Lewis Structures
For Covalent Compounds
NAB
Method
Information…
• Octet rule alone does not allow us to
create correct Lewis Structures
• Does not tell us where or how to place the
bonded electrons
• NAB method helps
N = “NEEDED”
1. Calculate the “N” as the sum of the
electrons necessary for all elements to
achieve an octet
–
Important exceptions (as usual)
•
•
•
H=2
Be = 4
B=6
NEEDED
Add together the number of valence
electrons for each atom in the molecule if
each had a full octet.
For example, CF4
There are 5 atoms in the molecule. If each
had a full octet, 5 x 8 = 40 e-s are needed
A “AVAILABLE”
2. Calculate “A” as the sum of all valence
electrons available to share (total
number of valence electrons for ALL
atoms involved)
AVAILABLE ELECTRONS
Add
together
the
number
of
valence
electrons for each atom in the molecule.
For example, CF4
Carbon has four valence electrons and each
fluorine has seven valence electrons
= 4 + 4(7) = 32 available e-
“B” BONDS
3. This is easy
– Shared is the difference between NEEDED
and AVAILABLE: N - A.
– Each bond requires 2 electrons
– So
B = (N – A) / 2
- B is the number of bonds that will
come off the central atom
BONDS
For example, CF4
The number of shared electrons is N – A
= 40 – 32 = 8 shared electrons
2 electrons per bond, so
8/2 = 4 bonds
Steps for Creating Lewis Structure
of Molecules
1. Create a “reasonable” skeleton structure
2. The least electronegative element will be
the central atom
3. Never H or any of the Halogens
•
WHY???
Write out the elements of the molecule so
that the least electronegative elements is
in the center surrounded by the other
elements. For example, CF4
F
F C F
F
How do you choose the central atom?
Many times it's readily apparent:
SiBr4
CH4
H2S
CO2
Place a covalent bond between the central
atom and the outside atoms. Remember
each covalent bond contains two
electrons.
F
F C F
F
The four covalent bonds use eight of the 32 valence
electrons in CF4
F
F
C F
F
• This uses 24
electrons. There are
no electrons left, so
this is The Lewis dot
structure for CF4
There are 24 valence electrons remaining. Add
electrons to the outer atoms as lone pairs to
satisfy the Octet Rule.
• Rule 1: NAB
• Rule 2: Place the least electronegative element at
the center, except for H which is always an outer
atom
• Rule 3: Add covalent bonds between the center atom
and the outer atoms
• Rule 4: Add lone pairs to the outer atoms
• Rule 5: Add lone pairs to the center atom
Number of Bonds





H: halogens and hydrogen can only have
1 bond
O: oxygen group can have 2 bonds
N: nitrogen group can have 3 bonds
C: carbon group can have 4 bonds
Be and B, Al: too small to hold a full
octet; Be forms only 2 bonds and B and
Al form only 3 bonds
RESONANCE
Consider the Lewis structure of the
carbonate ion, CO32-.
The Lewis structure for this ion has a carbon-oxygen double bond, and two
carbon-oxygen single bonds. But which of the three oxygens forms the
double bond? There are three possibilities:
When more than one Lewis structure can be
drawn, the molecule or ion is said to have
resonance
We use a double headed arrow to show
that individual structures are related by
resonance.
HYPERVALENCY
Larger atoms can exhibit hypervalency, in
which the octet is expanded and there are
more than eight electrons in the valence
orbitals.
Valence expansion (also known as
hypervalency) is allowed for atoms in the
3rd period and below of the periodic table.
Hypervalency