Transcript Chapter 10 - Suffolk County Community College

Chemistry: A Molecular Approach, 2nd Ed.
Nivaldo Tro
Chapter 10
Chemical
Bonding II
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
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Taste
• The taste of a food depends on the interaction
between the food molecules and taste cells on
your tongue
• The main factors that affect this interaction are
the shape of the molecule and charge
distribution within the molecule
• The food molecule must fit snugly into the
active site of specialized proteins on the
surface of taste cells
• When this happens, changes in the protein
structure cause a nerve signal to transmit
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Sugar & Artificial Sweeteners
• Sugar molecules fit into the active site of taste cell
•
•
•
receptors called Tlr3 receptor proteins
When the sugar molecule (the key) enters the
active site (the lock), the different subunits of the
T1r3 protein split apart
This split causes ion channels in the cell
membrane to open, resulting in nerve signal
transmission
Artificial sweeteners also fit into the Tlr3 receptor,
sometimes binding to it even stronger than sugar
 making them “sweeter” than sugar
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Structure Determines Properties!
• Properties of molecular substances depend on
•
the structure of the molecule
The structure includes many factors, such as:
the skeletal arrangement of the atoms
the kind of bonding between the atoms
ionic, polar covalent, or covalent
the shape of the molecule
• Bonding theory should allow you to predict the
shapes of molecules
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Molecular Geometry
• Molecules are 3-dimensional objects
• We often describe the shape of a molecule
•
•
with terms that relate to geometric figures
These geometric figures have characteristic
“corners” that indicate the positions of the
surrounding atoms around a central atom in
the center of the geometric figure
The geometric figures also have
characteristic angles that we call bond
angles
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Lewis Theory Predicts
Electron Groups
• Lewis theory predicts there are regions of
•
•
electrons in an atom
Some regions result from placing shared pairs
of valence electrons between bonding nuclei
Other regions result from placing unshared
valence electrons on a single nuclei
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Using Lewis Theory to Predict
Molecular Shapes
• Lewis theory says that these regions of
electron groups should repel each other
 because they are regions of negative charge
• This idea can then be extended to predict the
shapes of molecules
 the position of atoms surrounding a central atom will be
determined by where the bonding electron groups are
 the positions of the electron groups will be determined
by trying to minimize repulsions between them
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VSEPR Theory
• Electron groups around the central atom will
be most stable when they are as far apart as
possible – we call this valence shell electron
pair repulsion theory
because electrons are negatively charged, they
should be most stable when they are separated as
much as possible
• The resulting geometric arrangement will allow
us to predict the shapes and bond angles in
the molecule
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Electron Groups
• The Lewis structure predicts the number of valence
•
•
electron pairs around the central atom(s)
Each lone pair of electrons constitutes one electron
group on a central atom
Each bond constitutes one electron group on a
central atom
 regardless of whether it is single, double, or triple
••
•O
•
••
N
••
O ••
••
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there are three electron groups on N
three lone pair
one single bond
one double bond
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Electron Group Geometry
• There are five basic arrangements of electron
groups around a central atom
 based on a maximum of six bonding electron groups
 though there may be more than six on very large atoms, it is
very rare
• Each of these five basic arrangements results in
five different basic electron geometries
 in order for the molecular shape and bond angles to be
a “perfect” geometric figure, all the electron groups
must be bonds and all the bonds must be equivalent
• For molecules that exhibit resonance, it doesn’t
matter which resonance form you use – the
electron geometry will be the same
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Linear Electron Geometry
• When there are two electron groups around the
•
•
central atom, they will occupy positions on
opposite sides of the central atom
This results in the electron groups taking a linear
geometry
The bond angle is 180°
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Linear Geometry
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Trigonal Planar Electron Geometry
• When there are three electron groups around
•
•
the central atom, they will occupy positions in
the shape of a triangle around the central atom
This results in the electron groups taking a
trigonal planar geometry
The bond angle is 120°
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Trigonal Geometry
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Tetrahedral Electron Geometry
• When there are four electron groups around the
•
•
central atom, they will occupy positions in the
shape of a tetrahedron around the central atom
This results in the electron groups taking a
tetrahedral geometry
The bond angle is 109.5°
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Tetrahedral Geometry
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Trigonal Bipyramidal Electron Geometry
• When there are five electron groups around the central
•
•
•
•
•
atom, they will occupy positions in the shape of two
tetrahedra that are base-to-base with the central atom in the
center of the shared bases
This results in the electron groups taking a trigonal
bipyramidal geometry
The positions above and below the central atom are called
the axial positions
The positions in the same base plane as the central atom
are called the equatorial positions
The bond angle between equatorial positions is 120°
The bond angle between axial and equatorial positions is 90°
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Trigonal Bipyramid
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Trigonal Bipyramidal Geometry
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Octahedral Electron Geometry
• When there are six electron groups around the central
atom, they will occupy positions in the shape of two
square-base pyramids that are base-to-base with the
central atom in the center of the shared bases
• This results in the electron groups taking an octahedral
geometry
 it is called octahedral because the geometric figure has
eight sides
• All positions are equivalent
• The bond angle is 90°
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Octahedral Geometry
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Octahedral Geometry
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Molecular Geometry
• The actual geometry of the molecule may be
different from the electron geometry
• When the electron groups are attached to
atoms of different size, or when the bonding to
one atom is different than the bonding to
another, this will affect the molecular geometry
around the central atom
• Lone pairs also affect the molecular geometry
they occupy space on the central atom, but are not
“seen” as points on the molecular geometry
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Not Quite Perfect Geometry
Because the bonds and
atom sizes are not
identical in formaldehyde,
the observed angles are
slightly different from ideal
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The Effect of Lone Pairs
• Lone pair groups “occupy more space” on the
central atom
because their electron density is exclusively on the
central atom rather than shared like bonding electron
groups
• Relative sizes of repulsive force interactions is
Lone Pair – Lone Pair > Lone Pair – Bonding Pair > Bonding Pair – Bonding Pair
• This affects the bond angles, making the bonding
pair – bonding pair angles smaller than expected
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Effect of Lone Pairs
bonding electrons
are are
shared
by twoon
The nonbonding
electrons
localized
atoms,
so some
negative
charge
is
the
central
atom,ofsothe
area
of negative
charge
removed
from
the central atom
takes
more
space
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Bond Angle Distortion
from Lone Pairs
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Bond Angle Distortion
from Lone Pairs
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Bent Molecular Geometry:
Derivative of Trigonal Planar Electron Geometry
• When there are three electron groups around
•
the central atom, and one of them is a lone pair,
the resulting shape of the molecule is called a
trigonal planar — bent shape
The bond angle is less than 120°
 because the lone pair takes up more space
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Pyramidal & Bent Molecular Geometries:
Derivatives of Tetrahedral Electron Geometry
• When there are four electron groups around the
central atom, and one is a lone pair, the result is
called a pyramidal shape
 because it is a triangular-base pyramid with the
central atom at the apex
• When there are four electron groups around the
central atom, and two are lone pairs, the result is
called a tetrahedral—bent shape
 it is planar
 it looks similar to the trigonal planar—bent shape,
except the angles are smaller
• For both shapes, the bond angle is less than
109.5°
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Methane
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Pyramidal Shape
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Pyramidal Shape
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Tetrahedral–Bent Shape
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Tetrahedral–Bent Shape
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Derivatives of the
Trigonal Bipyramidal Electron Geometry
• When there are five electron groups around the central atom,
•
and some are lone pairs, they will occupy the equatorial
positions because there is more room
When there are five electron groups around the central atom,
and one is a lone pair, the result is called the seesaw shape
 aka distorted tetrahedron
• When there are five electron groups around the central atom,
•
•
•
and two are lone pairs, the result is called the T-shaped
When there are five electron groups around the central atom,
and three are lone pairs, the result is a linear shape
The bond angles between equatorial positions are less than
120°
The bond angles between axial and equatorial positions are
less than 90°
 linear = 180° axial–to–axial
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Replacing Atoms with Lone Pairs
in the Trigonal Bipyramid System
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Seesaw Shape
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T–Shape
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T–Shape
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Linear Shape
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Derivatives of the
Octahedral Geometry
• When there are six electron groups around the
•
central atom, and some are lone pairs, each even
number lone pair will take a position opposite the
previous lone pair
When there are six electron groups around the
central atom, and one is a lone pair, the result is
called a square pyramid shape
 the bond angles between axial and equatorial positions is
less than 90°
• When there are six electron groups around the
central atom, and two are lone pairs, the result is
called a square planar shape
 the bond angles between equatorial positions is 90°
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Square Pyramidal Shape
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Square Planar Shape
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Predicting the Shapes
Around Central Atoms
1. Draw the Lewis structure
2. Determine the number of electron groups
around the central atom
3. Classify each electron group as bonding or
lone pair, and count each type
 remember, multiple bonds count as one group
4. Use Table 10.1 to determine the shape and
bond angles
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Example 10.2: Predict the geometry and bond
angles of PCl3
1. Draw the Lewis
structure
a) 26 valence electrons
2. Determine the
Number of electron
groups around
central atom
a) four electron groups
around P
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Example 10.2: Predict the geometry and bond
angles of PCl3
3. Classify the electron groups
a) three bonding groups
b) one lone pair
4. Use Table 10.1 to determine the
shape and bond angles
a) four electron groups around P =
tetrahedral electron geometry
b) three bonding + one lone pair =
trigonal pyramidal molecular
geometry
c) trigonal pyramidal = bond angles
less than 109.5°
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Practice – Predict the molecular geometry
and bond angles in SiF5−
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Practice – Predict the molecular geometry
and bond angles in SiF5─
Si least electronegative
5 electron groups on Si
Si is central atom
5 bonding groups
0 lone pairs
Si = 4e─
F5 = 5(7e─) = 35e─
(─) = 1e─
total = 40e─
Shape = trigonal bipyramid
Bond angles
Feq–Si–Feq = 120°
Feq–Si–Fax = 90°
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Practice – Predict the molecular geometry
and bond angles in ClO2F
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Practice – Predict the molecular geometry
and bond angles in ClO2F
Cl least electronegative
4 electron groups on Cl
Cl is central atom
3 bonding groups
1 lone pair
Cl = 7e─
O2 = 2(6e─) = 12e─
F = 7e─
Total = 26e─
Shape = trigonal pyramidal
Bond angles
O–Cl–O < 109.5°
O–Cl–F < 109.5°
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Representing 3-Dimensional Shapes
on a 2-Dimensional Surface
• One of the problems with drawing molecules is
•
•
•
•
trying to show their dimensionality
By convention, the central atom is put in the plane
of the paper
Put as many other atoms as possible in the same
plane and indicate with a straight line
For atoms in front of the plane, use a solid wedge
For atoms behind the plane, use a hashed wedge
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SF6
F
F
F
S
F
F
F
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Multiple Central Atoms
• Many molecules have larger structures with many
•
•
interior atoms
We can think of them as having multiple central
atoms
When this occurs, we describe the shape around
each central atom in sequence
 
shape around left C is tetrahedral
H
O


|
||
 
shape around center C is trigonal planar H  C  C  O  H
shape around right O is tetrahedral-bent
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|
 
H
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Describing the Geometry
of Methanol
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Describing the Geometry
of Glycine
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Practice – Predict the molecular geometries
in H3BO3
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Practice – Predict the molecular geometries
in H3BO3
oxyacid, so H attached to O
34 electron
electron groups
groups on
on B
O
B least electronegative
O has
B
has
3 bonding groups
2
0 lone
2
ponepairs
pairs
B Is Central Atom
B = 3e─
O3 = 3(6e─) = 18e─
H3 = 3(1e─) = 3e─
Total = 24e─
Shape on B = trigonal planar
Shape on O = tetrahedral bent
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Polarity of Molecules
• For a molecule to be polar it must
1. have polar bonds


electronegativity difference - theory
bond dipole moments - measured
2. have an unsymmetrical shape

vector addition
• Polarity affects the intermolecular forces of
attraction
 therefore boiling points and solubilities

like dissolves like
• Nonbonding pairs affect molecular polarity,
strong pull in its direction
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Molecule Polarity
The H─Cl bond is polar. The bonding
electrons are pulled toward the Cl end of
the molecule. The net result is a polar
molecule.
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Vector Addition
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Molecule Polarity
The O─C bond is polar. The bonding
electrons are pulled equally toward both
O ends of the molecule. The net result is
a nonpolar molecule.
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Molecule Polarity
The H─O bond is polar. Both sets of
bonding electrons are pulled toward the
O end of the molecule. The net result
is a polar molecule.
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Predicting Polarity of Molecules
1. Draw the Lewis structure and determine the
molecular geometry
2. Determine whether the bonds in the molecule
are polar
a) if there are not polar bonds, the molecule is
nonpolar
3. Determine whether the polar bonds add
together to give a net dipole moment
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Example 10.5: Predict whether NH3 is a
polar molecule
1. Draw the Lewis
structure and
determine the
molecular geometry
a) eight valence electrons
b) three bonding + one lone
pair = trigonal pyramidal
molecular geometry
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Example 10.5: Predict whether NH3 is a
polar molecule
2. Determine if the bonds
are polar
a) electronegativity
difference
b) if the bonds are not polar,
we can stop here and
declare the molecule will
be nonpolar
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ENN = 3.0
ENH = 2.1
3.0 − 2.1 = 0.9
therefore the
bonds are
polar covalent
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Example 10.5: Predict whether NH3 is a
polar molecule
3) Determine whether
the polar bonds add
together to give a
net dipole moment
a) vector addition
b) generally, asymmetric
shapes result in
uncompensated
polarities and a net
dipole moment
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The H─N bond is polar. All
the sets of bonding
electrons are pulled toward
the N end of the molecule.
The net result is a polar
molecule.
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Practice – Decide whether the following
molecules are polar
EN
O = 3.5
N = 3.0
Cl = 3.0
S = 2.5
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Practice – Decide whether the following
molecules Are polar
Trigonal
Bent
Trigonal
Planar
2.5
1. polar bonds, N-O
2. asymmetrical shape
1. polar bonds, all S-O
2. symmetrical shape
nonpolar
polar
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Molecular Polarity
Affects
Solubility in Water
• Polar molecules are attracted
•
to other polar molecules
Because water is a polar
molecule, other polar
molecules dissolve well in
water
and ionic compounds as well
• Some molecules have both
polar and nonpolar parts
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Problems with Lewis Theory
• Lewis theory generally predicts trends in
properties, but does not give good numerical
predictions
 e.g. bond strength and bond length
• Lewis theory gives good first approximations of
•
•
the bond angles in molecules, but usually cannot
be used to get the actual angle
Lewis theory cannot write one correct structure for
many molecules where resonance is important
Lewis theory often does not predict the correct
magnetic behavior of molecules
 e.g. O2 is paramagnetic, though the Lewis structure
predicts it is diamagnetic
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Valence Bond Theory
• Linus Pauling and others applied the principles
•
•
of quantum mechanics to molecules
They reasoned that bonds between atoms
would occur when the orbitals on those atoms
interacted to make a bond
The kind of interaction depends on whether the
orbitals align along the axis between the
nuclei, or outside the axis
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Orbital Interaction
• As two atoms approached, the half-filled
valence atomic orbitals on each atom would
interact to form molecular orbitals
molecular orbtials are regions of high probability of
finding the shared electrons in the molecule
• The molecular orbitals would be more stable
than the separate atomic orbitals because they
would contain paired electrons shared by both
atoms
the potential energy is lowered when the molecular
orbitals contain a total of two paired electrons
compared to separate one electron atomic orbitals
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Orbital Diagram for the
Formation of H2S
H
1s
↑
+ ↑↓
1s
↑
3s
↑
↑ ↑↓ S
3p
↑↓
H─S bond
↑↓
H─S bond
H
Predicts bond angle = 90°
Actual bond angle = 92°
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Valence Bond Theory – Hybridization
• One of the issues that arises is that the number
of partially filled or empty atomic orbitals did
not predict the number of bonds or orientation
of bonds
C = 2s22px12py12pz0 would predict two or three
bonds that are 90° apart, rather than four bonds
that are 109.5° apart
• To adjust for these inconsistencies, it was
postulated that the valence atomic orbitals
could hybridize before bonding took place
one hybridization of C is to mix all the 2s and 2p
orbitals to get four orbitals that point at the corners
of a tetrahedron
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Unhybridized C Orbitals Predict the
Wrong Bonding & Geometry
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Valence Bond Theory
Main Concepts
1. The valence electrons of the atoms in a
molecule reside in quantum-mechanical atomic
orbitals. The orbitals can be the standard s, p,
d, and f orbitals, or they may be hybrid
combinations of these.
2. A chemical bond results when these atomic
orbitals interact and there is a total of two
electrons in the new molecular orbital
a) the electrons must be spin paired
3. The shape of the molecule is determined by the
geometry of the interacting orbitals
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Hybridization
• Some atoms hybridize their orbitals to
•
maximize bonding
• more bonds = more full orbitals = more stability
Hybridizing is mixing different types of orbitals
in the valence shell to make a new set of
degenerate orbitals
sp, sp2, sp3, sp3d, sp3d2
• Same type of atom can have different types of
hybridization
C = sp, sp2, sp3
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Hybrid Orbitals
• The number of standard atomic orbitals
combined = the number of hybrid orbitals formed
combining a 2s with a 2p gives two 2sp hybrid
orbitals
H cannot hybridize!!
its valence shell only has one orbital
• The number and type of standard atomic orbitals
•
combined determines the shape of the hybrid
orbitals
The particular kind of hybridization that occurs is
the one that yields the lowest overall energy for
the molecule
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Carbon Hybridizations
Unhybridized



2p
2s
sp hybridized



2sp

2p
sp2 hybridized

 
2sp2
sp3 hybridized




2p

2sp3
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sp3 Hybridization
• Atom with four electron groups around it
tetrahedral geometry
109.5° angles between hybrid orbitals
• Atom uses hybrid orbitals for all bonds and
lone pairs
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Orbital Diagram of the
sp3 Hybridization of C
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sp3 Hybridized Atoms
Orbital Diagrams
• Place electrons into hybrid and unhybridized valence orbitals
as if all the orbitals have equal energy
• Lone pairs generally occupy hybrid orbitals
sp3 hybridized atom
Unhybridized atom
2s

2s
 
2p
C

  
2p
N

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  
2sp3


  
2sp3
Copyright  2011 Pearson Education, Inc.
Practice – Draw the orbital diagram for
the sp3 hybridization of each atom
Unhybridized atom
3s

2s
  
3p
Cl
   
3sp3
  
2p
O
   
2sp3
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

sp3 hybridized atom
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Bonding with Valence Bond Theory
• According to valence bond theory, bonding
takes place between atoms when their atomic
or hybrid orbitals interact
“overlap”
• To interact, the orbitals must either be aligned
•
along the axis between the atoms, or
The orbitals must be parallel to each other and
perpendicular to the interatomic axis
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Methane Formation with sp3 C
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Ammonia Formation with sp3 N
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Types of Bonds
• A sigma (s) bond results when the interacting
atomic orbitals point along the axis connecting the
two bonding nuclei
 either standard atomic orbitals or hybrids
 s–to–s, p–to–p, hybrid–to–hybrid, s–to–hybrid, etc.
• A pi (p) bond results when the bonding atomic
orbitals are parallel to each other and perpendicular
to the axis connecting the two bonding nuclei
 between unhybridized parallel p orbitals
• The interaction between parallel orbitals is not as
strong as between orbitals that point at each other;
therefore s bonds are stronger than p bonds
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Orbital Diagrams of Bonding
• “Overlap” between a hybrid orbital on one atom
•
with a hybrid or nonhybridized orbital on
another atom results in a s bond
“Overlap” between unhybridized p orbitals on
bonded atoms results in a p bond
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CH3NH2 Orbital Diagram
H
s
·· s
s
H
C s N
H
s
s



s
s


1s H 1s H
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
s


s

C
s
1s H
98


sp3 N
s

sp3
H

H
1s H 1s H
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Formaldehyde, CH2O Orbital Diagram

pC


s

s
s

C

sp2
p

pO

 
sp2 O
1s H 1s H
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sp2
• Atom with three electron
groups around it
 trigonal planar system
 C = trigonal planar
 N = trigonal bent
 O = “linear”
 120° bond angles
 flat
• Atom uses hybrid orbitals for
s bonds and lone pairs, uses
nonhybridized p orbital for p
bond
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sp2 Hybridized Atoms
Orbital Diagrams

2s

2s
 
2p
  
2p
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sp2 hybridized atom
C
3s
1p

N
2s
1p

102

Unhybridized atom
 
2sp2

2p
 
2sp2

2p
Copyright  2011 Pearson Education, Inc.
Practice – Draw the orbital diagram for
the sp2 hybridization of each atom. How many s
and p bonds would you expect each to form?

2s

2s

2p
  
2p
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sp2 hybridized atom
B
3s
0p

 
2sp2
2p
O
1s
1p
  
2sp2

2p
103

Unhybridized atom
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Hybrid orbitals
overlap to form a
s bond.
Unhybridized p
orbitals overlap
to form a p bond.
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CH2NH Orbital Diagram

s

s

s

C
C
H
pN

H


sp2 N
s
N
H

1s H 1s H
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


pC
sp2
p
1s H
105
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Bond Rotation
• Because the orbitals that form the s bond point
•
along the internuclear axis, rotation around
that bond does not require breaking the
interaction between the orbitals
But the orbitals that form the p bond interact
above and below the internuclear axis, so
rotation around the axis requires the breaking
of the interaction between the orbitals
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sp
• Atom with two electron groups
linear shape
180° bond angle
• Atom uses hybrid orbitals for s bonds or lone
pairs, uses nonhybridized p orbitals for p bonds
p
s
p
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sp Hybridized Atoms
Orbital Diagrams

2s

2s
 
2p
  
2p
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C
2s
2p
N
1s
2p
112
sp hybridized atom
 
2sp
 
2p

2sp
 
2p

Unhybridized atom

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HCN Orbital Diagram

pC
sp C


2p
s



pN


sp N
s

1s H
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3
sp d
• Atom with five electron groups
around it
trigonal bipyramid electron
geometry
Seesaw, T–Shape, Linear
120° & 90° bond angles
• Use empty d orbitals from
•
valence shell
d orbitals can be used to
make p bonds
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sp3d Hybridized Atoms
Orbital Diagrams
sp3d hybridized atom
Unhybridized atom

3s

3s
  
3p
  
3p
P
3d

  
3sp3d
S  
3d
 


3sp3d
(non-hybridizing d orbitals not shown)
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SOF4 Orbital Diagram

dS
S 



s
s
s
s



sp3d
p
s


pO

 
sp2 O

2p F 2p F 2p F 2p F
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sp3d2
• Atom with six electron groups
around it
octahedral electron geometry
Square Pyramid, Square Planar
90° bond angles
• Use empty d orbitals from
•
valence shell to form hybrid
d orbitals can be used to make
p bonds
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sp3d2 Hybridized Atoms
Orbital Diagrams
sp3d2 hybridized atom
Unhybridized atom
↑↓
↑↓ ↑ ↑
3s
3p
↑↓
↑↓ ↑↓ ↑
5s
5p
S
↑ ↑ ↑ ↑ ↑ ↑
3sp3d2
3d
I
↑↓ ↑ ↑ ↑ ↑ ↑
5sp3d2
5d
(non-hybridizing d orbitals not shown)
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Predicting Hybridization and
Bonding Scheme
1. Start by drawing the Lewis structure
2. Use VSEPR Theory to predict the electron group
geometry around each central atom
3. Use Table 10.3 to select the hybridization
scheme that matches the electron group
geometry
4. Sketch the atomic and hybrid orbitals on the
atoms in the molecule, showing overlap of the
appropriate orbitals
5. Label the bonds as s or p
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Example 10.7: Predict the hybridization
and bonding scheme for CH3CHO
Draw the Lewis structure
Predict the electron group C1 = 4 electron areas
geometry around inside
 C1= tetrahedral
atoms
C2 = 3 electron areas
 C2 = trigonal planar
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Example 10.7: Predict the hybridization
and bonding scheme for CH3CHO
Determine the
hybridization of the
interior atoms
C1 = tetrahedral
 C1 = sp3
C2 = trigonal planar
 C2 = sp2
Sketch the molecule and
orbitals
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Example 10.7: Predict the hybridization
and bonding scheme for CH3CHO
Label the bonds
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Practice – Predict the hybridization of all
the atoms in H3BO3
H = can’t hybridize
B = 3 electron groups = sp2
O = 4 electron groups = sp3
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Practice – Predict the hybridization and
bonding scheme of all the atoms in NClO
••
•O
•
••
N
••
Cl ••
••
s:Osp2─Nsp2
↑↓
sp2
N = 3 electron groups =
O = 3 electron groups = sp2
Cl = 4 electron groups = sp3
O ↑↓ N
↑↓
↑↓
N
p:Op─Np
127
s:Nsp2─Clp
Cl
↑↓
O
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↑↓
Cl
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Problems with Valence Bond Theory
• VB theory predicts many properties better than
Lewis theory
bonding schemes, bond strengths, bond lengths,
bond rigidity
• However, there are still many properties of
molecules it doesn’t predict perfectly
magnetic behavior of O2
• In addition, VB theory presumes the electrons
are localized in orbitals on the atoms in the
molecule – it doesn’t account for delocalization
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Molecular Orbital Theory
• In MO theory, we apply Schrödinger’s wave
equation to the molecule to calculate a set of
molecular orbitals
 in practice, the equation solution is estimated
 we start with good guesses from our experience as to
what the orbital should look like
 then test and tweak the estimate until the energy of the
orbital is minimized
• In this treatment, the electrons belong to the
whole molecule – so the orbitals belong to the
whole molecule
 delocalization
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LCAO
• The simplest guess starts with the atomic
orbitals of the atoms adding together to make
molecular orbitals – this is called the Linear
Combination of Atomic Orbitals method
weighted sum
• Because the orbitals are wave functions, the
waves can combine either constructively or
destructively
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Molecular Orbitals
• When the wave functions combine constructively,
the resulting molecular orbital has less energy
than the original atomic orbitals – it is called a
Bonding Molecular Orbital
 s, p
 most of the electron density between the nuclei
• When the wave functions combine destructively,
the resulting molecular orbital has more energy
than the original atomic orbitals – it is called an
Antibonding Molecular Orbital
 s*, p*
 most of the electron density outside the nuclei
 nodes between nuclei
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Interaction of 1s Orbitals
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Molecular Orbital Theory
• Electrons in bonding MOs are stabilizing
lower energy than the atomic orbitals
• Electrons in antibonding MOs are
destabilizing
higher in energy than atomic orbitals
electron density located outside the
internuclear axis
electrons in antibonding orbitals cancel
stability gained by electrons in bonding orbitals
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Energy Comparisons of Atomic Orbitals
to Molecular Orbitals
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MO and Properties
• Bond Order = difference between number of
electrons in bonding and antibonding orbitals
 only need to consider valence electrons
 may be a fraction
 higher bond order = stronger and shorter bonds
 if bond order = 0, then bond is unstable compared to
individual atoms and no bond will form
• A substance will be paramagnetic if its MO
diagram has unpaired electrons
 if all electrons paired it is diamagnetic
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Hydrogen
Atomic
Orbital
Dihydrogen, H2
Molecular
Orbitals
s*
1s
Hydrogen
Atomic
Orbital
1s
s
Because more electrons are in
bonding orbitals than are in antibonding orbitals,
net bonding interaction
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H2
s* Antibonding MO
LUMO
s bonding MO
HOMO
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Helium
Atomic
Orbital
Dihelium, He2
Molecular
Orbitals
s*
1s
Helium
Atomic
Orbital
1s
s
BO = ½(2-2) = 0
Because there are as many electrons in
antibonding orbitals as in bonding orbitals,
there is no net bonding interaction
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Lithium
Atomic
Orbitals
Dilithium, Li2
Molecular
Orbitals
s*
2s
2s
s
s*
BO = ½(4-2) = 1
1s
Any fill energy level will
generate filled bonding
and antibonding MO’s;
therefore only need to
consider valence shell
1s
s
Because more electrons are
in bonding orbitals than are in
antibonding orbitals, there is a
net bonding interaction
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Lithium
Atomic
Orbitals
139
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Li2
s* Antibonding MO
s bonding MO
LUMO
HOMO
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Interaction of p Orbitals
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Interaction of p Orbitals
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O2
• Dioxygen is paramagnetic
• Paramagnetic material has unpaired electrons
• Neither Lewis theory nor valence bond theory
predict this result
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O2 as Described by Lewis and
VB Theory
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Oxygen
Atomic
Orbitals
2p
s*
p*
Oxygen
Atomic
Orbitals
O2 MO’s
2p
p
Because more electrons
are in bonding orbitals
than are in antibonding
orbitals, there is a net
bonding interaction
s
s*
BO = ½(8 be – 4 abe)
BO = 2
2s
Because there are unpaired
electrons in the
antibonding orbitals,
O2 is predicted to be
paramagnetic
2s
s
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Example 10.10: Draw a molecular orbital diagram
of N2− ion and predict its bond order and magnetic
properties
Write a MO
diagram for N2−
using N2 as a
base
Count the
number of
valence
electrons and
assign these to
the MOs
following the
aufbau
principle, Pauli
principle &
Hund’s rule
s*2p
N has 5 valence
electrons
2 N = 10e−
(−) = 1e−
total = 11e−
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p*2p
↑
↑↓
↑↓ ↑ ↓
s2p
p2p
↑↓ s*2s
↑↓
147
s2s
Copyright  2011 Pearson Education, Inc.
Example 10.10: Draw a molecular orbital diagram
of N2− ion and predict its bond order and magnetic
properties
Calculate the
bond order by
taking the
number of
bonding
electrons and
subtracting the
number of
antibonding
electrons, then
dividing by 2
Determine
whether the ion
is paramagnetic
or diamagnetic
BO = ½(8 be – 3 abe)
BO = 2.5
Because this is lower
than the bond order
in N2, the bond
should be weaker
Because there
are unpaired
electrons, this ion
is paramagnetic
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s*2p
p*2p
↑
↑↓
↑↓ ↑ ↓
p2p
↑↓ s*2s
↑↓
148
s2p
s2s
Copyright  2011 Pearson Education, Inc.
Practice – Draw a molecular orbital diagram of C2+
and predict its bond order and magnetic properties
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Practice – Draw a molecular orbital diagram of C2+
and predict its bond order and magnetic properties
C has 4 valence
electrons
s*2p
2 C = 8e−
(+) = −1e−
total = 7e−
p*2p
BO = ½(5 be – 2 abe)
BO = 1.5
Because there
are unpaired
electrons, this ion
is paramagnetic
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s2p
↑↓ ↑
p2p
↑↓ s*2s
↑↓
s2s
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Heteronuclear Diatomic Molecules & Ions
• When the combining atomic orbitals are
•
identical and equal energy, the contribution of
each atomic orbital to the molecular orbital is
equal
When the combining atomic orbitals are
different types and energies, the atomic
orbital closest in energy to the molecular
orbital contributes more to the molecular
orbital
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Heteronuclear Diatomic Molecules & Ions
• The more electronegative an atom is, the
•
•
•
lower in energy are its orbitals
Lower energy atomic orbitals contribute
more to the bonding MOs
Higher energy atomic orbitals contribute
more to the antibonding MOs
Nonbonding MOs remain localized on the
atom donating its atomic orbitals
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NO
a free radical
s2s Bonding MO
BO = ½(6 be – 1 abe)
BO = 2.5
shows more
electron density
near O because
it is mostly O’s
2s atomic orbital
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HF
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Polyatomic Molecules
• When many atoms are combined together, the
•
atomic orbitals of all the atoms are combined to
make a set of molecular orbitals, which are
delocalized over the entire molecule
Gives results that better match real molecule
properties than either Lewis or valence bond
theories
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Ozone, O3
MO Theory:
Delocalized
p bonding
orbital of O3
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