Chapter 10: The Shapes of
Molecules
• Lewis Structures for
Molecules and Ions
• Shapes of Molecules: a Certain Theory
• Polarity of Molecules
Using Octet Rule
molecular
formula
sum of
valence e-
place atom with
lowest EN in
center
draw single bonds.
subtract 2e- for
each bond
Lewis
Structure
atom
placement
remaining
valence e-
add Agroup
numbers
give each
atom 8e(2e- for H)
Using Octet Rule
Step 1. Place the atoms relative to each other. NF3; SOCl2
F
Cl
S
N
F
Cl
F
O
for compounds such as ABn, place the atom with lower group
number in the center.
Using Octet Rule
Step 1. Place the atoms relative to each other. NF3; SOCl2
F
Cl
S
N
F
Cl
F
O
lowest EN
for compounds such as ABn, place the atom with lower group
number in the center.
Using Octet Rule
Step 2. Determine the total number of valence electrons
available.
F
Cl
S
N
F
Cl
F
O
(for ions, add one e- for each negative charge and subtract one
each positive charge).
Using Octet Rule
Step 2. Determine the total number of valence electrons
available.
F
Cl
S
N
F
3x7 + 5 = 26
Cl
F
O
2x7 + 2x6 = 26
(for ions, add one e- for each negative charge and subtract one
each positive charge).
Using Octet Rule
Step 3. Draw a single bond from each surrounding atom to
the central atom.
F
F
N
Cl
F
3x7 + 5 = 26
S
Cl
O
2x7 + 2x6 = 26
Using Octet Rule
Step 3. Draw a single bond from each surrounding atom to
the central atom.
F
F
N
Cl
F
3x7 + 5 = 26 - 6 = 20
S
Cl
O
2x7 + 2x6 = 26- 6 = 20
Step 3a. Subtract 2e- for each single bond from the total number
of valence electrons.
Using Octet Rule
Step 4. Distribute the remaining electrons in pairs so that
each atom obtains eight electrons (or two for H).
F
F
N
Cl
F
3x7 + 5 = 26 - 6 = 20
S
Cl
O
2x7 + 2x6 = 26- 6 = 20
Start with more electronegative atoms.
Using Octet Rule
Step 4. Distribute the remaining electrons in pairs so that
each atom obtains eight electrons (or two for H).
F
F
N
Cl
F
3x7 + 5 = 26 - 6 = 20
- 18 = 2
S
Cl
O
2x7 + 2x6 = 26- 6 = 20
- 18 = 2
Start with more electronegative atoms.
Using Octet Rule
Step 4. Distribute the remaining electrons in pairs so that
each atom obtains eight electrons (or two for H).
F
F
N
Cl
F
3x7 + 5 = 26 - 6 = 20
- 18 = 2
S
Cl
O
2x7 + 2x6 = 26- 6 = 20
- 18 = 2
If any electrons remain, place them around the central atom
Using Octet Rule
Step 4. Distribute the remaining electrons in pairs so that
each atom obtains eight electrons (or two for H).
F
F
N
Cl
F
3x7 + 5 = 26 - 6 = 20
- 18 = 2
S
Cl
O
2x7 + 2x6 = 26- 6 = 20
- 18 = 2
If any electrons remain, place them around the central atom
Using Octet Rule
Step 1. Place the atoms relative to each other.
Step 2. Determine the total number of valence electrons
available.
Step 3. Draw a single bond from each surrounding atom to
the central atom
Step 4. Distribute the remaining electrons in pairs so that
each atom obtains eight electrons (or two for H).
Works for C, N, O as the central atom, single bonds only!
Using Octet Rule
BONUS!
Step 5. If, after step 4. The central atom still does not have
an octet, make a multiple bond by changing a lone pair
from one of the surrounding atoms into a bonding pair to
the central atom.
Let’s analyze CO2
Using Octet Rule
BONUS!
Step 5. If, after step 4. The central atom still does not have
an octet, make a multiple bond by changing a lone pair
from one of the surrounding atoms into a bonding pair to
the central atom.
Let’s analyze CO2
HCN, CO, N2
Resonance
Cl
Cl
Cl
Cl
S
S
O
O
O
C
O
O
C
O
thionyl chloride
O
O
O
O
O
ozone
O
Resonance
Cl
Cl
Cl
Cl
Cl
Cl
S
S
S
O
O
O
thionyl chloride
O
O
O
O
O
O
O
O
O
ozone
Resonance
general situation:
B
A
B
C
A
B
C
A
C
Resonance
general situation:
B
B
C
A
B
A
C
A
C
when symmetrical:
B
B
A
A
B
A
A
A
A
bond order is 1.5
Resonance
BOND ORDER =
O
O
C
O
carbonate
number of electron pairs
number of atom-to-atom linkages
2-
O
O
-
O
N
O
nitrate
3-
P
O
O
O
phosphate
Formal Charge
the charge an atom would have if the electrons
were shared equally
FC of atom = no. of v. e– – no. of unshared v. e– – 1/2 of shared v. e–
Formal Charge
the charge an atom would have if the electrons
were shared equally
FC of atom = no. of v. e– – no. of unshared v. e– – 1/2 of shared v. e–
Cl
Cl
Cl
Cl
S
S
O
O
thionyl chloride
Formal Charge
the charge an atom would have if the electrons
were shared equally
FC of atom = no. of v. e– – no. of unshared v. e– – 1/2 of shared v. e–
Cl
0
Cl
0
S
0
Cl
S
O
O
0
Cl
thionyl chloride
Formal Charge
the charge an atom would have if the electrons
were shared equally
FC of atom = no. of v. e– – no. of unshared v. e– – 1/2 of shared v. e–
Cl
0
Cl
0
S
0
Cl +1
S
0
0
O
O
0
Cl
thionyl chloride
-1
Formal Charge
the charge an atom would have if the electrons
were shared equally
FC of atom = no. of v. e– – no. of unshared v. e– – 1/2 of shared v. e–
Cl
0
Cl
0
S
0
Cl +1
S
0
0
O
O
0
Cl
thionyl chloride
-1
the structure with the lowest FC has a higher contribution
Formal Charge
the charge an atom would have if the electrons
were shared equally
Most important resonance structures will have
• smaller FC
• no like charges on adjacent atoms
• more negative FC on a more electronegative atom
Formal Charge
the charge an atom would have if the electrons
were shared equally
Most important resonance structures will have
• smaller FC
• no like charges on adjacent atoms
• more negative FC on a more electronegative atom
EXAMPLES: H2CN2 (diazomethane)
N3– (azide anion)
NCO– (cyanate anion) in the textbook
SO32– (sulfite anion)
Exceptions to the Octet Rule
we’ve seen them before…
• Electron-Deficient Molecules
• Odd-Electron Molecules
• Expanded Valence Shell
Exceptions to the Octet Rule
• Electron-Deficient Molecules
Typically, groups 2 and 3
AlCl3, BF3, BeCl2
Cl
Cl
Al
Cl
Exceptions to the Octet Rule
• Electron-Deficient Molecules
Typically, groups 2 and 3
AlCl3, BF3, BeCl2
Cl
Cl
Al
Cl
B
B
Cl
Al
Cl
Cl
Exceptions to the Octet Rule
• Odd-Electron Molecules, aka free radicals
Typically, group 5 elements (NITROGEN)
Example: NO (nitrogen monoxide)
O
N
Exceptions to the Octet Rule
• Odd-Electron Molecules, aka free radicals
Typically, group 5 elements (NITROGEN)
Example: NO (nitrogen monoxide)
O
N
N
O
O
N
N
O
Exceptions to the Octet Rule
• Expanded Valence Shell
ONLY period 3 or higher
Examples: SF6, PCl5
Cl
F
F
F
F
P
Cl
S
F
Cl
Cl
F
Cl
Using Bond Energies
To calculate heats of reactions:
!Horxn = !Horeactant bonds broken + !Hoproduct bonds formed
e
n
t
h
a
l
p
y
H
O
!Horxn
O
H
H
H
H
H
O
H
H
O
H
Molecular Shape
Valence-Shell Electron-Pair Repulsion Theory (VSEPR)
groups or unshared electron pairs around a central atom
are located as far away as possible from each others
A is the central atom:
A
X
(E)
A
A
AXnEm
X is a group; E is an unshared pair of electrons
Molecular Shape
Valence-Shell Electron-Pair Repulsion Theory (VSEPR)
non-bonding pair > double bond > single bond
repulsion effect decreases
A
X
(E)
A
A
AXnEm
X is a group; E is an unshared pair of electrons
Molecular Shape
Molecular Shape
Molecular Polarity
molecular polarity is a vector sum of bond polarity
dipole momentum (µ) = charge " distance
1 D (debye) = 3.34 " 10-30 C•m
H
Cl
!+
!"
Molecular Polarity
molecular polarity is a vector sum of bond polarity
dipole momentum (µ) = charge " distance
S
O
O
C
O
O
non-polar
O
H
H
Molecular Shape
Practice Problems
problem 10.4
What is required for an atom to expand its valence shell?
Which of the following atoms can expand its valence shell:
F, S, H, Cl, Al, Se, Ca
Molecular Shape
Practice Problems
problem 10.18 +
Draw a Lewis structure and calculate the formal charge of
each atom in
IF5
AlH4–
ClO–
CH3–
Molecular Shape
Practice Problems
problem 10.14 +
Draw a Lewis structure and the most important resonance
form for each ion, showing formal charges and oxidation
numbers of the atoms:
AsO4–
ClO2–
BrO3–
Molecular Shape
Practice Problems
follow-up problem 10.8
Draw molecular shapes and predict relative bond angles
of
(a) ICl2–
(b) ClF3
(c) SOF4
(d) IF4–
(e) ClO3–
Molecular Shape: Polyatomic
Practice Problems
problem 10.57 and 10.58
State ideal values for each of the bond angles in each
molecule, and note where you expect deviations:
H
H C N O H
H
H
H C O C H
H
H
H
H C C O
H H
O N O H
O
H
O
H O B O H
O
H C O H
Molecular Shape: Polyatomic
Practice Problems
problem 10.67 and 10.58
Which molecule in each pair has the greater dipole moment:
(a) ClO2 or SO2
(b) HBr or HCl
(c) BeCl2 or SCl2
(d) AsF3 or AsF5
Molecular Shape
Practice Problems
problem 10.67 and 10.58
Which molecule in each pair has the greater dipole moment:
(a) ClO2 or SO2
(b) HBr or HCl
(c) BeCl2 or SCl2
(d) AsF3 or AsF5
Molecular Shape
Practice Problems
problem 10.82
Like several other bonds, carbon-oxygen bonds have lengths
and strengths that depend on the bond order. Draw Lewis
structures for the following species, and arrange them in
order of increasing carbon-oxygen bond length.
(a) CO
(b) CO32–
(c) H2CO
(d) CH4O
(e) HCO3– (H is attached to O)
Molecular Shape
Practice Problems
problem 10.73
Both aluminum and iodine form chlorides with bridging Cl
atoms. Lewis structures not showing lone pairs are:
Cl
Cl
Al
Cl
Cl
Cl
Al
Cl
Cl
I
Cl
Cl
Cl
I
Cl
Cl
(a) What is the formal charge on each atom?
(b) Which of these molecules has a planar shape?
Molecular Shape
Practice Problems
problem 10.97
Some scientists speculate that many organic molecules
required for life on Earth arrived on meteorites. The
Murchison meteorite that landed in Australia in 1969
contained 92 different amino acids, including 21 found in
Earth organisms. A skeleton of one of these extraterrestrial
H
amino acids is
H3N
C
C
O
CH2 O
CH3
Draw a Lewis structure, and identify atoms with a nonzero
formal charge.