Lewis structures of compounds
Lewis structures are a convenient way of showing covalent
bonds in molecules or ions of non-transition elements
• The most important consideration is that the atoms attain a noble
gas configuration
• The trickiest part is determining the arrangement of the atoms in
the molecule or ion
-- usually, the single atom in the formula (e.g., C in CO2) will
be the central atom
• The Lewis structure does not give you information about the
shape of the molecule/ion (more about this later)
Rules for writing Lewis structures
1. Determine the number of valence electrons for all atoms in
the molecule or ion
• for ions, add one valence electron for each negative charge
or subtract one valence electron for each positive charge on
the ion
2. Write the skeletal arrangement of the atoms and connect
them with a single covalent bond (two dots or one dash)
• hydrogen and the halogens typically form only one covalent
bond
• oxygen normally forms two covalent bonds (one single bond
or two double bonds)
• nitrogen normally forms three covalent bonds (combination of
single, double and/or triple bonds)
• carbon normally forms four covalent bonds (combination of
single, double and/or triple bonds)
Rules for writing Lewis structures
3. Subtract two electrons for each single bond used in Step 2
from the total number of valence electrons
• this gives the number of remaining electrons available for
completing the structure
4. Distribute remaining electrons as pairs of electrons (pairs
of dots) around each atom to give each atom a noble gas
configuration (8 electrons)
• hydrogen needs only 2 electrons for a noble gas configuration
5. If there are not enough electrons to give each atom a noble
gas configuration, change single bonds to double or triple
bonds by shifting non-bonding pairs of electrons as needed
Lewis structure of water
Step 2. Write the skeletal arrangement of the atoms and
connect them with a single covalent bond (two dots or one
dash)
Lewis structures of compounds
Example: Write the Lewis structure for water (H2O)
Step 1. Determine the number of valence electrons for all
atoms in the molecule or ion
H: 1s1 -- one valence electron
O: [He] 2s22p4 -- six valence electrons
2H + O
total of 8 valence electrons
Lewis structure of water
Step 3. Subtract two electrons for each single bond used
in Step 2 from the total number of valence electrons
4 electrons
H:O:H
H:O:H
Place two dots between the hydrogen and oxygen atoms to
form the covalent bonds
8 total
valence
electrons
-
4 electrons
=
4 electrons
left to complete
the structure
Lewis structure of water
Step 4. Distribute remaining electrons as electron pairs
(pairs of dots) around each atom to give each atom a noble
gas configuration
hydrogen atoms have noble gas
configuration (2 electrons each)
: :
H:O:H
Lewis structures of compounds
Example: Write the Lewis structure for
carbon tetrachloride (CCl4)
Step 1. Determine the number of valence electrons for all
atoms in the molecule or ion
C: [He] 2s22p2 -- four valence electron
Cl: [Ne] 3s23p5 -- seven valence electrons
oxygen atom has noble gas
configuration (8 electrons)
Lewis structure of carbon tetrachloride
Step 2. Write the skeletal arrangement of the atoms and
connect them with a single covalent bond (two dots or one
dash)
C + 4 Cl
total of 32 valence electrons
Lewis structure of carbon tetrachloride
Step 3. Subtract two electrons for each single bond used
in Step 2 from the total number of valence electrons
8 electrons
Cl
Cl:C :Cl
Cl
Place two dots between the carbon and chlorine atoms to
form the covalent bonds
Lewis structure of carbon tetrachloride
Step 4. Distribute remaining electrons as electron pairs
(pairs of dots) around each atom to give each atom a noble
gas configuration
: :
: : :
: :
:
:Cl:
Cl:C :Cl
:Cl:
:
:
carbon atom has noble gas
configuration (8 electrons)
chlorine atoms have noble gas
configuration (8 electrons each)
: :
: :
Cl
Cl:C:Cl
Cl
32 total
valence
electrons
-
8 electrons
=
24 electrons
left to complete
the structure
Covalent bonds:
single, double, and triple
In a single bond
• one pair of electrons is shared.
In a double bond
• two pairs of electrons are shared
In a triple bond
• three pairs of electrons are shared
Write a Lewis structure for CO2
Step 1. Determine the total number of valence electrons
for the atoms in the compound
Write a Lewis structure for CO2
Step 2. The two O atoms are bonded to a central C atom.
Write the skeletal structure and place two electrons
between the C and each oxygen.
C: 4 valence electrons
O: 6 valence electrons
O:C:O
4 + ( 2 x 6 ) = 16 total valence electrons
Step 4. Distribute the 12 electrons (6 pairs) around the
carbon and oxygen atoms. Three possibilities exist.
: :
: :
O:C:O
:O:C:O:
II
:O:C:O:
4
electrons
16 valence electrons - 4 electrons = 12 electrons left
Write a Lewis structure for CO2
Step 5. Remove one pair of unbonded electrons from each O
atom in structure I and place one pair between each O and the C
atom forming two double bonds.
6 6
6
6
electrons
electrons
electronselectrons
Write a Lewis structure for CO2
Step 5. Remove one pair of unbonded electrons from each O
atom in Structure I and place one pair between each O and the C
atom forming two double bonds.
: :
: :
:O:C:O
Many of the atoms in these structures do not have eight
electrons around them.
:O:C:O:
Each atom now has 8
electrons around it
: :
O::C::O
: :
O::C::O
: :
Structure I
III
:
: :
::
I
double bond
: :
Step 3. Subtract the four electrons used in Step 2 from 16
(the total number of valence electrons) to obtain 12
electrons yet to be used.
Write a Lewis structure for CO2
:
:
: :
Write a Lewis structure for CO2
double bond
Carbon is sharing 4
electron pairs
Examples of multiple covalent bonds
O O
O C O
carbon dioxide
( CO2 )
H
H
C C
diatomic oxygen
( O2 )
H
Complex Lewis structures
( resonance structures )
There are some molecules and polyatomic
ions that can not be represented by a single
Lewis structure
N N
H
diatomic nitrogen
( N2 )
ethene
( C 2H 4 )
Write a Lewis structure for nitrate ( NO3- )
Write a Lewis structure for nitrate ( NO3- )
Step 1. Determine the total number of valence electrons
Step 2. The three O atoms are bonded to a central N
atom. Write the skeletal structure and place two
electrons between each pair of atoms
O O:N:O
:
• 5 from the nitrogen atom
• 6 from each O atom
• 1 from the –1 charge
The total number of valence electrons is 24
Since the extra electron present results in nitrate
having a –1 charge, the ion is enclosed in brackets
with a – charge.
Write a Lewis structure for nitrate ( NO3- )
Write a Lewis structure for nitrate ( NO3- )
Step 3. Subtract the 6 electrons used in Step 2
from 24, the total number of valence electrons,
to obtain 18 electrons yet to be placed.
Step 4. Distribute the 18 electrons around
the N and O atoms
:O
:O:N:O:
: :
: : :
: :
:
O
:
O N:O
-
electron deficient
Write a Lewis structure for nitrate ( NO3- )
Write a Lewis structure for nitrate ( NO3- )
Step 5. One pair of electrons is still needed to give all
the N and O atoms a noble gas structure
Step 5. One pair of electrons is still needed to give all
the N and O atoms a noble gas structure
• move the unbonded pair of electrons from the
N atom and place it between the N and the electrondeficient O atom, making a double bond
-
: :
:: :
: :
:O
:O:N:O:
Write a Lewis structure for nitrate ( NO3- )
Step 5. One pair of electrons is still needed to give all
the N and O atoms a noble gas structure
Step 5. There are three possible Lewis structures
:
:O
N O
:O
-
:O
N O:
:O:
-
O N O:
A molecule or ion that shows multiple correct Lewis
structures exhibits resonance
N O:
: :
: :
:O
-
: :
:
:
: :
-
:O:
: :
:
:
: :
• move the unbonded pair of electrons from the
N atom and place it between the N and the electrondeficient O atom, making a double bond
:O
-
Write a Lewis structure for nitrate ( NO3- )
: :
:
:
: :
: :
: : :
: :
:O
:O:N:O:
• move the unbonded pair of electrons from the
N atom and place it between the N and the electrondeficient O atom, making a double bond
• each Lewis structure is called a resonance structure
Write a Lewis structure for nitrate ( NO3- )
N O:
O N O:
Spectroscopy shows that all three of the N–O bonds in
the nitrate ion have identical lengths
• the N–O bonds are intermediate between single bonds
and double bonds
-
:O:
-
:O
N O
:O
-
:O
N O:
:O:
: :
:
:
: :
:O
:O:
: :
:
:
: :
N O
-
: :
:
:
: :
:O
:O
: :
:
:
: :
-
: :
:
:
: :
:O:
Step 5. There are three possible Lewis structures
: :
:
:
: :
Step 5. There are three possible Lewis structures
Write a Lewis structure for nitrate ( NO3- )
O N O:
The actual structure of the nitrate ion is a blend of all
three individual resonance structures
• this is referred to as a resonance hybrid
-
Compounds containing polyatomic ions
A polyatomic ion is a stable group of atoms that
has either a positive or negative charge and
behaves as a single unit in many chemical reactions
nitrate ion
sodium ion Na+
the polyatomic ion behaves as a unit because
its atoms are held together by covalent bonds
Na
Sodium nitrate has both ionic and covalent bonds
Ionic bonds exist between the sodium ions and
the nitrate ions
Na
+
:O
:O
N O:
Covalent bonds exist between the atoms of
nitrogen and oxygen atoms within the nitrate ion
ionic
bond
-
:O
-
Sodium nitrate has both ionic and covalent bonds
: :
:
:
: :
ionic
bond
+
:O
: :
:
:
: :
Example: Nitrate (NO3-)
Na
N O:
+
:O
-
: :
:
:
: :
•
Sodium nitrate ( NaNO3 ) comprises sodium ions
and nitrate ions in a 1:1 ratio
:O
N O:
covalent
bond
When sodium nitrate is dissolved in water,
the ionic bond breaks
The sodium ions and nitrate ions separate from
each other forming separate sodium and nitrate ions
-
O: N O:
N :O
:
:O
::
+
:
:O
:
Na
:
+
:O
:
:
::
:
:
Na
-
The nitrate ion, which is held together
by covalent bonds, remains as a unit
covalent
bond
covalent
bond
Exceptions to the octet rule
Octet Rule
Atoms tend to gain, lose, or share elections until they
attain a full valence shell
• 2 valence electrons for the first principal energy level
• 8 valence electrons for all other principal energy levels
The octet rule is a useful guideline for describing bonding,
but there are many exceptions:
•
Molecules and polyatomic ions containing an odd
number of electrons
•
Molecules and polyatomic ions in which an atom
has less than an octet of valence electrons
Molecules and polyatomic ions in which an atom
has more than an octet of valence electrons
•
Exceptions to the octet rule: Molecules and polyatomic
ions containing an odd number of electrons
Exceptions to the octet rule: Molecules/polyatomic ions
in which an atom has less than an octet of electrons
In most molecules and polyatomic ions, the total number of
valence electrons is even
This is relatively rare -- it is most commonly encountered
with compounds of boron and beryllium
• all electrons are paired up
Example: BF3
But there are some exceptions -- for molecules/ions with
an odd number of valence electrons, it is not possible to
pair up all electrons or attain a full octet around each atom
Example: NO
F
N O
These are the two most important Lewis structures for NO
Cl
Cl
F
Cl
Phosphorous atom has more
than 8 electrons around it
Cl
H
C
N
O
Hydrogen
Carbon
Nitrogen
Oxygen
1 covalent bond
4 covalent bonds
3 covalent bonds
1 lone pair
2 covalent bonds
2 lone pairs
Cl
Br
Halogens: 1 covalent bonds, 3 lone pairs
I
F
F
F
F
B
Group
8A
H
1 bond
He
Group
3A
Group
4A
Group
5A
Group
6A
Group
7A
0 bonds
B
C
N
O
F
Ne
3 bonds
4 bonds
3 bonds
2 bonds
1 bond
0 bonds
The octet rule is a useful guideline,
but there are numerous exceptions
P
B
Group
1A
5 + ( 5 x 7 ) = 40 valence electrons
Hydrogen, Carbon, Nitrogen, Oxygen, and Halogen
Atoms Usually Maintain Consistent Bonding Patterns
F
F
Numbers of covalent bonds typically formed by main
group elements to achieve noble gas configuration
Elements in Period 3 and beyond can expand their
valence shells to accommodate more than 8 electrons
• this is related to the d orbitals available starting with
principal energy level n = 3
Cl
B
F
B
F
This structure is most favorable even
though boron does not have a full octet
Exceptions to the octet rule: Molecules/polyatomic ions
in which an atom has more than an octet of electrons
Example: PCl5
F
F
5 + 6 = 11 valence electrons
N O
F
3 + ( 3 x 7 ) = 24 valence electrons
Si
P
S
Cl
Ar
4 bonds
3 bonds
(5)
2 bonds
(4, 6)
1 bond
(3, 5)
0 bonds
•
boron forms 3 covalent bonds and ends up with only
6 valence electrons
Br
Kr
1 bond
(3, 5)
0 bonds
•
elements in the third period and below can form additional
covalent bonds (this is because they have vacant
d orbitals that can hold additional bonding electrons)
I
Xe
1 bond
(3, 5, 7)
0 bonds
A quick note about formal charge
When drawing Lewis structures for a molecule or polyatomic ion,
a formal charge can be assigned to each atom in the structure
The formal charge of an atom is defined as the charge that the
atom would have if all of the atoms in the molecule/ion had the
same electronegativity
-- i.e., each bonding electron pair shared equally between the
two bonded atoms
Specific rules are followed to assign electrons to each atom in
the structure
-- the number of assigned electrons is compared with the
number of valence electrons in the isolated atom to
determine formal charge
A quick note about formal charge
Formal charge is used as an accounting device to determine
which Lewis structure for a molecule/ion is most favorable when
there exist several different structures that satisfy the octet rule
-- most favorable structures contain atoms with formal charges
closest to zero
-- most favorable structures have negative formal charges
assigned to the most electronegative atoms
Due to time constraints, we won’t be covering the topic of formal
charge in any detail
-- but if you are planning on taking further chemistry courses,
read through the section in Chapter 8 (Brown text) dealing
with formal charge and familiarize yourself with the concept