Transcript Chapter 9

Chapter 9
Molecular Geometry
And Bonding Theories
Molecular Shape
• A bond angle is the
angle defined by lines
joining the centers of
two atoms to a third
atom to which they are
covalently bonded
• The molecular
geometry or shape is
defined by the lowest
energy arrangement of
its atoms in threedimensional space.
Valence-Shell Electron-Pair
Repulsion Theory (VSEPR)
The geometric arrangement of
atoms bonded to a given atom is
determined principally by
minimizing electron pair repulsions
between the bonds to, and lonepairs on the atom.
Determining Molecular Shape
Steric number (SN) = (# of atoms bonded to
central atom) + (# of lone pairs on the central
atom)
Each Molecular
Shape has
characteristic
bond angles
Geometric Forms
Predicting a VSEPR Structure
1. Draw Lewis structure.
2. Determine the steric number of the
central atom.
3. Use the SN to determine the geometry
around the central atom.
4. The name for molecular structure is
determined by the number of lone
pairs and bonding pairs of electrons.
Examples
• What is the molecular geometry of BF3?
• What is the molecular geometry of CH4?
Central Atoms with Lone Pairs
• Electron-pair geometry describes the
arrangement of atoms and lone pairs of
electrons about a central atom.
 The electron-pair geometry will always be one of
the five geometries presented previously.
• The molecular geometry in these molecules
describes the shape of the atoms present (it
excludes the lone pairs).
Lone Pairs
• Lone pairs of electrons occupy more space
around a central atom than do bonding
electrons.
• Lone pair-lone pair repulsion is the largest.
• Lone pair-bonding pair repulsion is the next
largest.
• Bonding pair-bonding pair repulsion is the
smallest.
• In structures with lone pairs on the central
atom, the bond angles are a little smaller than
predicted based on the electron-pair
geometry.
Electron “cloud” geometry versus Molecular Geometry
SN = 4, Electron-pair Geometry =
Tetrahedral
No. of Bonded
Atoms
No. of Lone
Pairs
Molecular
Geometry
Bond Angles
4
0
Tetrahedral
109.5o
3
1
Trigonal
Pyramidal
<109.5o
2
2
Bent
<109.5o
Lone Pairs of Electrons Affect
the Bond Angles
SN = 5, Electron-pair Geometry = Trigonal
Bipyramidal
No. of Bonded
Atoms
No. of Lone
Pairs
Molecular
Geometry
Bond Angles
5
0
Trigonal
Bipyramidal
120o & 90o
4
1
Seesaw
<120o & 90o
3
2
T-shaped
<120o & 90o
2
3
Linear
180o
The lone pairs of electrons are always found in the trigonal
planar part of the structure to minimize repulsion.
SN = 6, Electron-pair Geometry = Octahedral
No. of Bonded
Atoms
No. of Lone
Pairs
Molecular
Geometry
Bond Angles
6
0
Octahedral
90o
5
1
Square
Pyramidal
<90o
4
2
Square Planar
90o
3
3
2
4
Although these arrangements
are possible, we will not
generally encounter any
molecules with these
arrangements.
Practice
What are the molecular geometries of
the ions: SCN- and NO2- ?
Polar Bonds and Polar Molecules
• Two covalently bonded atoms with different
electronegativities have partial electric charges
of opposite sign creating a bond dipole.
• A molecule is called a polar molecule when it
has polar bonds and a shape where the bond
dipole vector do no cancel each other. Nonpolar molecules (with polar bonds) are typically
high symmetric, e.g. BF3, CS2, CCl4
Fig. 7.30: Polar bonds can lead to polar molecules (but not
necessarily). A molecule will have a permanent dipole moment ( =
q۰r) when it has an asymmetric orientation of polar bonds.
1D = 3.336E-30 C-m (dipole moments are
measured in Debye units)
Measuring Polarity
• The permanent dipole moment () is a
measured value that defines the extent of
separation of positive and negative charge
centers in a covalently bonded molecule
 = q۰r
Dipole Moment ()
The dipole moment in CCl4 is zero
The calculated dipole moment
of CH2Cl2 is 2.05 Debye.
-0.33
+0.33
Dipole Moment ()
 = Qr where Q is the partial charge and r is the distance
between partial charges q+ and q-.
The dipole
moment of H2O
is 1.85 Debye.
-0.33
+0.33
Dipole Moment () Problem
The dipole moment on HBr has
been measured to be 0.78 D. If the
H-Br bond distance is 141 pm,
calculate the partial charges on the
H and Br atoms relative to the
charge on a single electron.
H
Br
141 pm
Atomic Orbitals and Bonds
•
•
A tetrahedral electron-pair geometry
(SN=4) requires that four orbitals of the
central atom must overlap with an
orbital of an outer atom to form a bond.
The central atom would use its s orbital
and its three p orbitals, but these
orbitals would not yield the 109° bond
angles observed in the tetrahedral
molecule.
Valence-Bond Theory
• Valence-bond theory assumes that
covalent bonds form when atomic
orbitals on different atoms overlap or
occupy the same region of space.
• Each pair of electrons (SN) requires an
orbital. e.g. for SN=4, four orbitals are
needed.
• A sigma () bond is a covalent bond in
which the greatest electron density lies
between the two atoms forming the
bond.
Sigma () bond examples
Valence Bond Theory
• Hybridization is the mixing of atomic orbitals
(AOs) to generate new sets of orbitals that
are then available to overlap and form
covalent bonds (with the assumed geometry
& based upon SN) with other atoms or lone
pair electrons.
• A hybrid atomic orbital is one of a set of
equivalent orbitals about an atom created
when specific atomic orbitals are “mixed”.
The # of AOs need is equal to the steric
number (SN).
Tetrahedral Geometry: sp3 Hybrid Orbitals
A tetrahedral orientation of valence electrons is
achieved by forming four sp3 hybrid orbitals form one
s and three p atomic orbitals (SN=4…required 4
hybrid orbitals).
Other sp3 Hybrid Examples (SN=4)
sp2 Hybridization (SN=3)
• In a covalent pi () bond, electron density is
greatest above and below the bonding axis.
sp Hybridization
• Pi bonds will not exist between two atoms
unless a sigma bond forms first.
The Bonding in Carbon Dioxide (SN=2)
The carbon atom
is sp hybridized
and these orbitals
form the two
sigma bonds. The
 bonds are form
from pure-p-type
orbital on the
carbon and are
thus rotated 90°
from one another.
dsp3 Hybridization (SN=5)
d2sp3 Hybridization (SN=6)
Practice
What are the hybridizations of the
central atoms of the ions: XeF4
and Cl-I-Cl ?
Practice
What are the hybridizations of the
central atoms of the molecules: XeF4
and HXeSH ?
Problems with Bonding Theories
• Lewis structure and valence bond theory help
us understand the bonding capacities of
elements.
• VSEPR and valence bond theories account
for the observed molecular geometries.
• None of these models enables us to explain
other properties of molecules, e.g why O2 is
attracted to a magnetic field while N2 is
repelled slightly.
Molecular oxygen (but not N2) is attracted to
the poles of a magnet.
Molecular Orbital (MO) Theory
• The wave functions() of atomic orbitals
(AOs) of atoms are combined to create
molecular orbitals (MOs) in molecules.
 Each MO is associated with an entire
molecule, not just a single atom. MOs are
spread out, or delocalized, over all the
atoms in a molecule.
MOs for H2
• The two wavefunctions (1s orbitals) may be
added or subtracted to yield two MOs, a 1s
bonding orbital and a *1s anti-bonding MO.
Bonding and Anti-bonding MOs
• Electrons in bonding orbitals serve to hold
atoms together in molecules by
increasing the electron density between
nuclear centers.
A sigma (2s) orbital
Bonding and Anti-bonding MOs
• Electrons in anti-bonding orbitals
decrease electron density between the
atoms
• a sigma(2s)
• antibonding
Bond Types
• A sigma () bond is a covalent bond in
which the highest electron density lies
along the bond axis.
• A pi () bond is formed by the mixing of
atomic orbitals that are not oriented
along the bonding axis in a molecule.
MO Guidelines
1.
The total number of MO formed equals the number
of atomic orbitals used in the mixing process.
2.
Orbitals with similar energy and shape mix more
effectively than do those that are different.
3.
Orbitals of different principal quantum numbers
have different sizes and energies, resulting in less
effective mixing.
4.
A MO can accommodate two electrons with
opposite spin.
5.
Electrons are placed in MO diagrams according to
Hund’s rule.
Bond Order
Bond Order = 1/2 (# bonding electrons - # antibonding
electrons)
The bond order is zero in He2 and the molecule is not stable.
Combinations of p-type Atomic Orbitals to
form MOs for N2 and O2
p-type AOs can also combine to form sigma
type MOs that can be bonding or antibonding
(2pz)
p-type AOs can also combine to form type MOs that can be bonding or antibonding
(2py)
*(2py)
*(2pz)
MO Diagrams for N2 and O2
MO Scheme for N2
Figure 9.27
• Electron configuration for
N2: 2s22s*22p22p4
• Bond order = 1/2 (8 - 2) = 3
 N2 has three bonds
 N2 has no unpaired
electrons
MO Scheme for O2
• Electron configuration
for O2: 2s22s*22p22p4
2p*2
• Bond order = 1/2 (8 - 4)
=2
 O2 has two bonds
 O2 has two unpaired
electrons in 2p*
Relative Energies of Bonding and Antibonding
MO’s in Second Row Diatomics
1.
The bonding and antibonding 2s orbitals are more stable
than any of the six molecular orbitals derived from the 2p orbitals.
This is because the 2s orbitals that give rise to 2s and *2s are
more stable than the 2p atomic orbitals.
2.
The two 2p bonding orbitals have identical energies,
because the atomic orbitals from which they are constructed have
identical energies. Likewise, the two *2p orbitals have identical
energies.
3.
The antibonding orbitals formed from the atomic 2p orbitals
are the least stable molecular orbitals, with the *2p orbital less
stable than the * orbitals.
4.
The relative energies of the 2p and 2p molecular orbitals
varies depending on several factors including electron-electron
repulsion.
Paramagnetism and
Diamagnetism
• Paramagnetism - atoms or molecules
having unpaired electrons are attracted
to magnetic fields
• Diamagnetism - atoms or molecules
having all paired electrons are repelled
by magnetic fields
Other Diatomic Molecules
• The 2s and 2p interactions are strong in Li2
through N2 but weaker in O2 through Ne2.
Use MO theory predict the bond orders for each of the
following molecular ions: N2+, N2-, He2+, Br2+
List each species that is paramagnetic.
Other Diatomic Molecules
• The MO diagram
illustrates how the
effective nuclear charge
alters the diagram.
• The odd electron is
more likely to be found
on nitrogen since it is in
an orbital closer in
energy to the atomic
orbitals of the nitrogen
atom.
Delocalization of Electrons
• The electrons in the
-system with
alternating single
and double bonds
can be delocalized
over several atoms
or even an entire
molecule.
Comparison of Theories
• MO theory may provide the most
complete picture of covalent bonding, but
it is also the most difficult to apply to large
molecules. Using Valence Bond theory,
the general shape of larger molecules can
be quickly estimated.
»MOs in CH4
Assigning Hybridization (VBT)
Morphine
"I think I can safely say that nobody
understands quantum mechanics.“
Richard Feyman
ChemTour: Partial Charges and
Bond Dipoles
Click to launch animation
PC | Mac
Students learn that covalent bonds often include unequal
distribution of electrons leading to partial charges on atoms,
bond dipole moments, and molecule polarity. Interactive
Practice Exercises ask students to calculate dipole
moments of polar molecule.
ChemTour: Greenhouse Effect
Click to launch animation
PC | Mac
This unit explores how excess carbon dioxide and CFCs in
the atmosphere contribute to global warming.
ChemTour: Vibrational Modes
Click to launch animation
PC | Mac
This tutorial illustrates the three vibrational modes: bending,
symmetric stretching, and asymmetric stretching. Students
learn that molecules can absorb specific wavelengths of
infrared radiation by converting this energy into molecular
vibrations.
ChemTour: Hybridization
Click to launch animation
PC | Mac
This tutorial animates the formation of hybrid orbitals from
individual s and p orbitals, shows examples of their
geometry, and describes how they can produce single,
double, and triple bonds. Includes Practice Exercises.
ChemTour: Chemistry of the
Upper Atmosphere
Click to launch animation
PC | Mac
This ChemTour examines how particles of the upper
atmosphere absorb and emit electromagnetic radiation.
ChemTour: Molecular Orbitals
Click to launch animation
PC | Mac
This animated tutorial offers a patient explanation of
molecular orbital theory, an alternative to the bonding theory
depicted by Lewis dot structures. Includes Practice
Exercises.
Ethylene, which has the molecular formula C2H4, is a
rigid molecule in which all 6 atoms lie in a plane.
Which of the following molecules also has a rigid planar
structure?
A) H2C=C=CH2
Planar Hydrocarbons
B) H2C=C=C=CH2
C) Neither
Consider the following arguments for each answer and
vote again:
A. A combination of 3 carbons and 4 hydrogens can
form the rigid planar molecule H2C=C=CH2.
B. The orientations of the π bonds in H2C=C=C=CH2
alternate in such a way as to create a planar structure.
C. The hybridization of the atomic orbitals on the
carbons prevents the retention of a planar structure in
molecules longer than C2H4.
Planar Hydrocarbons
Which of the following depicts a
molecule that is different from the
one shown to the left?
A)
B)
Molecular Structures of Pentane
C)
Please consider the following arguments for each
answer and vote again:
A. This molecule, known as isopentane, has a unique
T-shaped arrangement.
B. In this molecule, only two carbon atoms are bonded
to the second carbon.
C. This molecule is unbranched, whereas the one
pictured in the question is not.
Molecular Structures of Pentane
What is the bond order of the N-O bond in nitrate, NO3-?
A) 1
Bond Order of Nitrate
B) 11/3
C) 2
Consider the following arguments for each answer and
vote again:
A. The majority of the bonds in NO3- are single bonds,
so the bond order is 1.
B. The N-O bond is twice as likely to be a single bond as
it is to be a double bond, so the bond order should be
11/3.
C. The bond order is dictated by the strongest bond,
which in NO3- is a double bond.
Bond Order of Nitrate
According to Valence Shell Electron Pair
Repulsion (VSEPR) theory, 4 objects
around a central atom will have the
tetrahedral arrangement shown to the left
with bond angles of ~109.5º. Which of the
following compounds has a bond angle of
~109.5º?
A) SF2
B) SF3-
Molecular Geometry of SF , SF -, and SF
C) SF4
Please consider the following arguments for each answer and
vote again:
A. SF2 consists of a sulfur atom surrounded by 2 lone electron
pairs and bonded to 2 fluorine atoms, therefore, it has an
approximately tetrahedral bond angle.
B. The tetrahedral VSEPR arrangement of SF3- is formed by a
sulfur atom surrounded by 3 fluorine atoms and by the
additional electron (from the negative charge).
C. Sulfur tetrafluoride is the only molecule with a central atom
(sulfur) surrounded by 4 additional atoms (4 fluorine atoms)
and so is the only molecule with a bond angle of ~109.5º.
Molecular Geometry of SF , SF -, and SF
Which of the following is true of the
bond angle (θ1) in BrF2+ compared to
the bond angle (θ2) in ICl2-?
A) θ1 = θ2
B) θ1 > θ2
Bond Angles of BrF and ICl
C) θ1 < θ2
Please consider the following arguments for each
answer and vote again:
A. Both BrF2+ and ICl2- consist of a central halogen
atom bonded to two halogen atoms, and therefore
should have the same arrangement of atoms.
B. ICl2- has 1 more lone pair of electrons than BrF2+,
which forces the chlorine atoms closer together.
C. ICl2-, with 3 lone pairs, is linear whereas BrF2+, with
2 lone pairs, is bent.
Bond Angles of BrF and ICl
Boron trifluoride (BF3), which has the
structure shown to the left, is capable of
reacting with an unknown compound to
form a new compound without breaking
any bonds. Which of the following
could be the unknown compound?
A) BF3
Reaction of Boron Trifluoride
B) CH4
C) NH3
Please consider the following arguments for each
answer and vote again:
A. BF3 can dimerize to BF3-BF3 by forming a boronboron single bond.
B. By forming a boron-carbon bond, the carbon atom in
CH4 will increase its steric number to 5, thus
expanding its octet to compensate for boron's
incomplete octet.
C. The nitrogen lone electron pair can form a nitrogenboron bond yielding BF3-NH3, isoelectronic with
CH3-CH3.
Reaction of Boron Trifluoride
Pictured to the left is the planar
molecule ethylene, C2H4, which does
not have a permanent electric dipole
moment.
If chlorine atoms were substituted for two hydrogen
atoms, how many of the possible structures would also
not possess a dipole moment?
A) 0
B) 1
Dipole Moments of Dichloroethylene
C) 2
Consider the following arguments for each answer
and vote again:
A. Chlorine atoms always draw electron density away
from carbon atoms, so all possible structures will
possess a dipole moment.
B. Only if the chlorine atoms are diagonally opposite
will the two carbon-chlorine dipole moments cancel
each other.
C. So long as the two chlorine atoms are on different
carbon atoms, no permanent dipole moment will
form.
Dipole Moments of Dichloroethylene
For which central atom "X" does the
anion pictured to the left have a square
planar geometry?
A) C
Molecular Geometry of XF
22
B) S
C) Xe
Please consider the following arguments for each
answer and vote again:
A. CF42- forms a structure in which the 4 fluorine
atoms form a square plane with one negative charge
on either side of the plane.
B. With 2 lone electron pairs on the sulfur in SF42-, its
steric number is 6.
C. To maximize fluorine-fluorine distances, the 4
fluorine atoms in XeF42- will lie in a plane.
Molecular Geometry of XF
2-