CHEMISTRY 161
Chapter 10
Chemical Bonding & Molecular Structure
1
PREDICTING THE GEOMETRY
OF MOLECULES
1. derive Lewis structure of the molecule
2. discriminate between bonding and nonbonding electron pairs
H
O
H
3. VALENCE SHELL ELECTRON PAIR REPULSION
2
VALENCE SHELL ELECTRON PAIR REPULSION
VSEPR
1. identify in a compound the central atom
2. electrons repel each other
3. valence electron pairs stay as far apart as possible
4. non-bonding electrons repel more than bonding electrons
3
central atom
no non-bonding pairs
non-bonding pairs
4
AB2
BeCl2
Cl
Be
Cl
TWO ELECTRON PAIRS AROUND BERYLLIUM ATOM
5
Cl
Be
Cl
180°
270°
Be
Be
90°
180°
LINEAR ARRANGEMENT BEST
IT PUTS ELECTRON PAIRS FURTHEST APART
6
AB3
BF3
F
F
B
F
THREE ELECTRON PAIRS AROUND THE
BORON ATOM
7
F
F
B
F
THREE ELECTRON PAIRS AROUND THE
BORON ATOM
TRIGONAL PLANAR ARRANGEMENT BEST
120°
120°
B
120°
8
MOLECULAR SHAPE
F
B
F
F
THE SHAPE OF BF3 IS TRIGONAL PLANAR.
9
AB4
CH4
H
H
C
H
H
four electron pairs
expect square planar
90°
90°
C
90°
90°
10
better arrangement for four electron pairs
TETRAHEDRAL
109.5°
bigger than 90 ° in
square planar
C
4 electron pairs
tetrahedral
put on the H-atoms
11
TETRAHEDRAL
H
109.5°
C
C
H
H
H
shape of CH4 is tetrahedral
12
AB5
PF5
FIVE ELECTRON PAIRS AROUND PHOSPHORUS
F
F
P
P
F
5 electron pairs
F
F
trigonal bipyramidal
13
Bond angle
F
900
P
F
F
P
F
F
1200
shape of PF5 is trigonal bipyramidal
two of the F atoms different from the others
14
Bond angle
AXIAL
F
900
F
EQUATORIAL
F
P
F
F
1200
15
AB6
SF6
six electron pairs around the sulfur atom
F
F
S
F
S
F
F
F
6 electron pairs
octahedral
16
900
F
F
S
F
S
F
F
F
shape of SF6 is octahedral
17
central atom
no non-bonding pairs
non-bonding pairs
18
AB2E
SeO2
AB3
O
Se
O
19
O
Se
O
VSEPR treats double
bonds like a single bond
THREE ELECTRON PAIRS AROUND SELENIUM
ELECTRON PAIR GEOMETRY
Se
TRIGONAL PLANAR
20
Se
ADD OXYGENS
Se
O
O
SeO2 IS V-SHAPED (OR BENT)
THE MOLECULAR SHAPE IS THE POSITION OF
THE ATOMS
21
AB3E
NH3
AB4
H
N
H
H
electron pairs around the nitrogen atom
22
H
NH3
N
H
H
N
PUT ON THE 3 H ATOMS
N
H
H
NH3 is trigonal pyramidal
H
23
AB2E2
H
O
H
AB4
four electron pairs around the oxygen atom
PUT ON THE
2 H-ATOMS
O
O
H
H
shape of H2O is V-shaped or bent
24
AB4E
SF4
AB5
F
F
F
S
S
F
TRIGONAL BIPYRAMID
25
WHERE DOES LONE PAIR GO?
F
F
S
F
F
OR
F
F
S
F
F
lone pairs occupy the trigonal plane (the “equator”)
to minimize the number of 90° repulsions
26
AB4E
AB3E2
AB2E3
SF4
ClF3
XeF2
1 lone pair
2 lone pairs
3 lone pairs
F
F
F
Cl
Xe
F
S
F
F
F
F
F
F
See-saw
T-shaped
Linear
shaped
lone pairs occupy the trigonal plane (the “equator”)
first to minimize the number of 90° repulsions
27
AB5E
AB6
BrF5
F
F
F
F
Br
Br
F
Square pyramidal
28
AB6
AB4E2
XeF4
F
F
Xe
F
:
F
F
F
Xe
:
Xe
F
F
F
lone pairs MUST BE AT
1800
F
F
Xe
F
29
Summary of Molecular Shapes
Total valence
electron pairs
Electron Pair
Geometry
2
Linear
0
0
3
Trigonal
planar
4
Tetrahedral
Lone electron Shape of Molecule
pairs
Linear
Trigonal planar
1
V-shaped
0
Tetrahedral
1
Trigonal pyramid
2
V-shaped
30
Total valence
electron pairs
Electron Pair
Geometry
Lone electron Shape of Molecule
pairs
0
5
6
Trigonal
bipyramidal
Octahedral
Trig. bipyramid.
1
See-saw
2
T-shaped
3
Linear
0
Octahedral
1
Square pyramid
2
Square planar
31
POLYATOMICS
molecules with no single central atom
we apply our VSEPR rules to each atom in the chain
Example: ETHANOL
32
ETHANOL
C2H5OH
H
H
H
C
C
H
H
O
H
The atoms around the carbons form a.
tetrahedral arrangement
The atoms around the oxygen form a
V-shaped structure.
33
C
C
H
H
H
34
EXAMPLES
Cl2O
SO2Cl2
NH2OH
BF4-
ICl4Cl2CO
NH4+
Cl2SO
N2F2
35
1. Lewis structures
2. VSEPR model
WHY DO MOLECULES FORM?
36
simplest molecule
H2
two H-atoms
1s1
two H-atoms approach each other and the
electron waves interact
OVERLAP to form a region of increased
electron density between the atoms
37
38
39
chemical bond with electron density in between the
nuclei is called
 bond
40
VALENCE BOND THEORY
a covalent bond is formed by an
overlap of two valence atomic
orbitals that share an electron pair
the better the overlap the stronger the bond
the orbitals need to point along the bonds
41
H
CH4
What orbitals are used?
C
H
H
H
hydrogen atoms bond using their 1s orbitals
carbon needs four orbitals to bond with.
[He] 2s22p2
2s, 2px , 2py, 2pz
42
1. The electronic configuration of carbon is
[He] 2s22p2
The orbital diagram is:
43
1. The electronic configuration of carbon is
[He] 2s22p2
the orbital diagram is: [He]
the Lewis dot structure is
.
.C .
.
necessary to promote one 2s electron
44
PROMOTE AN ELECTRON
[He]
[He]
[He] 2s22p2
[He] 2s12p3
excited state (valence state)
Lewis dot structure
C
four unpaired electrons
we can use these to form chemical bonds
45
1. a covalent bond is formed by an overlap of two
valence atomic orbitals that share an electron pair
2. bonds formed with s orbitals will be different
to bonds formed with p orbitals
Experiment shows that all four bonds are identical
3. three p orbitals are mutually perpendicular,
suggesting 90° bond angles
experiment shows that methane has 109.5° bond
angles
combining the orbitals
46
we need four orbitals pointing to the vertices of
a tetrahedron
orbitals are just
mathematical functions
H
we can combine them
C
H
H
H
HYBRIDIZATION
47
COMBINING ORBITALS TO
FORM HYBRIDS
HYBRIDIZATION
number of atomic orbitals that are combined
IS EQUAL TO
the number of resulting hybrid orbitals
48
HYBRIDIZATION
Combine one s and one p
a sp- hybrid
+
+
ADD the orbitals
2s+ 2p
49
HYBRIDIZATION
Combine one s and one p
s+p
a sp- hybrid
2s+ 2p
+
+
What
do we
get?
The positive part cancels negative part
DESTRUCTIVE INTERFERENCE
The positive part adds to positive part
50
CONSTRUCTIVE INTERFERENCE
HYBRIDIZATION
Combine one s and one p to give
a sp- hybrid
2s+ 2p
s+p
Where is the nucleus?
+
REMEMBER IF WE MIX TWO WE MUST GET TWO BACK
The other combination is s - p
51
HYBRIDIZATION
Combine one s and one p
a sp- hybrid
+
+
2s- 2p
SUBTRACT the
orbitals
52
HYBRIDIZATION
Combine one s and one p
a sp- hybrid
SUBTRACTING THE p ORBITAL CHANGES ITS
PHASE
+
+
2s- 2p
SUBTRACT the
orbitals
53
HYBRIDIZATION
Combine one s and one p
a sp- hybrid
SUBTRACTING THE p ORBITAL CHANGES ITS
PHASE
+
+
2s- 2p
SUBTRACT the
orbitals
54
HYBRIDIZATION
Combine one s and one p
s-p
a sp- hybrid
2s- 2p
+
+
What
do we
get?
The positive part cancels negative part
DESTRUCTIVE INTERFERENCE
The positive part adds to positive part
55
CONSTRUCTIVE INTERFERENCE
HYBRIDIZATION
Combine one s and one p
a sp- hybrid
s-p
2s- 2p
Where is the nucleus?
+
The positive part cancels negative part
We get two equivalent sp orbitals
ORIENTED AT 1800
56
sp-HYBRIDIZATION
s and p orbitals
two sp-hybrids
57
COMBINE one s-orbital and two p-orbitals
Get three sp2 - orbitals oriented at 1200
s and p orbitals
three sp2-hybrids
directed at 1200
58
COMBINE one s-orbital and three p-orbitals
three sp3- orbitals oriented at 109.50
59
H
METHANE: CH4
four hybrid orbitals
needed to form four
C
bonds
H
H
s + px + py + pz
4 sp3 hybrids
H
an atom with sp3 hybrid orbitals is said to be
sp3 hybridized
The four sp3 hybrid orbitals form a tetrahedral
arrangement.
EPG of 4 pairs
sp3 hybridization
60
What happens to the energies of the orbitals?
What happens to orbital energies when the are
hybridized??
HYBRIDIZE
2p
E
2s
Orbitals in
free C atom
61
When orbitals are hybridized they have the
same energy:
HYBRIDIZE
2p
E
2s
Orbitals in
free C atom
E
sp3
Hybridized orbitals of
C atom in methane
The FOUR sp3 hybrids are DEGENERATE.
62
z
z
y
y
x
x
z Combine one s and three p orbitals…..
z
y
x
y
x
63
sp3 HYBRIDS
sp3 orbitals
C
Now form the bonds to the H-atoms……...
64
Each bond in methane results from the overlap of a
hydrogen 1s orbital and a carbon sp3 orbital.
Hydrogen
1s orbital
Carbon sp3
orbitals
H
Form a chemical bond by
sharing a pair of
electrons.
C
H
H
H
Each hybrid ready to overlap with H 1s orbitals
65
VALENCE BOND MODEL
Hybrid orbital model
Step 1:
Draw the Lewis structure(s)
Step 2:
Determine the geometry of the electron
pairs around each atom using VSEPR OR
preferably use the EXPERIMENTAL
GEOMETRY
Step 3:
Specify the hybrid orbitals needed to
accommodate the electron pairs on each
atom
66
OTHER MOLECULES USING sp3 HYBRIDS
sp3 hybrids are also employed in …...
all molecules that have a 4 pair EPG….
NH3, H2O, NH4+ , CCl4
AMMONIA…..
67
AMMONIA: NH3
VSEPR
N
H
H
Valence shell has four pairs
EPG is TETRAHEDRAL
H
Need sp3 hybrids
Nitrogen electronic configuration
N
[He]
2s
2p
HYBRIDIZE
68
When orbitals are hybridized they have the
same energy:
HYBRIDIZE
2p
E
2s
Orbitals in
free N atom
E
sp3
Hybridized orbitals of
N atom in ammonia
The FOUR sp3 hybrids are DEGENERATE.
sp3 hybridization…….
69
sp3 hybrids on N in AMMONIA
2s
2p
.
.N .
..
N
Now form a bond
Overlap H 1s…..
H
H
H
70
AMMONIA
2s
2p
.
.N .
..
Three  bonds
One lone pair in
an sp3 hybrid
N
H
H
H
71
AMMONIUM ION NH4
2s
+
+
H
2p
.
.N .
..
N
four  bonds.
ISOELECTRONIC WITH ?
H
H
H
CH4
72
WATER
H
O
H
FOUR PAIRS
2s
.
.O:
..
2p
EPG?
TETRAHEDRAL!!
O
O
HYBRIDIZATION?
sp3
73
WATER.
Overlap of two of
oxygen sp3 hybrids
with …..
H atom 1s orbitals.
To form
two 
bonds.
Lone pairs in two
of the sp3 hybrids.
O
H
H
Think about H3O+ !!!
74
HYDRONIUM ION.
Overlap of Oxygen
sp3 hybrids
containing a lone
pair
H+ ion empty
1s orbitals.
ISOELECTRONIC
WITH?
NH3
O
H+
H
H
75
QUESTION
Which of the following molecules is uses sp3
hybrids in the valence bond description of its
bonding?
1
C and D
A
CO2
B
C
D
E
NF3
O3
NO2
F2O
+
2
B and E
3
A and D
4
B and C
5
B and A
ANSWER…….
76
QUESTION
Which of the following molecules is uses sp3
hybrids in the valence bond description of its
bonding?
C
O
O
1
C and D
A
CO2
B
C
D
E
F
NF3
O3
NO2
F2O
N
F
O O
+
[O
F
N
O
F
O
+
O]
F
2
B and E
3
A and D
4
B and C
5
B and A
WHAT ABOUT OTHER EPG’S …….
77
VALENCE BOND THEORY FOR OTHER
ELECTRON PAIR GEOMETRIES
A four electron pair EPG uses sp3 hybrids
The three electron pair EPG uses sp2 hybrids
The two electron pair EPG uses sp hybrids
78
EPG’s
2
3
180°
120°
120°
X
X
180°
4
109.5°
X
120°
HYBRIDS
sp
sp2
sp3
lets look at a molecule that needs sp2
79
H
H
Ethylene:
VSEPR
C2H4
H
trigonal planar EPG
around each C-atom.
C
C
H
a HCH angle of 1200.
three hybrid orbitals on each carbon for the trigonal
planar EPG.
s + px + py
3 sp2 hybrids
The CARBON is sp2 hybridized
The 3 sp2 hybrid orbitals form a trigonal planar arrangement.
3 effective electron pairs
sp2 hybridization
80
FORMATION OF sp2 hybrids
VALENCE
STATE C atom
2p
E
2s
GROUND
STATE C atom
81
FORMATION OF sp2 hybrids
VALENCE
STATE C atom
2p
E
2s
GROUND
STATE C atom
2p
E
2s
HYBRIDIZE
82
FORMATION OF sp2 hybrids
VALENCE
STATE C atom
2p
E
2p
E
2s
GROUND
STATE C atom
2s
HYBRIDIZE
sp2 hybridized orbitals of C
E
2p
sp2
This leaves one p orbital unhybrized…….
83
An sp2 hydridized C atom
sp2 - hybrid orbital
z
UNHYBRIDIZED
p- orbital
y
x
The unhybridized p
orbital is
perpendicular to
sp2 plane.
Lets put it all together…….
84
H
H
DRAW TWO C-ATOMS
C
C
H
H
z
C
z
y
y
x
Now put the orbitals on…...
C
x
85
H
H
BONDING IN ETHYLENE
C
C
H
H
z
C
x
z
y
y
C
x
86
C
 bond
z
C
x
H
H
BONDING IN ETHYLENE
C
H
H
z
y
y
C
x
OVERLAP the sp2 hybrids from the two carbons
to form a sigma bond between them.
PUT THE ELECTRONS IN AND…..
87
overlap two sp2 hybrids on each carbon with hydrogen
H
H
1s orbitals to form sigma bonds and...
C
C
z
z
H
H
H
H
y
H
y
H
x
x
The two unhybridized p orbitals are left over to form…
88
The second part of the carbon-carbon double bond !
H
H
z
z
C
H
H
H
H
y
H
C
y
H
x
x
The two unhybridized p orbitals are left over to form a …..
pi bond (p bond)
89
The second part of the carbon-carbon double bond !
pi bond (p bond)
z
z
H
H
C
H
H
H
H
y
H
C
y
H
x
x
Electrons are shared between the unhybridized p
orbitals in an area above and below the line between
90
nuclei.
THE COMPLETE PICTURE!!!!!!!
C
pi bond (p bond)
z
z
H
H
H
H
H
H
y
H
C
y
H
x
x
sp2
SUMMARY...
sigma bonds ( bond)
91
p:C(2p)-C(2p)
H
:
H(1s)-C(sp2)
H
:C(sp2)-C(sp2)
H
C
: H(1s)-C(sp2)
C
H
 bonding
Now look at p bond
92
p:C(2p)-C(2p)
H
:
H(1s)-C(sp2)
H
:C(sp2)-C(sp2)
H
C
: H(1s)-C(sp2)
C
H
p bonding
93
Now look at ethyne (acetylene)
BONDING SCHEME IN
ETHYNE
p:C(2p)-C(2p) TWO OF THESE!!
: H(1s)-C(sp)
: H(1s)-C(sp)
H
C
C
H
:C(sp)-C(sp)
What does this look like????
94
DRAW TWO C-ATOMS
H
z
C
C
C
H
z
y
y
C
x
Now put the sp-orbitals on…...
x
95
OVERLAP the sp hybrids from the two carbons to
form a sigma bond between them.
z
C
x
z
y
y
C
x
Put in the unhybridized p orbitals
96
OVERLAP the sp hybrids from the two carbons to
form a sigma bond between them.
z
C
x
z
y
y
C
x
OVERLAP the hydrogen 1s orbitals
97
OVERLAP the C sp hybrids with H 1s to form sigma bonds
z
H
C
x
z
y
y
C
H
x
98
OVERLAP the sp hybrids from the two carbons to
form a sigma bond between them.
z
H
C
z
y
y
C
x
sigma framework of  bonds
H
x
pi bonding?
99
LATERAL OVERLAP of p orbitals to form pi bonds.
z
H
C
z
y
y
C
x
H
x
two pi bonds (p bonds)
100
LATERAL OVERLAP of p orbitals to form pi bonds.
z
H
C
z
y
y
C
x
H
x
SO…..
two pi bonds (p bonds)
101
SUMMARY
Single bond:
One  bond
Double bond:
One  bond,
Triple bond:
One  bond,
one p bond
two p bonds
102
VALENCE BOND THEORY
Step 1
Step 2
Lewis Dot Structure
Get Molecular Geometry
VSEPR
Step 3
Step 4
EXPERIMENTAL
Choose hybrids
Describe bonds…...
103
What about molecules with more than an octet
around the central atom?
Examples: PCl5, or SF4 or SiF62Four pairs needs
Four orbitals
Five pairs needs
Five orbitals
six pairs needs
six orbitals
104
PCl5
Cl
Cl
Cl
chlorine atoms
P
Cl
We ignore the
Cl
and just describe
central atom.
Need five hybrid orbitals on the phosphorus
to fit the trigonal bipyramidal EPG.
d + s + px + py + pz
5 dsp3 hybrids
5 effective electron pairs
dsp3 hybridization
Five equivalent orbitals……..
105
dsp3 - hybrid orbitals
TRIGONAL BIPYRAMID
EPG 5 PAIRS
z
900
y
1200
x
SIX PAIRS…..
overlap with orbitals on chlorine to form 5  bonds.
106
F
SF6
F
F
chlorine atoms
S
F
F
F
We ignore the
and just describe
central atom.
We need six hybrid orbitals on the sulfur to
allow for the octahedral EPG and six bonds.
d + d + s + px + py + pz
6 effective electron pairs
6 d2sp3 hybrids
d2sp3 hybridization
SIX equivalent orbitals……..
107
d2sp3 - hybrid orbitals
z
900
y
900
x
EXAMPLE
overlap with orbitals on flourine to form 6  bonds.
108
EXAMPLES
Describe the molecular structure and bonding in XeF2 and XeF4
F
Xe
F
EPG 5 pairs
Linear
dsp3 hybrids
Two axial 
bonds at 1800
Three lone pairs
in equatorial
hybrids
109
EXAMPLES
Describe the molecular structure and bonding in XeF2 and XeF4
F
Xe
F
F
Xe
F
EPG 5 pairs
Linear
dsp3 hybrids
Two axial 
bonds at 1800
Three lone pairs
in equatorial
hybrids
F
F
EPG 6 pairs
Square planar
d2sp3 hybrids
four  bonds at 900
in a plane
Two lone pairs
in axial hybrids
110
MOLECULAR ORBITAL THEORY
electrons occupy orbitals each of which spans
the entire molecule
molecular orbitals each hold up to two electrons
and obey Hund’s rule, just like atomic orbitals
111
H2 molecule:
1s orbital on Atom A
1s orbital on Atom B
the H2 molecule’s molecular orbitals can be
constructed from the two 1s atomic orbitals
1sA + 1sB = MO1
constructive interference
1sA – 1sB = MO2
destructive interference
112
 1s
1
 R( r )  2 
 a0 
3/2
r/a
e
0
113
ADDITION OF ORBITALS
builds up electron density in overlap region
1sA + 1sB = MO1
A
B
combine them by addition
114
ADDITION OF ORBITALS
builds up electron density in overlap region.
1sA + 1sB = MO1
A
B
what do we notice?
electron density between atoms
115
SUBTRACTION OF ORBITALS
results in low electron density in overlap region..
1sA – 1sB = MO2
A
B
subtract
116
SUBTRACTION OF ORBITALS
results in low electron density in overlap region..
1sA – 1sB = MO2
A
B
what do we notice?
no electron density between atoms
117
COMBINATION OF ORBITALS
1sA + 1sB = MO1
builds up electron density between nuclei
118
COMBINATION OF ORBITALS
1sA – 1sB = MO2 ANTI-BONDING
results in low electron density between nuclei
1sA + 1sB = MO1
BONDING
builds up electron density between nuclei
119
120
THE MO’s FORMED BY TWO
1s ORBITALS
121
1sA – 1sB = MO2
1s*
sigma anti-bonding = 1s*
1s
1sA + 1sB = MO1
sigma bonding = 1s
122
COMBINING TWO 1s ORBITALS
E
Energy of a
1s orbital in
a free atom
A
B
Energy of
a 1s orbital
in a free
atom
123
E
Energy of a
1s orbital in
a free atom
A
B
1s
Energy of
a 1s orbital
in a free
atom
1sA+1sB
MO
124
1sA-1sB
MO
E
Energy of a
1s orbital in
a free atom
A
1s*
1s
B
Energy of
a 1s orbital
in a free
atom
1sA+1sB
MO
125
COMBINING TWO 1s ORBITALS
1s*
E
1sA
A
B
1sB
1s
126
bonding in H2
H
H2
H
1s*
E
1s
1s
1s
127
H
H2
H
1s*
E
1s
1s
1s
the electrons are placed in the 1s molecular orbitals
128
H
H2
H
1s*
E
1s
1s
1s
H2: (1s)2
129
He2
atomic configuration of He
He
He2
1s2
He
1s*
E
1s
1s
1s
130
He2: (1s)2(1s*)2
He
He2
He
1s*
E
1s
1s
1s
bonding effect of the (1s)2 is cancelled by the
antibonding effect of (1s*)2
131
BOND ORDER
net number of bonds existing after the
cancellation of bonds by antibonds
He2
the electronic configuration is….
(1s)2(1s*)2
the two bonding electrons were cancelled out
by the two antibonding electrons
BOND ORDER = 0
132
BOND ORDER
measure of bond strength and molecular stability
If # of bonding electrons > # of antibonding electrons
the molecule is predicted to be stable
Bond
order
=
133
BOND ORDER
measure of bond strength and molecular stability
If # of bonding electrons > # of antibonding electrons
the molecule is predicted to be stable
Bond
order
# of bonding
–
electrons(nb)
= 1/2{
= 1/2 (n
b
# of antibonding
electrons (na)
}
- na)
high bond order indicates high bond
energy and short bond length
H2+,H2,He2+
134
H2
H2+
He2+
He2
1s*
E
1s
Magnetism
Bond order
Bond energy
(kJ/mol)
Bond length
(pm)
135
H2
H2+
He2+
He2
1s*
E
1s
Magnetism
Dia-
Bond order
1
Bond energy
(kJ/mol)
436
Bond length
(pm)
74
136
H2
H2+
Magnetism
Dia-
Para-
Bond order
1
½
Bond energy
(kJ/mol)
436
225
Bond length
(pm)
74
106
He2+
He2
1s*
E
1s
137
H2
H2+
He2+
Magnetism
Dia-
Para-
Para-
Bond order
1
½
½
Bond energy
(kJ/mol)
436
225
251
Bond length
(pm)
74
106
108
He2
1s*
E
1s
138
First row diatomic molecules and ions
H2
H2+
He2+
He2
Magnetism
Dia-
Para-
Para-
—
Bond order
1
½
½
0
Bond energy
(kJ/mol)
436
225
251
—
Bond length
(pm)
74
106
108
—
1s*
E
1s
139
second period
HOMONUCLEAR DIATOMICS
Li2
Li : 1s22s1
both the 1s and 2s overlap to produce 
bonding and anti-bonding orbitals
140
ENERGY LEVEL DIAGRAM FOR DILITHIUM
2s*
Li2
2s
2s
2s
E
1s*
1s
1s
1s
141
ELECTRONS FOR DILITHIUM
2s*
2s
Li2
2s
2s
E
1s*
1s
1s
1s
142
Electron configuration for DILITHIUM
2s*
Li2
(1s)2(1s*)2(2s)2
2s
2s
2s
E
Bond Order ?
1s
1s
1s
143
Electron configuration for DILITHIUM
2s*
Li2
(1s)2(1s*)2(2s)2
2s
2s
nb = 4
2s
E
1s
Bond Order = 1
1s
1s
na = 2
single bond.
144
Electron configuration for DILITHIUM
2s*
Li2
(1s)2(1s*)2(2s)2
2s
2s
2s
E
1s
the 1s and 1s* orbitals
can be ignored when
both are FILLED!
1s
1s
omit the inner shell
145
Li2
only valence orbitals contribute to
molecular bonding
(2s)2
Li
Li2
Li
2s*
E
2s
2s
2s
The complete configuration is: (1s)2(1s*)2 (2s)2
146
Be
Be2
Be
2s*
Be2
E
2s
2s
2s
147
Electron configuration for DIBERYLLIUM
Be2
Be
Be2
Be
2s*
E
2s
2s
2s
Configuration:
(2s)2(2s*)2
Bond order = 0
148
Electron configuration for DIBERYLLIUM
Be2
Be
Be2
Be
(2s)2(2s*)2
2s*
nb = 2
E
2s
2s
na = 2
2s
Bond Order = 1/2(nb - na) = 1/2(2 - 2) =0
No bond!!! The molecule is not stable!
Now B2...
149
B2
the Boron atomic configuration is
1s22s22p1
we expect B to use 2p orbitals to
form molecular orbitals
addition and subtraction
150
-molecular orbitals
151
p molecular orbitals
152
ENERGY LEVEL DIAGRAM
E
2s*
2s
2s
2s
153
2p*
p2p*
2p
E
2p
p2p
2p
154
expected orbital splitting
2p*
p2p*
2p
p2p
2p
2p
E
2s*
2s
2s
2s
This pushes the 2p up
155
MODIFIED ENERGY LEVEL DIAGRAM
2p*
p2p*
2p
E
2p
p2p
2s*
2s
2p
Notice that the 2p and p2p
have changed places!!!!
2s
2s
156
Electron configuration for B2
2p*
B is [He] 2s22p1
p2p*
2p
E
2p
p2p
2s*
2s
2p
Place electrons from 2s
into 2s and 2s*
2s
2s
157
2p*
p2p*
2p
E
2p
p2p
2s*
2s
2p
Place electrons from 2p
into p2p and p2p
2s
2s
Remember HUND’s RULE
158
ELECTRONS
ARE
UNPAIRED
2p*
Abbreviated configuration
p *
2p
(2s)2(2s*)2(p2p)2
2p
E
2p
p2p
2s*
2s
2p
Complete configuration
(1s)2(1s*)2(2s)2(2s*)2(p2p)2
2s
2s
159
Electron configuration for B2:
2p*
p2p*
(2s)2(2s*)2(p2p)2
na = 2
2p
E
2p
p2p
2s*
2s
2s
nb = 4
2p
Bond order 1/2(nb - na)
= 1/2(4 - 2) =1
2s Molecule is
predicted to be
stable and
160
paramagnetic.
A SUMMARY OF THE MO’s
Emphasizing nodal planes
161
ELECTRONIC CONFIGURATION OF THE
HOMONUCLEAR DIATOMICS
Li2
B2
C2
N2
O2
F2
162
Li2
E
2p
B2
C2
O2
N2
2p*
2p*
p2p*
2p
p2p
F2
p2p*
2p
2p
p2p
2p
2p
2s*
2s*
2s
2s
2s
2s
2s
2s
163
Second row diatomic molecules
B2
C2
N2
O2
F2
2p*
p2p*
E
2p
p2p
2s*
2s
Magnetism
Bond order
Bond E. (kJ/mol)
Bond length(pm)
164
Second row diatomic molecules
B2
C2
N2
O2
F2
2p*
p2p*
E
2p
p2p
2s*
2s
Magnetism
Para-
Bond order
1
Bond E. (kJ/mol)
290
Bond length(pm)
159
165
Second row diatomic molecules
B2
C2
Magnetism
Para-
Dia-
Bond order
1
2
Bond E. (kJ/mol)
290
620
Bond length(pm)
159
131
N2
O2
F2
2p*
p2p*
E
2p
p2p
2s*
2s
166
Second row diatomic molecules
B2
C2
N2
Magnetism
Para-
Dia-
Dia-
Bond order
1
2
3
Bond E. (kJ/mol)
290
620
942
Bond length(pm)
159
131
110
O2
F2
2p*
p2p*
E
2p
p2p
2s*
2s
167
Second row diatomic molecules
NOTE SWITCH
OF LABELS
B2
C2
N2
O2
Magnetism
Para-
Dia-
Dia-
Para-
Bond order
1
2
3
2
Bond E. (kJ/mol)
290
620
942
495
Bond length(pm)
159
131
110
121
F2
2p*
p2p*
E
p2p
2p
2s*
2s
168
Second row diatomic molecules
NOTE SWITCH
OF LABELS
B2
C2
N2
O2
F2
Magnetism
Para-
Dia-
Dia-
Para-
Dia-
Bond order
1
2
3
2
1
Bond E. (kJ/mol)
290
620
942
495
154
Bond length(pm)
159
131
110
121
143
169
2p*
p2p*
E
p2p
2p
2s*
2s
O2
O 2+
O 2–
O22-
2p*
E
p2p*
p2p
2p
2s*
2s
O2 :
O 2+ :
O 2– :
O22-:
170
O2
O 2+
O 2–
O22-
2p*
E
p2p*
p2p
2p
2s*
2s
171
O2
O 2+
O 2–
O22-
2p*
E
p2p*
p2p
2p
2s*
2s
172
O2
O 2+
O 2–
O22-
2p*
E
p2p*
p2p
2p
2s*
2s
173
O2
O 2+
O 2–
O22-
2p*
E
p2p*
p2p
2p
2s*
2s
174
O2
O 2+
O 2–
O22-
2p*
E
p2p*
p2p
2p
2s*
2s
O2 :
B.O. = (8 - 4)/2 = 2
O2+ : B.O. = (8 - 3)/2 = 2.5
O2– : B.O. = (8 - 5)/2 = 1.5
O22- : B.O. = (8 - 6)/2 = 1
175
O2
O 2+
O 2–
O22-
2p*
E
p2p*
p2p
2p
2s*
2s
O2 :
B.O. = 2
O2+ : B.O. = 2.5
BOND ENERGY ORDER
O2+ >O2 >O2– > O22-
O2– : B.O. = 1.5
O22- : B.O. = 1
176
OXYGEN
O
O
How does the Lewis dot picture correspond to MOT?
12 valence electrons
2p*
p2p*
E
2p
BO = 2 but PARAMAGNETIC
p2p
2s*
2s
177