Chemical
Bonding
Chapter 8
Each person
should get
out a small
WHITE
BOARD
1
Salt
•
•
•
•
NaCl
Ionic (metal+non)
Electrolyte
High MP (801ºC)
vs
Sugar
•
•
•
•
C6H12O6
Molecular (non+non)
Nonelectrolyte
Lower MP (185ºC)
2
Who gets with who?
• Metals + nonmetals
• IONIC
• Nonmetals + nonmetals
• MOLECULAR
• Metals + metals
• METALLIC BONDING (Alloys if a mixture.)
3
Valence Electrons
• Lewis Structures
– (aka: Lewis Symbols or Dot Diagrams or Lewis
Dot Diagrams)
• Phosphorus
– Electron configuration: [Ne] 3s23p3
– Lewis Structure
P
– Shows only the valence electrons
4
Electronic Nirvana
• The octet rule - 8 valence electrons
• The reasons that atoms come to the party to
bond and form molecules is to obtain an
octet
– H only acquires a “duet” 2 electrons
• Using Lewis Structures to account for all the
electrons and attempt to achieve “electronic
nirvana.” i.e stable.
5
Drawing Lewis Structures
1. Sum the valence electrons.
– for polyatomic ions assign an extra electron for
each negative charge.
– Subtract and electron for each positive charge
2. Connect all atoms with single bonds
3. Complete the octet of all atoms.
4. If there are not enough electrons to go
around, use multiple bonds (one at a time).
5. Put excess electrons on the central atom
– This is called an expanded octet.
6
Rules to guide you when deciding
what connects to what.
•
•
•
•
Sometimes the order listed may help.
Central atom is often written first.
H’s and F’s are terminal (on outside).
C’s love to hook together and do not like to be
terminal.
– Avoid unshared pairs on C’s.
• O’s do not bond together.
– Except in O2 and in peroxide molecules
• Avoid rings of 3 or less atoms.
7
Lewis Structures
1. H2
6. H2S
2. O2
7. CH2O
3. N2
8. (CH3)2CO
4. CH4
9. SO2
5. NF3
10.CO2
8
Formal Charge
Choosing between two
legitimate structures
9
You might be asked to decide between
more than one possible structure?
• First be sure you have followed the guidelines
above that would eliminate some possible
structures.
−
−
O＝C＝O
−
−
∣−
− C≡ O ∣
O
−
• Chemists have invented a tool called formal
charge that will provide information to help
eliminate some structures.
10
Applying Formal Charge
•
•
Count the number of valence electrons “assigned” to each atom.
– Unshared pairs belong entirely to the atom on which they reside.
– Shared pairs are divided evenly between the atoms on which they are
shared: half “assigned” to one atom and half belong to the other atom.
Subtract the number of valence electrons that are “assigned” to each atom
in the molecule from the number of valence electrons that each atom has as
an isolated atom
(FORMULA: valence electrons in the un-bonded atom – electrons that are
“assigned” to the atom in the molecule)
– If the number of electrons that are “assigned” to the atom in the molecule
is one more than the number of valence electrons in the elemental atom,
the formal charge is -1 (two more, the formal charge is −2)
– If the number of electrons that are “assigned” to the atom in the molecule
is one less than the number of valence electrons in the elemental atom,
the formal charge is +1 (two less, the formal charge is +2)
•
– If the number of electrons that are “assigned” to the atom in the molecule
is equal to the number of valence electrons in the elemental atom, the
formal charge is 0
−
−
−
∣
− C≡ O ∣
O＝
C
＝
O
O
The formula with the least amount
−
−
−
of formal charge will be prefered.
-1
0
+1
0
0 0
11
Charges, Formal and oxidation #
• Something to remember about formal
charge and oxidation number.
• Formal charge is the charge on an atom if
the bonds were 100% covalent.
• Oxidation number is the charge on an atom
if the bonds were 100% ionic.
• Of course, bonds are neither 100% ionic or
100% covalent and so oxidation numbers
and formal charges are not "real" charges,
just "constructs" that help us do chemistry.
12
Using formal charge to evaluate
the validity of structures
✓ The sum of the formal charges must equal the charge of
the species
✓ Molecules have a charge of 0
✓ A polyatomic ion’s charge should equal the formal
charge.
• Preferred structures have the lowest amount formal charge
‣ Atoms without any charge (or just the charge of the ion)
is ideal
‣ Lower charges are preferred
‣ Two charges of −1 and +1 would be better than −2 and 0
• All else being equal, any negative charge that must exist is
best placed on the most electronegative atom.
• Ultimately it may be necessary to have other information to
determine the “actual” structure.
13
Lewis Structures
−
1. NO3
−
2. NO2
2−
3. CO3
2−
5. SO3
−
6. ClO3
7.
−
ClO
2−
4. SO4
14
Expanded Octet
• Elements with three or more energy levels
have unused d-orbitals which can be
accessed to put electrons into for bonding to
allow for expanded octets.
• The maximum number of domains we will
study will be 6
– 7 and 8 do exist, but they are quite rare − and
go well beyond the scope of this course.
15
Lewis Structures
1. PCl5
5. SF6
2. SF4
6. BrF5
3. ClF3
7. XeF4
4. XeF2
16
Bond Length
Single.Double.Triple
17
Bonds − Length & Strength
• Single
• Overlap of one pair of electrons - Longest = weaker
• Double
• Overlap of two pair of electrons - Shorter = stronger
• Triple
• Overlap of three pair of electrons - Shortest = strongest
• more on why it is the shortest in the next chapter.
Length (A) / strength (kJ/mol)
Length (A) / strength (kJ/mol)
347
160
614
839
305
615
891
418
945
201
607
358
745
1070
204
498
18
Polarity
Unfair Sharing of Electrons
19
Bond Polarity
• Electronegativity
– A measure of the ability of an atom (that is
bonded to another atom) to attract electrons to
itself.
– Increases up and to the right on the periodic
chart.
• To determine polarity of bond
– compare electronegativity of the two atoms.
– Treat each bond individually.
20
Determining Polarity &
Symbolizing Polarity
H
F
δ+ δH F
2.1 - 4.0 = 1.9, very polar
F is the negative end.
or
H
F
21
Polarity is a Continuum
•
Nonpolar covalent
•
Polar covalent
•
Ionic
22
Nonpolar - Polar - Ionic
• Using shape and color as a visual
representation of the three types of bonds.
• Polarity is measured as a dipole moment
measure in debye units.
0D
1.92 D
23
Hydrogen-halide series.
• Larger halide atoms mean smaller electronegativity.
• Difference in electronegativity decreases.
• Therefore polarity decreases and measured dipole
moment is smaller.
24
C6H14 (nonpolar) H2O (polar)
• C6H14 has polar bonds symmetrically arranged,
and the molecule ends up nonpolar.
• H2O has polar bonds not symmetrically
arranged, and the molecule ends up polar.
25
The effect of a charged rod
• The stream of polar
water coming from a
buret is bent by the
charged rod.
• Nonpolar hexane is
not bent.
26
Of the 3 isomers that exist for
C2H2F2 how many are polar?
1.
2.
3.
4.
1 structure
2 structures
all 3 structures
none are polar
27
Of the 3 isomers that exist for
C2H2F2 how many are polar?
1.
2.
3.
4.
1 structure
2 structures
all 3 structures
none are polar
ciscis-dichloroethene
transtrans dichloroethene
1,1 dichloroethene
28
Which of these structures is most polar?
1.
2.
3.
4.
5.
6.
Ortho
ortho
meta
para
Meta
Para
Ortho and meta are equally polar
All three structures are equally polar
None of the structures are polar
29
Which of these structures is most polar?
1.
2.
3.
4.
5.
•
Ortho
ortho
meta
para
Meta
Para
Ortho and meta are equally polar
All three structures are polar
This is a planar molecule, and the
molecule will be most polar when the
polar C-Cl bond is oriented in the
same direction.
30
Select the non-polar bond(s) shown
below. Select all that apply.
1.
2.
3.
4.
5.
C−F
C−H
C−O
C−C
C=C
31
Select the least polar bond(s)
shown below
1.
2.
3.
4.
5.
•
C-F
C-H this bond is considered nonpolar.
However it is ever SO slightly polar.
C-O
C-C
C=C
Whether single or double, the polarity or
lack there of is calculate in the same
way.
32
Isomers
Same Chemical Formula
Different Arrangement of atoms
33
Isomers
• Isomers - Molecules with the same chemical
formula, but different arrangement of atoms.
• Draw a Lewis structure for C2H6
• Draw a Lewis structure for C2H4F2
– Draw isomers and name these isomers
– Free rotation around single
• Draw a Lewis structure for C2H2F2
– Draw isomers
– Let’s name these isomers
34
Isomers
• Sketch and build the molecule C4H10
• Sketch and build the molecule C4H8
• Draw as many isomers as you can for this
molecule.
35
Draw a Lewis Structure for C2H4F2
1.
Draw as many isomers as you can
36
Isomers are:
1.
2.
3.
4.
same chemical structure, different
number of neutrons.
same chemical structure, different
arrangement of electrons.
same chemical formula, but different
arrangement of atoms.
same chemical formula, different
number of electrons.
37
Isomers are:
1.
2.
3.
4.
same chemical structure, different
number of neutrons.
same chemical structure, different
arrangement of electrons.
same chemical formula, but different
arrangement of atoms.
same chemical formula, different
number of electrons.
38
Draw the structure C2H4F2
Select the number of different
isomers that can exist for C2H4F2
1.
2.
3.
4.
5.
1 structure
2 structures
3 structures
4 structures
5 structures
39
Select the number of different
isomers that can exist for C2H4F2
1.
2.
3.
4.
5.
1 structure
2 structures
3 structures
4 structures
5 structures
same
This structure is the same as the
one above it because of free
rotation around the single bond.
40
Draw the structure C2H2F2
Select the number of different
isomers that can exist for C2H2F2
1.
2.
3.
4.
5.
1 structure
2 structures
3 structures
4 structures
5 structures
41
Select the number of different
isomers that can exist for C2H2F2
1.
2.
3.
4.
5.
1 structure
2 structures
3 structures
4 structures
5 structures
ciscis-1,2 dichloroethene
transtrans 1,2 dichloroethene 1,1 dichloroethene
42
Resonance
Same arrangement of atoms
Movement of electrons
43
Resonance
• Two or more Lewis Structures in which;
– The connection and placement of the atoms is
the same
– But the arrangement of the electrons is different
• In contrast with an isomer which is a
difference arrangement of atoms
44
Carboxylic Acids: stable conjugate
After the H+ falls off, the C=O and C−O bonds can exhibit
resonance
•
✓
•
Resonance is the relocation electrons in an available nearby
location resulting in an “averaging” of the structures.
Resonance allows the electrons to “spread out,” distributing the
charge and stabilizing the conjugate base.
H
45
Resonance − we draw two (or three) separate Lewis
structures, but really, the molecule is a blend of the
various structures we can draw.
46
Resonance in Ozone: O3
• Draw a Lewis structure for ozone.
• Re-sketch the molecule with electrons in a different
arrangement - This is called a resonance structure.
• You wouldn’t ordinarily write two different versions of this, but
chemists noticed that the two bonds in ozone were the same
length AND that they both were shorter than a single bond,
but longer than a double bond.
• Chemists think that molecule does not actually flip between
the two structures, but is an average of the two.
47
Resonance in N2O
• Draw a Lewis structure for N2O
– Draw two isomers
• N−N−O
• N−O−N
– Can you draw resonance structures for these?
48
Resonance in N2O
• Draw four or five Lewis structures for N2O to
represent two different isomers.
– and draw at least two resonance structures for
each isomer
– How do we decide which is more likely to occur
• Formal Charge
49
Resonance in N2O
• Draw a Lewis structure for N2O
– Put one of the N’s in the middle.
– Putting the O in the middle leads to lousy formal
charge
• Draw two other resonance structures.
• Draw an isomers of the structure above and
it’s resonance structure.
−
−
N
＝
O
＝
N
−
−
∣−
N− O≡ N ∣
−
50
Using Formal Charge on N2O
• Apply formal charges to the 3 resonance
structures of N2O.
0
+1
-1
-2
−
−
N ＝O＝N
−
−
-1 +2 -1
+1
+1
-1
+1
0
∣−
− O≡ N ∣
N
−
-2 +2 0
• Remember the rule:
– If you must choose between two equal amounts
of formal charge, the negative formal charges
should reside on the more electronegative
atoms.
51
Resonance in Benzene: C6H6
• Draw a Lewis Structure C6H6 with 30 valence electrons make a ring structure.
• Resonance
52
Resonance in Benzene: C6H6
• C6H6 with 30 valence electrons - make a ring structure double bonds will be necessary.
• The molecule does not flip between the two structures, but
is an average of the two.
• Bond length studies tell us that all 6 C-C bonds have the
same length - intermediate between single and double.
53
Representing Benzene: C6H6
• This is an alternative way of representing carbon
structures.
• At each corner the appropriate number of hydrogens are
attached.
• The circle structure on the right emphasizes the
delocalization of the electrons.
54
Which structure(s) below exhibit(s)
resonance structures:
A) CO2 B) NO2− C) SO2
1.
2.
3.
4.
5.
B
C
A & C only
B & C only
A, B, & C
55
Which structure(s) below exhibit(s)
resonance structures:
A) CO2 B) NO2− C) SO2
−
−
−
−
−
−
−
−
−
−
−
−
1. B
− − − −
− − −
O − N＝O
O − S＝O
2. C
−
−
−
−
−
−
− − −
3. A & C only
O＝N− O
O＝S− O
−
−
4. B & C only
5. A, B, & C
− C − While you can write a resonance structure
O ＝ ＝O for CO2, with a triple bond and a single
bond, the formal charge is very poor, thus
− C≡
O−
O the resonance structure is not valid.
−
56
The End
57
Exceptions to the
Octet Rule
Other than the
Expanded Octet
58
Exceptions to the Octet Rule
•
•
Odd number of electrons.
Sketch a Lewis Structure for
– NO, NO2
59
Odd Number of Electrons
Free Radicals
• Free radicals are very unstable and react quickly
with other compounds, trying to capture the
needed electron to gain stability.
• Generally, free radicals attack the nearest stable
molecule, "stealing" its electron.
• When the "attacked" molecule loses its electron, it
becomes a free radical itself, beginning a chain
reaction.
• Once the process is started, it can cascade, finally
resulting in the disruption of a living cell.
60
Exceptions to the Octet Rule
•
Odd number of electrons.
– NO, NO2
•
Less than an octet.
– BF3
61
Less Than an Octet
• Sketch a Lewis Structure for BF3
• BF3 gas is very reactive because of its
incomplete octet.
• It reacts with molecules such as water
and ammonia with their available
unshared pair of electrons.
BF3 + NH3 → H3NBF3
• In H3NBF3 the boron would now have
an octet.
62
Exceptions to the Octet Rule
•
Odd number of electrons.
– NO, NO2
•
Less than an octet.
– BF3
•
More than an octet.
– Expanded octet on a central atom
– only atoms that have unused “d” orbitals
– Period 3 and beyond
63
Silly Chemistry Jokes
Early chemists describe the first dirt molecule.
65
Hybridization
Since one bonding model
may not be enough!
66
Why another model?
• VSEPR was useful to predict the 3-D shape of
molecules, with little explanation as to why the bond
exists.
• We have said that a bond is an overlap of orbitals.
• In polyatomic molecules we need to blend the
concept of overlap of orbitals with the observed
geometry.
67
Hybrid Orbitals
• Atomic orbitals have particular shapes.
• When an atom is about to become part of a
molecule, that atom’s atomic orbitals must
morph into a new set of orbitals with different
shapes and orientations than the original
atomic orbitals. These new morphed orbitals
are called hybrid orbitals.
• The orientation of hybrid orbitals is the same
orientation of the shapes you learned from the
VSEPR model.
68
Summary of Hybrid Orbitals
Atomic orbitals
that morph
together
To make hybrid
orbitals
The resulting
shape of the hybrid
orbitals
s+p
(px py or pz)
Two equivalent
sp orbitals
Linear
s+p+p
Three equivalent
sp2 orbitals
Trigonal planar
s+p+p+p
Four equivalent
sp3 orbitals
Tetrahedral
s+p+p+p+
d
Five equivalent
sp3d orbitals
Trigonal
bipyramidal
s+p+p+p+
d+d
Six equivalent
sp3d2 orbitals
Octahedral
69
sp hybrid
Linear
s
p
sp sp
sp
sp
next.... 70
2
sp
s
hybrid
p
p
sp2
sp2
Trigonal Planar
sp2
sp2
sp2
sp2
next.... 71
s
p
p
p
sp3 hybrid
sp3
sp3
Tetrahedral
sp3
sp3
next.... 72
Bonding is Overlap of Orbitals
NH3
• So when N gets ready to bond in the ammonia molecule, its
5 valence electrons in the s and p orbitals are morphed into
4 orbitals called sp3 hybrid orbitals.
• One of the sp3 orbitals will have 2 electrons and the other
three sp3 orbitals each have 1 electron.
• Hydrogen only has 1 electron in a single s orbital, so there
is no morphing to be done.
• Three of the hybrid sp3 orbitals in
the N overlap with the single
sp3
atomic s orbital in each of the three
3
sp
H’s to make the three bonds.
sp3
sp3
73
So What about Multiple Bonds?
- Double and Triple Bonds Sigma (σ) and Pi (π) Bonds
• Overlap region that occurs directly between
two nuclei on the internuclear axis is called a
sigma bond - these are single bonds.
• Overlap of two p orbitals oriented
perpendicular to the internuclear axis is called
a pi bond - the other part(s) of the double and
triple bonds.
74
Pi (π) Bonds - The “other” part of
Double and Triple Bonds
• Side to side overlap of
the p orbitals.
• Above and below the
internuclear axis.
• Total overlap covers
less area causing the
pi (part) bond to be
weaker.
• The “hot dog bun”
bonds.
75
Draw a Lewis structure for C2H4
• Let’s look at what’s going on around each C
 How many domains around one of the C’s?
 3 domains
 What is the domain shape?
 Trigonal planar
 What is the name of the hybrid orbitals?
 The trigonal planar geometry around each carbon indicates sp2
hybridorbitals.
H
H
C
C
H
C
H
76
Hybridization in C2H4
• The view below looks down on the molecule. The
unhybridized p orbitals are not showing. They
would be projecting above and below this picture.
• The sp2 hybrids on the carbons are overlapping
with the s orbitals of the hydrogens.
H
H
C
C
H
C
H
77
Hybridization in C2H4
• Let’s consider the unhybridized p orbitals on the
carbon again and look at the side view.
• Overlap of those p orbitals completes the pi bond
(the double bond)
 Note that the pi bond has two parts (one part above the
sigma bond and one part below the sigma bond).Only 1
part is labeled.
C
σ
σ
σ
C
σ
σ
π
78
Sigma (σ) and Pi (π) Bonds in C2H4
• Look at your Lewis structure of C2H4
• Indicate the Sigma (σ) and Pi (π) Bonds
Between the C’s 1 σ bond
and one π bond
H
H
C
C
H
H
1 σ bond in each C-H bond
79
Hybridization in CH2O
• Sketch Lewis Structure CH2O
• What is the # of domains around the C and the O?
 3 domains around each
• What does this tell us about the hybridization around the C
and O?
 sp2 hybridization on both C and O
• This leaves one p orbital unhybridized on each atom. This
unhybridized p overlaps to make the π bond
H
H
C
O
π
sp2
sp2
sp2
p
sp2
sp2
sp2
p
80
Draw a Lewis Structure for CO2
• What is # number of domains around the C?
 2 domains
• What does this tell us about the hybridization
around the C?
sp hybridization on the C
• This leaves two p orbitals unhybridized on the C.
These unhybridized p’s overlap to make the π bonds
O
C
O
81
More about hybridization in CO2
• What are the number of domains around one of the O’s
 3 domains
• What would be the hybridization orbitals?
 sp2 hybridization.
• This would leave one unhybridized p orbital around the
oxygen.
O
C
O
82
Hybridization Overlap in CO2
• The view below looks down on the molecule. The
unhybridized p orbitals are not showing. They
would be projecting above and below this picture.
• The carbon’s sp hybrid orbitals overlap with the
oxygen’s sp2 hybrid orbitals.
• Note how the oxygen orbitals are oriented in
opposite planes.
83
Bonding in CO2
• Put the sigma bonds together with the pi
bonds for the entire molecule.
84
Draw a Lewis structure for C6H6
•
What’s the number of domains and their shape around each C?
 3, trigonal planar
•
What’s the hybridization around each C?
 Trigonal planar geometry around each carbon indicates sp2 hybridization
•
•
•
This leaves a single p orbital on each carbon projecting above and below
the planar ring.
The p orbitals overlap for pi bonding and because of the alternating
double bonds you can write resonance structures
These electrons are said to be “delocalized.”
tip this ring flat
85
Draw a Lewis structure for N2
• The presence of a triple bond tells us that there are how
many unhybridized p orbitals?
 Two p orbitals on each N must be unhybridized.
• What is the hybridization of each N?
 2 p’s unhybridized indicates sp hybridization
• The last diagram shows the sigma overlap of the sp hybrid
orbitals to make up one third of the triple bond.
N
N
86
Bonding in N2
• Pi orbitals overlap above and below the
internuclear axis.
• The “hot dog bun” bonds !
87
Draw a Lewis structure for PF5
•
How many domains and the
shape around the P?
 5 domains, trigonal bipyramid.
•
What is the hybridization of
around the P?
 sp3d hybridization
•
How many domains and the
shape around each F?
sp3
sp3
sp3
sp3
F
F
sp3d
 4 domains, tetrahedral shape
•
What is the hybridization around
each F?
•
sp3 hybridization
F
sp3d
P
F
sp3d
sp3d
sp3d
F
88
Geometry, Hybridization & σ, π bonds
Around the left O
• Name the electron
domain geometry.
 tetrahedral
• Name the molecular
geometry (shape).
 bent
• Give the bond angle
 <<109.5º
• Name the hybridization
orbitals.
Around the Cl
•
Name the electron domain geometry.
 tetrahedral
•
Name the molecular geometry (shape).
 Trigonal pyramid
•
Give the bond angle
 <109.5º
•
Name the hybridization orbitals.
 sp3
•
List the number of σ and
π bonds.
 3 σ bonds, no π bonds
 sp3
• List the number of σ and
π bonds.
 2 σ bonds, no π bonds
89
Geometry, Hybridization & σ π bonds
Around the C
• Name the electron
domain geometry.
 tetrahedral
• Name the molecular
geometry (shape).
 tetrahedral
Around the O
•
 tetrahedral
•
• Name the hybridization
orbitals.
Name the molecular geometry (shape).
 bent
•
Give the bond angle
 <<109.5º
•
• Give the bond angle
 109.5º
Name the electron domain geometry.
Name the hybridization orbitals.
 sp3
•
List the number of σ and
π bonds.
 2 σ bonds, no π bonds
 sp3
• List the number of σ and
π bonds.
 4 σ bonds, no π bonds
90
Geometry, Hybridization & σ π bonds
Around the left C
• Name the electron domain
geometry.
 linear
• Name the molecular geometry
(shape).
 linear
• Give the bond angle
 180º
• Name the hybridization orbitals.
 sp
• List the number of σ and π
bonds.
 2 σ bonds, 2 π bonds
91
Geometry, Hybridization & σ π bonds
Around the left C
Around the right O
• Name the electron
domain geometry.
•
 Trigonal planar
•
• Name the molecular
geometry (shape).
 trigonal planar
• Give the bond angle (6)
 >120º
• Name the hybridization
orbitals.
Name the electron domain geometry.
 tetrahedral
Name the molecular geometry (shape).
 bent
•
Give the bond angle (7)
 <<109.5º
•
Name the hybridization orbitals.
 sp3
•
List the number of σ and
π bonds.
 2 σ bonds, no π bonds
 sp2
• List the number of σ and
π bonds.
 3 σ bonds, 1 π bonds
92
Bonding and Geometry
Review
Chapter 8 & 9
True / False
93
The Lewis structure of N2 is:
94
The Lewis structure of N2 is:
• False
• There is a triple bond and one unshared pair
of electrons on each N for a total of 10 (not
12) valence electrons in the molecule.
95
The product in the reaction:
BF3(g) + NH3(g) → F3BNH3(g)
contains a single covalent bond between
boron and nitrogen
96
The product in the reaction:
BF3(g) + NH3(g) → F3B:NH3(g)
contains a single covalent bond between
boron and nitrogen
• True
• This bond forms because nitrogen can donate
its two non-bonded electrons to form what is
sometimes called a coordinate covalent bond.
Ultimately it’s a bond - made of two shared
electrons - just like any other bond.
• This is a Lewis acid/base reaction.
• LA - accepts electrons, LB - donates electrons
97
The bond between Ca and O is covalent
in character.
98
The bond between Ca and O is covalent
in nature.
• False:
• It is ionic. Oxides of metals are always ionic
as it is made of a metal and a nonmetal.
99
For the following hydrogen halides,
bond polarity increases in the order
listed below.
HCl < HBr < HI
100
For the following hydrogen halides,
bond polarity increases in the order
listed below.
HCl < HBr < HI
• False.
• Polarity increases with electronegativity
on the halogen. Thus the order is HI <
HBr < HCl
101
Review:
For the following hydrogen halides,
acidity increases in the order listed
below.
HCl < HBr < HI
102
Review:
For the following hydrogen halides,
acidity increases in the order listed
below.
HCl < HBr < HI
• True.
• The Acidity increase is NOT due to the
polarity of the bond, but instead
associated with the length of the bond.
• While they are all strong acids with
very large Ka values HI is the
strongest.
103
In MgCl2 there is both an ionic
Mg-Cl bond and a covalent Cl-Cl bond.
104
In MgCl2 there is both an ionic
Mg-Cl bond and a covalent Cl-Cl bond.
• False:
• Mg is ionic bonded to both Cl’s which are not
bonded to each other.
105
The lattice energy of MgO is greater than
that of NaCl.
106
The lattice energy of MgO is greater
than that of NaCl.
• True.
• Lattice energy increases with increasing ionic
charge. (The most important of the two factors.) The
charge on Mg and O is 2 (not 1 as in Na and Cl) thus
the lattice energy in MgO is greater.
• The second factor is that lattice energy increases
with decreasing size of the ions. For this compound
the size of O2− is significantly smaller than Cl− (In
addition, Mg2+ is slightly smaller than Na+) making
MgO an even tighter lattice.
107
In H3CCH3 there exists both sigma and
pi bonds.
108
In H3CCH3 there exists a both sigma
and pi bonds.
• False.
• There are no double bonds in this
molecule, thus there are no pi bonds.
H
H C
H
H
C H
H
109
In HCCH, acetylene, there exists two
pi bonds.
110
In HCCH, acetylene, there exists two
pi bonds.
• True.
• There is a triple bond between the two
C’s and and a triple bond always
contains two pi bonds.
H
C
C
H
111
Fe can form a 2+ ion because it loses its
two 4s valence electrons.
112
Fe can form a 2+ ion when it loses its
two 4s valence electrons.
• True.
• Fe 4s2 3d6
•
⊗ ⊗∅∅∅∅
• The 4s electrons are furthest away from the
nucleus and get ripped off by other atoms.
• One of the 3d6 electron is also quite willing
to leave and thus Fe often forms 3+ ion as
well. Fe+3 O ∅ ∅ ∅ ∅ ∅
113
In the BF3 molecule, the differing
electronegativities of B and F cause the
bonds and the molecule to be polar.
114
In the BF3 molecule, the differing
electronegativities of B and F and cause
the bonds and the molecule to be polar.
• False
• The B-F bonds are polar, but because the
molecule is trigonal planar and all three
bonds are the same, the polarity cancels out.
115
There are three isomers each for the
following molecules:
C2H4F2 and C2H2F2
116
H H
There are three isomers each
for the following molecules: H C C F
H H
C2H2F2 and C2H4F2
H F
F C C F
H H
• False there are only two for C2H4F4 because
there IS rotation around a single bond, but there
ARE three for C2H2F2 because there is NO
rotation around a double bond
ciscis-1,2 dichloroethene
transtrans 1,2 dichloroethene 1,1 dichloroethene117
The dissociation of Cl2 to form two Cl
atoms in the gas phase is endothermic,
and this energy is termed bond enthalpy.
118
The dissociation of Cl2 to form two Cl
atoms in the gas phase is endothermic,
and this energy is termed bond enthalpy.
• True
• It is called bond enthalpy and it is
always endothermic. Bond breaking is
always endothermic and bond forming
is exothermic.
119
The H2Se and H2O are analogous
molecules. The bonds in both are polar.
Electronegativity values:
H=2.2, O=3.5 Se=2.4
120
The bonds in H2Se are analogous to
the bonds in H2O and are polar.
Electronegativity values:
H=2.2, O=3.5 Se=2.4
• False.
• They are analogous molecules (comparable in
certain respects, in particular their valence
electrons and Lewis structures), but the H2Se
bonds are nonpolar.
121
The reaction of calcium with Cl to form
CaCl2 illustrates the octet rule.
122
The reaction of calcium with Cl to
form CaCl2 illustrates the octet rule.
• True.
• The calcium achieves an octet by losing
its two outer electrons. The chlorines
complete their octets by each gaining
one electron.
• All ionic bonds form in order to satisfy
their octets
123
The oxidation state (or number) of
iodine in ICl is -1.
124
The oxidation state (or number) of
iodine in ICl is -1.
• False.
• Because Cl is more electronegative, it
carries a -1 oxidation and the I is +1
125
One of the three
resonance structures
of the sulfite ion is:
126
One of the three
resonance structures
of the sulfite ion is:
• False.
• There are not enough electrons in this structure.
There should be 4(6) + 2 = 26, 13 lines.
• This is a resonance structure for SO3, the
molecule, not SO3-2 the polyatomic ion. The sulfite
ion does not have a double bond and thus does
not have resonance structures
127
FeCl3 contains a metallic bond.
128
FeCl3 contains a metallic bond.
• False.
• FeCl3 has ionic bonds.
• Metallic bonds are the name given to bonds
between metal atoms - a subject we will take
up later in Chap 23.
129
CO2 has polar bonds and is a polar
molecule.
Electronegativity values:
C=2.5 O=3.5
130
CO2 has polar bonds and is a polar
molecule.
Electronegativity values:
C=2.5 O=3.5
• False.
• The C=O bond is polar, but they are
symmetrically oriented causing the polarity
to cancel out and the molecule to be
nonpolar.
131
The bond enthalpy of a CN triple bond is
larger than that for a CN double bond.
132
The bond enthalpy of a CN triple bond
is larger than that for a CN double
bond.
• True.
• The enthalpy for a triple bond is always
higher than for an analogous double
bond.
133
A sigma bond can exist between
hybridized orbitals and atomic orbitals.
134
A sigma bond can exist between
hybridized orbitals and atomic
orbitals.
• True.
• Remember that a bond is overlap of any
two orbitals, which can be atomic or hybrid
orbitals.
135
According to the VSEPR model the
H2S molecule is linear.
136
According to the VSEPR model the
H2S molecule is linear.
• False.
• There are 4 domains around the central S. It is
bent due to the two unshared pairs on the S atom.
137
Hybridization of the s and p orbitals
named sp3 hybrids in carbon in CH4 is
responsible for its tetrahedral structure.
138
Hybridization of the s and p orbitals
named sp3 hybrids in carbon in CH4 is
responsible for its tetrahedral structure.
• True.
139
Benzene, C6H6 is an example of a
molecule in which electrons are
delocalized over the s orbitals of carbon.
140
Benzene, C6H6 is an example of a
molecule in which electrons are
delocalized over the s orbitals of carbon.
• False.
• It is the p orbitals that contain the
delocalized electrons.
141
In NH3 the electron domain geometry
is tetrahedral.
142
In NH3 the electron domain geometry
is tetrahedral.
• True.
• The electron domain geometry is tetrahedral.
• The molecular geometry is trigonal pyramidal
due to the unshared pair on the N.
143
•
All three of the AB3 molecules below contain
unshared pairs on the central A atom.
144
•
All three of the AB3 molecules below contain
unshared pairs.
• False.
• Molecule i is trigonal planar.
• Trigonal planar never has unshared pairs.
145
Molecules ii and iii each contain only one
nonbonded pair of electrons.
146
Molecules ii and iii each contain one nonbonded
pair of electrons.
• False.
• Structure ii does have 1 unshared pair causing
trigonal pyramid
• But structure iii is a (laying sideways) trigonal
bipyramid with two nonbonded pairs on iii causing the
t-shape.
147
The End
148