O
H O S O
O