Molecular Geometry
Chapter 10: Molecular Geometry
and Chemical Bonding
Topics
•
•
•
•
•
•
• Describes the general shape of a molecule, as
determined by the relative positions of the
atomic nuclei
The Valence-Shell Electron-Pair Repulsion
(VSEPR) Model
Dipole Moment
Molecular Geometry
Valence Bond Theory
Molecular Orbital Theory
Electron Configurations of Diatomic Molecules
1
• May be predicted using the VSEPR model
– Valence-Shell Electron-Pair Repulsion
– Assumes that the valence-shell electron pairs are
arranged about each atom so that they are kept ____
_________ from one another as possible,
• Motive: to minimize electron pair _____________.
• Must consider electron groups around an atom
– Bonding electrons
– Non-bonding electrons (lone electron pairs)
2
VSEPR Models 1
VSEPR Models 2
• Two electron pairs are 180° apart (a linear arrangement).
• Five electron pairs are arranged with
• Three electron pairs are 120° apart in one plane (a trigonal planar
arrangement).
– three pairs in a plane 120° apart and
– two pairs at 90°to the plane and 180° to each other (a trigonal bipyramidal
arrangement).
• Four electron pairs are 109.5° apart in three dimensions (a tetrahedral
arrangement).
3
• Six electron pairs are 90° apart (an octahedral arrangement).
4
Determining Molecular Geometry
These arrangements are illustrated below with
balloons and models of molecules for each.
• Based on the arrangement of electron pairs
• It only takes into consideration ____________
electrons
– Non-bonding electrons are only considered to
determine the arrangement of the molecule, NOT the
geometry
• The direction in space of the bonding pairs gives
the molecular geometry.
• Consider the following compounds and draw
their Lewis Structures:
– CO2, SO2, BF3
5
6
1
Molecular Geometry:
2 and 3 Electron Pairs
Determining Molecular Geometry
• CO2
O
C
O
– Two double bonds around C
– The VSEPR model considers a double or triple bond as though it
were one lone pair
• Thus two electron groups
• SO2
O
S
O
O
S
O
O
S
O
– There are three structures
– There are three electron groups around S
F
• BF3
– B is bonded to three F
– There are three electron groups around B
F
B
F
7
8
Molecular Geometry: 5 Electron Pairs
Molecular Geometry: 4 Electron Pairs
9
10
Predicting Molecular Geometry
Molecular Geometry: 6 Electron Pairs
• Write the electron-dot formula
– from the chemical formula.
• Based on the electron-dot formula:
– Determine the number of electron pairs around the
central atom (including bonding and nonbonding pairs)
– Determine the arrangement of the electron pairs about
the central atom (Figure 10.3)
– Obtain the molecular geometry from the directions of
the bonding pairs for this arrangement (Figure 10.4)
11
12
2
Example 1: AsF3
Example 1: Molecular Geometry
• AsF3 has 1(5) + 3(7) = 26 valence
electrons;
• As is the central atom.
• Use the VSEPR model to predict the
geometries of the following molecules:
a. AsF3
b. PH4+
c. BCl3
F
As
F
F
• There are four regions of electrons
around As
•three bonding pairs
•one lone pair
• The electron arrangement is:_____________
• It has one lone pair
•The molecular geometry is _______________.
13
14
Example 1: PH4+
Example 1: BCl3
• PH4+ has 1(5) + 4(1) – 1 = 9 valence electrons;
• P is the central atom.
• BCl3 has 1(3) + 3(7) = 24
valence electrons;
• B is the central atom.
• There are four regions of
electrons around P:
•four bonding electron
pairs.
B
• There are three regions of
electrons around B; all are
bonding.
• The electron-pairs arrangement is __________
• All regions are bonding, so the molecular
geometry is __________________.
15
Cl
Cl
Cl
• The electron-pair arrangement is
_____________.
16
Example 2: ICl3
Examples 2: Molecular Geometry
• ICl3 has 1(7) + 3(7) = 28 valence
electrons.
• I is the central atom.
• Using the VSEPR model, predict the
geometry of the following species:
a. ICl3
b. ICl4-
Cl
Cl
• There are five regions:
• three bonding
• two lone pairs.
I
Cl
• The electron-pair arrangement is
_________________.
• The geometry is ____________.
17
18
3
Example 2: ICl4• ICl4- has 1(7) + 4(7) + 1 = 36
valence electrons.
• I is the central element.
Predicting Bond Angles
-
Cl
Cl
• There are six regions around I
• four bonding
• two lone pairs.
I
• 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 type of
atom.
Cl
• When this is not the case, the bond angles
deviate from these values in sometimes
predictable ways.
Cl
• The electron-pair arrangement
is _______________.
• The geometry is ___________.
• Because a lone pair tends to require more space
than a bonding pair, it tends to reduce the bond
angles.
19
20
Angles for Tetrahedral Arrangements
Effect of Multiple Bonds on Angles
Multiple bonds require more space than single
bonds and, therefore, constrict the bond angle.
21
22
Dipole Moment
Determining Polarity
• A quantitative measure of the polarity in a
molecule
• Depends on the atoms present and the polarity of
the bonds in the molecule
• First, find the electron-dot formula and the
molecular geometry.
• Second, use vectors to represent the charge
separation of each bond
– A polar bond is characterized by separation of
electrical charge.
– They begin at δ+ atoms and go to δ- atoms.
– Recall that vectors have both magnitude and direction.
• For HCl, we can represent the charge separation
using δ+ and δ- to indicate partial charges.
• Third, sum the vectors.
– If the sum of the vectors is zero, the dipole moment is
__________.
– Cl is more electronegative: it has the δ- charge
– H has the δ+ charge.
23
δ+
δ-
H
Cl
• Vectors of equal magnitude but opposite direction cancel out
– If there is a net vector, the molecule is ________.
24
4
Determining Polarity Example
Example 3: Determining Polarity
• To illustrate polarity, arrows with a ‘+’ on one end of
the arrow are used
• Which of these molecules is polar?
Cl
– You must always consider the geometry of the molecule
• Consider CO2 and H2O.
– CO2 is linear, and H2O is bent
O
C
H
O
O
The vectors add to zero
(cancel) for CO2.
Its dipole moment is zero.
F
H
C
Cl
H
H
H
H
25
S
F
Cl
Cl
B
Cl
F
F
Cl
Cl
For H2O, a net vector
points up.
Water has a dipole
moment.
I
H
C
H
Cl
26
Properties of Polar Compounds
• Polar molecules experience attractive forces
between molecules;
– Thus, they orient themselves in a δ+ to δ- 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.
– Consider the following compound, which has two
conformations
• 1,2-dichloroethene
27
28
Properties of Polar Compounds
cis-1,2-dichloroethene
+
H
H
C
+
+
Cl
+
C
Cl
• Must make use of quantum mechanics
• Two theories explain electronic structures
of bonds
Cl
+
C
+
+
Cl
H
trans-1,2-dichloroethene
+
+
C
Understanding Bonding Electronics
– Valence bond theory
H
• Overlap of atoms’ Hybrid orbitals result in bonds
29
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.
– Molecular Orbital Theory
• New Molecular orbitals are formed from the
orbitals of individual atoms
30
5
Valence Bond Theory
How Many Bonds can be Formed?
• A bond forms between two atoms
when:
• Chlorine
– A portion of two orbitals occupy the
same region (they overlap)
– __ unpaired electron
– forms __ bond (HCl)
• The greater the orbital overlap, the ______ the bond
• Orbitals (except s orbitals) bond in the direction in
which they protrude or point, so as to obtain maximum
overlap
• Oxygen
– __ unpaired electrons
– forms __ bonds (H2O)
– The total number of electrons in both orbitals is
no more than _________.
• Carbon
31
32
i d l
The Four Bonds of Carbon
t
Hybrid Orbitals
• An electron from the 2s orbital is _______ to the
vacant 2p orbital
• Upon promotion of an electron, atomic orbitals combine
to form ‘new’ orbitals
– Process is called ______________
– The new orbitals are called hybrid orbitals
• In the case of carbon,
– 1 s orbital and 3 p orbitals make 4 sp3 orbitals
33
34
Hybrid Orbitals 2
Spatial arrangement of sp3 hybrid orbitals
s-orbital:
• Any number and type of orbital is capable of
hybridizing
p-orbital:
– The number of hybrid orbitals formed always equals the
number of atomic orbitals used.
• Hybrid orbitals are named by using the atomic
orbitals that combined:
A. One sp3-orbital
B. Four sp3-orbital
around a nucleus
C. Bonding in CH4
• One carbon sp3orbital bonds to
one hydrogen sorbital
35
–
–
–
–
–
1 s orbital + 1 p orbital = _______ orbitals
1 s orbital + 2 p orbitals = _______ orbitals
1 s orbital + 3 p orbitals = _______ orbitals
1 s orbital + 3 p orbitals + 1 d orbital = _______ orbitals
1 s orbital + 3 p orbitals + 2 d orbitals = ______ orbitals
36
6
Hybrid Orbitals Diagrams
Hybrid Orbitals Characteristics
37
38
Obtaining Bond Description
In General
1. Write the Lewis electron-dot formula.
2. Use VSEPR to determine the electron
arrangement about the atom.
Orbital
Type
sp
Orbitals
formed
Electron groups
around atom
2
2
_______
3. From the arrangement, deduce the hybrid orbitals.
sp2
3
3
_______
4. Assign the valence electrons to the hybrid orbitals
one at a time, pairing only when necessary.
sp3
4
4
_______
sp3d
5
5
_______
sp3d2
6
6
_______
5. Form bonds by overlapping singly occupied hybrid
orbitals with singly occupied orbitals of another
atom.
39
Electron
arrangement
40
• The orbital diagram of the ground-state N atom is
Example 4: Hybridization
• Use valence bond theory to describe the
bonding about an N atom in N2H4.
F
N
N
F
F
1s
F
2s
2p
• The sp3 hybridized N atom is
1s
sp3
• Consider one N in N2F4:
• The Lewis electron-dot structure shows
three bonds and one lone pair around
each N atom.
– 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.
– They have a tetrahedral arrangement.
41
42
7
Example 5: Hybridization
Example 5: Hybridization
• Hybridization of the s, three p, and one d orbital occurs
next
• Use valence bond theory to describe the bonding in the
ClF2- ion.
– Five sp3d hybridized orbitals form, four 3d orbitals remain:
– The chlorine must be able to form two bonds. How does it do it?
• The valence orbital diagram for the Cl- ion is
sp3d
3s
3p
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.
3d
• First promotion of one 3p electron occurs to a d orbital:
– These use the axial positions of the trigonal bipyramid.
– Shape of molecule: linear
3s
3p
• Three lone pairs occupy three sp3d orbitals. These are in
the equatorial position of the trigonal bipyramid.
3d
43
44
Multiple Bonds
Double Bond
• One hybrid orbital is required for
• Consider ethylene
– each bond (whether a single or a multiple bond)
– and for each lone pair.
H
H
C
C
H
• Multiple bonding involves the overlap of
– one hybrid orbital
– and one or two nonhybridized ___ orbitals
H
• There is one double bond between carbons
– Must have one hybridized orbital and one
nonhybridized p orbital
• For a double bond
– Requires bonding via __ nonhybridized p orbital from each atom
• There are 4 single bonds with hydrogens
• For a triple bond
– Formed from hybridized orbitals in carbon
– Requires bonding via __ nonhybridized p orbitals from each atom
45
46
Orbital Diagram for C in Ethylene
Triple Bond
C—C bond between
hybridized orbitals (sp2)
• Consider acetylene
H
C
C
H
• There is one triple bond between carbons
– Must have one hybridized orbital and ___
nonhybridized p orbital
C—C bond between
nonhybridized p orbitals
• There are 2 single bonds with hydrogens
– Formed from hybridized orbitals in carbon
47
48
8
σ-Bonds
Orbital Diagram for C in Acetylene
• Multiple bonds may be identified as either of these
C—C bond between
hybridized orbitals (sp)
• A σ−bond (sigma)
– Has a ____________ shape about the bond axis.
– Formed when orbitals overlap along their _______
sp
• either when two s orbitals overlap or with directional orbitals (p or
hybrid)
2 C—C bonds between
nonhybridized p orbitals
49
50
π-Bonds
Ethylene and Acetylene
• A π−bond (pi)
– Has an electron distribution above and below the bond
axis.
– It is formed by the _____________ overlap of two
parallel p orbitals.
51
Ethylene
Acetylene
____ σ-bonds
____ σ-bonds
____ π-bond
___ π-bonds
52
O
Example 6: Bond Description
Example 6
C
H
• Describe the bonding about the C atom in
formaldehyde, CH2O, using valence bond
theory.
H
O
• After promotion, the orbital diagram is
C
H
1s
• The electron arrangement is ______________
using sp2 hybrid orbitals.
• The ground-state orbital diagram for C is
1s
53
2s
H
2s
2p
• After hybridization, the orbital diagram is
1s
2p
sp2
2p
54
9
O
Example 6
Molecular Orbital (MO) Theory
C
H
H
• There are _____ σ bonds arising from two C—H
bonds
• A second theory that explains electronic
structure of molecules
– For each bond one C sp2 hybrid orbital overlaps with the
1s orbital on the H atoms.
– Used to explain bonds between two atoms
• As atoms approach one another, their atomic
orbitals overlap and form molecular orbitals.
• A third ___ bond forms from the C—O bond
– Results from the overlap of one sp2 hybrid orbital and
one O half-filled p orbital.
• The electrons in a bond can be found either
– in a region between the nuclei
– or as far away from each other as possible
• One ___ bond forms for the second C—O bond
– from the sideways overlap of the C 2p orbital and an O
2p orbital.
55
56
Bonding and Antibonding Orbitals
Applying MO Theory
• Once a molecular orbital is formed, it can be
occupied in the same way as are atomic orbitals.
• If the electrons are
between nuclei,
– _______ molecular
orbitals form
– They are obtained
by adding 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
σ−molecular orbitals
– One is a bonding orbital (represented by σ1s)
– One is an antibonding orbital (represented by σ∗1s)
• If the electrons are in a region other than between nuclei
– _______________ orbitals form
– Result from subtraction of atomic orbitals.
57
58
Molecular Orbitals for H2
Molecular Orbitals for H2
• The bonding molecular orbital is lower in energy than the
antibonding molecular orbital.
• The two electrons are placed in the lower-energy
bonding molecular orbital with __________ spin.
• The ground-state molecular orbital electron configuration
is _________.
– Thus bonding MOs fill before antibonding MOs
• Each MO is able to hold ______ electrons
• We have a total of two electrons in the bond: one from
each hydrogen atom
59
– Excited states are also possible
60
10
Molecular Orbitals for He2
Bond Order
• Draw the MO diagram for He2
• The number of bonds that exist between two
atoms
• Each atom has 2 electrons, for a total of 4 electrons
He
atom
σ*1s
1s
– Can be calculated using the equation
He
atom
1s
– nb = number of electrons in bonding orbitals
– na = number of electrons in antibonding orbitals
σ1s
• Bond order also tells us about the stability of the
bond
• The energy decrease from the bonding electrons is offset
by the energy increase of antibonding electrons.
– If BO = ____, no stable bond forms
– Thus, He2 is not a stable molecule
• Electron configuration: ______________.
61
62
Bond Order Calculations
Conditions for Forming MO
• For H2 (two electrons),
• The strength of the interaction between two
atomic orbitals to form MOs is determined by
two factors:
– the molecular electron configuration is ______.
– Bond order = ½ (2 - 0) = ____.
– The energy difference between the interacting orbitals
– The magnitude of the overlap
• For H2+ (one electron),
– the molecular electron configuration is ______.
– Bond order = ½ (1 - 0) = ____.
• For the interaction to be strong, the energies of
the two orbitals must be approximately equal
and the overlap must be large
• For He2 (four electrons),
– the molecular electron configuration is __________.
– Bond order = ½(2 - 2) = ____.
• This means no bond forms.
63
64
Conditions for Forming MO - Example
Li2
• Consider the diatomic molecule Li2
• The two 1s atomic orbitals combine to form two
σ MOs:
– Each atom has an electron configuration of 1s2 2s1
– ____________.
• When the molecule forms, only like atomic
orbitals will interact
• The two 2s atomic orbitals combine to form two
σ MOs:
– The ___ orbitals will interact with each other
– The ___ orbitals will interact with each other
– A 1s orbital of one atom will ____ interact with the 2s
orbital of another
– _______________.
• Each atom has 3 electrons, thus
• The energies of each are different
65
– Six electrons occupy the orbitals in order of
increasing energy:
66
11
MOs of Be2
Li2 MO Diagram
• Determine the molecular electron configuration
of Be2
σ 1s
σ 1∗s
• Be2 has 8 electrons.
σ ∗2s
σ 2s
– The molecular orbital diagram is the same as for Li2.
– The eight electrons occupy the orbitals in order of increasing
energy:
The molecular electron configuration is
(σ1s)2(σ1s*)2(σ2s)2
or
σ 1s
________
σ 1∗s
σ ∗2 s
σ 2s
• The molecular electron configuration is
KK is the abbreviation for (σ1s)2(σ*1s)2 .
– (σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2
Bond order = _______________.
or
__________________
• The bond order is ___________________.
– No bond forms between Be atoms
67
68
MOs from p-Orbitals
MOs from p-Orbitals
• When atomic p orbitals interact two types of
molecular orbitals are formed
• When atomic p orbitals interact two types of
molecular orbitals are formed
– ________ π molecular orbitals form when they overlap
sideways
– ______ σ molecular orbitals form when they overlap
end-to-end
• π2p and π*2p
• Since there are two sets of p orbitals interacting, each will give
two π molecular orbitals for a total of ___ molecular orbitals
• σ2p and σ*2p.
69
70
Example 7: MO Diagram of O2
Energetic Diagram of Molecular Orbitals
• Relative energies of the
molecular orbitals formed
from second period atomic
orbitals.
• Give the orbital diagram and electron
configuration of the O2 molecule.
• Is the molecular substance diamagnetic or
paramagnetic?
• What is the order of the bond in this
molecule?
• The arrows are for N2.
• Because there are two
bonding π orbitals, they can
hold four electrons.
– They are filled one electron per
orbital
– Then a second electron is
added to the same orbital with
opposite spin.
71
72
12
Example 7: Electron Configuration of O2
Molecular Electron Configuration Shortcut
• Order of filling up MOs
• O2 has 16 electrons. The KK shell holds 4
electrons so 12 remain.
– σ1s σ*1s σ2s σ*2s π2p σ2p π*2p σ*2p
– Each MO holds 2 electrons
– Thus, each σ subshell holds ___ electrons
– And, each π subshell holds ___ electrons
The molecular electron configuration is
_________________________
The bond order is _______________.
• Give the electron configuration of the F2
molecule.
• Is the molecular substance diamagnetic or
paramagnetic?
The molecule is ________________.
73
74
σ*2p
Heteronuclear Diatomic Molecules
π*2p
• So far we have looked only at homonuclear
(same atom) diatomic nucleus.
• What happens when two different atoms
combine?
2p
Electron Configuration:
π2p
σ*2s
• Consider the molecule NO.
• Draw the Molecular Orbitals of this molecule
2s
2s
σ2s
σ*1s
N atom
(7 atoms)
75
2p
σ2p
76
Heteronuclear Diatomic Molecules
1s
O atom
1s
(8 atoms)
σ1s
Summary of Second Period Diatomic Molecules
• What if the heteronuclear
diatomic molecule forms from
two very different atoms?
• Consider HF
• One p orbital from F bonds with
the 1s orbital in H
• The other two p orbitals do not
bond.
• Difference in energies between
orbitals is due to the F electrons
being more tightly held.
77
78
13