Transcript No Slide Title

Chemical Bonding I:
The Covalent Bond
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Valence electrons are the outer shell electrons of an
atom. The valence electrons are the electrons that
particpate in chemical bonding.
Group
e- configuration
# of valence e-
1A
ns1
1
2A
ns2
2
3A
ns2np1
3
4A
ns2np2
4
5A
ns2np3
5
6A
ns2np4
6
7A
ns2np5
7
2
Lewis Dot Symbols for the Representative Elements &
Noble Gases
3
The Ionic Bond
Ionic bond: the electrostatic force that holds ions together in
an ionic compound.
Li + F
1 22s22p5
1s22s1s
LiF
e- +
Li+ +
Li+ F 1s2 1s22s22p6
[He]
[Ne]
Li
Li+ + e-
F
F -
F -
Li+ F -
4
Electrostatic (Lattice) Energy
Lattice energy (U) (LE) of an ionic compound is the energy
required to break apart the ions in their lattice arrangement
into the ions in the gas phase.
For example the energy to take NaCl from its lattice arrangement
into the gas phase ions would be represented by NaCl(s) --->
Na+(g) + Cl-(g)
5
Born-Haber Cycle for Determining Lattice Energy
°
DHoverall
= DH1°+ DH2 °+ DH3°+ DH4 °+ DH5 °
6
7
A covalent bond is a chemical bond in which two or more
electrons are shared by two atoms.
Why should two atoms share electrons?
F
+
7e-
F
F F
7e-
8e- 8e-
Lewis structure of F2
single covalent bond
lone pairs
F
F
lone pairs
single covalent bond
lone pairs
F F
lone pairs
8
Sigma (s) and Pi Bonds (p)
1 sigma bond
Single bond
Double bond
1 sigma bond and 1 pi bond
Triple bond
1 sigma bond and 2 pi bonds
How many s and p bonds are in the acetic acid (vinegar)
molecule CH3COOH?
H
C
H
O
H
C
O
H
s bonds = 6 + 1 = 7
p bonds = 1
9
Lewis structure of water
H
+
O +
H
single covalent bonds
H O H
or
H
O
H
2e- 8e- 2eDouble bond – two atoms share two pairs of electrons
O C O
or
8e- 8e- 8e-
O
O
C
double bonds
double bonds
Triple bond – two atoms share three pairs of electrons
N N
8e- 8e-
triple bond
or
N
N
triple bond
10
Lengths of Covalent Bonds
Bond Lengths
Triple bond < Double Bond < Single Bond
11
Polar covalent bond or polar bond is a covalent
bond with greater electron density around one of the
two atoms
electron poor
region
H
electron rich
region
F
e- poor
H
d+
e- rich
F
d-
12
Electronegativity is the ability of an atom to attract
toward itself the electrons in a chemical bond.
Electron Affinity - measurable, Cl is highest
X (g) + e-
X-(g)
Electronegativity - relative, F is highest
13
Variation of Electronegativity with Atomic Number
14
Classification of bonds by difference in electronegativity
Difference
Bond Type
0
Covalent
2
0 < and <2
Ionic
Polar Covalent
Increasing difference in electronegativity
Covalent
share e-
Polar Covalent
partial transfer of e-
Ionic
transfer e15
Classify the following bonds as ionic, polar covalent, or
covalent: The bond in CsCl; the bond in H2S; and the NN
bond in H2NNH2.
Cs – 0.7
Cl – 3.0
3.0 – 0.7 = 2.3
Ionic
H – 2.1
S – 2.5
2.5 – 2.1 = 0.4
Polar Covalent
N – 3.0
N – 3.0
3.0 – 3.0 = 0
Covalent
16
Writing Lewis Structures
1. Draw skeletal structure of compound showing
what atoms are bonded to each other. Put least
electronegative element in the center.
2. Count total number of valence e-. Add 1 for
each negative charge. Subtract 1 for each
positive charge.
3. Complete an octet for all atoms except
hydrogen
4. If structure contains too many electrons, form
double and triple bonds on central atom as
needed.
17
Write the Lewis structure of nitrogen trifluoride (NF3).
Step 1 – N is less electronegative than F, put N in center
Step 2 – Count valence electrons N - 5 (2s22p3) and F - 7 (2s22p5)
5 + (3 x 7) = 26 valence electrons
Step 3 – Draw single bonds between N and F atoms and complete
octets on N and F atoms.
Step 4 - Check, are # of e- in structure equal to number of valence e- ?
3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons
F
N
F
F
18
Write the Lewis structure of the carbonate ion (CO32-).
Step 1 – C is less electronegative than O, put C in center
Step 2 – Count valence electrons C - 4 (2s22p2) and O - 6 (2s22p4)
-2 charge – 2e4 + (3 x 6) + 2 = 24 valence electrons
Step 3 – Draw single bonds between C and O atoms and complete
octet on C and O atoms.
Step 4 - Check, are # of e- in structure equal to number of valence e- ?
3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons
Step 5 - Too many electrons, form double bond and re-check # of e-
O
C
O
O
2 single bonds (2x2) = 4
1 double bond = 4
8 lone pairs (8x2) = 16
Total = 24
19
A resonance structure is one of two or more Lewis structures
for a single molecule that cannot be represented accurately by
only one Lewis structure.
O
O
+
-
-
O
O
+
O
O
What are the resonance structures of the
carbonate (CO32-) ion?
-
O
C
O
O
-
O
C
O
O
-
-
-
O
C
O
O
-
20
Exceptions to the Octet Rule
The Incomplete Octet
BeH2
BF3
B – 3e3F – 3x7e24e-
Be – 2e2H – 2x1e4e-
F
B
H
F
Be
H
3 single bonds (3x2) = 6
9 lone pairs (9x2) = 18
Total = 24
F
21
Exceptions to the Octet Rule
Odd-Electron Molecules
NO
N – 5eO – 6e11e-
N
O
The Expanded Octet (central atom with principal quantum number n > 2)
SF6
S – 6e6F – 42e48e-
F
F
F
S
F
F
F
6 single bonds (6x2) = 12
18 lone pairs (18x2) = 36
Total = 48
22
Chemical Bonding II
Molecular Geometry and
Hybridization of Atomic Orbitals
23
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Valence shell electron pair repulsion (VSEPR) model:
Predict the geometry of the molecule from the electrostatic
repulsions between the electron (bonding and nonbonding) pairs.
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
B
B
24
0 lone pairs on central atom
Cl
Be
Cl
2 atoms bonded to central atom
25
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
0
trigonal
planar
trigonal
planar
AB3
3
26
27
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
trigonal
planar
tetrahedral
AB3
3
0
trigonal
planar
AB4
4
0
tetrahedral
28
29
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
trigonal
planar
AB3
3
0
trigonal
planar
AB4
4
0
tetrahedral
tetrahedral
AB5
5
0
trigonal
bipyramidal
trigonal
bipyramidal
30
31
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
trigonal
planar
AB3
3
0
trigonal
planar
AB4
4
0
tetrahedral
tetrahedral
AB5
5
0
trigonal
bipyramidal
trigonal
bipyramidal
AB6
6
0
octahedral
octahedral
32
33
Predicting Molecular Geometry
1. Draw Lewis structure for molecule.
2. Count number of lone pairs on the central atom and
number of atoms bonded to the central atom.
3. Use VSEPR to predict the geometry of the molecule.
What are the molecular geometries of SO2 and SF4?
O
S
AB2E
bent
F
O
F
S
F
AB4E
F
distorted
tetrahedron
34
Dipole Moments and Polar Molecules
electron poor
region
electron rich
region
H
F
d+
d-
Dipole moment is defined as magnitude of charge (Q) times distance (r)
between the charges.
u = (Q)(r)
Q charge in coulombs (C)
r distance between charges in meters (m)
35
Behavior of Polar Molecules
field off
field on
36
Which of the following molecules have a dipole moment?
H2O, CO2, SO2, and CF4
O
S
dipole moment
polar molecule
dipole moment
polar molecule
F
O
C
C
O
F
no dipole moment
nonpolar molecule
F
F
no dipole moment
nonpolar molecule
37
Does CH2Cl2 have a
dipole moment?
38
39
How does Lewis theory explain the bonds in H2 and F2?
Sharing of two electrons between the two atoms.
Bond Enthalpy
Bond Length
Overlap Of
H2
436.4 kJ/mol
74 pm
2 1s
F2
150.6 kJ/mol
142 pm
2 2p
Valence bond theory – bonds are formed by sharing
of e- from overlapping atomic orbitals.
40
Change in electron density as two hydrogen atoms
approach each other.
41
Valence Bond Theory and NH3
N – 1s22s22p3
3 H – 1s1
If the bonds form from overlap of 3 2p orbitals on nitrogen with
the 1s orbital on each hydrogen atom, what would the molecular
geometry of NH3 be?
If use the
3 2p orbitals
predict 90o
Actual H-N-H
bond angle is
107.3o
42
Hybridization – mixing of two or more atomic
orbitals to form a new set of hybrid orbitals.
1. Mix at least 2 nonequivalent atomic orbitals (e.g. s
and p). Hybrid orbitals have very different shape
from original atomic orbitals.
2. Number of hybrid orbitals is equal to number of
pure atomic orbitals used in the hybridization
process.
3. Covalent bonds are formed by:
a. Overlap of hybrid orbitals with atomic orbitals
b. Overlap of hybrid orbitals with other hybrid
orbitals
43
Formation of sp3 Hybrid Orbitals
44
Formation of Covalent Bonds in CH4
45
Formation of sp Hybrid Orbitals
46
Formation of sp2 Hybrid Orbitals
47
How do I predict the hybridization of the central atom?
1. Draw the Lewis structure of the molecule.
2. Count the number of lone pairs AND the number of
atoms bonded to the central atom
# of Lone Pairs
+
# of Bonded Atoms
Hybridization
Examples
2
sp
BeCl2
3
sp2
BF3
4
sp3
CH4, NH3, H2O
5
sp3d
PCl5
6
sp3d2
SF6
48
Stoichiometry
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Micro World
atoms & molecules
Macro World
grams
Atomic mass is the mass of an atom in
atomic mass units (amu)
By definition:
1 atom 12C “weighs” 12 amu
On this scale
1H
= 1.008 amu
16O
= 16.00 amu
50
The average atomic mass is the weighted
average of all of the naturally occurring
isotopes of the element.
51
Naturally occurring lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
Average atomic mass of lithium:
(7.42 x 6.015) + (92.58 x 7.016)
= 6.941 amu
100
52
Average atomic mass (6.941)
53
The Mole (mol): A unit to count numbers of particles
Dozen = 12
Pair = 2
The mole (mol) is the amount of a substance that
contains as many elementary entities as there
are atoms in exactly 12.00 grams of 12C
1 mol = NA = 6.0221367 x 1023
Avogadro’s number (NA)
54
eggs
Molar mass is the mass of 1 mole of shoes in grams
marbles
atoms
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms = 6.941 g of Li
For any element
atomic mass (amu) = molar mass (grams)
55
One Mole of:
S
C
Hg
Cu
Fe
56
How many atoms are in 0.551 g of potassium (K) ?
1 mol K = 39.10 g K
1 mol K = 6.022 x 1023 atoms K
1 mol K
6.022 x 1023 atoms K
0.551 g K x
x
=
1 mol K
39.10 g K
8.49 x 1021 atoms K
57
Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
1S
SO2
2O
SO2
32.07 amu
+ 2 x 16.00 amu
64.07 amu
For any molecule
molecular mass (amu) = molar mass (grams)
1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2
58
How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
1 mol C3H8O molecules = 8 mol H atoms
1 mol H = 6.022 x 1023 atoms H
1 mol C3H8O 8 mol H atoms 6.022 x 1023 H atoms
72.5 g C3H8O x
x
x
=
1 mol C3H8O
1 mol H atoms
60 g C3H8O
5.82 x 1024 atoms H
59
Formula mass is the sum of the atomic masses
(in amu) in a formula unit of an ionic compound.
1Na
NaCl
22.99 amu
1Cl + 35.45 amu
NaCl
58.44 amu
For any ionic compound
formula mass (amu) = molar mass (grams)
1 formula unit NaCl = 58.44 amu
1 mole NaCl = 58.44 g NaCl
60
What is the formula mass of Ca3(PO4)2 ?
1 formula unit of Ca3(PO4)2
3 Ca
3 x 40.08
2P
2 x 30.97
8O
+ 8 x 16.00
310.18 amu
61
Percent composition of an element in a compound =
n x molar mass of element
x 100%
molar mass of compound
n is the number of moles of the element in 1 mole
of the compound
2 x (12.01 g)
x 100% = 52.14%
46.07 g
6 x (1.008 g)
%H =
x 100% = 13.13%
46.07 g
1 x (16.00 g)
%O =
x 100% = 34.73%
46.07 g
%C =
C2H6O
52.14% + 13.13% + 34.73% = 100.00%
62
Percent Composition and Empirical Formulas
Determine the empirical formula of a
compound that has the following
percent composition by mass:
K 24.75, Mn 34.77, O 40.51 percent.
1 mol K
nK = 24.75 g K x
= 0.6330 mol K
39.10 g K
nMn = 34.77 g Mn x
1 mol Mn
= 0.6329 mol Mn
54.94 g Mn
nO = 40.51 g O x
1 mol O
= 2.532 mol O
16.00 g O
63
Percent Composition and Empirical Formulas
nK = 0.6330, nMn = 0.6329, nO = 2.532
0.6330 ~
K:
~ 1.0
0.6329
Mn :
0.6329
= 1.0
0.6329
2.532 ~
O:
~ 4.0
0.6329
KMnO4
64
Q. What
is the
Empirical
formula of
Ethanol
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
g CO2
mol CO2
mol C
gC
6.0 g C = 0.5 mol C
g H2O
mol H2O
mol H
gH
1.5 g H = 1.5 mol H
g of O = g of sample – (g of C + g of H)
4.0 g O = 0.25 mol O
Combusted 11.5 g ethanol
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
65
A process in which one or more substances is changed into one
or more new substances is a chemical reaction
A chemical equation uses chemical symbols to show what
happens during a chemical reaction
reactants
products
3 ways of representing the reaction of H2 with O2 to form H2O
66
How to “Read” Chemical Equations
2 Mg + O2
2 MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
67
Balancing Chemical Equations
1. Write the correct formula(s) for the reactants on
the left side and the correct formula(s) for the
product(s) on the right side of the equation.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2
CO2 + H2O
2. Change the numbers in front of the formulas
(coefficients) to make the number of atoms of
each element the same on both sides of the
equation. Do not change the subscripts.
2C2H6
NOT
C4H12
68
Balancing Chemical Equations
3. Start by balancing those elements that appear in
only one reactant and one product.
C2H6 + O2
2 carbon
on left
C2H6 + O2
6 hydrogen
on left
C2H6 + O2
CO2 + H2O
start with C or H but not O
1 carbon
on right
multiply CO2 by 2
2CO2 + H2O
2 hydrogen
on right
2CO2 + 3H2O
multiply H2O by 3
69
Balancing Chemical Equations
4. Balance those elements that appear in two or
more reactants or products.
C2H6 + O2
2 oxygen
on left
2CO2 + 3H2O
multiply O2 by 7
2
4 oxygen + 3 oxygen = 7 oxygen
(3x1)
on right
(2x2)
C2H6 + 7 O2
2
2CO2 + 3H2O
2C2H6 + 7O2
4CO2 + 6H2O
remove fraction
multiply both sides by 2
70
Balancing Chemical Equations
5. Check to make sure that you have the same
number of each type of atom on both sides of the
equation.
2C2H6 + 7O2
4CO2 + 6H2O
4 C (2 x 2)
4C
12 H (2 x 6)
12 H (6 x 2)
14 O (7 x 2)
14 O (4 x 2 + 6)
Reactants
4C
12 H
14 O
Products
4C
12 H
14 O
71
Amounts of Reactants and Products
1. Write balanced chemical equation
2. Convert quantities of known substances into moles
3. Use coefficients in balanced equation to calculate the
number of moles of the sought quantity
4. Convert moles of sought quantity into desired units
72
Methanol burns in air according to the equation
2CH3OH + 3O2
2CO2 + 4H2O
If 209 g of methanol are used up in the combustion, what
mass of water is produced?
grams CH3OH
moles CH3OH
molar mass
CH3OH
209 g CH3OH x
moles H2O
grams H2O
molar mass
coefficients
H2O
chemical equation
4 mol H2O
18.0 g H2O
1 mol CH3OH
=
x
x
32.0 g CH3OH
2 mol CH3OH
1 mol H2O
235 g H2O
73
Limiting Reagent:
Reactant used up first in
the reaction.
2NO + O2
2NO2
NO is the limiting reagent
O2 is the excess reagent
74
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3
Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
g Al
mol Al
mol Fe2O3 needed
g Fe2O3 needed
OR
g Fe2O3
124 g Al x
mol Fe2O3
1 mol Al
27.0 g Al
x
mol Al needed
1 mol Fe2O3
2 mol Al
Start with 124 g Al
160. g Fe2O3
=
x
1 mol Fe2O3
g Al needed
367 g Fe2O3
need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent
75
Use limiting reagent (Al) to calculate amount of product that
can be formed.
g Al
mol Al
mol Al2O3
2Al + Fe2O3
124 g Al x
1 mol Al
27.0 g Al
x
1 mol Al2O3
2 mol Al
g Al2O3
Al2O3 + 2Fe
102. g Al2O3
=
x
1 mol Al2O3
234 g Al2O3
At this point, all the Al is consumed
and Fe2O3 remains in excess.
76
Reaction Yield
Theoretical Yield is the amount of product that would
result if all the limiting reagent reacted.
Actual Yield is the amount of product actually obtained
from a reaction.
% Yield =
Actual Yield
x 100%
Theoretical Yield
77
1.
What is the character of the bond in
carbon monoxide?
a. ionic
b. polar covalent
c. non-polar covalent
d. coordinate covalent
78
2. Which of the following is the bet explanation of
the phenomenon of hydrogen bonding?
a. Hydrogen has a strong affinity for holding onto
valence electrons
b. Hydrogen can only hold two valence electrons
c. Electron negative atoms disproportionately
carry shared pairs when bonded to hydrogen
d. Hydrogen bonds have ionic character
79
3. Which of the following has a formula
weight between 74 and 75 grams per
mole?
a. KCl
b. C4H10O
c. (LiCl)2
d. BF3
80
4. In the reactoin shown, if 39.03 g of Na2S are reacted
with 113.3 g of AgNO3, how much, if any, of either
reagent will be left over once the reaction has gone
to completion?
Na2S+ 2 AgNO3Ag2S + 2 NaNO3
a.
b.
c.
d.
41.37 g AgNO3
13.00 g Na2S
26.00 g Na2S
74.27 g AgNO3
81
What is the percent by mass of carbon in CO2?
a. 12%
b. 27%
c. 33 %
d. 44 %
82
Sulfur dioxide oxidizes in the presence of O2 gas
as per the reaction
2 SO2 + O2 2 SO3
Approximately how many grams of sulfur trioxide
are produced by the complete oxidation of 1
mole of sulfur dioxide?
a. 1g
b. 2g
c. 80g
d. 160 g
83
How many carbon atoms exist in 12 amu of 12C?
a. 1
b. 12
c. 6.02 x 1023
d. 7.22x1024
84
When an electron moves from a 2p to a 3s orbital,
the atom containng that electron:
a. becomes a new isotope
b. becomes a new element
c. absorbs energy
d. releases energy
85
86
5. One way to test for the presence of iron in
solution is by adding potassium thiocyanate to
the solution. The product when this reagent
reacts with iron is FeSCN2+, which creates a
dark red color in solution via the following net
ionic equation.
Fe 3+ (aq) + SCN-FeSCN2+
How many grams of iron sulfate would be
needed to produce 2 moles of
a. 400 g
b. 800 g
c. 200 g
d. 500 g
87