Chemistry 1A
Chapters 8 and 9
Molecular Orbital
Theory
ƒ Goal: to generate molecular orbital
diagrams that describe molecules,
like the atomic orbital diagrams that
describe atoms.
ƒ Molecular Orbital: a volume within
which a high percentage of the
negative charge for an electron in a
molecule is found.
Atomic Orbitals and
Molecular Orbitals
Molecular Orbital Diagram for O2
Like Atomic Orbital Diagram for O
Goal: To determine the MO
Diagram for Diatomic
Molecules
• We assume that the electrons would fill
the molecular orbitals of molecules like
electrons fill atomic orbitals in atoms.
– The molecular orbitals are filled in a way that
yields the lowest potential energy for the
molecule.
– The maximum number of electrons in each
molecular orbital is two. (We follow the Pauli
exclusion principle.)
– Orbitals of equal energy are half filled with
parallel spin before they begin to pair up. (We
follow Hund's Rule.)
Linear Combination of
Atomic Orbitals - LCAO
• Atomic orbitals (AOs) of about the
same energy overlap to form
molecular orbitals (MOs).
• Pairs of AOs overlap in two
extreme ways.
– In-phase to bonding MO
– out-of-phase to antibonding MO
Electron and Light Waves
is like
Describes the
variation in the
intensity of negative
charge created by an
electron.
Describes the variation
in the electric and
magnetic fields created
by light.
Wave Interference
In-Phase and Out-ofPhase Interactions
• Light waves can interact in-phase and
out-of-phase.
– In-phase leads to an increase in the intensity
of the light (brighter).
– Out-of-phase leads to a decrease in the
intensity of the light (less bright).
• Electron waves can interact in-phase and
out-of-phase.
– In-phase leads to an increase in the intensity
of the negative charge.
– Out-of-phase leads to a decrease in the
intensity of the negative charge.
MOs from
1s Orbitals
Bonding Molecular
Orbital
• From in-phase interaction of two atomic
orbitals.
• Leads to an increase in negative charge
between two nuclei where the atomic
orbitals overlap.
• Leads to more +/- attraction between the
negative charge generated by the
electrons and the nuclei.
• Thus there is a decrease in PE.
• Energy would be required to separate the
atoms, so electrons in this type of
molecular orbital tend to keep atoms
together.
Antibonding Molecular
Orbital
• From out-of-phase interaction of two
atomic orbitals.
• Leads to a decrease in negative charge
between two nuclei where the atomic
orbitals overlap.
• Leads to less +/− attraction between the
negative charge generated by the
electrons and the nuclei.
• Thus there is an increase in PE.
• Electrons are more stable in separate
atomic orbitals of separate atoms, so
electrons in this type of molecular orbital
tend to keep atoms separate.
MOs from 1s Orbitals (cont.)
Molecular Orbitals from
2px Atomic Orbitals
Molecular Orbitals
from 2py Atomic
Orbitals
Why σ2p lower PE than
π2p?
• The π2p molecular orbital comes from
parallel overlap of p orbitals, and σ2p
comes from end-on overlap.
• Greater overlap between end-on p
orbitals leads to a greater increase
in negative charge between the
nuclei, a greater increase in +/attraction, a greater stabilization of
the electrons, and greater decrease
in PE of these electrons.
Why π2p* lower PE than
σ2p*?
• The π2p molecular orbital comes from
parallel overlap of p orbitals, and σ2p
comes from end-on overlap.
• Less overlap between parallel p
orbitals leads to a lesser decrease in
negative charge between the nuclei,
a lesser decrease in +/- attraction, a
lesser destabilization of the
electrons, and lesser increase in PE
of these electrons.
MO Diagram
for O2, F2,
Ne2, CO,
and NO
Procedure for Drawing
MO Diagrams
• Determine the number of electrons in
the molecule. We get the number of
electrons per atom from their atomic
number on the periodic table.
• Fill the molecular orbitals from bottom to
top until all the electrons are added.
Describe the electrons with arrows. Put
two arrows in each molecular orbital,
with the first arrow pointing up and the
second pointing down.
• Orbitals of equal energy are half filled
with parallel spin before they begin to
pair up.
Bond Order
1/2(e− bonding MO's − e− antibonding MO's)
• We use bond orders to predict the
stability of molecules.
– If the bond order for a molecule is
equal to zero, the molecule is
unstable.
– A bond order of greater than zero
suggests a stable molecule.
– The higher the bond order is, the more
stable the bond.
Paramagnetic or
Diamagnetic
• If all the electrons are paired, the
molecule is diamagnetic.
• If one or more electrons are
unpaired, the molecule is
paramagnetic.
Hydrogen, H2
• The bond order is 1.
Bond Order = 1/2(2 − 0) = 1
• The bond order above zero
suggests that H2 is stable.
• Because there are no unpaired
electrons, H2 is diamagnetic.
Helium, He2
• The bond order is 0 for He2.
Bond Order = 1/2(2 − 2) = 0
• The zero bond order for He2
suggests that He2 is unstable.
• If He2 did form, it would be
diamagnetic.
Oxygen, O2
• O2 has a bond order
of 2.
Bond Order
= 1/2(10 − 6) = 2
• The bond order of
two suggests that the
oxygen molecule is
stable.
• The two unpaired
electrons show that
O2 is paramagnetic.
Fluorine, F2
• F2 has a bond order
of 1.
Bond Order
= 1/2(10 − 8) = 1
• The bond order of
one suggests that
the fluorine
molecule is stable.
• Because all of the
electrons are paired
F2 is diamagnetic.
Neon, Ne2
• Ne2 has a bond
order of 0.
Bond Order
= 1/2(10 − 10) = 0
• The bond order zero
for Ne2 suggests that
Ne2 is unstable.
• If Ne2 did form, it
would be
diamagnetic.
Carbon
Monoxide, CO
Nitrogen
Monoxide, NO
Molecular Orbital
Theory Issues
• Problem: At the level described in
general chemistry, the molecular
orbital theory is too limited in its
application.
• Solution: Switch to a simpler
model that incorporates as much of
the molecular orbital theory as
possible but that has broader
application. The new model is
called the valence bond model.
Models – Advantages
and Disadvantages (1)
• They help us to visualize, explain, and
predict chemical changes.
• Because a model is a simplified version of
what we think is true, the processes it
depicts are sometimes described using
the phrase as if. When you read, “It is as
if an electron were promoted from one
orbital to another,” the phrase is a
reminder that we do not necessarily think
this is what really happens. We merely
find it useful to talk about the process as if
this is the way it happens.
Models – Advantages
and Disadvantages (2)
• One characteristic of models is that
they change with time. Because our
models are simplifications of what
we think is real, we are not surprised
when they sometimes fail to explain
experimental observations. When
this happens, the model is altered to
fit the new observations.
Assumptions of the
Valence-Bond Model
• Only the highest energy
electrons participate in bonding.
• Covalent bonds usually form to
pair unpaired electrons.
• Covalent bonds arise when
atomic orbitals on adjacent
atoms overlap to form
molecular orbitals.
Valence Electrons
• Valence electrons are the highestenergy s and p electrons in an
atom.
Fluorine
F
1s2 2s2 2p5
Valence electrons
Atomic Orbital
Overlap for F2 (1)
• One way to describe how the covalent
bond between fluorine atoms forms is to
say that a 2p orbital from each atom
overlaps to form a σ2p molecular orbital.
• The two unpaired electrons from the 2p
orbitals of the two fluorine atoms are
then in the more stable σ2p molecular
orbital, making the F2 molecule more
stable than the separate F atoms.
Atomic Orbital Overlap for F2 (2)
• Because the bond arises due to the
formation of a sigma molecular orbital, it is
called a sigma bond.
H2 Formation (1)
• 1s atomic orbitals, one from each
hydrogen atom, overlap to form a σ1s
molecular orbital. The two electrons in H2
are found in the σ1s, leading to a more
stable H2 molecule.
H2 Formation (2)
• Again, because the bond arises due to the
formation of a sigma molecular orbital, it is
called a sigma bond.
Things to explain for the bonding
in a methane molecule, CH4
• Why are there four C-H covalent bonds?
• Why do all of the bonds have the same
length and strength?
• Why is the molecule tetrahedral with
bond angles of 109.5°?
Carbon – 4 bonds
Formation of
sp3 hybrid
orbitals
Orbital Overlap in Methane, CH4
Bonding in CH4
• Four equivalent sigma C-H covalent
bonds arise due to the overlap of
four sp3 hybrid orbitals on the
carbon atom with 1s atomic orbitals
on the hydrogen atoms. Hybrid
orbitals always form sigma bonds.
• Because the sp3 hybrid orbitals are
arranged in a tetrahedral geometry,
the molecule has tetrahedral
geometry.
Methane, CH4
Explanation of Bonding
Patterns - Valence Bond
Model – Part 1
• Step 1: Write, “Only the highest energy
electrons participate in bonding.”
• Step 2: Draw the orbital diagram for the
valence electrons. Include the empty “d”
orbitals for the third period nonmetals
and below.
• Step 3: If there is a formal charge, add
or subtract electrons.
– Add one electron for -1.
– Subtract one electron for +1.
Explanation of Bonding
Patterns - Valence Bond
Model – Part 2
• Step 4: If necessary, promote one or more
electrons from a pair to an empty orbital to get
the number of unpaired electrons equal to the
number of bonds to be explained. Rewrite the
orbital diagram if electrons have been promoted.
• Step 5: Predict the hybridization, and rewrite
the orbital diagram, showing the predicted hybrid
orbitals.
• Step 6: Write, “Covalent bonds form in order to
pair unpaired electrons.”
• Step 7: Indicate that the unpaired electrons
form the bonds.
• Step 8: Indicate that the paired electrons are
lone pairs.
Descriptions of Molecules
• You will be given a Lewis structure, and
you will be asked the following.
– What is the hybridization for each atom in
the structure?
– Write a description of the formation of each
bond in terms of the overlap of atomic
orbitals.
– Describe the electron group geometry
around each atom that has two or more
atoms attached to it.
– Draw the geometric sketch of the molecule,
including bond angles.
– Describe the molecular geometry around
each atom that has two or more atoms
attached to it.
Steps for Hybridization
– Part 1
• Step 1: Count the number of
electron groups around each atom.
– An electron group is any one of the
following.
• single bond
• double or triple bond (each counts as
one group)
• lone pair
Steps for Hybridization
– Part 1
• Step 2 Apply the following
guidelines.
– 1 group - No hybridization
– 2 groups - sp hybridization
– 3 groups - sp2 hybridization
– 4 groups - sp3 hybridization
– 5 groups - sp3d hybridization
– 6 groups - sp3d2 hybridization
Describing the bonds in terms of
the overlap of atomic orbitals
• Step 1: Describe single bonds that do not
involve hydrogen as due to the overlapping of
two hybrid orbitals, one from each atom.
• Step 2: Describe single bonds that involve
hydrogen as due to the overlapping a hybrid
orbital with a 1s orbital for the hydrogen atom.
• Step 3: Describe double bonds as one bond
forming from the overlap of two hybrid orbitals,
one from each atom. Describe the second bond
as due to the overlap of two p orbitals, one from
each atom.
• Step 4: Describe triple bonds as one bond
forming from the overlap of two hybrid orbitals,
one from each atom. Describe each of the other
two bonds as due to the overlap of two p orbitals,
one from each atom.
Steps for Molecular
Geometry
• Step 1: To determine the name of the
electron group geometry around each
atom that is attached to two or more
atoms, count the number of electron
groups around each atom and apply the
guidelines found on the following slides.
• Step 2: Use one or more of the
geometric sketches shown on the
following slides for the geometric sketch
of your molecule.
Steps for Molecular
Geometry (cont.)
• Step 3: To determine the name of
the molecular geometry around
each atom that has two or more
atoms attached to it, count the
number of bond groups and lone
pairs, and then apply the guidelines
found on the following slides.
Electron Group and
Molecular Geometry –
Part 1
• 2 e- groups – linear electron
group geometry - 180° angles.
– 2 bond groups and 0 lone pairs –
linear molecular geometry
Electron Group and
Molecular Geometry –
Part 2
• 3 e- groups – trigonal
planar electron group
geometry - 120° angles.
– 3 bond groups and 0 lone
pairs – trigonal planar
molecular geometry
– 2 bond groups and 1 lone
pair – bent molecular
geometry
Electron Group and
Molecular Geometry –
Part 3
• 4 e- groups – tetrahedral
electron group
geometry – 109.5° angles.
– 4 bond groups and 0 lone
pairs – tetrahedral molecular
geometry
– 3 bond groups and 1 lone
pair – trigonal pyramid
molecular geometry
– 2 bond groups and 2 lone
pairs – bent molecular
geometry
Things to explain for the
bonding in a ethene (ethylene)
molecule, C2H4
• Why are there four equivalent C-H covalent bonds?
• Why are all the bond angles about 120°?
• According to our model, how does the double bond
form?
• Why is one bond in the double bond weaker than the
other?
Carbon – 2 single bonds and
one double bond
Formation
of sp2
hybrid
orbitals
Orbital Overlap for Ethene
(Ethylene) Molecule, C2H4
Bonding in C2H4
• Four equivalent sigma C-H covalent bonds
arise due to the overlap of sp2 hybrid orbitals
on the carbon atoms with 1s atomic orbitals
on the hydrogen atoms.
• The double bond between the carbon atoms
consists of
– one sigma bond formed due to the overlap of two
sp2 hybrid orbitals, one from each carbon.
– One pi bond due to the overlap of unhybridized 2p
orbitals, one from each carbon.
• Because the sp2 hybrid orbitals are arranged
in a trigonal planar geometry, the geometry
about each carbon atom is trigonal planar with
angles of 120°.
Ethene (ethylene)
Pi bonds are generally
weaker than sigma bonds.
• When p orbitals overlap to form pi bonds
there is less overlap than for the overlap
between hybrid orbitals to form sigma
bonds.
• Greater overlap between hybrid orbitals
leads to a greater increase in the
negative charge between the nuclei for
the sigma MO compared to the pi MO.
This leads to a greater +/- attraction,
greater stabilization, and lower PE.
Therefore, sigma bonds are generally
stronger than pi bonds.
Things to explain for the
bonding in an ethyne
(acetylene) molecule, C2H2
•
•
•
•
Why are there two equivalent C-H covalent bonds?
Why are all the bond angles 180°?
How does the triple bond between carbons atoms form?
Why are two bonds in the triple bond weaker than the
other one?
Carbon – 1 single bond and 1
triple bond
Formation
of sp
hybrid
orbitals
Orbital Overlap
for Ethyne
(Acetylene)
Molecule, C2H2
Bonding in C2H2
• Two equivalent sigma C-H covalent bonds arise
due to the overlap of sp hybrid orbitals on the
carbon atoms with 1s atomic orbitals on the
hydrogen atoms.
• The triple bond between the carbon atoms
consists of
– one sigma bond formed due to the overlap of two sp
hybrid orbitals, one from each carbon.
– Two pi bonds, each of which arises due to the overlap
of unhybridized 2p orbitals, one from each carbon.
• Because the sp hybrid orbitals are arranged in
a linear geometry, the geometry about each
carbon atom is linear with angles of 180°.
Ethyne (acetylene) , C2H2
Things to explain for the
bonding in an ammonia
molecule, NH3
• Why are there three equivalent N-H
covalent bonds?
• Why are the bond angles about 107°?
Nitrogen – 3 bonds & 1 lone pair
Orbital Overlap for Ammonia, NH3
Bonding in NH3
• Three equivalent covalent bonds arise due to
the overlap of three sp3 hybrid orbitals on the
nitrogen atom with 1s atomic orbitals on the
hydrogen atoms. The lone pair can be viewed
as in an sp3 hybrid orbital.
• Because the sp3 hybrid orbitals are arranged in
a tetrahedral geometry, the molecule has
tetrahedral electron group geometry with
bond angles of about 109.5° (actually about
107°).
• A molecule that has three bonds and one lone
pair around its central atom is said to have
trigonal pyramid molecular geometry.
Electron Group
Geometry and
Molecular Geometry
• Electron group geometry
describes all of the electron
groups around the central atom,
including lone pairs.
• Molecular geometry just
describes the bond groups.
Ammonia, NH3, Geometry
Ammonia molecules have
tetrahedral electron group geometry
and trigonal pyramid molecular
geometry.
Ammonia, NH3
Things to explain for the
bonding in an ammonium
ion, NH4+
• Why are there four equivalent N-H
covalent bonds?
• Why does the polyatomic ion have
tetrahedral geometry with bond angles of
109.5°?
Formal Charge
• A measure of the stability of an atom in a
molecule
• Formal charge = number of valence electrons
needed to be uncharged − approximate number
of valence electrons atom has
– The A-group number tells us the number of valence
electrons needed to be uncharged.
– We assume that electrons in covalent bonds are
shared equally, so each bond contributes one electron
to each atom.
– Lone pairs contribute two electrons to the atom on
which they are found.
• Formal charge = A-group number − number of
lines − number of dots
Nitrogen – 4 bonds
Orbital Overlap for ammonium, NH4+
Bonding in NH4+
• Four equivalent sigma covalent
bonds arise due to the overlap of
four sp3 hybrid orbitals on the
nitrogen atom with 1s atomic
orbitals on the hydrogen atoms.
• Because the sp3 hybrid orbitals are
arranged in a tetrahedral geometry,
the molecule has tetrahedral
electron group geometry with bond
angles of about 109.5°.
Things to explain for the
bonding in a water
molecule, H2O
• Why are there two equivalent O-H
covalent bonds?
• Why are the bond angles about 105°?
Oxygen – 2 bonds & 2 lone
pairs
Orbital Overlap for Water, H2O
Bonding in H2O
• Two equivalent sigma covalent bonds arise
due to the overlap of two sp3 hybrid orbitals
on the oxygen atom with 1s atomic orbitals on
the hydrogen atoms. The lone pairs can be
viewed as in sp3 hybrid orbitals.
• Because the sp3 hybrid orbitals are arranged
in a tetrahedral geometry, the molecule has
tetrahedral electron group geometry with bond
angles of about 109.5° (actually about 105°).
• A molecule that has two bonds and two lone
pairs around its central atom is said to have
bent molecular geometry.
Water, H2O, Geometry
Water molecules have tetrahedral
electron group geometry and bent
molecular geometry.
Water, H2O
Oxygen – 1 bond & 3 lone pairs
Carbon – 3 bonds & 1 lone pair
Oxygen – 3 bonds & 1 lone pair
Bonding in CO
• The triple bond between the carbon
and oxygen atoms consists of
– one sigma bond formed due to the
overlap of two sp hybrid orbitals, one
from the carbon atom and one from
the oxygen atom.
– Two pi bonds, each of which arises
due to the overlap of unhybridized 2p
orbitals, one from the carbon atom
and one from the oxygen atom.
Things to explain for the
bonding in a boron
trifluoride molecule, BF3
• Why are there three equivalent B-F
covalent bonds?
• Why is the molecule trigonal planar with
bond angles of 120°?
Boron – 3 bonds
Orbital Overlap in BF3
Trigonal Planar Geometry – BF3
Bonding in BF3
• Three equivalent sigma covalent
bonds arise due to the overlap of
three sp2 hybrid orbitals on the
boron atom with sp3 atomic
orbitals on the fluorine atoms.
• Because the sp2 hybrid orbitals
are arranged in a trigonal planar
geometry, the molecule has
trigonal planar geometry.
Halogens – 1 bond & 3 lone
pairs
Things to explain for the
bonding in a phosphorus
pentafluoride molecule, PF5
• Why are there five equivalent P-F covalent
bonds?
• Why is the molecule trigonal planar with
bond angles of 90 °, 120 °, and 180°?
Phosphorus – 5 bonds
Formation
of sp3d
Hybrid
Orbitals
Orbital Overlap in PF5
Bonding in PF5
• Five equivalent sigma covalent
bonds arise due to the overlap of
five sp3d hybrid orbitals on the
phosphorus atom with sp3 atomic
orbitals on the fluorine atoms.
• Because the sp3d hybrid orbitals
are arranged in a trigonal
bipyramid geometry, the molecule
has trigonal bipyramid geometry.
Bonding in SF4
• Four equivalent sigma
covalent bonds arise
due to the overlap of
sp3d hybrid orbitals on
the sulfur atom with sp3
atomic orbitals on the
fluorine atoms.
• The molecule has
trigonal bipyramid
electron group
geometry and see-saw
molecular geometry.
Orbital Overlap in SF4
Bonding in IF3
• Three equivalent sigma
covalent bonds arise due to
the overlap of sp3d hybrid
orbitals on the iodine atom
with sp3 atomic orbitals on
the fluorine atoms.
• The molecule has trigonal
bipyramid electron group
geometry and T-shaped
molecular geometry.
Orbital Overlap in IF3
Bonding in XeF2
• Two equivalent sigma
covalent bonds arise due
to the overlap of sp3d
hybrid orbitals on the
xenon atom with sp3
atomic orbitals on the
fluorine atoms.
• The molecule has trigonal
bipyramid electron group
geometry and linear
molecular geometry.
Orbital Overlap in XeF2
Electron Group and
Molecular Geometry – Part 4
• 5 e- groups – trigonal bipyramid electron
group geometry – 90 °, 120 °, and 180°
angles.
– 5 bond groups and 0 lone pairs – trigonal
bipyramid molecular geometry
– 4 bond groups and 1 lone pair – see-saw molecular
geometry
– 3 bond groups and 2 lone pairs – T-shaped
molecular geometry
– 2 bond groups and 3 lone pairs – linear molecular
geometry
Things to explain for the
bonding in a sulfur
hexafluoride molecule, SF6
• Why are there six
equivalent S-F
covalent bonds?
• Why is the
molecule
octahedral with
bond angles of
90 ° and 180°?
Sulfur - 6 bonds
Formation
of sp3d2
Hybrid
Orbitals
Orbital Overlap in SF6
Bonding in SF6
• Six equivalent sigma covalent
bonds arise due to the overlap of
sp3d2 hybrid orbitals on the sulfur
atom with sp3 atomic orbitals on
the fluorine atoms.
• Because the sp3d2 hybrid orbitals
are arranged in an octahedral
geometry, the molecule has
octahedral geometry.
Bonding in IF5
• Five equivalent sigma
covalent bonds arise due
to the overlap of sp3d2
hybrid orbitals on the
iodine atom with sp3
atomic orbitals on the
fluorine atoms.
• The molecule has
octahedral electron group
geometry and square
pyramid molecular
geometry.
Orbital Overlap in IF5
Bonding in XeF4
• Four equivalent sigma
covalent bonds arise due
to the overlap of sp3d2
hybrid orbitals on the
xenon atom with sp3
atomic orbitals on the
fluorine atoms.
• The molecule has
octahedral electron group
geometry and square
planar molecular
geometry.
Orbital Overlap in XeF4
Electron Group and
Molecular Geometry – Part 5
• 6 e- groups – octahedral electron group
geometry – 90 ° and 180° angles.
– 6 bond groups and 0 lone pairs – octahedral
molecular geometry
– 5 bond groups and 1 lone pair – square pyramid
molecular geometry
– 4 bond groups and 2 lone pairs – square planar
molecular geometry
Drawing Lewis
Structures (1)
•
If the formula represents an
uncharged molecule, try to draw
a Lewis structure in which all of
the atoms have their most
common bonding pattern.
–
–
–
–
–
Hydrogen – 1 bond 0 lone pairs
Halogens – 1 bond 3 lone pairs
O, S, Se – 2 bonds 2 lone pairs
N, P – 3 bonds 1 lone pair
C – 4 bonds 0 lone pairs
Drawing Lewis Structures (2)
If the shortcut doesn’t work,
•
Step 1: Determine the total number of
valence electrons for the molecule or
polyatomic ion.
–
–
–
For uncharged molecules, the total number
of valence electrons is the sum of the
valence electrons of each atom.
For polyatomic cations, the total number of
valence electrons is the sum of the valence
electrons for each atom minus the charge.
For polyatomic anions, the total number of
valence electrons is the sum of the valence
electrons for each atom plus the charge.
Drawing Lewis
Structures (3)
• Step 2: Draw a reasonable skeletal structure,
using single bonds to join all the atoms.
– Try to arrange the atoms to yield the most typical
number of bonds for each atom.
– Apply the following guidelines in deciding what
element belongs in the center of your structure.
• Hydrogen and fluorine atoms are never in the center.
• Oxygen atoms are rarely in the center.
• The element with the fewest atoms in the formula is
often in the center.
• The atom that is capable of making the most bonds is
often in the center.
– Oxygen atoms rarely bond to other oxygen atoms.
– The molecular formula often reflects the molecular
structure.
– Carbon atoms commonly bond to other carbon atoms.
Drawing Lewis
Structures (4)
• Step 3: Subtract 2 electrons
from the total for each of the
single bonds (lines) described in
Step 2.
Drawing Lewis
Structures (5)
• Step 4: Try to distribute the remaining electrons
as lone pairs to obtain a total of eight electrons
around each atom except hydrogen and boron.
– In a reasonable Lewis structure, carbon, nitrogen,
oxygen, and fluorine always have eight electrons
around them.
– Hydrogen will always have a total of two electrons from
its one bond.
– Boron can have fewer than eight electrons but never
more than eight.
– The nonmetallic elements in periods beyond the
second period (P, S, Cl, Se, Br, and I) usually have
eight electrons around them, but they can have more.
– The bonding properties of the metalloids arsenic, As,
and tellurium, Te, are similar to those of phosphorus, P,
and sulfur, S, so they usually have eight electrons
around them but can have more.
Drawing Lewis
Structures (6)
•
Step 5: Do one of the following.
–
–
–
If in Step 4 you were able to obtain an octet of
electrons around each atom other than hydrogen
and boron, and if you used all of the remaining
valence electrons, go to Step 6.
If you have electrons remaining after each of the
atoms other than hydrogen and boron have their
octet, you can put more than eight electrons around
elements in periods beyond the second period.
If you do not have enough electrons to obtain octets
of electrons around each atom (other than hydrogen
and boron), convert one lone pair into a multiple
bond for each two electrons that you are short.
Drawing Lewis
Structures (7)
• Step 6: Check your structure.
– See if all of the atoms have their most
common bonding pattern.
– Determine the formal charge for each
atom that does not have its most
common bonding pattern.
• Formal charge = A-group # - # lines - #
dots
• Formal charge = 0 means stable
Drawing Lewis
Structures (8)
• Step 7: If necessary, try to rearrange
your structure to give each atom its
most common bonding pattern and a
minimum of formal charges.
– One way to do this is to return to Step 2
and try another skeleton.
– If your central atom is below the second
period, you can bring a lone pair in from
an outer atom with a negative formal
charge to form a new bond to a central
atom with a positive formal charge.
Lewis
Structure
Drawing
Summary
Nitrate Resonance
Resonance
• We can view certain molecules
and polyatomic ions as if they
were able to resonate—to switch
back and forth—between two or
more different structures. Each of
these structures is called a
resonance structure. The
hypothetical switching from one
resonance structure to another is
called resonance.
3 Criteria for Resonance
• There must be a double or triple
bond.
• There must be an adjacent atom
with at least one lone pair.
• The adjacent atom cannot be a
fluorine atom or an oxygen with 2
bonds and two lone pairs.
To draw Resonance Forms
Nitrate Resonance
Steps for Drawing
Resonance Hybrid
•
•
•
•
•
Draw the skeletal structure with solid lines for
the bonds that are found in all of the
resonance forms.
Where there is sometimes a multiple bond
and sometimes not, draw a dotted line.
Draw in all of the lone pairs that are found on
every one of the resonance forms. (Leave off
the lone pairs that are on one or more
resonance form but not on all of them.)
Put on full formal charges on those atoms
that have the formal charge in all of the
resonance forms.
Put on partial formal charges on the atoms
that have formal charges in one or more of
the resonance forms but not in all of them.
NO3- Resonance Hybrid
Delocalized Pi System for NO3-
PO43- Controversy – Minimize
Formal Charges
PO43- Controversy –
Follow Octet Rule
PO43- Geometry
• Both possible Lewis structures predict
the same geometry, and both show
that the four bonds are equivalent.
Triglycerides (Fats and Oils)
Saturated Triglyceride Tristearin
Tristearin – Line Drawing
Unsaturated Triglyceride
Cis and Trans
• When there is a double bond between two
carbons and when like groups are on different
carbons and the same side of the double bond
the arrangement is called cis.
• When the like groups are on opposite sides of
the double bond the arrangement is called
trans.
Hydrogenation
Hydrogenation
- Example
Trans Fats
• Hydrogenation is reversible.
• When the double bond is reformed, it is
more likely to form the more stable trans
form than the less stable cis form.
• Therefore, partial hydrogenated vegetable
oils contain trans fats, which are
considered to be damaging to your health.
Olestra - a Fat Substitute
Predicting Molecular
Polarity
• Step 1: Draw a reasonable Lewis
structure for the substance.
• Step 2: Identify each bond as either
polar or nonpolar.
– If there are no polar bonds, the
molecule is nonpolar.
– If the molecule has polar bonds, move
on to Step 3.
Electronegativities
Bond Type Prediction
Predicting Molecular
Polarity (cont.)
• Step 3: If there is only one central atom,
examine the electron groups around it.
– If there are no lone pairs on the central atom,
and if all the bonds to the central atom are
the same, the molecule is nonpolar.
– If the central atom has at least one polar
bond and if the groups bonded to the central
atom are not all identical, the molecule is
probably polar. Move on to Step 4.
• Step 4: Draw a geometric sketch of the
molecule.
Predicting Molecular
Polarity (cont.)
• Step 5: Determine the symmetry of the
molecule using the following steps.
– Describe the polar bonds with arrows
pointing toward the more electronegative
element. Use the length of the arrow to
show the relative polarities of the different
bonds.
– Decide whether the arrangement of arrows
is symmetrical or asymmetrical
• If the arrangement is symmetrical and the arrows
are of equal length, the molecule is nonpolar.
• If the arrows are of different lengths, and if they
do not balance each other, the molecule is polar.
• If the arrangement is asymmetrical, the molecule
is polar.
Shortcuts for Predicting
Molecular Polarity
• If the central atom has no lone
pairs and if all the bonds are
identical (all single or all double and
all to an atom of the same
element), the molecule is nonpolar.
• If there’s at least one polar bond,
and if the groups are not identical,
the molecule is probably polar.
• Molecules with the O-H or N-H
bond are polar.
Condensation
(Gas to Liquid)
Dipole-Dipole Attractions
Dipole-Dipole Attractions in a Liquid
Connections Between Topics
Lewis structures
↓
Molecular geometry
Molecular polarity
Electronegativity
↓
Bond polarity
Other factors
Types of attractions between particles
↓
Relative strengths of these attractions
↓
Relative condensation temperatures
Hydrogen Bonds in HF
Hydrogen Bonds in Water
Hydrogen
Bonds
in
Methanol
Hydrogen
Bonds
in
Ammonia
London
Forces
London Forces
in Polar
Molecules
Why Larger Molecules Have
Stronger London Forces
Predicting Types of
Attractions
Predicting Types of
Attractions
• Step 1: Determine the type of attraction
between the particles using the following
steps.
– Step 1(a): Classify each substance as either a
metallic element, an element composed of atoms, an
element composed of molecules, an ionic compound,
or a molecular compound.
• Metallic elements have metallic bonds.
• The particles in elements that are composed of
atoms (the noble gases) and molecules (the other
nonmetallic elements) are held together by
London forces.
• The ions in ionic compounds are held together by
ionic bonds.
• Continue to Step 1(b) for molecular compounds.
Predicting Types of
Attractions (cont.)
• Step 1(b): For molecular
compounds, draw the Lewis
structure for the molecule.
– If the Lewis structure contains an
O-H, N-H, or H-F bond, the
attractions that form when the
substance condenses are hydrogen
bonds enhanced by London forces.
– For other molecular compounds, do
Step 1(c).
Predicting Types of
Attractions (cont.)
• Step 1(c): If there are no O-H,
N-H, or H-F bonds, determine the
polarity of the bonds.
– If there are no polar bonds, the
molecules are nonpolar and form
London forces when they condense.
– If there is at least one polar bond, do
Step 1(d).
Predicting Types of
Attractions (cont.)
• Step 1(d): Predict whether the polar
bonds are symmetrically or
asymmetrically arranged.
– If the distribution of polar bonds is
symmetrical and their dipoles equal, the
molecules are nonpolar and form London
forces when they condense.
– If the distribution of polar bonds is
asymmetrical, or symmetrical with unequal
dipoles, the molecules are polar and form
dipole-dipole attractions enhanced by
London forces when they condense.
Predicting Strengths of
Attractions
• Step 2: Although we cannot predict the
relative strengths of attractions between
all particles, we can apply one of the
following guidelines to predict the relative
strengths of attractions between some
particles.
– Chemical bonds are generally stronger
than intermolecular attractions.
– Hydrogen bonds are generally stronger
than dipole-dipole attractions, which are
generally stronger than London forces.
– Larger molecules tend to have stronger
attractions.
Properties Affected by
Strengths of Attractions
• Stronger attractions lead to
– Higher melting and boiling point
temperatures.
– Greater viscosity for liquids.
– Greater surface tension for liquids.
– Lower rate of evaporation for
liquids.
– Lower vapor pressures above a
liquid in a closed container.
Water Solubility
• If we are comparing the water
solubility of two similar molecules,
the one with the higher percentage of
the molecule that is polar
(hydrophilic) is expected to have
higher water solubility.
• We predict that the molecule with
the higher percentage of its structure
that is nonpolar (hydrophobic) to be
less soluble in water.
Hydrophobic and Hydrophilic
Methamphetamine