CHEM1310 Lecture

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Chapter 6. Bonding
6.1 Types of Chemical Bonds
6.2 Electronegativity
6.3 Bond Polarity and Dipole Moments
6.4 Ions: Electron Configurations and Sizes
6.5 Formation of Binary Ionic Compounds
6.6 Partial Ionic Character of Covalent Bonds
6.7 The Covalent Chemical Bond: A Model
6.8 Covalent Bond Energies and Chemical Reactions
6.9 The Localized Electron Bonding Model
6.10 Lewis Structure
6.11 Resonance
6.12 Exceptions to the Octet Rule
6.13 Molecular Structure: The VSEPR Model
What is a Chemical Bond? (1)
• Bonding is the force of attraction that holds
atoms together in an element (N2) or
compound (CO2 or NaCl).
• The distances between bonded atoms are
less than those between non-bonded atoms.
• The forces between bonded atoms are
greater than those between non-bonded
atoms.
• The principal types of bonding are ionic,
covalent, and metallic.
What is a Chemical Bond? (2)
• A chemical bond links two atoms or
groups of atoms when the forces acting
between them are sufficient to lead to the
formation of an aggregate (a molecule)
with sufficient stability to make it
convenient for the chemist to consider it as
an independent "molecular species”
Paraphrased from Linus Pauling (1967).
Types of Chemical Bonds (1)
• Ionic Bonds
– Ionic substances are formed when an atom
that loses electrons easily reacts with an atom
that gains electrons easily.
Na ·
e- + Cl
Na+
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Na+ + e.. → : Cl :
..
..
+
: Cl : →
..
→
Loss of an electron
Gain of an electron
NaCl
Combination to form
the compound NaCl
Types of Chemical Bonds (2)
For ionic bonds, the energy of interaction
between a pair of ions can be
calculated by using Coulomb's law. The
energy depends only on distance.
Q1Q2
V
4 r
Types of Chemical Bonds (3)
• Covalent Bond
– is the sharing of pairs of electrons between
atoms. Covalent bonding does not require
atoms be the same elements but that they be
of comparable electronegativity. Covalent
bonds give an the angular relation between
the atoms (in polyatomic molecules, does not
apply to molecules like H2).
H·
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·H
→
H:H
or
H―H
Electronegativity (EN)
• the tendency of an atom in a molecule to
attract shared electrons. Shared electrons are
closer to atoms with greater electronegativity.
• Trends in EN
– In a group (column): EN decreases w/
increasing Z (# of protons)
– In a period (row): EN increases w/
increasing Z
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Polar covalent
bonds
Dipole Moments
•
•
•
Bonded atoms share electrons unequally, whenever they differ in
Electronegativity
HF: The F atom carries negative charge and the H atom positive charge
of equal magnitude. The molecules align in an electric field.
Polar molecules posses a dipole moment, μ
Covalent/Polar/Ionic
Covalent
e.g., H2, Cl2 N2
Polar Covalent
e.g., HF, H2O
Ionic Bond
e.g., LiF, NaCl
Percent Ionic Character
Ionic vs. Covalent Bonding
Electronegativity
The Process of Bond Formation (1).
•
Two atoms with (i) unfilled electron shells or (ii) opposite charge, are separated by
a great distance.
The initial potential energy is zero. The atoms do not sense each other.
•
Via some random process the atoms approach each other.
There is an attractive force between the atoms. The PE goes negative.
•
The distance between the atoms reaches the ideal bond distance.
The PE reaches a minimum, attraction equals the repulsion. There is no net force.
•
The distance continues to decrease
Repulsion goes up sharply. PE goes positive. The atoms are forced back to the
ideal bond distance.
The Process of Bond Formation (2).
The Process of Bond
Formation (3)
When two atoms (or
molecules) approach each
other, the result depends on
the atom (or molecule) type.
1)Cl + Cl both have unfilled
shells. A bond can form.
2)K+ + Cl- have opposing
charges. A bond can form.
3)Ar + Ar both have filled
shells. A bond cannot form.
Core Electrons vs Valence Electrons
• Electrons can be divided into:
– Valence electrons (e- in unfilled shells,
outermost electrons). Valence electrons
participate in bonding.
– Core electrons (e- in a filled shells). Core
electrons do not participate in bonding.
Valence electrons in molecules are usually
distributed in such a way that each maingroup element is surrounded by eight
electrons (an octet of electrons). Hydrogen is
surrounded by two valence electrons.
Sizes of Ions
Cations (+):
smaller than
parent atom
Anions (-): larger
than parent atom
Isoelectronic
series:
O2FNa+
Mg2+
Ionic Radii
In picometers
Al3+
As Z, the
nuclear
charge,
increases
the atomic
radii
decreases
Ions: Predicting Forumlae of Salts
K:
[Ar]4s1
Can lose 1 electron
Na+: [Ar]
Ca:
[Ar]4s2
Can lose 2 electrons
Ca+2: [Ar]
O:
[He]
2s22p4
Can gain 2 electrons
O-2: [Ne]
Dot Structures: closed shell elements
He
Ne
2
2
8
2
Ar
8
8
2
Kr
8
8
18
Xe 2, 8, 8, 18, 18
Rn 2, 8, 8, 18, 18, 32
Dot Structures: steps in converting a molecular formula into a
Step 1
Molecular
formula
Lewis Dot structure
Draw the individual atoms/ions - with their valence
electrons. H has 1, Carbon has 4, N has 5, O has 6, etc.
Determine the total number of valence electrons. Account
for the net charge.
Make the molecule. Place the atom with
Step 2
lowest EN in center (CO2, SO4 ).
Atom
placement
Step 3
Make single
bonds
Make the bonds. Add 2 e- to each bond
Follow the octet rule. Add
Step 4 remaining e- (double, triple bonds,
account for net charge)
Impose octet rule
Step 4
to C, N, O, F,
Lewis
structure
Check the formal and
total charges, and
octet rule.
Lewis Dot Structure NF3
Example
NF3
Lewis Dot Structure: NF3
Molecular
formula
count
valence eAtom
placement
Add in
valence edone
Dot Structures: NF3
Dot Structures
Write Lewis Structures for Molecules with One
Central Atom
PROBLEM:
Write a Lewis structure for CCl2F2, one of the compounds
responsible for the depletion of stratospheric ozone.
Dot Structures
The Lewis Structures for Molecules with One
Central Atom
PROBLEM:
Write a Lewis structure for CCl2F2, one of the compounds
responsible for the depletion of stratospheric ozone.
:
SOLUTION:
:
:
: Cl :
F :
:
C
:
:Cl
:
: F:
Dot Structures
The Lewis Structure for Molecules with More
than One Central Atom
PROBLEM:
Write the Lewis structure for methanol (molecular formula
CH4O, i.e., CH3OH), an important industrial alcohol that is used
as a gasoline alternative in car engines. If you drink it you get
drunk and then go blind and die.
Hydrogen can have only one bond. C and O are bonded. H
fills in the rest of the bonds.
Dot Structures
The Lewis Structure for Molecules with More
than One Central Atom
Write the Lewis structure for methanol (molecular formula
CH4O, i.e., CH3OH), an important industrial alcohol that is used
as a gasoline alternative in car engines. If you drink it you get
drunk and then go blind and die.
Hydrogen can have only one bond. C and O are bonded. H
fills in the rest of the bonds.
There are 4(1) + 4 + 6 = 14 valence e-.
C forms 4 bonds, O forms 2. Each H forms 1 bond. O has 2
paisr of nonbonding e-.
:
H
H
C
O
:
PROBLEM:
H
H
Dot Structures
Writing Lewis Structures for Molecules with
Multiple Bonds.
PROBLEM:
Write Lewis structures for the following:
(a) Ethylene (C2H4), the most important
reactant in the manufacture of
polymers.
(b) Nitrogen (N2), the most abundant
atmospheric gas. Try O2, H2.
Dot Structures
PROBLEM:
The Lewis Structures for Molecules with
double or triple bond.
Write Lewis structures for:
(a) Ethylene (C2H4), the most important reactant in the
manufacture of polymers
(b) Nitrogen (N2), the most abundant atmospheric gas
If a central atom does not have 8e-, an octet, then a pair of e- can be
moved from an adjacent atom to form a multiple bond.
PLAN:
Ethylene
SOLUTION:
H
:
H
(a) There are 2(4) + 4(1) = 12 valence e-. H can can form
only one bond.
C
C
H
H
H
H
H
C
C
H
Dot Structures
The Lewis Structures for Molecules with
double or triple bond.
Write Lewis structures for:
(a) Ethylene (C2H4), the most important reactant in the
manufacture of polymers
(b) Nitrogen (N2), the most abundant atmospheric gas
PROBLEM:
PLAN:
If a central atom does not have 8e-, an octet, then a pair of e- can be
moved from an adjacent atom to form a multiple bond.
N2
SOLUTION:
(b) N2 has 2(5) = 10 valence e-. Therefore a triple bond is required to make
the octet around each N.
N
.
:
N
.
:
:
:.
N
.
N
N
:
.:
N
.
Dot Structures
Write the Lewis Structure for O3 (Ozone)
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Resonance
Delocalized Electron-Pair Bonding
O3 can be drawn in 2 ways -
O
O
O
O
O
O
Neither structure is acurate. Reality is hybrid of the two.
B
B
O
O
O
A
O
C
O
O
O
O
A
O
C
Resonance structures have the same relative atom placement but a
difference in the locations of bonding and nonbonding electron pairs.
is used to indicate that resonance occurs.
Formal Charge
Selecting the Best Resonance Structure
An atom “owns” all of its nonbonding electrons and half of its bonding electrons.
Formal charge is the charge an atom would have if the bonding electrons were
shared equally.
Formal charge of atom =
# valence e- - (# unshared electrons + 1/2 # shared electrons)
B
# valence
e-
=6
O
e-
A
# nonbonding
# bonding
For OC
O
For OA
e-
=4
= 4 X 1/2 = 2
O
C
For OB
# valence e- = 6
Formal charge = 0
# valence e- = 6
# nonbonding e- = 6
# bonding e- = 2 X 1/2 = 1
Formal charge = -1
# nonbonding e- = 2
# bonding e- = 6 X 1/2 = 3
Formal charge = +1
Resonance
SAMPLE PROBLEM 10.4
PROBLEM:
Writing Resonance Structures
-
Write resonance structures for the nitrate ion, NO3 .
Resonance
SAMPLE PROBLEM 10.4
PROBLEM:
PLAN:
-
Write resonance structures for the nitrate ion, NO3 .
Use 5+(3x6)+1=24 valence e-. After you write out the dot
structure see if other structures can be drawn in which the
electrons can be delocalized over more than two atoms.
SOLUTION:
O
Writing Resonance Structures
Nitrate has 1(5) + 3(6) + 1 = 24 valence e-
O
O
O
N
N
N
O
O
O
O
O
Something is
wrong. N does not
have an octet; shift
a pair of e- to form a
double bond
O
O
O
O
N
N
N
O
O
O
O
O
Resonance
O
O
O
O
N
N
N
O
O
O
O
O
None of these are the ‘true’ structure. The
true structure is a average of all three,
and cannot be represented by a simple
dot structure.
Resonance and Formal Charge
Problem: Draw all the resonance structures of NCO-.
Determine the relative weights of the resonance
structures.
Resonance and Formal Charge
Problem: Draw all the resonance structures of NCO-.
N
C
O
N C
A
formal charges
-2
N
N
O
C
B
0
+1
C
O
-1
0
N C
O
C
0
O
0
N
0
C
-1
O
The true structure is a weighted average of Forms A, B and C.
Form A: Formal charge of -2 on N and +1 on O. O is the most electronegative atom in the molecule. Low weight.
Form B: Formal charge on the N is more negative than that of O. O is the
most electro-negative atom in the molecule. Low weight.
Form C: Formal charge on O is -1. High weight.
Resonance and Formal Charge
Three criteria for estimating the weights of resonance structures:
Smaller formal charges (absolute value) give higher weight than
larger charges.
Opposing charges on adjacent atoms gives a low weight.
A negative formal charge is on an electronegative atom (O and
sometimes N and S) gives a high weight.
Resonance and Weighted Averages
EXAMPLE: NCO- has 3 possible resonance forms -
N
C
A
O
N C
O
N
B
C
O
C
None of these are the ‘true’ structure.
The true structure is a weighted average of all
three.
The weight of C is greater than the weights of A
and B.
Carbon Monoxide
42
Dot Structures - Octet Expansion & Contraction
PROBLEM:
PLAN:
Write Lewis structures for (a) H3PO4 (pick the most heavily
weighted structure); (b) BFCl2.
Draw the Lewis structures for the molecule. Note that these dare
exceptiosn to the octet rule. P is a Period-3 element and can have
an expanded valence shell.
SOLUTION:
(a) H3PO4 has two resonance forms and formal charges
indicate the more important form.
-1
0
O
0 H O
P
O
0
H
0
+1
O H 0
0
0
0 H O
0
O
0
P
O H 0
(b) BFCl2 has only 1 Lewis
structure.
0
0 O
H more weight
0
lower formal charges
F
B
Cl
Cl
VSEPR - Valence Shell Electron Pair Repulsion Theory
Each electron pair (bonded and non-bonded) around a central
atom is located as far away as possible from the others, to minimize
electrostatic repulsions.
But a double bond counts as one bond. A triple bond counts as one
bond.
These repulsions maximize the space allowed for each electron
pair.
The result is five electron-group arrangements of minimum
energy seen in a large majority of molecules and polyatomic
ions.
Linear, trigonal planer, tetrahedral, trigonal bipyramidal,
octahedral.
Electron-pair repulsions and the five basic molecular shapes.
linear
trigonal bipyramidal
tetrahedral
trigonal planar
octahedral
The single molecular shape of the linear electron-group
arrangement.
Examples:
CS2, HCN, BeF2
Figure 10.4
The two molecular shapes of the trigonal planar electrongroup arrangement.
Class
Examples:
SO2, O3, PbCl2, SnBr2
Shape
Examples:
SO3, BF3, NO3-, CO32-
Factors Affecting Actual Bond Angles
Bond angles are consistent with theoretical angles when the atoms
attached to the central atom are the same and when all electrons are
bonding electrons of the same order.
H
Effect of Double Bonds
1200
ideal
1200
larger EN
C
H
O
greater
electron
density
Factors Affecting Actual Bond Angles
Bond angles are consistent with theoretical angles when the atoms
attached to the central atom are the same and when all electrons are
bonding electrons of the same order.
H
Effect of Double Bonds
1200
ideal
1200
H
larger EN
C
H
1220
O
1160
C
H
greater
electron
density
real
Effect of Nonbonding(Lone) Pairs
Lone pairs take more space than
bonding pairs..
Sn
Cl
Cl
950
O
Factors Affecting Actual Bond Angles
Bond angles are consistent with theoretical angles when the atoms
attached to the central atom are the same and when all electrons are
bonding electrons of the same order.
H
Effect of Double Bonds
1200
ideal
1200
H
larger EN
C
H
1220
O
greater
electron
density
1160
C
H
real
O
Factors Affecting Actual Bond Angles
Bond angles are consistent with theoretical angles when the atoms
attached to the central atom are the same and when all electrons are
bonding electrons of the same order.
H
Effect of Double Bonds
1200
ideal
1200
O
1160
C
H
greater
electron
density
real
Effect of Nonbonding(Lone) Pairs
Lone pairs repel bonding pairs
more strongly than bonding pairs
repel each other.
H
larger EN
C
H
1220
Sn
Cl
Cl
950
O
Figure 10.5
The three molecular shapes of the tetrahedral electrongroup arrangement.
Examples:
CH4, SiCl4,
SO42-, ClO4-
NH3
H 2O
PF3
OF2
ClO3
SCl2
H 3 O+
Figure 10.6
Lewis structures and molecular shapes.
Figure 10.7
The four molecular shapes of the trigonal bipyramidal
electron-group arrangement.
PF5
SF4
AsF5
XeO2F2
SOF4
IF4+
IO2F2-
ClF3
XeF2
BrF3
I3 -
IF2-
Figure 10.8
The three molecular shapes of the octahedral electrongroup arrangement.
SF6
IOF5
BrF5
TeF5
-
XeOF4
XeF4
ICl4-
A summary of common molecular shapes with two to six
electron groups.
SAMPLE PROBLEM
PROBLEM:
Predicting Molecular Shapes with Five or Six
Electron Groups
Determine the molecular shape and predict the bond angles
(relative to the ideal bond angles) of (a) SbF5 and (b) BrF5.
SAMPLE PROBLEM 10.7
PROBLEM:
SOLUTION:
Predicting Molecular Shapes with Five or Six
Electron Groups
Determine the molecular shape and predict the bond angles
(relative to the ideal bond angles) of (a) SbF5 and (b) BrF5.
(a) SbF5 - 40 valence e-; all electrons around central
atom will be in bonding pairs; shape is AX5 - trigonal
bipyramidal.
F
F
F
F
Sb
F
F
F
Sb
F
F
F
(b) BrF5 - 42 valence e-; 5 bonding pairs and 1 nonbonding pair on central
atom. Shape is AX5E, square pyramidal.
F
F
F
Br
F
F
SAMPLE PROBLEM
PROBLEM:
PLAN:
Predicting Molecular Shapes with More Than
One Central Atom
Determine the shape around each of the central atoms in
acetone, (CH3)2C=O.
Find the shape of one atom at a time after writing the Lewis
structure.
SOLUTION:
tetrahedral
H
H C
H
O
C
H
C H
H
tetrahedral
trigonal planar
O
>1200
H
C
H
H C
C
H
HH
<1200
Figure 10.11
The tetrahedral centers of ethane and ethanol.
ethane
ethanol
CH3CH3
CH3CH2OH
Figure 10.12
The orientation of polar molecules in an electric field.
Electric field OFF
Electric field ON
Dipole moment = μ = QR
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Non-Polar
covalent bonding
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SAMPLE PROBLEM 10.9
PROBLEM:
Predicting the Polarity of Molecules
From electronegativity (EN) values (button) and their periodic
trends, predict whether each of the following molecules is polar
and show the direction of bond dipoles and the overall
molecular dipole when applicable:
(a) Ammonia, NH3
(b) Boron trifluoride, BF3
(c) Carbonyl sulfide, COS (atom sequence SCO)
SAMPLE PROBLEM
Predicting the Polarity of Molecules
continued
(b) BF3 has 24 valence e- and all electrons around the B will be involved in
bonds. The shape is AX3, trigonal planar.
F
B
F
F
1200
F (EN 4.0) is more electronegative
than B (EN 2.0) and all of the dipoles
will be directed from B to F. Because
all are at the same angle and of the
same magnitude, the molecule is
nonpolar.
(c) COS is linear. C and S have the same EN (2.0) but the C=O bond is
quite polar(DEN) so the molecule is polar overall.
S
C
O
SAMPLE PROBLEM
Predicting the Polarity of Molecules
continued
(b) BF3 has 24 valence e- and all electrons around the B will be involved in
bonds. The shape is AX3, trigonal planar.
F
B
F
F
1200
F (EN 4.0) is more electronegative
than B (EN 2.0) and all of the dipoles
will be directed from B to F. Because
all are at the same angle and of the
same magnitude, the molecule is
nonpolar.
(c) COS is linear. C and S have the same EN (2.0) but the C=O bond is
quite polar(DEN) so the molecule is polar overall.
S
C
O
Carbonate, CO3-2
Thiocyanate ion, NCS-1
N2O, laughing gas
The VSEPR Theory
SN = 5 (bonded atoms) + 0 (lone pairs)
5: Trigonal bipyramidal
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The VSEPR Theory
SN = 6 (bonded atoms) + 0 (lone pairs)
6: Octahedral
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The VSEPR Theory
Pyramidal
SN = 3 (bonded atoms) + 1 (lone pairs)
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Bent
SN = 2 (bonded atoms) + 2 (lone pairs)
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