Drawing Lewis Structures
The following step-by step instructions are designed to produce a correct Lewis
Structure for any molecular assembly (neutral molecule or polyatomic ion) which
contains only Main Group elements joined by two-atom shared-pair covalent bonds.
1.
2.
3.
4.
5.
6.
7.
Connectivity - arrange bonded atoms
VSE - count valence shell electrons
BP - assign bond pairs
PLP - assign lone pairs to peripheral atoms
CLP - assign lone pairs to central atoms
Rearrange - find best Lewis Structure(s) [formal charge]
Extra - assigning oxidation numbers
These rules are not appropriate for free radicals or molecular assemblies which
contain multicenter bonds or transition metals.
Step 1: Connectivity - arrange the atomic symbols so that covalently bonded atoms
are contiguous.
Commentary
Some general rules and definitions:
1. the number of covalent bonds an atom forms is
called its valence.
2. Some atoms have fixed valence. For example:
H = 1, C = 4, F = 1.
3. Some atoms have variable valence. For
example:
O = 2 (sometimes 3), B, N = 3 (sometimes 4).
4. an atom bonded to only one other atom is
peripheral (monovalent atoms such as H and F
are always peripheral).
5. an atom bonded to two or more other atoms is
central.
Often, the formula is written to indicate connectivity.
For example:
HCN = H bonded to C, C bonded to N, H and N are not
bonded.
CH3OCH3 = three H bonded to C1, C1 bonded to O, O
bonded to C2, C2 bonded to three H.
CH3CH2OH = three H bonded to C1, C1 bonded to C2,
C2 bonded to two H and O, O bonded to H.
Otherwise, as a general rule, the least electronegative
elements (if not monovalent) are central, the most
electronegative elements are peripheral. For example:
CO32- = 3 peripheral O bonded to central C
PO43- = 4 peripheral O bonded to central P
H3O+ = 3 monovalent (therefore peripheral) H bonded to
central O
Note that the order or geometric arrangement of symbols
written on the page is irrelevant as long as bonded atom
pairs are contiguous. The final Lewis Structure
represents onlythe approximate electronic
disribution,not the actual 3-dimensional arrangement
of the atoms.
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Step 2: VSE - Count the total number of Valence Shell Electrons; divide these VSE
into pairs.
Commentary
1. Sum the number of valence shell electrons of each atom;
the number of valence shell electrons for an atom is equal to its group
number
2. Subtract the charge on the assembly
3. The total should be even (these rules do no apply to assemblies with an odd
number of electrons)
Examples:
HCN = 1+4+5-0 = 10 VSE = 5 VSE pairs
C2H6O = 2(4)+6(1)+6-0 = 20 VSE = 10 VSE pairs
CO32- = 4+3(6)-(-2) = 24 VSE = 12 VSE pairs
PO43- = 5+4(6)-(-3) = 32 VSE = 16 VSE pairs
H3O+ = 3(1)+6-(+1) = 8 VSE = 4 VSE pairs
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Step 3: Assign BP - Place one VSE pair of electrons between each bonded pair of
atoms.
Commentary
Connect each contiguous pair of atoms with
one of the VSE pairs; each Bond Pair is shown
as a line. Examples:
HCN = 5 VSE pairs - 2 BP = 3 pairs remaining
C2H6O = 10 VSE pairs - 8 BP = 2 pairs
remaining
CO32- = 12 VSE pairs - 3 BP = 9 pairs
remaining
PO43- = 16 VSE pairs - 4 BP = 12 pairs
remaining
H3O+ = 4 VSE pairs - 3 BP = 1 pair remaining
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Step 4: Assign Peripheral LP - Place up to three VSE pairs on each peripheral atom.
Commentary
Distribute the VSE pairs remaining after Step 3
among the peripheral atoms as Lone Pairs (H cannot
accept lone pairs; see the Rule of Orbitals below). At
this stage, no peripheral atom may have more than 4
VSE pairs (1 BP + 3 LP). Examples
HCN : of 3 VSE pairs, all are assigned to N; no VSE
pairs remain.
C2H6O : of 2 VSE pairs, none are assigned (since all
peripheral atoms are H); 2 VSE pairs remain.
CO32- : of 9 VSE pairs, 3 are assigned to each O; no
VSE pairs remain.
PO43- : of 12 VSE pairs, 3 are assigned to each O; no
VSE pairs remain.
H3O+ : 1 pair, not assigned (since all peripheral
atoms are H); 1 VSE pair remains.
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Step 5: Assign Central LP - Place any remaining VSE pairs as Lone Pairs on central
atom(s) according to the Rule of Orbitals.
Commentary
The Rule of Orbitals: the total number of lone pairs and bond
pairs (LP+BP) associated with an atom cannot exceed the
number of Valence Shell Orbitals (VSO = n2, where n is the
row of the Periodic Table in which that atom resides).
n = 1 (H): maximum VSE pairs (LP+BP) = VSO = 1;
n = 2 (B, C, N, O, F): maximum VSE pairs (LP+BP) = VSO =
4 ("octet rule")
n = 3 ((Al, Si, P, S, Cl): maximum VSE pairs (LP+BP) = VSO
= 9; etc.
Examples
C2H6O = 2 pairs, both assigned to O since each C already has
4 BP.
H3O+ = 1 pair, assigned to O.
While C,N,O and F always fill their four valence shell
orbitals with Lone Pairs and/or Bond Pairs, B often does not.
Thus, these four atoms always obey the "octet rule" (the only
atoms on the periodic chart which always do!), but some
compounds with B have an empty valence shell orbital and are
called "electron deficient". Furthermore, third row elements
(e.g., Al, Si, P, S, Cl) often have more than four valence shell
orbitals filled with Lone Pairs and/or Bond Pairs; this is called
(illogically) "expanded valence". Obviously, elements from
the fourth and higher rows can also exhibit "expanded
valence". The tendency of most main group elements (except
H) is to form molecular assemblies which fill at least the first
four orbitals (s and p subshells); this tendency, plus the
ubiquity of assemblies which contain C, N and O, has led to
the (over) emphasis on the "octet rule".
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Step 6: Rearrange VSE Pairs - If necessary, push electron pairs according to the
Rule of Orbitals and the Principle of Electroneutrality.
Commentary
Principle of Electroneutrality: each atom in a covalent molecular assembly has a
formal charge close to zero.
Formal Charge: FC = (Group Number) - (Bond Pairs) - 2(Lone Pairs)
Electron Pushing: formally changing a lone pair into a bond pair, or vice versa, while
retaining association with the atom.
Examples
HCN
Original Lewis Structure
H: FC = 1-1-2(0) = 0;
H: rule of orbitals satisfied (1 orbital, 1
VSE pair);
C: FC = 4-2-2(0) = +2;
C: rule of orbitals not satisfied (4
orbitals, only 2 VSE pairs; it must be
associated with 4 VSE pairs);
N: FC = 5-1-2(3) = -2;
N: rule of orbitals satisfied (4 orbitals,
4 VSE pairs)
Rearranged Lewis Structure
Push two lone pairs on N into C-N
bonding position, creating a C-N triple
covalent bond.
H: FC = 0;
C: FC = 4-4-2(0) = 0;
N: FC = 5-3-2(1) = 0;
All atoms: rule of orbitals satisfied.
C2H6O (both isomers)
Original Lewis Structures
H (all): FC = 1-1-2(0) = 0;
H (all): rule of orbitals satisfied;
C (all): FC = 4-4-2(0) = 0;
C (all): rule of orbitals satisfied;
O: FC = 6-2-2(2) = 0;
O: rule of orbitals satisfied;
the original Lewis Diagrams
(produced with rules 1-5) are correct,
no rearrangement necessary.
CO32Original Lewis Structure
O (all): FC = 6-1-2(3) = -1;
O (all): rule of orbitals satisfied;
C: FC = 4-3-2(0) = +1;
C: rule of orbitals not satisfied; it must
be associated with 4 VSE pairs.
Rearranged Lewis Structure
Push a lone pair on any of the three O
into bonding position, creating one CO double bond. For each of the three
Lewis Structures thus produced:
O (single bond): FC = -1;
O (double bond): FC = 6-2-2(2) = 0
(average formal charge on each
oxygen = -2/3);
C: FC = 4-4-2(0) = 0;
all atoms: rule of orbitals satisfied;
The set of three rearranged Lewis
Structures correctly depicts the
resonance in this ion.
PO43O (all): FC = 6-1-2(3) = -1;
O (all): rule of orbitals satisfied;
P: FC = 5-4-2(0) = +1;
P: rule of orbitals satisfied (P can
accomodate up to 9 VSE pairs);
Push a lone pair on any of the four O
into bonding position, creating one PO double bond. For each of the four
Lewis Diagrams thus produced:
O (single bond): FC = -1;
O (double bond): FC = 0
(average formal charge on each
oxygen = -3/4);
P: FC = 5-5-2(0) = 0;
all atoms: rule of orbitals satisfied;
The set of four rearranged Lewis
Diagrams correctly depicts the
resonance structures of this ion.
H3 O+
H (all): FC = 0;
H (all) rule of orbitals satisfied;
O: FC = 6-3-2(1) = +1;
O: rule of orbitals satisfied.
The original Lewis Structure is
correct. The formal charge of +1 on
oxygen is reduced by electronegative
induction, so each H atom attains a
slight positive charge while the charge
on the O atom approaches 0.
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Extra: Oxidation Numbers
General Rules
1. Each atom in an element has ON = 0
Examples: He(g), H2(g), Br2(l), P4(g), S8(s), Fe(s)
2. In ionic compounds, molecules and polyatomic ions
a. Oxygen: ON(O) = -2 usually
Examples: Na2SO3, SO3, SO42-; but in Na2O2 and H2O2, ON(O) = -1
b. Hydrogen with a metal: ON(H) = -1
Examples: NaH, CaH2
c. Hydrogen with a non-metal: ON(H) = +1
Examples: CH4, H2O, HPO42d. Fluorine: ON(F) = -1 always
3. The sum of all oxidation numbers is equal to the charge on the compound,
molecule or ion
4. For any Main Group atom, maximum ON = Group Number
Examples: C (+4), P (+5), Se (+6), Br (+7), Xe (+8)
5. For any Main Group atom, minimum ON = Group Number - 8
Examples: C (-4), P (-3), Se (-2), Br (-1), Xe (0)
Lewis Diagrams
The oxidation number for an individual atom in a Lewis Diagram is calculated as
follows:






Assign both electrons of a lone pair to the valence shell of the atom.
Assign both electrons of a bond pair to the valence shell of the most
electronegative atom.
Examples
o H-F => H :F
o C-O => C :O
o C-H => C: H
If both bonded atoms are identical (and therefore have the same
electronegativity), assign one bonding electron to the valence shell of each
atom.
Examples
o C-C => C. .C
o C=C => C: :C
VSE = the total number of electrons (lone pairs, bond pairs, and single
electrons) thus assigned to the valence shell of the atom.
ON = Group Number - VSE
Check: the sum of the oxidation numbers of all atoms in the Lewis
Structure = the total charge on the Lewis Structure.
Examples:
HCN: C is more electronegative
than H, so it is assigned the H-C
bond pair; N is more
electronegative than C, so it is
assigned all three C-N bond
pairs. Then




ON(H) = 1-0 = +1
ON(C) = 4-2 = +2
ON(N) = 5-8 = -3
(+1) + (+2) + (-3) = 0
CH3OCH3: O is more
electronegative than C, so it is
assigned both C-O bond pairs; C
is more electronegative than H,
so both C(1) (the left-most C
atom) and C(2) (the right-most C
atom) are assigned every C-H
bond pair (both C atoms are
identical). Then




ON(H) = 1-0 = +1
ON(C(1)) = ON(C(2)) =
4-6 = -2
ON(O) = 6-8 = -2
6(+1) + 2(-2) + (-2) = 0
Note that the oxidation number of
both C atoms is equal to the value
obtained by applying the General
Rules.
CH3CH2OH: O is more
electronegative than either C or
H, so O is assigned both the C-O
and O-H bond pairs; C is more
electronegative than H, so both
C(1) (the left-most C atom) and
C(2) are assigned every C-H
bond pair. The C-C bond pair is
split between the two C atoms.
Then
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



ON(H) = 1-0 = +1
ON(C(1)) = 4-7 = -3
ON(C(2)) = 4-5 = -1
ON(O) = 6-8 = -2
6(+1) + (-3) + (-1) + (-2)
=0
Note that the average oxidation
number of the two C atoms is
equal to the value obtained by
applying the General Rules.
CO32-: O is more electronegative
than C, so each O is assigned the
C-O bond pair. Then



ON(O) = 6-8 = -2
ON(C) = 4-0 = +4
3(-2) + (+4) = -2
It is left as an exercise for the
reader to perform these
calculations on the resonance
structures.
PO43-: O is more electronegative
than P, so each O is assigned the
P-O bond pair. Then



ON(O) = 6-8 = -2
ON(P) = 5-0 = +5
4(-2) + (+5) = -3
It is left as an exercise for the
reader to perform these
calculations on the resonance
structures.
H3O+: O is more electronegative
than H, so O is assigned all of the
H-O bond pairs. Then



ON(H) = 1-0 = +1
OH(O) = 6-8 = -2
(+1)+(+1)+(+1)+(-2) = +1