Molecular
Structure
Part I: Lewis Diagrams
CH. 6 – MOLECULAR STRUCTURE
I
II
III
I. Lewis Diagrams
(p. 170 – 175)
Molecular Compounds:
Are
made of nonmetals
 Nonmetals
have high
electronegativity, so they do NOT
release their electrons.
 Two
nonmetals share some of their
valence electrons (in bonds) to
Shared Valence
achieve full octets.
 The
atoms are CO-valent-ly bonded!
I. Lewis Dot Structures
Are designed to show the
placement of valence
electrons in covalently
bonded compounds.

Covalent
compounds
= Molecules
= Compounds with shared electrons
I. Lewis Dot Structures
Step
1: Add up all valence
electrons in the compound.
This
is the total number of
electrons available to bond the
molecule.
(Remember:
inner electrons do
not participate in bonding.)
I. Lewis Dot Structures
Step
1: Add up all valence
electrons in the compound.
NO2
(e- tally)
1 N × 5e- =
5e2 O × 6e- = + 12e17e-
If
the formula carries a CHARGE,
add or subtract electrons
accordingly. NO2- = 17e- +1e = 18e-
I. Lewis Dot Structures
Step
1: Add up all valence
electrons in the compound.
If
the formula carries a CHARGE,
add or subtract electrons
accordingly:
Negative
charge: ADD electrons
Positive charge: SUBTRACT
electrons
I. Lewis Dot Structures
Step
1: Add up all valence
electrons in the compound.
(This
1 2
is the “electron tally.”)
1 C × 4e- = 4eCF4 4 F × 7e- = +28e32e
3 4 5 6 7 8
I. Lewis Dot Structures
Step
1: Add up all valence
electrons in the compound.
NH3
1 2
3 4 5 6 7
1 N × 5e- = 5e3 H × 1e- = + 3e8e
8
I. Lewis Dot Structures
Step
1 2
1: Add up all valence
electrons in the compound.
- = 6e21
S
×
6e
SO2 2 O × 6e- = 12eCharge = +2e3 4 5 6 7 8
20e-
I. Lewis Dot Structures
Step
1: Add up all valence
electrons in the compound.
NOTE: Charged dot structures
2SO2
(like SO22- or PO43-) will
always be drawn inside
square brackets!
1 2
3 4 5 6 7 8
I. Lewis Dot Structures
Step
2: Draw a skeleton
structure.
 A)The
first element in the
formula goes in the center.*
 B)The
second element goes
around the first element.
Left,
right, top, bottom
 (Don’t
choose weird angles.)
F
F C F
F
I. Lewis Dot Structures
A)
The first element in the
formula goes in the center.*
*BUT NOT HYDROGEN
H
cannot be the center atom.
 If H is first in the formula, skip it.
the second element in the
center.
 Treat H as the second element.
H2S
 Put
H S H
I. Lewis Dot Structures
Step
2: Draw a skeleton
structure.
C)
Draw a bond (line)
connecting each
secondary atom to the
center.
 (Do
not connect secondary atoms to
each other.)
F
F C F
F
I. Lewis Dot Structures
Step
3: Calculate
remaining electrons.
A)
Each bond (line) represents
two electrons that are shared
between two atoms.
 Number
B)
of bonds x 2e- = # e- in bonds
Subtract the bonded e- from
the total.
I. Lewis Dot Structures
Step
3: Calculate
remaining electrons.
CF4
= 32e4 bonds × 2e- = - 8e24eThese electrons
will appear as dots.
F
F C F
F
I. Lewis Dot Structures
Step
4: Distribute remaining
electrons.
A)
Octet rule: every atom needs
8 electrons in its valence.*
A
bond is two valence electrons
that count as valence for both
elements involved (at the same
time).
CO-valence!
I. Lewis Dot Structures
A)
Octet rule: every
atom needs 8 electrons
in its valence.*
*BUT NOT HYDROGEN.
H2S
can only have two
electrons (because 1s is its
only orbital)
H S H
When it has a bond, H is
“full.”
Both H’s are full.
H
No dots for H!
I. Lewis Dot Structures
A)
Octet rule: every atom needs
8 electrons in its valence.*
*But not hydrogen.
B)
Add electron dots to atoms
as needed:
You
must use up all the e-’s
available.
You may NOT use more e-’s than
that!
I. Lewis Dot Structures
Step
4: Distribute remaining
electrons.
CH4
= 32e4 bonds × 2e- = 8eF
24eF C F
These electrons
will appear as dots.
F
I. Lewis Dot Structures
Step
4: Distribute remaining
electrons.
24e- dots:
C: 4 bonds x 2e- = 8 v.e-.
No dots needed on C
F: 1 bond x 2e- = 2 v.e-.
6 dots needed on each F
F
F C F
F
I. Lewis Dot Structures
Step
4: Distribute remaining
electrons.
24e- dots:
C: 4 bonds x 2e- = 8 v.e-.
No dots needed on C
F: 1 bond x 2e- = 2 v.e-.
6 dots needed on each F
F
F C F
F
I. Lewis Dot Structures
Step
4: Distribute remaining
electrons.
24e- dots:
We used exactly 24e-!
 Octet
check:
Do all atoms* have 8 v.e-.?
YES! We win. 
F
F C F
F
I. Lewis Dot Structures
Step
4: Distribute remaining
electrons.
NH3
= 8e3 bonds × 2e- = -6eH
2eN
H
H
These electrons
will appear as dots.
I. Lewis Dot Structures
Step
4: Distribute remaining
electrons.
We used exactly 2e-!
 Octet
check:
H
Do all atoms* have 8 v.e-.?
H N H
 N has 8
H
has 2
 We
win. 
I. Lewis Dot Structures
Step
4: Distribute remaining
electrons.
C)
If you run out of electrons:
Share
more!
2
e- short = 1 double bond
4
e- short = 2 double bonds
or 1 triple bond
I. Lewis Dot Structures
 Let’s
try CO2
(e- tally)
1 C × 4e- =
4e2 O × 6e- = + 12e16e-
16e- (e- tally)
- 4e- (in 2 bonds)
12e- (as dots)
But 12 dots won’t be enough.
We’re 2 epairs short.
That means
we need 2
more bonds.
(4 bonds total)
O C O

I. Lewis Dot Structures
 Let’s
try CO2
(e- tally)
1 C × 4e- =
4e2 O × 6e- = + 12e16eHere we go again:
16e- (e- tally)
- 8e- (in 4 bonds)
8e- (as dots)
We’re 2 epairs short.
That means
we need 2
more bonds.
(4 bonds total)
O C O
I. Lewis Dot Structures
 Let’s
try CO2
(e- tally)
1 C × 4e- =
4e2 O × 6e- = + 12e16eHere we go again:
16e- (e- tally)
- 8e- (in 4 bonds)
8e- (as dots)
Octet check!
 Remember,
each bond
counts as 2e-
O C O
 4 dots, 2 bonds
 4 bonds
 4 dots, 2 bonds
WIN 
I. Lewis Dot Structures
Step
4: Distribute remaining
electrons.
 D)
If you have left-over electrons:
Make an Expanded Octet
 This
is a fancy name for “more than 8
electrons on the central atom.”
Central
atom must be at an energy
level ≥ 3. Expanded octets cannot
exist in the 1st or 2nd E.L.
4. Distributing Electrons
Make
an Expanded Octet
1)
If there are simply left-over
electrons, put them on the central
atom.
 No
multiple bonds allowed!
 The
limit for expanded octets is 12
electrons total.
 ONLY
the central atom gets more than
8 electrons.
I. Lewis Dot Structures
 Let’s
try SeCl4
(e- tally)
1 Se × 6e- =
6e4 Cl × 7e- = + 28e34e-
34e- (e- tally)
- 8e- (in 4 bonds)
26e- (as dots)
But 26 dots won’t fit!
We’ve got 1 extra
e- pair.
That pair goes on Se.
(Se gets 10 e- total)
Each Cl gets an octet.
Cl
Cl Se Cl
Cl
4. Distributing Electrons
Make
an Expanded Octet
If
there are more than 4 atoms of
the secondary element, they still
bond to the central atom.
 Still
no multiple bonds allowed!
 The
limit is 6 secondary atoms bonded
to the central atom (Still 12 e- total)
 Still
ONLY the central atom gets more
than 8 electrons.
I. Lewis Dot Structures
Let’s
try PF5
What’s your e- tally?
1 P × 5e- =
5e5 F × 7e- = + 35e40eHow many e- in bonds and
how many in dots?
40e- (e- tally)
- 10e- (in 5 bonds)
30e- (as dots)
Even before doing a tally
we can tell that P must
expand its octet in order
to bond 5 F atoms.
The 5 Fs are
distributed in a
pentagon.
(If there were six, it
would be a hexagon.)
Octet check:
 P gets 10 e- total
 Each F gets an octet.
F
F P F
F F
5. Finishing:
A)
If your dot structure was
for an ion:
Place
square brackets around
the ion and write the charge
outside the brackets:
ClO4
1 Cl × 7e- = 7e4 O × 6e- = 24eCharge
+ 1e32e32e- 8e24e-
O
O Cl O
O
5. Finishing
B)
Resonance Structures
 Some
molecules with double bonds
can’t be correctly represented by a
single Lewis diagram.
 The actual structure is an average of
all the possibilities.
B) Resonance Structures
Truth:
the electrons are evenly
distributed, but there isn’t a way to
draw that in a Lewis dot structure.
(Half a bond? Half a dot?)
Show
all possible structures
separated by a double-headed
arrow.
D. Resonance Structures
 SO3 has
1 double bond which could be
in any of these three places:
O
O S O
O
O S O
O
O S O
HOMEWORK:
Molecular
1st
Geometry WS
two columns ONLY!
(We’ll
learn the others next time.)
Skyward!
Do
the worksheet first– it will give
you the answers to the Skyward
problems.