Lewis Structures
Lewis structures are used to help visualize what molecules look like. They are 2dimansional representations of molecular structures, based on the arrangement of valence
electrons in the formation of chemical bonds. It is assumed that most elements require an
octet of valence electrons, to attain a noble gas configuration. There are exceptions to
this rule, which we will discuss. An important exception is hydrogen, which fills its
valence shell (1s) with only 2 electrons. There is a series of steps that can be followed to
simplify the determination of a Lewis Structure.
1. Examine the atoms in the molecule and determine how many valence electrons they
each contribute. For polyatomic anions, add enough extra electrons to account for the
negative charge. For polyatomic cations, remove enough electrons to account for the
positive charge.
2. Place the least electronegative element in the center of the molecule (never
hydrogen).
3. Using pair of electrons, make bonds between the central atom and each surrounding
atom.
4. Complete the octets of electrons for the atoms bonded to the central atom. Show all
lone pairs.
5. Place all remaining electrons on the central atom. If the octet rule is not met for the
central atom, move lone pairs of electrons to make multiple bonds to the central atom.
This sounds more confusing than it is.
What is the structure of nitrogen trifluoride, NF3?
N = group 5 = 5 valence electrons
F = group 7 = 7 valence electrons
Total electrons = 5 + (3x7) = 26 electrons
The least electronegative element is nitrogen which will be the central atom.
Now, make bonds between N and three F.
F N F
F
This uses up 6 of the 26 electrons. Now, add an octet of valence electrons around each of
the fluorine atoms.
F N F
F
We have now used 24 of the 26 electrons. Add the remaining 2 electrons to complete the
octet on N.
F N F
F
This accounts for all 26 electrons and gives every atom in the molecule a share in 8
valence electrons, an octet.
There are 3 bonding pairs of electrons and 10 lone pairs of electrons.
What is the Lewis Structure for nitric acid, HNO3?
N = Group 5 = 5 valence electrons
O = Group 6 = 6 valence electrons
H = Group 1 = 1 valence electron
Total electrons = 5 + (3x6) + 1 = 24 electrons
The least electronegative element is nitrogen , the central atom. Hydrogen is always at
the edge of a molecule, since it can only form one bond.
Make bonds between N and three O. The hydrogen should be bonded to one of the
oxygens. Nitric acid is one of the "xoy-acids", like sulfuric, phosphoric and choric acids.
The acidic hydrogen, which forms H+ in aqueous solutions, is always bonded to an
oxygen in these acids.
O N O H
O
This uses up 8 of the 24 electrons. Now, create octets of electrons around the oxygens.
O N O H
O
This uses up all 24 electrons. Each oxygen is surrounded by 8electrons and the hydrogen
has 2 electrons. However, nitrogen only has a share in 6 electrons. To give nitrogen a
chare in 8 electrons, move one of the lone pairs of electrons from an oxygen, and make a
second bonding pair between the oxygen and nitrogen.
O N O H
O
O N O H
O
This now gives N and all of the O a share in 8 electrons. With two pairs of bonding
electrons, the nitrogen and oxygen are joined by a double bond, which would be
represented as N=O.
What is the Lewis Structure of the carbonate ion, CO32-?
First, decide on the number of electrons involved.
C = Group 4 = 4 valence electrons
O = Group 6 = 6 valence electrons
Total electrons = 4 + (3x6) = 22 electrons, but the ion carries a charge of -2. This
indicates that the ion has gained 2 additional electrons, so add 2 more electrons to the
total. We have 24 electrons to work with.
The least electronegative atom is carbon, so it will be the central atom. Draw bonds
between the central C and the three O atoms.
O C O
O
This uses up 6 of the 24 electrons. Next, complete the octets of electrons around the
oxygen atoms.
O C O
O
This brings the electrons used to 24. But, there is a problem. The C has a share in only 6
electrons. Again, move a lone pair from oxygen to make a double bond with the carbons.
This gives both atoms a share in a complete octet.
O C O
O
O C O
O
The double bond could be between the carbon and any one of the equivalent oxygen
atoms.
O C O
O
or
O C O
O
or
O C O
O
A C=O bond will be stronger and shorter than a C-O bond. There is more electron
density between the nuclei. However, it has been shown that all three carbon-oxygen
bonds are identical. This can be explained if we use all three configurations to explain
the structure. When two or more Lewis Structures can be used to describe the same
molecule, each is called a resonance structure. Each has an identical atomic structure,
only the position of the electrons has been changed. A double arrow is used to represent
resonance structures.
O C O
O
O C O
O
O C O
O
Another way of representing these structures is:
O
C
O
O
O
C
O
O
O
C
O
O
Or, more compactly as
O C
O
O
which shows the partial double bond character of all three bonds.
What is the Lewis Structure of formaldehyde, CH2O?
Carbon - 4 valence electrons
Oxygen = 6 valence electrons
Hydrogen = 1 valence electron
Total = 4 + 6 + (2x1) = 12 valence electrons
The atoms can be arranged 2 different ways:
H C O H
or
H C H
O
This leaves 6 more electrons to complete the octets around C and O.
H C O H
or
H C H
O
Again, we have to move lone pair electrons to create double bonds.
H C O H or
H C H
O
Which of these two structures represents formaldehyde? To determine this, we need to
look at the concept of formal charge. This is a comparison of the number of valence
electrons on an isolated atom versus the number of electrons assigned to an atom in a
given Lewis Structure.
The number of electrons in an isolated atom is simply the Group number of the
element.
The number of electrons assigned by a Lewis Structure is the number of lone pair
electrons (unshared) plus one-half of the shared, bonding electrons.
In the figure below, the electrons "belonging" to carbon are shown in red, and the
electrons "belonging" to oxygen are shown in blue.
H C O H or
H C H
O
For the structure on the left, the formal charges will be:
valence e- - lone pair e- - ½ shared e- = formal charge
1
0
1
0
4
2
3
-1
6
2
3
+1
element
H
C
O
For the structure on the right, the formal charges will be:
element
H
C
O
valence e- - lone pair e- - ½ shared e- = formal charge
1
0
1
0
4
0
4
0
6
4
2
0
For a neutral molecule, a Lewis Structure with no formal charges is preferred.
Small formal charges are preferable to large formal charges.
Like charges on adjacent atoms are undesirable.
A more negative formal charge should reside on a more electronegative element.
The structure on the left puts a positive formal charge on oxygen, an electronegative
element.
The structure on the right has no formal charges and is the best structure for
formaldehyde.
The Lewis dot structures of the individual, non-metal atoms give a good indication of the
bonding possibilities for the atoms.
C
N
O
F
Carbon tends to form 4 bonds and have no lone pairs.
Nitrogen tends to form three bonds and have on e lone pair.
Oxygen tends to form two bonds and have two lone pairs.
Fluorine (and all halogens) tends to form one bond and have 3 lone pairs.
With these electron configurations, none of these atoms will have any formal charge.
This is how they are most stable.
In the correct (right) structure for formaldehyde, CH2O, carbon has four bonds and no
lone pairs, and oxygen has two bonds and two lone pairs.
In the incorrect (left) structure, carbon has three bonds an a lone pair, and oxygen has
three bonds and a lone pair, leading to non-zero formal charges.
Exceptions to the octet rule
1. Molecules with an odd number of electrons.
These species are called free radicals because they contain an unpaired electron. An
example would be nitrogen monoxide, NO. Because of the odd number of electrons,
Lewis structures can not really be used to describe it.
What is the structure of nitrogen monoxide, NO?
N = Group 5 = 5 valence electrons
O = Group 6 = 6 valence electrons
Total electrons = 5 + 6 = 11 electrons
Add another electron in order to come up with possible Lewis structures. So, assume
there are 12 electrons.
Bond the O and N.
N O
Next, create an octet on the oxygen, and place remaining electrons on the nitrogen.
N O
To complete the octet on N, move a lone pair from O to give N a share in 8 electrons.
N O
To make a free radical, with only 11 electrons, we can remove an electron from either N
or O.
The two resulting structures are shown below, with their formal charges.
N O
0 0
N O
-1 +1
The first has no formal charge and is the correct representation for NO.
2. Incomplete Octets (fewer than 8 electrons on central atom)
These compounds involve Be (2 valence e-), B and sometimes Al (3 valence e-). For
example, B forms compounds with halogens with the formula, BX3, like BF3.
What is the structure of boron trifluoride, BF3?
B = Group 3 = 3 valence electrons
F = Group 7 = 7 valence electrons
Total electrons = 3 + (3x7) = 24 electrons
F B F
F
A resonance structure could be drawn with a double bond between a F and B. But, this
would put a formal positive charge on F, the most electronegative element.
3. Expanded Octet (more than 8 valence electrons on the central atom)
This can occur in atoms that possess d orbitals, in the 3rd period and beyond. An
example of this would be phosphorus pentafluoride, PF5.
What is the structure of phosphorus pentafluoride?
P = Group 5 = 5 valence electrons
F = Group 7 = 7 valence electrons
Total = 5 + (5x7) = 40 electrons
Bond the 5 F to a central P, and complete the octets on the surrounding atoms.
F
F
F
P
F
F
This uses all 40 electrons, and P, the central atom, has a share in 10 valence electrons.