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CHEMISTRY 161
Chapter 10
Chemical Bonding & Molecular Structure
2
PREDICTING THE GEOMETRY OF MOLECULES
1. derive Lewis structure of the molecule
2. discriminate between bonding and non-bonding electron pairs
O HH
3. VALENCE SHELL ELECTRON PAIR REPULSION
3
VALENCE SHELL ELECTRON PAIR REPULSION
3. valence electron pairs stay as far apart as possible
VSEPR
1. identify in a compound the central atom
2. electrons repel each other
4. non-bonding electrons repel more than bonding electrons
4
central atom
no non-bonding pairs non-bonding pairs
5
Cl Be Cl
BeCl2
TWO ELECTRON PAIRS AROUND BERYLLIUM ATOM
AB2
6
Be
180°
180°
Be90°
270°
LINEAR ARRANGEMENT BEST
Cl Be Cl
IT PUTS ELECTRON PAIRS FURTHEST APART
7
BF3
F B F
F
THREE ELECTRON PAIRS AROUND THE BORON ATOM
AB3
8
B
120°
120°
120°
F B F
F
THREE ELECTRON PAIRS AROUND THE BORON ATOM
TRIGONAL PLANAR ARRANGEMENT BEST
9
F F
F
B
THE SHAPE OF BF3 IS TRIGONAL PLANAR.
MOLECULAR SHAPE
10
CH4
H C H
H
H
four electron pairs
90°
C
90° 90°
90°
expect square planar
AB4
11
better arrangement for four electron pairs
109.5°
C
TETRAHEDRAL
4 electron pairs tetrahedral
bigger than 90 ° in square planar
put on the H-atoms
12
109.5°
C
TETRAHEDRAL
C
H
HH
H
shape of CH4 is tetrahedral
13
PF5 FIVE ELECTRON PAIRS AROUND PHOSPHORUS
P
5 electron pairs trigonal bipyramidal
F
PFF
F F
AB5
14
PPF
F
F
F
F
Bond angle
900
1200
shape of PF5 is trigonal bipyramidal
two of the F atoms different from the others
15
PFF
F
F
FBond angle
900
1200
AXIAL
EQUATORIAL
16
F
S
F
F
F
F
F
S
six electron pairs around the sulfur atom
6 electron pairs
SF6
octahedral
AB6
17
S S
F
F
F
F
F
F
shape of SF6 is octahedral
900
18
central atom
no non-bonding pairs non-bonding pairs
19
SeO2 O Se O
AB2E AB3
20
Se
VSEPR treats double bonds like a single bondO Se O
THREE ELECTRON PAIRS AROUND SELENIUM
ELECTRON PAIR GEOMETRY
TRIGONAL PLANAR
21
Se
SeO2 IS V-SHAPED (OR BENT)
ADD OXYGENSSe
O O
THE MOLECULAR SHAPE IS THE POSITION OF
THE ATOMS
22
NH3
H N H
H
electron pairs around the nitrogen atom
AB3E AB4
23
NH3H N H
H
NPUT ON THE 3 H ATOMS
N
H
H
H
NH3 is trigonal pyramidal
24
AB2E2 AB4
H O H four electron pairs around the oxygen atom
O
PUT ON THE 2 H-ATOMS
O
HH
shape of H2O is V-shaped or bent
25
F
SFF
F
SF4
AB4E AB5
TRIGONAL BIPYRAMID
S
26
SF
F
F
F
SFF
F
F
WHERE DOES LONE PAIR GO?
OR
lone pairs occupy the trigonal plane (the “equator”) to minimize the number of 90° repulsions
27
SF4
1 lone pair
See-saw
shaped
ClF3
2 lone pairs
T-shaped
SF
F
F
F
ClF
F
F
lone pairs occupy the trigonal plane (the “equator”)
first to minimize the number of 90° repulsions
AB4E AB3E2
F
F
F
Xe
XeF2
3 lone pairs
Linear
AB2E3
28
BrF5
Br
F
F
F
F
F
Square pyramidal
Br
AB5E AB6
29
XeF4
XeF
F
F
F
XeXe
F
F
F
F
:
lone pairs MUST BE AT 1800
XeF
F
F
F
:
AB4E2AB6
30
Summary of Molecular ShapesTotal valence electron pairs
Electron Pair Geometry
Lone electron pairs
Shape of Molecule
2
3
4
Linear
Trigonal planar
Tetrahedral
0
0
1
0
1
2
Linear
Trigonal planar
V-shaped
Tetrahedral
Trigonal pyramid
V-shaped
31
Total valence electron pairs
Electron Pair Geometry
Lone electron pairs
Shape of Molecule
5
6
Trigonal bipyramidal
Octahedral
1
2
3
0
1
2
See-saw
T-shaped
Linear
Octahedral
Square pyramid
Square planar
0 Trig. bipyramid.
32
molecules with no single central atom
we apply our VSEPR rules to each atom in the chain
POLYATOMICS
Example: ETHANOL
33
ETHANOL
The atoms around the carbons form a.
The atoms around the oxygen form a
OCH H
H
H
C
H
H
tetrahedral arrangement
V-shaped structure.
C2H5OH
34
HH
H
CH
H
C
H
O
35
EXAMPLES
Cl2O BF4-
ICl4-
SO2Cl2 Cl2CO Cl2SO
NH2OH NH4+ N2F2
36
1. Lewis structures
2. VSEPR model
WHY DO MOLECULES FORM?
37
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
38
39
40
chemical bond with electron density in between the nuclei is called
bond
41
a covalent bond is formed by an
overlap of two valence atomic
orbitals that share an electron pair
VALENCE BOND THEORY
the better the overlap the stronger the bond
the orbitals need to point along the bonds
42
C
H
H
H
H
What orbitals are used?
hydrogen atoms bond using their 1s orbitals
carbon needs four orbitals to bond with.
2s, 2px , 2py, 2pz
[He] 2s22p2
CH4
43
[He] 2s22p2
1. The electronic configuration of carbon is
The orbital diagram is:
44
[He] 2s22p2
1. The electronic configuration of carbon is
the orbital diagram is: [He]
the Lewis dot structure is C ...
.
necessary to promote one 2s electron
45
[He] 2s12p3
[He]
PROMOTE AN ELECTRON
[He]
[He] 2s22p2
excited state (valence state)
Lewis dot structure
four unpaired electrons
we can use these to form chemical bonds
C
46
2. bonds formed with s orbitals will be different
to bonds formed with p orbitals
3. three p orbitals are mutually perpendicular,
suggesting 90° bond angles
combining the orbitals
1. a covalent bond is formed by an overlap of two
valence atomic orbitals that share an electron pair
Experiment shows that all four bonds are identical
experiment shows that methane has 109.5° bond
angles
47
we need four orbitals pointing to the vertices of a tetrahedron
orbitals are just mathematical functions
HYBRIDIZATION
C
H
H
H
Hwe can combine them
48
COMBINING ORBITALS TO FORM HYBRIDS
HYBRIDIZATION
number of atomic orbitals that are combined
the number of resulting hybrid orbitals
IS EQUAL TO
49
Combine one s and one p a sp- hybrid
+
+ ADD the orbitals
2s+ 2p
HYBRIDIZATION
50
Combine one s and one p a sp- hybrid
++
s + p
The positive part cancels negative part
2s+ 2p
HYBRIDIZATION
The positive part adds to positive part
CONSTRUCTIVE INTERFERENCE
DESTRUCTIVE INTERFERENCE
What do we get?
51
Combine one s and one p to give a sp- hybrid
+
s + p
REMEMBER IF WE MIX TWO WE MUST GET TWO BACK
The other combination is s - p
HYBRIDIZATION
2s+ 2p
Where is the nucleus?
52
HYBRIDIZATIONCombine one s and one p a sp- hybrid
+
+SUBTRACT the
orbitals2s- 2p
53
HYBRIDIZATIONCombine one s and one p a sp- hybrid
+
+SUBTRACT the
orbitals2s- 2p
SUBTRACTING THE p ORBITAL CHANGES ITS PHASE
54
HYBRIDIZATIONCombine one s and one p a sp- hybrid
+
+SUBTRACT the
orbitals2s- 2p
SUBTRACTING THE p ORBITAL CHANGES ITS PHASE
55
HYBRIDIZATIONCombine one s and one p a sp- hybrid
+ +
s - p2s- 2p
The positive part cancels negative part
The positive part adds to positive part
CONSTRUCTIVE INTERFERENCE
DESTRUCTIVE INTERFERENCE
What do we get?
56
HYBRIDIZATIONCombine one s and one p a sp- hybrid
+
s - p
The positive part cancels negative part
We get two equivalent sp orbitals
ORIENTED AT 1800
2s- 2p
Where is the nucleus?
57
sp-HYBRIDIZATION
s and p orbitals
two sp-hybrids
58
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
59
COMBINE one s-orbital and three p-orbitals
three sp3- orbitals oriented at 109.50
60
four hybrid orbitals
needed to form four
bonds
s + px + py + pz
an atom with sp3 hybrid orbitals is said to be
4 sp3 hybrids
The four sp3 hybrid orbitals form a tetrahedral
arrangement.
sp3 hybridizationEPG of 4 pairs
METHANE: CH4
sp3 hybridized
What happens to the energies of the orbitals?
C
H
H
H
H
61
E2p
2s
Orbitals in free C atom
What happens to orbital energies when the are hybridized??
HYBRIDIZE
62
E2p
2s
Orbitals in free C atom
E sp3
Hybridized orbitals of C atom in methane
When orbitals are hybridized they have the same energy:
The FOUR sp3 hybrids are DEGENERATE.
HYBRIDIZE
63
x
y
z
x
y
z
x
y
z
x
y
z
Combine one s and three p orbitals…..
64
C
sp3 HYBRIDS
Now form the bonds to the H-atoms……...
sp3 orbitals
65
C
Each bond in methane results from the overlap of a
hydrogen 1s orbital and a carbon sp3 orbital.
H
HH
H
Hydrogen 1s orbital
Carbon sp3 orbitals
Each hybrid ready to overlap with H 1s orbitals
Form a chemical bond by sharing a pair of electrons.
66
VALENCE BOND 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
Hybrid orbital model
67
sp3 hybrids are also employed in …...
all molecules that have a 4 pair EPG….
NH3, H2O, NH4+ , CCl4
OTHER MOLECULES USING sp3 HYBRIDS
AMMONIA…..
68
AMMONIA: NH3
2s 2p
N [He]
NVSEPR
Valence shell has four pairs
EPG is TETRAHEDRAL
Nitrogen electronic configuration
Need sp3 hybrids
HYBRIDIZE
H
H
H
69
E2p
2s
Orbitals in free N atom
E sp3
Hybridized orbitals of N atom in ammonia
When orbitals are hybridized they have the same energy:
sp3 hybridization…….
The FOUR sp3 hybrids are DEGENERATE.
HYBRIDIZE
70
N
N... .
.
2s 2p
sp3 hybrids on N in AMMONIA
Now form a bond
Overlap H 1s….. H
H
H
71
Three bonds
One lone pair in an sp3 hybrid
N
HH
H
N... .
.
2s 2p
AMMONIA
72
four bonds.N
HH
H
N... .
.
2s 2p
AMMONIUM ION NH4+
H+
ISOELECTRONIC WITH ?
CH4
73
O
2s 2p
WATER
O:.. ..
H O H
O
FOUR PAIRS
EPG?
TETRAHEDRAL!!
HYBRIDIZATION? sp3
74
Overlap of two of oxygen sp3 hybrids with …..
H atom 1s orbitals.
WATER.
Lone pairs in two of the sp3 hybrids.
To form two bonds.
Think about H3O+ !!!
O
H
H
75
Overlap of Oxygen sp3 hybrids containing a lone pair
H+ ion empty 1s orbitals.
HYDRONIUM ION.
O
H
H
H+
ISOELECTRONIC WITH?
NH3
76
QUESTIONWhich of the following molecules is uses sp3 hybrids in the valence bond description of its bonding?
A CO2
B NF3
C O3
D NO2+
E F2O
ANSWER…….
1 C and D
2 B and E
3 A and D
4 B and C
5 B and A
77
QUESTIONWhich of the following molecules is uses sp3 hybrids in the valence bond description of its bonding?
A CO2
B NF3
C O3
D NO2+
E F2O
WHAT ABOUT OTHER EPG’S …….
1 C and D
2 B and E
3 A and D
4 B and C
5 B and AOF F
C OO
O N O[ ]+
NF F
F
O O O
78
A four electron pair EPG uses sp3 hybrids
The three electron pair EPG uses sp2 hybrids
VALENCE BOND THEORY FOR OTHER ELECTRON PAIR GEOMETRIES
The two electron pair EPG uses sp hybrids
79
X
180°
180°
X
120°
120°
120°
X
109.5°
EPG’s2 3
4
HYBRIDS
sp sp2 sp3
lets look at a molecule that needs sp2
80
Ethylene: C2H4 C C
H
H
H
H
The CARBON is sp2 hybridized
3 effective electron pairs
3 sp2 hybridss + px + py
The 3 sp2 hybrid orbitals form a trigonal planar arrangement.
three hybrid orbitals on each carbon for the trigonal
planar EPG.
VSEPR trigonal planar EPG around each C-atom. a HCH angle of 1200.
sp2 hybridization
81
GROUND STATE C atom
E2p
2s
FORMATION OF sp2 hybrids
VALENCE STATE C atom
82
GROUND STATE C atom
E2p
2s
FORMATION OF sp2 hybrids
E2p
2s
VALENCE STATE C atom
HYBRIDIZE
83
GROUND STATE C atom
E2p
2s
sp2 hybridized orbitals of C
FORMATION OF sp2 hybrids
E sp2
2p
E2p
2s
VALENCE STATE C atom
HYBRIDIZE
This leaves one p orbital unhybrized…….
84
x
y
z
The unhybridized p orbital is perpendicular to sp2 plane.
sp2 - hybrid orbitalUNHYBRIDIZEDp- orbital
An sp2 hydridized C atom
Lets put it all together…….
85
x
z
y y
x
z
C C
H
H
H
H
DRAW TWO C-ATOMS
C C
Now put the orbitals on…...
86
x
y
z
x
y
z
C C
BONDING IN ETHYLENEC C
H
H
H
H
87
x
y
z
x
y
z
OVERLAP the sp2 hybrids from the two carbons to form a sigma bond between them.
C C
BONDING IN ETHYLENE
bond
PUT THE ELECTRONS IN AND…..
C C
H
H
H
H
88
x
y
z
x
y
z
overlap two sp2 hybrids on each carbon with hydrogen 1s orbitals to form sigma bonds and...
The two unhybridized p orbitals are left over to form…..
H
HH
H
C C
H
H
H
H
89
x
y
z
x
y
z
The two unhybridized p orbitals are left over to form a …..
H
HH
H
pi bond ( bond)
The second part of the carbon-carbon double bond !
C C
H
H
H
H
90
x
y
z
x
y
z
H
HH
H
pi bond ( bond)The second part of the carbon-carbon double bond !
Electrons are shared between the unhybridized p
orbitals in an area above and below the line between
nuclei.
C C
H
H
H
H
91
x
y
z
x
y
z
H
HH
H
pi bond ( bond)
sp2
sigma bonds ( bond)
THE COMPLETE PICTURE!!!!!!!
C C
H
H
H
H
SUMMARY...
92
C C
H
H
H
H
H(1s)-C(sp2)
:C(sp2)-C(sp2)C(2p)-C(2p)
H(1s)-C(sp2)
Now look at bond
bonding
93
C C
H
H
H
H
H(1s)-C(sp2)
:C(sp2)-C(sp2)C(2p)-C(2p)
H(1s)-C(sp2)
Now look at ethyne (acetylene)
bonding
94
H(1s)-C(sp)
:C(sp)-C(sp)
C(2p)-C(2p) TWO OF THESE!!
H(1s)-C(sp)
BONDING SCHEME IN ETHYNE
C CH H
What does this look like????
95
x
z
y y
x
z
DRAW TWO C-ATOMS
C C
Now put the sp-orbitals on…...
C CH H
96
x
y
z
x
y
z
OVERLAP the sp hybrids from the two carbons to form a sigma bond between them.
C C
Put in the unhybridized p orbitals
97
x
y
z
x
y
z
OVERLAP the sp hybrids from the two carbons to form a sigma bond between them.
C C
OVERLAP the hydrogen 1s orbitals
98
x
y
z
x
y
z
OVERLAP the C sp hybrids with H 1s to form sigma bonds
C C HH
99
x
y
z
x
y
z
OVERLAP the sp hybrids from the two carbons to form a sigma bond between them.
C C HH
sigma framework of bonds pi bonding?
100
x
y
z
x
y
z
C C HH
two pi bonds ( bonds)
LATERAL OVERLAP of p orbitals to form pi bonds.
101
x
y
z
x
y
z
LATERAL OVERLAP of p orbitals to form pi bonds.
C C HH
two pi bonds ( bonds)
SO…..
102
Single bond: One bond
SUMMARY
Double bond: One bond, one bond
Triple bond: One bond, two bonds
103
VALENCE BOND THEORY
Step 1 Lewis Dot Structure
Step 2 Get Molecular Geometry
VSEPR EXPERIMENTAL
Step 3 Choose hybrids
Step 4 Describe bonds…...
104
What about molecules with more than an octet around the central atom?
Examples: PCl5, or SF4 or SiF62-
Four pairs needs Four orbitals
Five pairs needs Five orbitals
six pairs needs six orbitals
105
PCl5
Cl
PClCl
Cl Cl
We ignore the
chlorine atoms
and just describe
central atom.
Need five hybrid orbitals on the phosphorus
d + s + px + py + pz
5 effective electron pairs dsp3 hybridization
5 dsp3 hybrids
to fit the trigonal bipyramidal EPG.
Five equivalent orbitals……..
106
dsp3 - hybrid orbitals
x
y
zTRIGONAL BIPYRAMID EPG 5 PAIRS
overlap with orbitals on chlorine to form 5 bonds.
1200
900
SIX PAIRS…..
107
SF6
F
S
F
F
F
F
F
We need six hybrid orbitals on the sulfur to
allow for the octahedral EPG and six bonds.
6 d2sp3 hybrids
d2sp3 hybridization
d + d + s + px + py + pz
6 effective electron pairs
SIX equivalent orbitals……..
We ignore the
chlorine atoms
and just describe
central atom.
108overlap with orbitals on flourine to form 6 bonds.
d2sp3 - hybrid orbitals
x
y
z 900
900
EXAMPLE
109
EXAMPLES
Xe FF
EPG 5 pairs
dsp3 hybrids
Two axial bonds at 1800
Three lone pairs in equatorial
hybrids
Describe the molecular structure and bonding in XeF2 and XeF4
Linear
110
EXAMPLES
Xe FF
EPG 5 pairs
dsp3 hybrids
Two axial bonds at 1800
Three lone pairs in equatorial
hybrids
EPG 6 pairs
d2sp3 hybrids
four bonds at 900 in a plane
Two lone pairs in axial hybrids
Describe the molecular structure and bonding in XeF2 and XeF4
Linear
XeF
F
F
F
Square planar
111
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
112
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
1sA – 1sB = MO2
constructive interference
destructive interference
113
0r/a-3/2
01 e
12)(
a
rRs
114
ADDITION OF ORBITALSbuilds up electron density in overlap region
1sA + 1sB = MO1
combine them by addition
A B
115
ADDITION OF ORBITALSbuilds up electron density in overlap region.
1sA + 1sB = MO1
A Bwhat do we notice?
electron density between atoms
116
SUBTRACTION OF ORBITALSresults in low electron density in overlap region..
1sA – 1sB = MO2
A B
subtract
117
SUBTRACTION OF ORBITALSresults in low electron density in overlap region..
1sA – 1sB = MO2
A Bwhat do we notice?
no electron density between atoms
118
COMBINATION OF ORBITALS
1sA + 1sB = MO1
builds up electron density between nuclei
119
COMBINATION OF ORBITALS
1sA + 1sB = MO1
builds up electron density between nuclei
1sA – 1sB = MO2
results in low electron density between nuclei
BONDING
ANTI-BONDING
120
121
THE MO’s FORMED BY TWO 1s ORBITALS
122
1sA + 1sB = MO1
1sA – 1sB = MO2
sigma anti-bonding = 1s*
sigma bonding = 1s
1s
1s*
123
E
Energy of a 1s orbital in a free atom
Energy of a 1s orbital in a free atom
A B
COMBINING TWO 1s ORBITALS
124
E
Energy of a 1s orbital in a free atom
Energy of a 1s orbital in a free atom
A B
1sA+1sB
MO
1s
125
E
Energy of a 1s orbital in a free atom
Energy of a 1s orbital in a free atom
A B
1sA-1sB
MO
1sA+1sB
MO
1s
1s*
126
E 1sAA B
1s
1s*
1sB
COMBINING TWO 1s ORBITALS
127
E1s
1s*
1s
1s
H HH2
bonding in H2
128
E1s
1s*
1s
1s
H HH2
the electrons are placed in the 1s molecular orbitals
129
E1s
1s*
1s
1s
H2: (1s)2
H HH2
130
E1s
1s*
1s
1s
He2
He HeHe2
atomic configuration of He 1s2
131
E1s
1s*
1s
1s
He2: (1s)2(1s*)2
He HeHe2
bonding effect of the (1s)2 is cancelled by the
antibonding effect of (1s*)2
132
BOND ORDER
net number of bonds existing after the cancellation of bonds by antibonds
the two bonding electrons were cancelled out by the two antibonding electrons
He2
(1s)2(1s*)2
the electronic configuration is….
BOND ORDER = 0
133
BOND ORDER
=
measure of bond strength and molecular stability
If # of bonding electrons > # of antibonding electrons
Bondorder
the molecule is predicted to be stable
134
BOND ORDER
= {
high bond order indicates high bond energy and short bond length
# of bonding electrons(nb)
# of antibonding electrons (na)
– 1/2 }
measure of bond strength and molecular stability
If # of bonding electrons > # of antibonding electrons
Bondorder
the molecule is predicted to be stable
H2+,H2,He2
+
= 1/2 (nb - na)
135
1s*
1s
Magnetism
Bond order
Bond energy (kJ/mol)
Bond length (pm)
H2+
E
He2+ He2H2
136
1s*
1s
Magnetism
Bond order
Bond energy (kJ/mol)
Bond length (pm)
H2+
E
He2+ He2H2
Dia-
1
436
74
137
1s*
1s
Magnetism
Bond order
Bond energy (kJ/mol)
Bond length (pm)
H2+
Para-
½
225
106
E
He2+ He2H2
Dia-
1
436
74
138
1s*
1s
Magnetism
Bond order
Bond energy (kJ/mol)
Bond length (pm)
H2+
Para-
½
225
106
E
He2+
Para-
½
251
108
He2H2
Dia-
1
436
74
139
1s*
1s
Magnetism
Bond order
Bond energy (kJ/mol)
Bond length (pm)
First row diatomic molecules and ions
H2+
Para-
½
225
106
E
He2+
Para-
½
251
108
He2
—
0
—
—
H2
Dia-
1
436
74
140
HOMONUCLEAR DIATOMICS
Li2 Li : 1s22s1
both the 1s and 2s overlap to produce bonding and anti-bonding orbitals
second period
141
E
1s
1s*
1s
1s
2s
2s*
2s
2s
ENERGY LEVEL DIAGRAM FOR DILITHIUM
Li2
142
E
1s
1s*
1s
1s
2s
2s*
2s
2s
Li2
ELECTRONS FOR DILITHIUM
143
E
1s 1s
1s
Electron configuration for DILITHIUM
2s
2s*
2s
2s
(1s)2(1s*)2(2s)2
Li2
Bond Order ?
144
E
1s 1s
1s
Electron configuration for DILITHIUM
2s
2s*
2s
2s
(1s)2(1s*)2(2s)2
Li2
nb = 4 na = 2
Bond Order = 1
single bond.
145
E
1s 1s
1s
Electron configuration for DILITHIUM
2s
2s*
2s
2s
(1s)2(1s*)2(2s)2
the 1s and 1s* orbitals can be ignored when
both are FILLED!
Li2
omit the inner shell
146
E2s
2s*
2s
2s
Li LiLi2
The complete configuration is: (1s)2(1s*)2 (2s)2
Li2 (2s)2 only valence orbitals contribute to molecular bonding
147
E2s
2s*
2s
2s
Be BeBe2 Be2
148
E2s
2s*
2s
2s
Be2Be BeBe2
Electron configuration for DIBERYLLIUM
Configuration: (2s)2(2s*)2 Bond order = 0
149
E2s
2s*
2s
2s
(2s)2(2s*)2Be BeBe2
Be2
Electron configuration for DIBERYLLIUM
nb = 2
na = 2
Bond Order = 1/2(nb - na) = 1/2(2 - 2) =0
No bond!!! The molecule is not stable! Now B2...
150
B2
the Boron atomic configuration is
1s22s22p1
form molecular orbitals
we expect B to use 2p orbitals to
addition and subtraction
151
-molecular orbitals
152
molecular orbitals
153
ENERGY LEVEL DIAGRAM
E
2s
2s*
2s
2s
154
2p*
2p
2p
2p*
E2p 2p
155
E
expected orbital splitting
2s
2s*
2s
2s
2p
2p*
2p
2p
2p
2p*
This pushes the 2p up
156
E
MODIFIED ENERGY LEVEL DIAGRAM
2s
2s*
2s
2s
2p
2p*
2p2p
2p
2p*
Notice that the 2p and 2p
have changed places!!!!
157
E
2s
2s*
2s
2s
Electron configuration for B2
2p
2p*
2p2p
2p
2p*
Place electrons from 2s into 2s and 2s*
B is [He] 2s22p1
158
E
2s
2s*
2s
2s
2p
2p*
2p2p
2p
2p*
Place electrons from 2p into 2p and 2p
Remember HUND’s RULE
159
E
2s
2s*
2s
2s
2p
2p*
2p2p
2p
2p*(2s)2(2s*)2(2p)2
Abbreviated configuration
Complete configuration
(1s)2(1s*)2(2s)2(2s*)2(2p)2
ELECTRONS ARE UNPAIRED
160
E
2s
2s*
2s
2s
Electron configuration for B2:
Bond order
2p
2p*
2p2p
2p
2p*(2s)2(2s*)2(2p)2
Molecule is predicted to be stable and paramagnetic.
na = 2
nb = 4
1/2(nb - na)
= 1/2(4 - 2) =1
161
A SUMMARY OF THE MO’s
Emphasizing nodal planes
162
ELECTRONIC CONFIGURATION OF THE HOMONUCLEAR DIATOMICS
B2 C2 N2 O2 F2Li2
163
B2 C2 N2
O2 F2
E
2s
2s*
2s
2s
2p
2p*
2p2p
2p
2p*
2s
2s*
2s
2s
2p*
2p
2p
2p
2p*
2p
Li2
164
2p*
2p*
2p
2p
2s*
2s
Magnetism
Bond order
Bond E. (kJ/mol)
Bond length(pm)
Second row diatomic molecules
B2 C2 N2 O2 F2
E
165
2p*
2p*
2p
2p
2s*
2s
Magnetism
Bond order
Bond E. (kJ/mol)
Bond length(pm)
Second row diatomic molecules
B2
Para-
1
290
159
C2 N2 O2 F2
E
166
2p*
2p*
2p
2p
2s*
2s
Magnetism
Bond order
Bond E. (kJ/mol)
Bond length(pm)
Second row diatomic molecules
B2
Para-
1
290
159
C2
Dia-
2
620
131
N2 O2 F2
E
167
2p*
2p*
2p
2p
2s*
2s
Magnetism
Bond order
Bond E. (kJ/mol)
Bond length(pm)
Second row diatomic molecules
B2
Para-
1
290
159
C2
Dia-
2
620
131
N2
Dia-
3
942
110
O2 F2
E
168
2p*
2p*
2p
2p
2s*
2s
Magnetism
Bond order
Bond E. (kJ/mol)
Bond length(pm)
Second row diatomic molecules
B2
Para-
1
290
159
C2
Dia-
2
620
131
N2
Dia-
3
942
110
O2
Para-
2
495
121
F2
E
NOTE SWITCH OF LABELS
169
2p*
2p*
2p
2p
2s*
2s
Magnetism
Bond order
Bond E. (kJ/mol)
Bond length(pm)
Second row diatomic molecules
B2
Para-
1
290
159
C2
Dia-
2
620
131
N2
Dia-
3
942
110
O2
Para-
2
495
121
F2
Dia-
1
154
143
E
NOTE SWITCH OF LABELS
170
2p*
2p*
2p
2p
2s*
2s
E
O2 O2+ O2
–
O2 :
O2+ :
O2– :
O22-:
O22-
171
2p*
2p*
2p
2p
2s*
2s
E
O2 O2+ O2
– O22-
172
2p*
2p*
2p
2p
2s*
2s
E
O2 O2+ O2
– O22-
173
2p*
2p*
2p
2p
2s*
2s
E
O2 O2+ O2
– O22-
174
2p*
2p*
2p
2p
2s*
2s
E
O2 O2+ O2
– O22-
175
2p*
2p*
2p
2p
2s*
2s
E
O2 O2+ O2
– O22-
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
176
2p*
2p*
2p
2p
2s*
2s
E
O2 O2+ O2
– O22-
O2 : B.O. = 2
O2+ : B.O. = 2.5
O2– : B.O. = 1.5
O22- : B.O. = 1
O2+ >O2 >O2
– > O22-
BOND ENERGY ORDER
177
O O
OXYGEN
How does the Lewis dot picture correspond to MOT?
2p*
2p*
2p
2p
2s*
2s
E
12 valence electrons
BO = 2 but PARAMAGNETIC