CHEMISTRY 115
LEWIS DOT DIAGRAMS AND ASSIGNING FORMAL CHARGES
Bonds and Lone Pairs:
Each covalent bond between atoms usually contains 2 electrons. Between 2 atoms we can have
any of the following:
a single bond (2 electrons)
a double bond (4 electrons – i.e., 2 2-electron bonds)
a triple bond (6 electrons – i.e., 3 2-electron bonds)
In addition, we can have electrons on atoms which are not shared in bonds:
a lone pair of electrons (2 electrons, one with spin up and one with spin down)
a lone electron (a single electron – happens less often, usually only when necessary)
The Octet Rule:
Noble gas configurations are especially stable. Ignoring any d-type electrons, (we will focus on the
non-metals, which have either no d-electrons or a completely filled set of d-electrons – in either
case, the d electrons will usually not participate in bonding.) However, as we will see, when an
empty set of d orbitals is close in energy to the levels containing the valence electrons, then these
orbitals may be used to accept electrons and thus participate in bonding. The distinction is
important – d orbitals are not the same thing as d-electrons. The latter are electrons that already
occupy d orbitals, d orbitals, (as is the case also for s, p, and f orbitals), are regions of space that
have particular shapes that show the highest probabilities of where the electrons will be, if
electrons occupy that orbital.
The special stability of noble gas configurations means we would expect either 2 valence electrons,
(a 1s2 – He configuration), or 8 valence electrons, (a ns2 np6 – Ne, Ar, Kr, Xe, or Rn noble gas
configuration), to be especially stable. [Note again that we do not include the 10 electrons that
occupy filled d orbitals.]
When a bond is formed, both atoms can benefit by sharing the electrons in the bond. For example,
H2O would look like this –
H O H
Each H atom brought only 1 electron to the molecule, but each H atom “feels” the presence of 2
electrons in its vicinity – the 2 it shares with the O atom – and thus ends up with the He
configuration, because it shares its 1 electron with one from oxygen. Each O atom brings only 6
valence electrons to the molecule, but “feels” the presence of a total of 8 electrons in its vicinity –
i.e., the Ne configuration. It “feels” the presence of 2 electrons that are shared with the left H atom,
2 which are shared with the right H atom and 4 more which are unshared that occur as 2 sets of
lone pairs. Lone pair electrons are shared with no other atom and thus “belong” solely to the atom
on which they reside. Note that both H atoms and the O atom have achieved noble gas
configurations, without the necessity of forming ions in order to do so.
-1-
Raynor: 2/6/2016
Preferred “octets”:
For elements in groups 14 (or IVA in older tables), or higher, an s2p6 configuration is preferred.
Thus these elements need to feel the presence of 8 electrons in their vicinity, either shared or in
lone pairs. For elements in lower numbered groups, too many electrons would need to be added to
reach a total of 8. For these elements, a maximum of two times the number of valence electrons is
desired, no more.
I.e., # e– in preferred “octet” = 2  # valence electrons; or 8, (whichever value is smaller)
Table 3-1: # Valence Electrons and Preferred Octet by Group Number
Group
1
2
3 (13)
4 (14)
5 (15)
6 (16)
–
1
2
3
4
5
6
# valence e
2
4
6
8
8
8
“octet”
7 (17)
7
8
8 (18)
8
8
Note that boron has only 6 electrons in its preferred octet, instead of 8, since 2  the number of its
valence electrons, (i.e., 23) < 8. However, oxygen has a full 8 in its octet, since 2 the number of
its valence electrons, (i.e., 26) > 8.
It is easy to understand why some atoms tend to have fewer than 8 in their octets. For example,
when boron forms bonds, it has 3 valence electrons available to share. Thus it can donate 1 each to
3 other atoms bringing the total number to 6 electrons in its vicinity. In order for it to have a total
of 8 in its vicinity, it would need to form a 4th bond without sharing – i.e., both electrons in that
bond would need to come from the other atom in the bond. This gives no benefit for the other atom
– it is providing all the electrons in that bond – and thus boron is limited to 2 its valence
electrons.
Determining the Number of Bonding and Lone Pair Electrons:
The following simple formula is pretty much foolproof, as long as it is applied properly. [There is
no point in using it if you don’t know what the value you get from it is used for.] When the
formula fails, it still provides important information – it alerts us that one or more of the atoms will
end up with more than or less than an octet.
S=N–A
where
N = total # electrons Needed by all atoms in order for them to each have a Noble gas
configuration; (2 ONLY for each H atom, usually 8 for each other atom, but see table above.)
A = total # valence electrons Actually Available from the atoms, (including the charge, if
any. Add 1 electron for each negative charge on the whole molecular unit, subtract 1 for each
positive charge on the molecular unit. Note that A is identical to the total number of electrons
which must be shown in the final diagram.
S = total # Shared electrons in the molecule, (i.e., it is identical to the number of electrons
that must appear in bonds between atoms);
Important Note: Even if you ignore this formula and use a different method the create your
diagram, your final diagram must have a total of A electrons in it. I.e., when you add up all the
electrons in bonds with all the electrons in lone pairs, the sum must equal A.
-2-
Raynor: 2/6/2016
So, for H2O
N = 2 (for 1st H) + 2 (for 2nd H) + 8 (for O) = 12
A = 1 (for 1st H) + 1 (for 2nd H) + 6 (for O) = 8
S = 12 – 8 = 4
This means that we need a total of 4 electrons occurring in bonds between the atoms. 2
electrons in the O-H bond on the left and 2 electrons in the O-H bond on the right. The
remaining electrons, which is given by A – S = 8 – 4 = 4 will occur only in lone pair
positions. Thus water will have a total of 2 lone pairs. Both lone pairs must be on the O atom
in order for it to achieve its desired octet of 8.
Note that we can also use the values of A and S to determine the number of lone pair electrons.
Since S is the number of electrons in bonds and A is the total number of electrons in the diagram,
the number of lone pair electrons must be given be U = A – S, where U is the number of unpaired
electrons. Dividing that value by 2 gives the number of pairs to place in as lone pairs. [Note: in the
cases where S must be adjusted because one or more atoms either exceed their desired octets or
have less than a full octet, use S = actual number of electrons that occur in bonds in the final
diagram.]
Important Note:
Since H atoms can only have octets containing 2 electrons, H atoms can NEVER form more than 1
bond to another atom. The electron in H occupies the 1s level, when it is in its ground state. The 1s
level can accept a maximum of 2 electrons – one with spin up and one with spin down. Any
addition electrons would have to occupy levels much higher in energy – e.g., the 2s level. Thus H
atoms form only 1 bond each.
Rules for Forming Lewis Dot Pictures:
1)
Choose a symmetric skeleton for the molecule or ion..
a)
The least electronegative element is usually in the center (exception – H atoms are
almost never in the center). Since oxygen is the 2nd-most electronegative element, (with
F being the most electronegative), it is rare for O atoms to be in the center. Exceptions
are H2O, OF2, O3, etc. Once you’ve placed the center atom, distribute other atoms
symmetrically around it. E.g., for OF2: F O F is a more symmetric structure than
O F F.
b)
O atoms usually do not bond together unless to form O2, O3 or the ions O2–2 (peroxide)
or O2–1 (superoxide). So for the ion SO3–2, we would arrange the atoms this way:
O S O
O
c)
In acids containing oxy-anions, the acidic H atoms usually bond to O atoms first, then
to the central atom, if any H atoms are left over. [In oxy-acids where H atoms bond to
the central atom, instead of to available O atoms, you will be advised how to place
them.
H2SO4:
O
H O S O H
O
-3-
Raynor: 2/6/2016
d)
If a charge is present on the unit, place square brackets around the diagram and place
the charge at the upper right corner of the right-hand bracket:
–2
SO32–:
O S O
O
e)
If there is more than 1 central atom, then choose the most symmetrical skeleton:
C2H2:
2)
H H
C C
H H
and
P2O74–
O
O


O P O P O 




O
O
4
Calculate N, The number of valence electrons needed by all atoms in the molecule or ion to
achieve noble gas configurations. DO NOT INCLUDE CHARGES HERE. This value is the
value needed for every atom in the structure to leave without bonding to any others, but still
have a noble gas configuration
For H2SO4: N  1 8 (S atom)  4  8 (O atoms)  2  2 (H atoms)  8  32  4  44 e needed.
For SO32–: N  1 8 (S atom)  3  8 (O atoms)  8  24  32 e needed.
3)
Calculate A, the number of valence electrons actually available. For atoms, the number of
valence electrons available = its Group number, (if it is in Groups 1 or 2), or its Group
number – 10, (if it is in Groups 13 - 18) If you use an older-style periodic table, the number
of valence electrons is the same as the Group Number, (which is in Roman numerals). Since
this gives the actual number of electrons in the diagram, we must adjust our value by adding
the relevant number of electrons if the structure is negatively charged and by subtracting the
relevant number if the structure is positively charged. Thus you need to add 1 electron for
each negative charge on the overall chemical formula or subtract 1 electron for each positive
charge on the formula.
For H 2SO 4 : A  1 6 (S atom)  4  6 (O atoms)  2 1 (H atoms)  6  24  2
 32 e  available.
For SO32 : A  1 6 (S atom)  3  6 (O atoms)  2 (Since there is a  2 charge)
 6  18  2  26 e available.
NOTE: Whether you use this method or not, you MUST make sure to have this number
of electrons in your final diagram. I.e., you must have a total of 32 e– in the H2SO4
diagram and 26 e– in the diagram of SO32–.
4)
Calculate S, the total number of electrons shared in the molecule or ion, using the equation
S = N – A.
For H2SO4: S = N – A = 44 – 32 = 12 e– in shared positions (= 6 pairs of e– in bonds)
For SO32–: S = N – A = 32 – 26 = 6 e– in shared positions (= 3 pairs of e– in bonds)
-4-
Raynor: 2/6/2016
5)
Place all bonding electrons in the skeleton, as shared pairs. Note, each atom must have at
least one bond to all its nearest neighbors in the skeleton. Use double or triple bonds only
when absolutely necessary. Bonding electrons can be shown using either two dots per bond or
one long dash, —, where the dash indicates a pair of electrons.
O
O
H O S O H
For H2SO4: H O S O H
or
O
O
O S O 


O


2–
For SO3 :
6)
2
O S O 


O


or
2
Add the rest of the available electrons as unshared electrons. Add these in as lone pairs as
much as possible. Use these electrons to ensure that, as much as possible, every atom will
complete its desired octet. (I.e., 2 for H, 6 for B atoms, 8 for most others, etc.) Again, you
can either use dots to represent individual electrons, or dashes to indicate pairs, or a mix of
the two. (Lewis pictures often use dashes for bonds and dots for lone pairs.)
For H2SO4:
2–
For SO3 :
O
H O S O H or
O
O S O


O


2
or
O
H O S O H or
O
O S O


O


2
or
O
H O S O H
O
O S O 


O


2
Note — when putting in your bonding and lone pair electrons, if there is more than one way
of placing the electrons, use the most symmetrical arrangements of them:
For CO2: N = 8 + 8 + 8 = 24; A = 4 + 6 + 6 = 16; S = 24 – 16 = 8, (4 bonding pairs);
U = A – # in bonds = 16 – 8 = 8, (4 sets of lone pairs):
7)
NO – (unsymmetrical)
O C
O
YES – (symmetrical)
O C
O
Some compounds or ions contain an odd number of electrons. It is usually best to place the
odd electron as a lone electron, not as a bonding electron. With an odd number of electrons,
we usually have at least one atom which ends up with less than an octet of electrons – the less
electronegative element should be the one with less than an octet. I.e., the more
electronegative atoms will fill their octets first, since they have stronger affinities of
electrons.
Nitrous Oxide: NO – N = 16; A = 11; S = 5; U = 6 – this would predict this structure:
N
O (Note that this puts 9 electrons around O and 7 around N – not acceptable!)
N
O (Note that this puts 7 electrons around O and 9 around N – not acceptable!)
or
-5-
Raynor: 2/6/2016
[Oxygen and Nitrogen can NEVER exceed their octets! The next level above the valence
level would be the 3s level, which involves an increase in the principle quantum number, n,
and thus is at a much higher energy.]
Instead, when S is an odd number, reduce it by one to ensure that only an even number
of electrons is used in each bond. Thus, we will reduce S from 5 to become S = 4. This
forces the odd electron to become a lone pair electron, the number of lone pair (or lone)
electrons becomes U = 11 – 4 = 7. Thus, the lone electron will be placed in a lone pair
position, instead of in a bonding one.
We have 2 possible sites for the 7th lone pair electron: on the N atom or on the O atom.
Possible structures then become
N
O
(Note that this puts 8 electrons around O and 7 around N – acceptable)
N
O
(Note that this puts 7 electrons around O and 8 around N – acceptable)
or
Note that the first of these two structures is most probable – the more electronegative
element, (O), has its desired full octet of 8 electrons and the less electronegative element, (N),
has 1 fewer than a complete octet. Thus, the 1st structure will be our choice.
8)
Some elements, (besides H), will tend to have less than an octet of electrons, as indicated
earlier. These are the ones in Groups 1, 2 and 13, (or 3), which have 1, 2 and 3 valence
electrons, respectively. These will tend to form exactly 1, 2 and 3 bonds, respectively, with
other atoms. I.e., when calculating N, (the number needed for an “octet”), use 2 for atoms in
Group 1, 4 for atoms in Group 2, and 6 for atoms in Group 13.
BF3: N = 16 (for B) + 38 (for F) = 30; A = 13 + 37 = 24; S = N – A = 6; U = 24 – 6 = 18
F B F
F
9)
Some compounds must exceed their octet – i.e., have more than 8 electrons around them.
This can only happen if there is an empty d orbital just above the valence p orbital – i.e., only
for atoms in the 3rd or higher period. (P, S, Cl, Br, etc.) This cannot happen with 2nd row
atoms – i.e., never for C, N, O, F… The closest empty orbital for the atoms in the 2nd period
is the 3s orbital, which is much higher in energy than the 2s and 2p orbitals. Thus, elements
in the 1st and 2nd periods will never exceed a total of 8 electrons, (i.e., never go beyond an
s2p6 configuration). You can tell when it is necessary for an atom to exceed its octet, because
the calculated number of shared electrons, (S, in the formula), is too little for all the single
bonds needed just to “glue” together the atoms.
PH5: N = 18 + 52 = 18; A = 15 + 51 = 10; S = 18 – 10 = 8, (4 bonding pairs) – this is
not enough electrons to bind 5 H atoms to 1 P atom. Therefore, P must exceed its octet:
H H
H P H
H
[Note that P has 10 electrons around it – more than an octet.]
-6-
Raynor: 2/6/2016
I 3 : N = 38 = 24; A = 37 + 1 = 22; S = 24 – 22 = 2 (1 bonding pair) – this is not enough
to bind together the 3 I atoms – one of them must exceed its octet:
1
I I I
[Note that the central I has 10 electrons around it, more than an octet, while the outer I atoms
each have 8, perfect octets.] [Note that F3 cannot exist since F cannot exceed its octet.]
Determining Formal Charges:
When more than one Lewis Dot picture is possible, formal charges can be used to determine which
picture is most likely. The concept is simple – each atom in a bond is assumed to “own” half of all
bonding electrons and is assumed to “own” all electrons in lone pairs on that atom. The formal
charge is then the difference between the effective nuclear charge on the atom, (= the number of
valence electrons in the atom) and the number of electrons the atom “owns” in the diagram. I.e., we
will assume that every bond is a perfect covalent bond – one with equal sharing of all electrons.
Formal charge = # valence e— of atom — ½  [# e— shared by atom] — 1  [# e— in lone pairs]
Note – due to the definition of formal charge, the sum of all formal charges on all atoms must be
equal to the overall charge on the molecule or ion.
Example 1: There are 3 different Lewis diagrams for the thiocyanate anion: SCN— ion, shown
below. The formal charges on each atom are shown under it.
[S
C
N ]—
0
0
–1
[ S
C
N ]—
–1
0
0
[ S
+1
C
N ]—
0
–2
To demonstrate, let’s do the formal charges in the 1st diagram: For S: FC = 6 – ½(4) – (4) = 0.
For C: FC = 4 – ½(8) – 0 = 0. For N: FC = 5 – ½(4) – (4) = –1.
Using Formal Charges to Select Best Lewis Diagram:
Once the formal charges are determined for each diagram, the best diagram is chosen, by applying
the rules listed below, in order. The best structure is one with all 0 formal charges, since that
indicates that all atoms are involved with perfectly covalent bonds. (I.e., we stop after step 1
below). However, if no such perfect structure exists, we then look for the worst structures next.
(I.e., we apply steps 2-4 below, stopping once we have only 1 structure remaining. That then is our
best structure for the molecule or ion.)
1)
If any structure has all of its formal charges = 0, it is a perfect structure.
However, if no perfect structure exists, we apply the next 3 steps, in order, eliminating poorer
structures, until the best structure remains. Keep in mind that we are looking for structures
containing covalent bonds between its atoms – with no charges on the atoms.
2)
First, we reject all structures which have large magnitudes formal charges. [I.e., formal
charges of 2 or higher.]
3)
If more than 1 structure remains, we reject all structures where large charge differences,
(2 or higher), exist between atoms, especially if those atoms are bonded to one
another. [I.e., if one atom has a formal charge of -1 and it is bonded to an atom with a
-7-
Raynor: 2/6/2016
formal charge of +1, it has a difference of 2 and is unfavored.] High charge
differences between formal charges on bonded atoms imply that the bonding is no
longer covalent and is thus unfavorable.
4)
Finally, if more than 1 structure still remains, we look to see which types of atoms have
non-zero formal charges in each structure. A structure which has negative formal
charges on more electronegative elements, or positive formal charges on more
electropositive elements, is more favored. [Remember – oxygen is the 2nd most
electronegative element, (fluorine is the 1st). Many compounds contain oxygen. If the
structure contains oxygen and in one Lewis diagram an oxygen carries a negative
formal charge while in another structure a different element has a negative formal
charge, the first structure would be preferred. For other elements, use the rules we
established for eletronegativity.]
Rules 1 – 3 rules arise from the idea that true covalent bonds should involve completely uncharged
atoms.
Consider the three thiocyanate anions shown in Example 1. None of these structures is perfect –
they all have at least one non-zero formal charge. Using Rule 2, Structure 3 is found to be the least
likely structure, since one of its atoms it has a formal charge of –2. Rule 3 does not help us
distinguish between the remaining structures, Structures 1 and 2, since in both structures; one of
the bonds has a charge difference of 1. However, when we apply Rule 4, we find that Structure 1
should be the more likely structure, since N is somewhat more electronegative than S. Note,
however, that the real structure will be a blend of all three structures, with the largest weighting
from Structure 1, some weighting from Structure 2, and little contribution from Structure 3.
Notice that we can also use formal charges to decide between the two final candidates we had for
NO. The two possible structures are redrawn below, with their formal charges shown as well:
N
0
O
0
N
O
–1
+1
Clearly, the first structure is more likely – it is a perfect structure, (i.e., all its formal charges are 0).
The second structure shows both a large charge difference along the NO bond, (a difference of 2),
and also puts the negative charge on the less electronegative element. Thus we expect that the
unpaired electron will reside with the N atom and not with the O atom, in agreement with our
earlier conclusions.
Formal Charges and Expanded Octets:
For elements that are allowed to expand their octets, a more likely structure may occur with an
expanded octet, even if an acceptable one is found with the usual octet on all atoms. For example,
consider the structures shown below for ClO3—.
O
O
Cl
Cl = +2
–O = –1
O
O
O
Cl
O
O
Cl = +1
–O = –1
=O = 0
O
Cl
Cl = 0
–O = –1
=O = 0
-8-
O
O
O
Cl
O
Cl = –1
=O = 0
Raynor: 2/6/2016
The first drawing is the one we would predict using the rules for Lewis structures – all its atoms
have complete octets. However, it has very unlikely formal charges, due to the large magnitude
formal charge on Cl, Thus, we would reject this structure using Rule 2. When we apply Rule 3, we
find that Structure 2 is also rejected, due to large charge differences along the Cl—O bonds, (a
difference of 2). Structures 3 and 4 are equivalent to one another, except as to which element
carries the –1 formal charge. Since O is more electronegative than Cl, Structure 3 is most likely,
(and Structure 4 is next most likely). We would expect that the bond lengths in ClO3— should be
approximately 2/3 Cl-O double bond and 1/3 Cl-O single bond, once all 3 resonance forms for
Structure 3 are considered.
Let us take another look at the sulfate anion, SO4–2. The formal charges for the Lewis diagram we
derived earlier are as follows: +2 on S and –1 on each O. This gives a very unfavorable set of
formal charges. Since S can exceed its octet, let’s consider what happens to the formal charges as
we move a lone pair off of an O atom into the bond with S. Note that O will then still possess 8
electrons in its vicinity, but S will possess 2 more for each double bond formed:
For SO42-:
FC’s:
O
L
M
O S
M
M
NO
O
OP
P
P
Q
2
O: all –1
S: +2
O
L
M
O S
M
M
NO
O
OP
P
P
Q
2
O–: –1
O=: 0
S: +1
O
L
M
O S
M
M
NO
O
OP
P
P
Q
2
O–: –1
O=: 0
S: 0
O
L
M
O S
M
M
NO
O
O
L
P
M
O PM
O S
P
QM
NO
O–: –1
O=: 0
S: +1
2
O
OP
P
P
Q
2
O=: 0
S: +2
Note that the best structure is the middle one – the one with 2 double bonds and 2 single bonds.
The 1st and 5th are rejected using Rule 2 and the 2nd and 4th are rejected by Rule 3.
Resonance Structures:
When a structure contains 2 or more identical atoms involved in different bonding environments,
then more than one equivalent Lewis drawing can be made. For example, consider our chosen
structure for ClO3– from above. There are 3 different drawings we can make, all of which involve 2
C=O type bonds and 1 C–O type bond. These are called resonance structures and we represent
the final structure by showing all three with double-headed arrows between them:
O
O
Cl
O
O 
O
Cl
O
O 
O
Cl
O
The actual structure of this ion involves 3 identical Cl–O bonds, each of which is 2/3 of a double
bond and 1/3 of a single bond, i.e., a 5/3 bond. What this means in reality is that the expected
length of the bond is approximately 2/3 of a double bond plus 1/3 of a single bond. [Note: to
calculate the average size of a bond simply do the following: choose any of the bonds involved in
resonance, (i.e., the left-most O-Cl bond, or the top one or the right-hand one), and add up the
number of bonds it contains in each structure. Then, divide by the total number of resonance
structures. In the example above, if we choose the left-hand O-Cl bond, we find it has 2 bonds in
the 1st structure, 1 in the 2nd and 2 in the third. This adds up to 5 bonds total in 3 structures, so we
divide 5 by 3 predicting that each bond is a 5/3 bond. We get the same outcome regardless of
which bond we follow through the structures.
-9-
Raynor: 2/6/2016
Note that the sulfate anion will have 6 resonance structures, each involving a different pair of
bonds with the SO double bond – i.e., with double bonds in positions 1 and 2, 1 and 3, 1 and 4, 2
and 3, 2 and 4, and 3 and 4:
O
L
M
O S
M
M
NO
O
O
L
P
M
O PM
O S
P
QM
NO
2
O
O
L
P
M
O PM
O S
P
QM
NO
2
O
O
L
P
M
O PM
O S
P
QM
NO
2
O
O
L
P
M
O PM
O S
P
QM
NO
2
O
O
L
P
M
O PM
O S
P
QM
NO
2
O
OP
P
P
Q
2
We thus expect that each SO bond will average out to be ½ of a double bond and ½ of a single
bond, i.e., a 1½ bond. [Out of 6 structures, each bond is a double bond in 3 of the structures and a
single bond in 3 of them, thus each bond is 1/6[32 + 31] = 3/2 bond.]
A classical example of resonance is given by benzene. The two resonance structures of benzene are
shown below. Note that the expected bond length is approximately the average of a C-C single
bond and a C-C double bond, a 3/2 bond. The third structure indicates the resonance using a circle
to indicate totally delocalized electrons. [In all 3 structures, there are H atoms bonded to each
carbon atom. These H atoms are not shown in the drawings below.]


- 10 -
Raynor: 2/6/2016