Drawing Lewis Structures
Drawing the "best" Lewis structure for a molecule is a great skill to have for your chemistry class! This concept will come in handy in the upcoming chapters as well. We will follow the steps below each time.
Octet Rule: atoms need to have 8 electrons in the valence shell
Examples: C, N, O, F all need 8 electrons in their valence shell to have the "best" Lewis structure
Example: CF4 → the C atom has 8 valence electrons and each F atom has 8 valence electrons in the Lewis Structure
Exceptions to the Octet Rule
Some elements are octet deficient (require less than 8 electrons in their valence shell)
Examples:
H and He can only have 2 electrons
Be can only have 4 electrons
In group 13, Al and B can only have 6 electrons
Some elements have an expanded octet (can have more than 8 electrons in their valence shell)
Examples:
Non-metals in period 3 and onwards can have expanded octet.
P can have 10 electrons
S can have 12 electrons
Some molecules like radicals have an odd number of electrons (these are rare!)
Example: NO
First let’s count the number of valence electrons (VE) in both atoms.
- N →5 VEs
- O → 6 VEs
- So in total there are 11 VEs which is an odd number. We would be left with a molecule that looks like this:
Note: There is an unpaired electron on the N. This is called a radical molecule and is highly reactive so it is very short-lived.
Formal Charges
The last thing we need to check to ensure we have the BEST Lewis Structure is to count formal charges.
FC is the formal charge of an atom
VE are the valence electrons the atom has
Example:
OF2
FC for F= 7 VE - 1 bond connected - 6 electrons on it
FC for F=7-1-6
FC for F=0
FC for O= 6 VE - 2 bonds connected - 4 electrons on it
FC for O= 6 - 2 - 4
FC for O=0
Therefore, we know this is the best possible Lewis structure!
Example: Lewis Structures
Draw Lewis structures of the following molecules.
a) PCl3
b) NH4+
c) SeF4
Example: Lewis Structure
Draw the Lewis Structure for SO42–
Determine the central atom. The central atom is usually the atom which has the lowest electronegativity. In this case it is S.
Count the total # of valence electrons.
6 (S) + [4 X 6 (O)] + 2= 32
Arrange the terminal atoms around the central atom.
Use e- to fill octets of terminal atoms.
Evaluate and rearrange electrons to find the best structure.
In our example all octets are satisfied but we have a 2+ on sulfur and negative charges on all oxygen atoms. To solve this we should move a l.p on oxygen (the negative element) to a double bond. Remember sulfur can expand its octet.