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Chapter 7 -- Covalent Bonds
and Molecular Structure
Covalent Bonds
Electronegativity
Electron-Dot Structures
Resonance
VESPR Theory
Hybridization
What are the trends?
Electronegativity increases
1
1A
2.1
H
1.0
Li
0.9
Na
0.8
K
0.8
Rb
0.7
Cs
18
8A
2
2A
13
3A
1.5
Be
1.2
Mg
1.0
Ca
1.0
Sr
0.9
Ba
3
3B
1.3
Sc
1.2
Y
1.0
La
<= 1.0
4
4B
1.5
5
5B
1.6
Ti
V
1.4
1.6
Zr
1.3
Hf
Nb
1.5
Ta
6
6B
1.6
Cr
1.8
Mo
1.7
W
7
7B
1.5
Mn
1.9
Tc
1.9
Re
1.0 - 2.0
8
9
10
|-------8B-------|
1.8
Fe
2.2
Ru
2.2
Os
1.9
Co
2.2
Rh
2.2
Ir
1.9
Ni
2.2
Pd
2.2
Pt
11
1B
1.9
Cu
1.9
Ag
2.4
Au
2.0 - 3.0
12
2B
1.6
Zn
Cd
1.9
16
6A
17
7A
2.5
3.0
3.5
4.0
B
C
N
O
F
1.5
1.8
2.1
2.5
3.0
Al
Si
P
S
1.8
2.0
2.4
Ga
Ge
1.7
1.8
In
1.8
Hg
15
5A
2.0
1.6
1.7
14
4A
Tl
Sn
1.9
Pb
3.0 - 4.0
As
1.9
Sb
1.9
Bi
Se
2.1
Te
2.0
Po
Cl
He
Ne
Ar
2.8
Br
Kr
2.5
I
Xe
2.1
At
Rn
Electronegativity increases
Electronegativity: The ability of an atom to attract
electrons towards itself (unitless value)
>= 4.0
Electronegativity Increases Left to Right
• There are more protons in Fluorine’s nucleus
F
C
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Electronegativity Increases Bottom to Top
•  2nd
shell electrons closer to the nucleus compared to 3rd
shell electrons
Cl
F
Bond Characterization
•  Bonds can be categorized as ionic, covalent,
or polar covalent
•  This is a “loose definition”
Difference in
Electronegativity
Type of Bond
<0.5
Covalent
≥0.5 to <2.00
Polar Covalent
≥2.0
Ionic
Bond Characterization
Molecule
Value 1 Value 2 Difference
Type of
Bond
Cl2
3.0
3.0
0
Covalent
HCl
2.2
3.0
0.8
“Polar”
Covalent
NaCl
0.9
3.0
2.1
Ionic
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Ionic, Polar Covalent, and
Covalent Bonds
δ+
Na+ Cl
H
δ+
M+ X:-
Cl
Y:X
δ-
Cl
δ−
X:X
Polar covalent
with partial
charges
Ionic with
full charges
Cl
Covalent with
no charge!
What type of bond are each of the following? For
the polar covalent bonds, assign “partial
charges” to the atoms.
K—Br
H—OH
Br—Br
H—NH2
H—BH2
Cl—PCl2
H—CH3
Cl—AlCl2
Na—OH
Covalent Bonds
Attractive Forces
Repulsive Forces
Attractive Forces
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H H (Too close)
+
(Too far)
H
Energy
H
0
(Just right)
H H
-
Bond Length (74 pm)
Internuclear Distance
Making and Breaking
Covalent Bonds
•  One hydrogen molecule is lower in
energy than two hydrogen atoms
436 kJ/mol released
Higher
Energy
H· + ·H
H—H
Lower
Energy
436 kJ/mol absorbed
Average Bond Dissociation Energies, D (kJ/mol)
H-H
436
C-H
410
N-H
390
O-H
460
F-F
159
H-C
410
C-C
350
N-C
300
O-C
350
Cl-Cl
243
H-F
570
C-F
450
N-F
270
O-F
180
Br-Br
193
H-Cl
432
C-Cl
330
N-Cl
200
O-Cl
200
I-I
151
H-Br
366
C-Br
270
N-Br
240
O-Br
210
S-F
310
H-I
298
C-I
240
N-I
—
O-I
220
S-Cl
250
H-N
390
C-N
300
N-N
240
O-N
200
S-Br
210
H-O
460
C-O
350
N-O
200
O-O
180
S-S
225
H-S
340
C-S
260
N-S
—
O-S
—
C-O
732
O-O
498
N-N
945
Multiple covalent bonds
C=C
611
CC
835
Example: H—H ! H· + H· D = 436 kJ/mol
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How much energy is required to break
all four bonds of 1.4 g of methane
(CH4)?
1.4 g×
1 mol CH4
16.04 g
×
410. kJ 4 mol C-H
×
=1.4×102 kJ
1 mol C-H 1 mol CH4
Covalent Bonds
•  Formed by the sharing of electrons between
two elements
•  Covalent bonds are best illustrated with the
use of “Lewis dot structures”
•  Each dot in a Lewis dot structure represents one
valence electron
Br
+
7 valence
electrons
Br
7 valence
electrons
Br Br
octet rule is
satisfied for
both bromines
Lewis Dot Structures
•  Lines are shorthand for two electrons,
especially to emphasize covalent
bonds:
Br
+
Br
Br
+
Br
Br
Br
Br
+
Br
Br
Br
Br Br
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Covalent Molecules
•  Like ionic compounds, the octet rule
must be satisfied for covalent molecules
•  Carbon, for example, has four valence
electrons
•  Carbon shares four more electrons with
other elements to satisfy the octet rule
C
H
H
H
H
H C H
H
H
Other Examples
N
H
H
H
C
H
H
H
O
H
H N H
H
F
H
H C H
H
H
H
“Lone pair”
electrons
H O H
H
F H
Double and Triple Bonds
Carbon (and other elements) can satisfy the octet rule
by forming double or triple bonds to other elements
C
C
C
C
H
H
H
H
H
H
H C
H
H
H
H
C
C H
C
H
C
H
C
H
H
C
H
C
H
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More Examples
PBr3
HI
C 2H 6
C 2H 4
C 2H 2
N(CH3)3
BH3
AlCl3
NaCl
Third Period Elements
•  Third period (and below) elements can
adopt an “expanded octet”
•  It’s wise to follow “the method” for
determining Lewis structures
Determining the Lewis Dot Structure...
Start
Count the total number of valence
electrons. Don’t forget to add or subtract
electrons for the overall charge.
Determine and write the central atom first.
Use single bonds to connect all of the
other atoms to the central atom. Hydrogen
may be attached to oxygen and not
necessarily to the central atom.
•  Second period elements prefer
eight electrons around them; they
can’t have more than eight and it’s
generally bad to have less than eight
•  The third period and below
elements sometimes have more than
eight electrons around them
•  If there are several structures
possible, choose the one with the
minimum amount of formal charges
Distribute the remaining electrons
around the peripheral atoms.
Too many
Put the excess electrons
on the central atom.
Ran out
Start creating double and triple
bonds between the central and
peripheral atoms.
Did you run out of electrons or do
you still have more electrons?
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Examples
NH3
CH4
HCN
NO2
C 2H 6
C 2H 4
C 2H 2
SF6
PCl5
Formal Charge
•  Atoms in a covalent molecule can have a
“formal charge”
–  “Electron bookkeeping”
•  Formal charge is different than the oxidation
number (and computed differently)
⎛ Number of ⎞
⎛ Number of⎞ ⎛ Number of ⎞
1
Formal Charge = ⎜ valence electrons⎟ − ⎜ bonding ⎟ − ⎜ nonbonding⎟
⎜
⎟ 2⎜
⎟ ⎜
⎟
⎝ in free atom ⎠
⎝ electrons ⎠ ⎝ electrons ⎠
Examples
SO42-
PO43-
O3
I31-
CO
AlCl3
ClO31-
CH4
PCl3
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Formal Charges, Quick and Dirty Method
(Optional, but effective!)
If (and only if) an atom has an octet of electrons and it
is a second period element or halogen, then the atom is
neutral if it is making the “preferred” number of bonds:
C (Si)
N (P)
O (S)
F,Cl,Br,I
4
3
2
1
If the element is making one less bond: -1
If the element is making one more bond: +1
(Triple bonds count as three bonds, double bonds
count as two bonds)
Determining the formal charge...
Is the element C, N, O,
F, Cl, Br, I, Si, S, or P?
Start
No
Yes
Is the octet rule
satisfied?
(
)
Is the element
making the preferred
number of bonds?
No
⎛ # of bonding e- s⎞
Formal Charge = # of valence e s − ⎜
⎟ − # of non - bonding e s
2
⎝
⎠
-
Yes
(
)
No
Yes
The element has no
formal charge, done!
Preferred Number of Bonds
C, Si
N, P
O, S
F, Cl, Br, I
4
3
2
1
Is the element
making one more or
one less than
preferred?
One less
-1
One more
+1
Molecular Shape
•  Molecules are 3D structures
•  (V)alence (S)hell (E)lectron (P)air (R)epulsion
•  Electrons repel one another, we must space the epairs out as much as possible!
•  To determine “molecular shape”:
•  Step 1: Count the number of “charge clouds”
around each element in a molecule
•  Step 2: Count the number of “lone pairs” around
an element
•  Step 3: Assign a molecular shape based upon the
number of lone pairs and charge clouds
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Charge Clouds
H
H
H
C
H
Each single bond counts as one
charge cloud
C
N
2 charge clouds
around carbon
H
Triple bonds are counted as
one charge cloud
4 charge clouds
around carbon
O
Double bonds are counted as
one charge cloud
C
H
H
Each lone pair of electrons
counts as one charge cloud
O
H
H
4 charge clouds
around oxygen
3 charge clouds
around carbon
How many charge clouds surround
the following blue elements?
O
H
O
C
C
H
C
H
O
C
H
H
O
H
H
H
O
H
O
C
C
H
H
C
C
O
H
O
H
H
H
H
Two Charge Clouds
Molecular shape is “linear”
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Three Charge Clouds
Molecular shape is “trigonal planar”
Four Charge Clouds
Molecular shape is “tetrahedral”
Five Charge Clouds
Molecular shape is “trigonal bipyramidal”
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Six Charge Clouds
Molecular shape is “octahedral”
Electron Pairs
•  The presence of lone pairs on an atom
complicate the shape (a little)
•  X-ray crystallography, a method for
finding the three dimensional structure
of molecules, only “sees” nuclei, not
lone pair electrons...
Three Charge Clouds, One Lone Pair
Molecular shape is “bent”
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Four Charge Clouds, One Lone Pair
Molecular shape is “trigonal pyramidal”
Four Charge Clouds, Two Lone Pairs
Molecular shape is “bent”
Five Charge Clouds, One Lone Pair
Molecular shape is “seesaw”
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Five Charge Clouds, Two Lone Pairs
Molecular shape is “T-shaped”
Five Charge Clouds, Three Lone Pairs
Molecular shape is “linear”
Six Charge Clouds, One Lone Pair
Molecular shape is “square pyramidal”
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Six Charge Clouds, Two Lone Pairs
Molecular shape is “square planar”
Start
Determining the
Molecular Shape...
Count the number of
charge clouds
2
3
4
5
6
# of lone
pairs?
linear
0
# of lone
pairs?
1
trigonal
planar
2
0
bent
1
octahedral
square
planar
# of lone
pairs?
square
pyramidal
0
tetrahedral
# of lone
pairs?
1
trigonal
pyramidal
2
0
bent
trigonal
bipyramidal
2
1
3
see-saw
t-shaped
linear
5
6
Determining the bond angle...
Start
Count the number of
charge clouds
2
180°
3
120°
4
109.5°
90°, 120°,180°
90°,180°
Note: the presence of triple bonds, double bonds, and lone
pair electrons may “tweak” these bond angles from ideal.
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What is the molecular shape of
the following blue elements?
O
H
O
C
C
H
C
H
O
H
C
H
O
H
H
H
O
H
O
C
C
H
H
C
H
O
C
O
H
H
H
H
What is the molecular shape of
the following blue elements?
F
F
F
F
Xe
S
F
F
F
F
-2
F
Cl
Cl
Cl
Cl
Cl
F
F
Sb
Cl
Valence Bond Theory
•  How do atoms “bond”?
•  The 1s atomic orbital from each hydrogen atom
combine to form one “molecular” orbital
•  Both electrons now occupy the molecular orbital
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Valence Bond Theory
•  Bonds are created by the overlap of orbitals
•  The higher the overlap of the orbitals, the
“stronger” the bond
•  A bond that forms directly between the atoms
is called a sigma bond (σ)
Hybridization
•  Elements will often rearrange atomic
orbitals to form “hybrid” orbitals
•  The driving force for hybridization is to
maximize orbital overlap when making
new bonds
Hybridization (sp3)
Four atomic orbitals
Four sp3 hybrid orbitals
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Hybridization
Tetrahedral!
Energies of Hybrid Orbitals
Energies of Hybrid Orbitals
“degenerate”
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sp2 Hybridization
sp Hybridization
sp3d Hybridization
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sp3d2 Hybridization
Determining the hybridization of an atom...
Start
Count the number of
charge clouds
2
sp
3
sp2
4
5
sp3
sp3d
6
sp3d2
State the hybridization of each
non-hydrogen element:
CH4
C 2H 4
acetone
NH3
PCl5
PtCl4
C 2H 2
HC2H3O2
allene
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Pi (π) Bonds
•  Bonding above and below the bond axis
•  Occurs between p orbitals from two atoms
Two π-Bonds (Triple Bond)
•  Bonding between two p orbitals of two separate atoms
• The two p orbitals are perpendicular to one another
Rotatable Bonds
•  Single bonds are fully rotatable
•  Double and triple bonds are not
rotatable
H
H
H
C
H
C
H
H
Does occur
H
H
C
C
H
H
Doesn’t occur
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Double Bonds are Shorter than Single
0.134 nm?
0.154 nm?
0.154 nm
H
C
0.134 nm
H3 C
C
H
CH
HC
HC
CH2
CH
C
H
Kekule benzene is
not observed!
Kekule benzene
Resonance
HC
H
C
HC
CH Each of the
six bonds
CH is 0.139 nm!
C
H
H
C
HC
HC
H
C
CH Resonance
HC
CH
HC
C
H
CH
CH
C
H
Resonance Structures
There are three alternative ways to write the Lewis dot
structure for nitrate ion:
O
O
N
O
O
N
O
O
N
O
O
O
• All are identical in energy
• All are equally valid
• All N-O bonds are identical in length
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Resonance Structures
H
H
H
C
C
C
C
C
H
H
H
H
C
H
H
C
C
C
C
C
C
C
C
H
C
H
H
H
H
C
C
H
H
H
All of the bonds of
napthalene are
identical in length
C
C
C
C
H
C
C
H
C
C
H
C
H
C
C
C
C
H
H
H
Resonance Structures
OCN1- Cyanate ion
-1
N
C
-2
O
-2
+1
N
C
+1
N
O
-2
-1
N
C
O
O
+2
N
Oxygen is more
electronegative
than nitrogen...
C
-1
N
C
-1
O
+1
C
-1
O
•  Which resonance structure is the most representative
of the overall picture?
More Resonance Examples
O
O
O
S
O
H
N
O
O
O
H
H
C
C
H
H
O
N
N
N
O
C
-2
H
H
C
H
H
C
CH 3
CH3
NH
23