Transcript Chapter 10 Molecular Geometry and Chemical Bonding Theory

Chapter 10
Molecular Geometry
and Chemical
Bonding Theory
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Contents and Concepts
Molecular Geometry and Directional Bonding
We can predict the molecular geometry of a molecule—that is,
its general shape as determined by the relative positions of
atomic nuclei—with a simple model: the valence-shell
electron-pair repulsion model.
After exploring molecular geometry, we explain chemical
bonding by means of valence bond theory, which gives us
insights into why bonds form and why they have definite
directions in space, giving particular molecular geometries.
1. The Valence-Shell Electron-Pair Repulsion (VSEPR) Model
2. Dipole Moment and Molecular Geometry
3. Valence Bond Theory
4. Description of Multiple Bonding
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Molecular Orbital Theory
Although valence bond theory satisfactorily describes most
molecules, it has trouble explaining the bonding in
molecules such as oxygen, O2, which has even numbers of
electrons but is paramagnetic.
Molecular orbital theory is an alternative theory that views
the electronic structure of molecules much the way we
think of atoms—in terms of orbitals that are successively
occupied by electrons.
5. Principles of Molecular Orbital Theory
6. Electron Configurations of Diatomic Molecules of the
Second-Period Elements
7. Molecular Orbitals and Delocalized Bonding
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Learning Objectives
Molecular Geometry and Directional Bonding
1. The Valence-Shell Electron-Pair Repulsion
(VSEPR) Model
a. Define molecular geometry.
b. Define valence-shell electron-pair repulsion
model.
c. Note the difference between the
arrangement of electron pairs about a
central atom and molecular geometry.
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d. Note the four steps in the prediction of
geometry by the VSEPR model.
e. Predict the molecular geometry for a
molecule with two, three, or four electron
pairs.
f. Note that a lone pair tends to require more
space than a corresponding bonding pair
and that a multiple bond requires more
space than a single bond.
g. Predict the molecular geometry for a
molecule with five or six electron pairs.
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2. Dipole Moment and Molecular Geometry
a. Define dipole moment.
b. Explain the relationship between the dipole
moment and molecular geometry.
c. Note that the polarity of a molecule can
affect certain properties, such as boiling
point.
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3. Valence Bond Theory
a. Define valence bond theory.
b. State the two conditions needed for bond
formation according to valence bond theory.
c. Define hybrid orbitals.
d. State the five steps in describing bonding,
following the valence bond theory.
e. Apply valence bond theory to a molecule
with two, three, or four electron pairs.
f. Apply valence bond theory to a molecule
with five or six electron pairs.
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4. Description of Multiple Bonding
a. Define an s (sigma) bond.
b. Define a p (pi) bond.
c. Apply valence bond theory (multiple
bonding).
d. Explain geometric, or cis-trans, isomers in
terms of the p-bond description of a double
bond.
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Molecular Orbital Theory
5. Principles of Molecular Orbital Theory
a. Define molecular orbital theory.
b. Define bonding orbitals and antibonding
orbitals.
c. Define bond order.
d. State the two factors that determine the
strength of interaction between two atomic
orbitals.
e. Describe the electron configurations of H2,
He2, Li2, and Be2.
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6. Electron Configurations of Diatomic Molecules
of the Second-Period Elements
a. Define homonuclear diatomic molecules
and heteronuclear diatomic molecules.
b. Describe molecular orbital configurations
(homonuclear diatomic molecules).
c. Describe molecular orbital configurations
(heteronuclear diatomic molecules).
7. Molecular Orbitals and Delocalized Bonding
a. Describe the delocalized bonding in
molecules such as O3.
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In this chapter, we discuss how to explain the
geometries of molecules in terms of their electronic
structures.
We also explore two theories of chemical bonding:
valence bond theory and molecular orbital theory.
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Molecular geometry is the general shape of a
molecule, as determined by the relative positions
of the atomic nuclei.
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The valence-shell electron-pair repulsion (VSEPR)
model predicts the shapes of molecules and ions
by assuming that the valence-shell electron pairs
are arranged about each atom so that electron
pairs are kept as far away from one another as
possible, thereby minimizing electron pair
repulsions.
The diagram on the next slide illustrates this.
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Two electron pairs are 180° apart (a linear
arrangement).
Three electron pairs are 120° apart in one plane
(a trigonal planar arrangement).
Four electron pairs are 109.5° apart in three
dimensions (a tetrahedral arrangment).
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Five electron pairs are arranged with three pairs in
a plane 120° apart and two pairs at 90°to the
plane and 180° to each other (a trigonal
bipyramidal arrangement).
Six electron pairs are 90° apart (an octahedral
arrangement).
This is illustrated on the next slide.
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These arrangements are illustrated below with
balloons and models of molecules for each.
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To describe the molecular geometry, we describe
the relative positions of the atoms, not the lone
pairs. The direction in space of the bonding pairs
gives the molecular geometry.
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The diagrams below illustrate molecular geometry
and the impact of lone pairs on it for linear and
trigonal planar electron-pair arrangements.
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Molecular geometries with a tetrahedral electronpair arrangement are illustrated below.
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Molecular geometries for the trigonal bipyramidal
electron-pair arrangement are shown on the next
slide.
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Molecular geometries for the octahedral electronpair arrangement are shown below.
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The VSEPR model considers a double or triple
bond as though it were one lone pair.
When resonance structures are required for the
electron-dot diagram, you may choose any one to
determine the electron-pair arrangement and the
molecular geometry.
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Predicting Molecular Geometry Using VSEPR
1. Write the electron-dot formula from the formula.
2. Based on the electron-dot formula, determine
the number of electron pairs around the central
atom (including bonding and nonbonding
pairs).
3. Determine the arrangement of the electron
pairs about the central atom (Figure 10.3).
4. Obtain the molecular geometry from the
directions of the bonding pairs for this
arrangement (Figure 10.4).
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?
Use the VSEPR model to predict the
geometries of the following molecules:
a. AsF3
b. PH4+
c. BCl3
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AsF3 has 1(5) + 3(7) = 26 valence electrons; As is
the central atom. The electron-dot formula is
F
As
F
F
There are four regions of electrons around As: three
bonds and one lone pair.
The electron regions are arranged tetrahedrally.
One of these regions is a lone pair, so the molecular
geometry is trigonal pyramidal.
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PH4+ has 1(5) + 4(1) – 1 = 8 valence electrons; P
is the central atom. The electron-dot formula is
+
H
H
P
H
H
There are four regions of electrons around P:
four bonding electron pairs.
The electron-pairs arrangement is tetrahedral.
All regions are bonding, so the molecular geometry
is tetrahedral.
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BCl3 has 1(3) + 3(7) = 24 valence electrons;
B is the central atom.
The electron-dot formula is
Cl
Cl
B
Cl
There are three regions of electrons around B; all
are bonding.
The electron-pair arrangement is trigonal planar.
All of these regions are bonding, so the molecular
geometry is trigonal planar.
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Concept Check 10.1
An atom in a molecule is surrounded by four pairs of
electrons: one lone pair and three bonding pairs. Describe
how the four electron pairs are arranged about the atom.
How are any three of these pairs arranged in space?
What is the geometry about this central atom, taking into
account just the bonded atoms?
The electron-pair arrangement is tetrahedral.
Any three pairs are arranged as a trigonal pyramid.
When one pair of the four is a lone pair, the
geometry is trigonal pyramidal.
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?
Using the VSEPR model, predict the
geometry of the following species:
a. ICl3
b. ICl4-
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ICl3 has 1(7) + 3(7) = 28 valence electrons. I is the
central atom. The electron-dot formula is
Cl
Cl
I
Cl
There are five regions: three bonding and two lone
pairs.
The electron-pair arrangement is trigonal
bipyramidal.
The geometry is T-shaped.
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ICl4- has 1(7) + 4(7) + 1 = 36 valence electrons. I is
the central element. The electron-dot formula is
-
Cl
Cl
I
Cl
Cl
There are six regions around I: four bonding and
two lone pairs.
The electron-pair arrangement is octahedral.
The geometry is square planar.
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Predicting Bond Angles
The angles 180°, 120°, 109.5°, and so on are
the bond angles when the central atom has no
lone pair and all bonds are with the same other
atom.
When this is not the case, the bond angles deviate
from these values in sometimes predictable ways.
Because a lone pair tends to require more space
than a bonding pair, it tends to reduce the bond
angles.
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The impact of lone pair(s) on bond angle for
tetrahedral electron-pair arrangements has been
experimentally determined.
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Multiple bonds require more space than single
bonds and, therefore, constrict the bond angle.
This situation is illustrated below, again with
experimentally determined bond angles.
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Dipole Moment
A quantitative measure of the degree of charge
separation in a molecule.
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Measurements are based on the fact that polar
molecules are oriented by an electric field. This
orientation affects the capacitance of the charged
plates that create the electric field.
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In part A, there is no
electric field; molecules
are oriented randomly.
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In part B, there is an
electric field; molecules
align themselves
against the field.
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A polar bond is characterized by separation of
electrical charge. Polar molecules, therefore, have
nonzero dipole moments.
For HCl, we can represent the charge separation
using + and - to indicate partial charges.
Because Cl is more electronegative than H, it has
the - charge, while H has the + charge.
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+
-
H
Cl
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The figure below shows the orbitals involved in HCl
bond: the H 1s and the Cl 3p.
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To determine whether a molecule is polar, we
need to determine the electron-dot formula and the
molecular geometry. We then use vectors to
represent the charge separation. They begin at +
atoms and go to - atoms. Vectors have both
magnitude and direction.
We then sum the vectors. If the sum of the vectors
is zero, the dipole moment is zero. If there is a net
vector, the molecule is polar.
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To illustrate this process, we use arrows with a +
on one end of the arrow. We’ll look at CO2 and
H2O. CO2 is linear, and H2O is bent.
+
O
C
O
The vectors add to zero
(cancel) for CO2.
Its dipole moment is
zero.
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+
+ +
O
H
+
H
For H2O, a net vector
points up.
Water has a dipole
moment.
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The relationship between molecular geometry and
dipole moment is summarized in Table 10.1.
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Polar molecules experience attractive forces
between molecules; in response, they orient
themselves in a + to - manner. This has an
impact on molecular properties such as boiling
point. The attractive forces due to the polarity lead
the molecule to have a higher boiling point.
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We can see this illustrated with two compounds:
cis-1,2-dichloroethene
H
H
C
Cl
Cl
C
+
+
+
+
Cl
+
Cl
+
+
C
H
+
+
C
trans-1,2-dichloroethene
H
The net polarity is
down; this is a
polar molecule.
There is no net
polarity; this is a
nonpolar molecule.
Boiling point 60°C.
Boiling point 48°C.
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Concept Check 10.2
Two molecules, each with the general formula AX3,
have different dipole moments. Molecule Y has a
dipole moment of zero, whereas molecule Z has a
nonzero dipole moment. From this information, what
can you say about the geometries of Y and Z?
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The formula AX3 can have the following
molecular geometries and dipole moments:
Trigonal planar (zero)
Trigonal pyramidal (can be nonzero)
T-shaped (can be nonzero)
Molecule Y is likely to be trigonal planar, but might be
trigonal pyramidal or T-shaped.
Molecule Z must be either trigonal pyramidal or Tshaped.
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?
Which of the following molecules would be
expected to have a zero dipole moment?
a. GeF4
b. SF2
c. XeF2
d. AsF3
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GeF4: 1(4) + 4(7) = 32 valence electrons.
Ge is the central atom.
8 electrons are bonding; 24 are nonbonding.
Tetrahedral molecular geometry.
F
GeF4 is
nonpolar and has
a zero dipole moment.
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F
Ge
F
F
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SF2: 1(6) + 2(7) = 20 valence electrons.
S is the central atom.
4 electrons are bonding; 16 are nonbonding.
Bent molecular geometry.
F
S
F
SF2 is
polar and has
a nonzero dipole moment.
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XeF2: 1(8) + 2(7) = 22 valence electrons.
Xe is the central atom.
4 electrons are bonding; 18 are nonbonding.
Linear molecular geometry.
F
Xe
F
XeF2 is
nonpolar and has
a zero dipole moment.
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AsF3: 1(5) + 3(7) = 26 valence electrons.
As is the central atom.
6 electrons are bonding; 20 are nonbonding.
Trigonal pyramidal molecular geometry.
F
AsF3 is
polar and has
a nonzero dipole moment.
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As
F
F
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Which of the following molecules would be expected
to have a zero dipole moment?
a. GeF4
tetrahedral molecular geometry
zero dipole moment
b. SF2
bent molecular geometry
nonzero dipole moment
c. XeF2
linear molecular geometry
zero dipole moment
d. AsF3
trigonal pyramidal molecular geometry
nonzero dipole moment
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Valence bond theory is an approximate theory
put forth to explain the electron pair or covalent
bond by quantum mechanics.
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A bond forms when
• An orbital on one atom comes to occupy a
portion of the same region of space as an orbital
on the other atom. The two orbitals are said to
overlap.
• The total number of electrons in both orbitals is
no more than two.
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The greater the orbital overlap, the stronger the
bond.
Orbitals (except s orbitals) bond in the direction in
which they protrude or point, so as to obtain
maximum overlap.
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Hybrid orbitals are orbitals used to describe the
bonding that is obtained by taking combinations of
the atomic orbitals of the isolated atoms.
The number of hybrid orbitals formed always
equals the number of atomic orbitals used.
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Hybrid orbitals are named by using the atomic
orbitals that combined:
• one s orbital + one p orbital gives two sp
orbitals
• one s orbital + two p orbitals gives three sp2
orbitals
• one s orbital + three p orbitals gives four sp3
orbitals
• one s orbital + three p orbitals + one d orbital
gives five sp3d orbitals
• one s orbital + three p orbitals + two d orbitals
gives six sp3d2 orbitals
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Hybrid orbitals have definite directional
characteristics, as described in Table 10.2.
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To obtain the bonding description about any atom
in a molecule:
1. Write the Lewis electron-dot formula.
2. Use VSEPR to determine the electron
arrangement about the atom.
3. From the arrangement, deduce the hybrid
orbitals.
4. Assign the valence electrons to the hybrid
orbitals one at a time, pairing only when
necessary.
5. Form bonds by overlapping singly occupied
hybrid orbitals with singly occupied orbitals of
another atom.
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Let’s look at the methane molecule, CH4. Simply
using the atomic orbital diagram, it is difficult to
explain its four identical C—H bonds.
The valence bond theory allows us to explain this
in two steps: promotion and hybridization.
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First, the paired 2s electron is promoted to the
unfilled orbital. Now each orbital has one electron.
Second, these orbitals are hybridized, giving four
sp3 hybrid orbitals.
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?
Use valence bond theory to describe the
bonding about an N atom in N2F4.
F
N
N
F
F
F
The Lewis electron-dot structure shows three
bonds and one lone pair around each N atom.
They have a tetrahedral arrangement.
A tetrahedral arrangement has sp3 hybrid orbitals.
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The orbital diagram of the ground-state N atom is
1s
2s
2p
The sp3 hybridized N atom is
sp3
1s
Consider one N in N2F4: the two N—F bonds are
formed by the overlap of a half-filled sp3 orbital
with a half-filled 2p orbital on F. The N—N bond
forms from the overlap of a half-filled sp3 orbital on
each. The lone pair occupies one sp3 orbital.
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?
Use valence bond theory to describe the
bonding in the ClF2- ion.
The valence orbital diagram for the Cl- ion is
3s
3p
3d
After the promotion to get two half-filled orbitals,
the orbital diagram is
3s
3p
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3d
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The sp3d hybridized orbital diagram is
sp3d
3d
Two Cl—F bonds are formed from the overlap of
two half-filled sp3d orbitals with half-filled 2p
orbitals on the F atom. These use the axial
positions of the trigonal bipyramid.
Three lone pairs occupy three sp3d orbitals. These
are in the equatorial position of the trigonal
bipyramid.
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One hybrid orbital is required for each bond
(whether a single or a multiple bond) and for each
lone pair.
Multiple bonding involves the overlap of one hybrid
orbital and one (for a double bond) or two (for a
triple bond) nonhybridized p orbitals.
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To describe a multiple bond, we need to
distinguish between two kinds of bonds.
A s bond (sigma) has a cylindrical shape about
the bond axis. It is formed either when two s
orbitals overlap or with directional orbitals (p or
hybrid), when they overlap along their axis.
A p bond (pi) has an electron distribution above
and below the bond axis. It is formed by the
sideways overlap of two parallel p orbitals. This
overlap occurs when two parallel half-filled p
orbitals are available after s bonds have formed.
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Figure A illustrates the s bonds in C2H4.
The top of Figure B shows the p orbital on each
carbon at 90° to each other, with no overlap.
The bottom of Figure B shows parallel p orbitals
overlapping to form a p bond.
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In acetylene, C2H2, each C has two s bonds and
two p bonds.
The s bonds form using the sp hybrid orbital on C.
This is shown in part A.
The two p bonds form from the overlap of two sets
of parallel p orbitals. This is illustrated in Figure B.
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The description of a p bond helps to explain the cistrans isomers of 1,2-dichloroethene.
The overlap of the parallel p orbitals restricts the
rotation around the C=C bond. This fixes the
geometric positions of Cl: either on the same side
(cis) or on different sides (trans) of the C=C bond.
H
H
C
Cl
H
C
C
cis
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Cl
Cl
Cl
trans
C
H
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Concept Check 10.3
An atom in a molecule has one single bond and
one triple bond to other atoms. What hybrid orbitals
do you expect for this atom? Describe how you
arrive at your answer.
One single bond and one triple bond requires two
hybrid orbitals and two sets of two parallel p
orbitals. That requires sp hybridization.
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?
Describe the bonding about the C atom
in formaldehyde, CH2O, using valence
bond theory.
O
C
H
H
The electron arrangement is trigonal pyramidal
using sp2 hybrid orbitals.
The ground-state orbital diagram for C is
1s
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2s
2p
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After promotion, the orbital diagram is
1s
2s
2p
After hybridization, the orbital diagram is
1s
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sp2
2p
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The C—H s bonds are formed from the overlap of
two C sp2 hybrid orbitals with the 1s orbital on the H
atoms.
The C—O s bond is formed from the overlap of one
sp2 hybrid orbital and one O half-filled p orbital.
The C—O p bond is formed from the sideways
overlap of the C 2p orbital and an O 2p orbital.
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As atoms approach one another, their atomic
orbitals overlap and form molecular orbitals.
Molecular orbitals concentrated in regions between
nuclei are called bonding orbitals. They are
obtained by adding atomic orbitals.
Molecular orbitals having zero values in regions
between nuclei (and are in other regions) are
called antibonding orbitals. They are obtained by
subtracting atomic orbitals.
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The figure on the next slide illustrates the
combination of two 1s orbitals.
The top shows the formation of the bonding
molecular orbital; the bottom shows the formation
of the antibonding molecular orbital.
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Once a molecular orbital is formed, it can be
occupied in the same way as are atomic orbitals.
As an example, we’ll study hydrogen, H2.
Each hydrogen atom has one electron in a 1s
orbital, for a total of two electrons.
When the two 1s orbitals combine, they form two s
molecular orbitals.
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The bonding molecular orbital is lower in energy
than the antibonding molecular orbital.
The notation for a molecular orbital has the original
atomic orbitals as subscript.
The * indicates the antibonding molecular orbital.
s 1s
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s 1s
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The two electrons are placed in the lower-energy
bonding molecular orbital with opposite spin.
s 1s
s 1s
The ground-state molecular orbital electron
configuration is (σ1s)2. Excited states are also
possible.
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Bond order—the number of bonds that exist
between two atoms—can be calculated using the
equation
1
Bond order  nb  na 
2
nb = number of electrons in bonding orbitals
na = number of electrons in antibonding orbitals
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For H2 (two electrons), the molecular electron
configuration is (s1s)2.
The bond order is ½(2 - 0) = 1.
For H2+ (one electron), the molecular electron
configuration is (s1s)1.
The bond order is ½(1 - 0) = ½.
For He2 (four electrons), the molecular electron
configuration is (s1s)2 (s1s)2.
The bond order is ½(2 - 2) = 0. This means no
bond forms.
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The strength of the interaction between two atomic
orbitals to form molecular orbitals is determined by
two factors:
•
•
The energy difference between the interacting
orbitals
The magnitude of the overlap
For the interaction to be strong, the energies of the
two orbitals must be approximately equal and the
overlap must be large.
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Second-Period Homonuclear Diatomic
Molecules
In Li2, the two 1s atomic orbitals combine to form
two s molecular orbitals: s1s and s1s.
The two 2s atomic orbitals combine to form two s
molecular orbitals: s2s and s2s.
Six electrons occupy the orbitals in order of
increasing energy:
s 1s
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s 1s
s 2s
s 2s
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s 1s

s 1s
s 2s

s 2s
The molecular electron configuration is
(s1s)2(s1s*)2(s2s)2
or
KK(s2s)2
KK is the abbreviation for (s1s)2(s*1s)2 .
The bond order is ½(4 - 2) = 1.
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Be2 has 8 electrons. The molecular orbital diagram
is the same as for Li2. The eight electrons occupy
the orbitals in order of increasing energy:
s 1s

s 1s
s 2s

s 2s
The molecular electron configuration is
(s1s)2(s1s)2(s2s)2(s2s)2 or KK(s2s)2(s2s)2
The bond order is ½(4 - 4) = 0.
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Molecular orbitals formed by the overlap of atomic
p orbitals are of two types.
When two p orbitals overlap end-to-end, two s
molecular orbitals form: s2p and s*2p.
When two p orbitals overlap sideways, two p
molecular orbitals form: p2p and p*2p. Because two
p orbitals on each atom overlap with two p orbitals
on another atom, a total of four molecular orbitals
form: two bonding and two antibonding.
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The relative
energies of the
molecular
orbitals formed
from second
period atomic
orbitals are
shown here.
The arrows are
for N2.
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Because there are two bonding p orbitals, together
they can hold four electrons. They are filled one
electron per orbital before a second electron is
added to the same orbital with opposite spin.
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?
Give the orbital diagram and electron
configuration of the F2 molecule.
Is the molecular substance diamagnetic or
paramagnetic?
What is the order of the bond in F2?
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F2 has 18 electrons. The KK shell holds 4
electrons so 14 remain.
s 2s

s 2s
p 2p
s 2p
p 2* p
s 2 p
The molecular electron configuration is
KK(s2s)2(s2s)2(p2p)4(s2p)2 (p*2p)4
The bond order is ½(8 - 6) = 1.
The molecule is diamagnetic.
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?
A number of compounds of the nitrosonium
ion, NO+, are known, including nitrosonium
hydrogen sulfate, (NO+)(HSO4-). Use the
molecular orbitals similar to those of
homonuclear diatomic molecules and obtain
the orbital diagram, electron configuration,
bond order, and magnetic characteristics of
the NO+ ion.
Note: The stability of the ion results from the
loss of an antibonding electron from NO.
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NO+ has 14 electrons. The KK shell holds 4
electrons, leaving 10 electrons for bonding.
s 2s

s 2s
p 2p
s 2p
p 2* p
s 2 p
The molecular electron configuration is
KK(s2s)2(s2s)2(p2p)4(s2p)2
The bond order is ½(8 - 2) = 3.
The ion has a diamagnetic molecular orbital
electron configuration.
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Table 10.3 (next slide) compares the theoretical
bond order to the experimental bond length, bond
dissociation energy, and magnetic character for the
second-period homonuclear diatomic molecules.
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Molecular orbital theory can describe delocalized
bonding in terms of a single electron configuration.
Besides having the conventional s bonds between
two atoms, these molecules have molecular
orbitals formed by more than two atomic orbitals.
Ozone, O3, is one example.
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O
O
O ↔
O
O
O
Each oxygen has three localized electron pairs
around it, suggesting that each atom uses sp2
hybridized orbitals.
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The overlap of one hybrid orbital on the central O
atom with a hybrid orbital on each terminal O
forms the two O—O s bonds.
This leaves one hybrid orbital on the central O
atom and two hybrid orbitals on the terminal O
atoms.
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This leaves one unhybridized 2p orbital on each of
the three O atoms.
Combining these three atomic orbitals forms three
molecular orbitals: p antibonding, nonbonding, and
p bonding.
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All three of the molecular orbitals span the entire
molecule. Both the bonding orbitals and the
nonbonding orbitals (same energy as isolated
atoms) are doubly occupied. This agrees with the
resonance structure in which one delocalized pair
is bonding and one is a lone pair (nonbonding) on
the terminal atom.
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