1. Chapter 10 Chemical Bonding II

2. 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

3. 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

4. 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

5. 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

6. 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

7. 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

8. 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

10. 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 •
there are three electron groups on N
three lone pair
one single bond
one double bond

11. 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
Tro: Chemistry: A Molecular Approach, 2/e

12. Linear
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°

13. Trigonal Planar
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°

14. Tetrahedral
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°

15. Trigonal Bipyramidal
5 electron groups around a central atom occupy positions in the shape of two tetrahedra base-to-base with the central atom in the center of the shared bases
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°

16. Octahedral
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°

17. 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

18. Not Quite Perfect Geometry
Because the bonds and atom sizes are not identical in formaldehyde, the observed angles are slightly different from ideal

19. 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

20. Effect of Lone Pairs
The nonbonding electrons are localized on the central atom, so area of negative charge takes more space
The bonding electrons are shared by two atoms, so some of the negative charge is removed from the central atom

21. Bond Angle Distortion from Lone Pairs

22. Bond Angle Distortion from Lone Pairs

23. 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

24. Pyramidal Geometry: A Derivative 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
Bond angle is less than 109.5°

25. Bent Tetrahedral Molecular Geometry: A Derivative of Tetrahedral Electron Geometry
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
the bond angle is less than 109.5°
Tro: Chemistry: A Molecular Approach, 2/e

26. 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
Seesaw Shape - five electron groups around the central atom, and one is a lone pair (aka distorted tetrahedron)
T-shaped – 5 electron groups around the central atom, and two are lone pairs
Linear shape – 5 electron groups around the central atom, and three are lone pairs
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

27. Replacing Atoms with Lone Pairs in the Trigonal Bipyramid System

28. Seesaw Shape

29. T–Shape

30. T–Shape

31. Linear Shape

32. 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
Square Pyramid Shape – 6 electron groups around the central atom, and one is a lone pair, the result is called a
the bond angles between axial and equatorial positions is less than 90°
Square Planar Shape – 6 electron groups around the central atom, and two are lone pairs, the result is called a
the bond angles between equatorial positions is 90°

33. Square Pyramidal Shape

34. Square Planar Shape

36. 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

37. 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

38. 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°

39. Practice – Predict the molecular geometry and bond angles in SiF5−

40. 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°

41. Practice – Predict the molecular geometry and bond angles in ClO2F

42. 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°

43. 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

45. S F
SF6

46. 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
shape around center C is trigonal planar
shape around right O is tetrahedral-bent

47. Describing the Geometry of Methanol

48. Describing the Geometry of Glycine

49. Practice – Predict the molecular geometries in H3BO3

50. Practice – Predict the molecular geometries in H3BO3
oxyacid, so H attached to O
3 electron groups on B
4 electron groups on O
B least electronegative
O has
2 bonding groups
2 lone pairs
B has
3 bonding groups
0 pone 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

51. Polarity of Molecules For a molecule to be polar it must
have polar bonds
electronegativity difference - theory
bond dipole moments - measured
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

52. 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.

53. Vector Addition

55. 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.

56. 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.

57. 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

58. 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

59. 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
ENN = 3.0
ENH = 2.1
3.0 − 2.1 = 0.9
therefore the bonds are polar covalent

60. 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
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.

61. Practice – Decide whether the following molecules are polarEN
O = 3.5
N = 3.0
Cl = 3.0
S = 2.5

62. 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
polar
nonpolar

63. 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

64. 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

65. 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

66. 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

67. Orbital Diagram for the Formation of H2S↑ H 1s ↑↓
H─S bond + S ↑ ↑↓ 3s 3p ↑ H 1s ↑↓
H─S bond
Predicts bond angle = 90°
Actual bond angle = 92°

68. 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

69. Unhybridized C Orbitals Predict the Wrong Bonding & Geometry

70. Valence Bond Theory Main Concepts
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.
A chemical bond results when these atomic orbitals interact and there is a total of two electrons in the new molecular orbital
the electrons must be spin paired
The shape of the molecule is determined by the geometry of the interacting orbitals

71. 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

72. 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

73. Carbon Hybridizations
Unhybridized    2s 2p
sp hybridized    
2sp 2p
sp2 hybridized    
2sp2 2p
sp3 hybridized    
2sp3

74. 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

76. Orbital Diagram of the sp3 Hybridization of C

77. 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
Unhybridized atom
sp3 hybridized atom C        2s 2p
2sp3 N         2s 2p
2sp3

78. Practice – Draw the orbital diagram for the sp3 hybridization of each atom
Unhybridized atom
sp3 hybridized atom Cl         3s 3p
3sp3 O         2s 2p
2sp3

79. 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

80. Methane Formation with sp3 C

81. Ammonia Formation with sp3 N

82. 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

84. 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

85. CH3NH2 Orbital Diagram s s sp3 C         sp3 N s s s s s  
1s H
1s H
1s H
1s H
1s H

86. Formaldehyde, CH2O Orbital Diagramp  
p C
p O s
sp2 C      
sp2 O s s  
1s H
1s H

87. 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

89. sp2 Hybridized Atoms Orbital Diagrams
Unhybridized atom
sp2 hybridized atom C
3 s
1 p        2s 2p
2sp2 2p N
2 s
1 p         2s 2p
2sp2 2p

90.      2s 2p 2sp2 2p         2s 2p 2sp2 2p B 3 s 0 p O
Practice – Draw the orbital diagram for the sp2 hybridization of each atom. How many s and p bonds would you expect each to form?
Unhybridized atom
sp2 hybridized atom B
3 s
0 p      2s 2p
2sp2 2p O
1 s
1 p         2s 2p
2sp2 2p

91. Hybrid orbitals overlap to form a s bond.
Unhybridized p orbitals overlap to form a p bond.

92. CH2NH Orbital Diagram ･ p   p C p N s sp2 C       sp2 N s s s
1s H
1s H
1s H

93. Bond Rotation
Orbitals form the s bond point along the internuclear axis
rotation around that bond does not require breaking the interaction between the orbitals
Orbitals that form the p bond interact above and below the internuclear axis
rotation around the axis requires the breaking of the interaction between the orbitals

96. 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

99. sp Hybridized Atoms Orbital Diagrams
Unhybridized atom
sp hybridized atom C 2s 2p        2s 2p
2sp 2p N 1s 2p         2s 2p
2sp 2p

100. HCN Orbital Diagram 2p    
p C
p N s
sp C    
sp N s 
1s H

101. sp3d 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

103. sp3d Hybridized Atoms Orbital Diagrams
Unhybridized atom
sp3d hybridized atom     P      3s 3p
3sp3d 3d     S      3s 3p 3d
3sp3d
(non-hybridizing d orbitals not shown)

104. SOF4 Orbital Diagram p   d S p O s sp3d S         sp2 O s
2p F
2p F
2p F
2p F

105. 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

107. sp3d2 Hybridized Atoms Orbital Diagrams
Unhybridized atom
sp3d2 hybridized atom ↑↓ ↑↓ ↑ ↑ S ↑ ↑ ↑ ↑ ↑ ↑ 3s 3p 3d
3sp3d2 5s 5p ↑↓ 5d ↑ I ↑↓ ↑ ↑ ↑ ↑ ↑
5sp3d2
(non-hybridizing d orbitals not shown)

109. 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

110. Example 10.7: Predict the hybridization and bonding scheme for CH3CHO
Draw the Lewis structure
Predict the electron group geometry around inside atoms
C1 = 4 electron areas
 C1= tetrahedral
C2 = 3 electron areas
 C2 = trigonal planar

111. 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

112. Example 10.7: Predict the hybridization and bonding scheme for CH3CHO
Label the bonds

113. Practice – Predict the hybridization of all the atoms in H3BO3
H = can’t hybridize
B = 3 electron groups = sp2
O = 4 electron groups = sp3

114. Practice – Predict the hybridization and bonding scheme of all the atoms in NClO• • • • • • • O N C l • • • •
s:Osp2─Nsp2 ↑↓ ↑↓ ↑↓
N = 3 electron groups = sp2
O = 3 electron groups = sp2
Cl = 4 electron groups = sp3 ↑↓
s:Nsp2─Clp ↑↓ ↑↓
p:Op─Np

115. 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

116. 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

117. 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

118. 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

119. Interaction of 1s Orbitals

120. 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

121. Energy Comparisons of Atomic Orbitals to Molecular Orbitals

122. 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

123. Because more electrons are in
Dihydrogen, H2
Molecular
Orbitals
Hydrogen
Atomic
Orbital
Hydrogen
Atomic
Orbital s* 1s 1s s
Because more electrons are in
bonding orbitals than are in antibonding orbitals,
net bonding interaction

124. H2
s bonding MO
HOMO
s* Antibonding MO
LUMO

125. Because there are as many electrons in
Dihelium, He2
Molecular
Orbitals
Helium
Atomic
Orbital
Helium
Atomic
Orbital s* 1s 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

126. therefore only need to consider valence shell
Dilithium, Li2
Molecular
Orbitals
Lithium
Atomic
Orbitals
Lithium
Atomic
Orbitals s* 2s 2s s
Any fill energy level will generate filled bonding and antibonding MO’s;
therefore only need to consider valence shell s*
BO = ½(4-2) = 1 1s 1s s
Because more electrons are in bonding orbitals than are in antibonding orbitals, there is a net bonding interaction

127. Li2
s* Antibonding MO
LUMO
s bonding MO
HOMO

128. Interaction of p Orbitals

129. Interaction of p Orbitals

131. O2 Dioxygen is paramagnetic
Paramagnetic material has unpaired electrons
Neither Lewis theory nor valence bond theory predict this result

132. O2 as Described by Lewis and VB Theory

133. s* p* 2p O2 MO’s 2p p s s* 2s 2s s BO = ½(8 be – 4 abe) BO = 2 Oxygen
Atomic
Orbitals
Oxygen
Atomic
Orbitals p* 2p
O2 MO’s 2p
Because more electrons are in bonding orbitals than are in antibonding orbitals, there is a net bonding interaction p s
Because there are unpaired electrons in the
antibonding orbitals,
O2 is predicted to be paramagnetic s*
BO = ½(8 be – 4 abe)
BO = 2 2s 2s s

134. 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
s2s
s*2s
p2p
s2p
p*2p
s*2p
N has 5 valence electrons
2 N = 10e−
(−) = 1e−
total = 11e− ↑ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓

135. 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
s2s
s*2s
p2p
s2p
p*2p
s*2p ↑ ↑ ↓ ↑ ↓ ↑ ↓
Because there are unpaired electrons, this ion is paramagnetic ↑ ↓ ↑ ↓

136. Practice – Draw a molecular orbital diagram of C2+ and predict its bond order and magnetic properties

137. Practice – Draw a molecular orbital diagram of C2+ and predict its bond order and magnetic properties
C has 4 valence electrons
2 C = 8e−
(+) = −1e−
total = 7e−
s2s
s*2s
p2p
s2p
p*2p
s*2p
BO = ½(5 be – 2 abe)
BO = 1.5 ↑ ↓ ↑
Because there are unpaired electrons, this ion is paramagnetic ↑ ↓ ↑ ↓

138. Heteronuclear Diatomic Molecules & Ions
If the combining atomic orbitals are identical and equal energy  the contribution of each atomic orbital to the molecular orbital is equal
If 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

139. 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

140. NO a free radical s2s Bonding MO
shows more electron density near O because it is mostly O’s 2s atomic orbital
BO = ½(6 be – 1 abe)
BO = 2.5

141. HF

142. 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

143. Ozone, O3
MO Theory: Delocalized p bonding orbital of O3