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
Transcript
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).
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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 · → Na+ + e- Loss of an electron
e- + Cl → Gain of an electron: Cl :. .
. .-
: Cl :. .
. .-+Na+ → NaCl Combination to form
the compound NaCl
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.
1 2
4
Q QV
rπε=
o
Types of Chemical Bonds (2)
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• 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:HH· ·H → or H―H
Types of Chemical Bonds (3)
• 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
04/19/23Zumdahl Chapter 13 7
Electronegativity (EN)
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
Polar covalent bonds
Covalent
e.g., H2, Cl2 N2
Polar Covalent
e.g., HF, H2O
Ionic Bond
e.g., LiF, NaCl
Percent Ionic CharacterCovalent/Polar/Ionic
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 decreaseRepulsion 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.
• 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 main-group element is surrounded by eight electrons (an octet of electrons). Hydrogen is surrounded by two valence electrons.
Core Electrons vs Valence Electrons
Cations (+): smaller than parent atom
Anions (-): larger than parent atom
Isoelectronic series:
O2-
F-
Na+
Mg2+
Al3+Ionic Radii
In picometers
As Z, the nuclear charge, increases the atomic radii decreases
Sizes of Ions
K: [Ar]4s1
O: [He] 2s22p4
Can lose 1 electron
Can gain 2 electrons
Na+: [Ar]
O-2: [Ne]
Ions: Predicting Forumlae of Salts
Ca: [Ar]4s2 Can lose 2 electrons Ca+2: [Ar]
He 2
Ne 2
8
Ar2
8
8
Kr2
8
8
18
Xe 2, 8, 8, 18, 18
Rn 2, 8, 8, 18, 18, 32
Dot Structures: closed shell elements
Molecular formula
Atom placement
Make single bonds
Impose octet rule
to C, N, O, F,
Lewis structure
Make the molecule. Place the atom with lowest EN in center (CO2, SO4 ).
Make the bonds. Add 2 e- to each bond
Check the formal and total charges, and
octet rule.
Step 2
Step 3
Step 4
Dot Structures: steps in converting a molecular formula into a Lewis Dot structure
Step 4
Step 1Draw 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.
Follow the octet rule. Add remaining e- (double, triple bonds,
account for net charge)
NF3
Lewis Dot Structure NF3 Example
Lewis Dot Structure: NF3
Molecular formula
Atom placement
count valence e-
Add in valence e-
done
Dot Structures: NF3
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
SOLUTION:
PROBLEM: Write a Lewis structure for CCl2F2, one of the compounds responsible for the depletion of stratospheric ozone.
C
Cl
Cl F
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
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
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-.
C O H
H
H
H
::
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
The Lewis Structures for Molecules with double or triple bond.
PLAN:
Ethylene SOLUTION:
PROBLEM: 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.
(a) There are 2(4) + 4(1) = 12 valence e-. H can can form only one bond.
CCH
H H
H
:
CCH
H H
H
Dot Structures
The Lewis Structures for Molecules with double or triple bond.
PLAN:
N2
SOLUTION:
PROBLEM: 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.
(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
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Write the Lewis Structure for O3 (Ozone)
Dot Structures
Delocalized Electron-Pair Bonding
Resonance structures have the same relative atom placement but a difference in the locations of bonding and nonbonding electron pairs.
OO O
A
B
C
OO O
A
B
C
O3 can be drawn in 2 ways - OO O
OO O
Neither structure is acurate. Reality is hybrid of the two.
is used to indicate that resonance occurs.
O
O O
Resonance
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
SAMPLE PROBLEM 10.4 Writing Resonance Structures
PROBLEM: Write resonance structures for the nitrate ion, NO3-.
Resonance
SAMPLE PROBLEM 10.4 Writing Resonance Structures
PLAN:
SOLUTION:
PROBLEM: 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.
Nitrate has 1(5) + 3(6) + 1 = 24 valence e-
N
O
O O
N
O
O O
N
O
O O
Something is wrong. N does not have an octet; shift a pair of e- to form a double bond
N
O
O O
N
O
O O
N
O
O O
Resonance
N
O
O O
N
O
O O
N
O
O O
Resonance
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.
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
A
N C O
B
N C O
C
N C O N C O N C O
formal charges
-2 0 +1 -1 0 0 0 0 -1
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 electro-negative 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
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.
Three criteria for estimating the weights of resonance structures:
Resonance and Formal Charge
EXAMPLE: NCO- has 3 possible resonance forms -
N C O
A
N C O
B
N 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.
Resonance and Weighted Averages
42
Carbon Monoxide
PLAN:
SOLUTION:
PROBLEM: 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.
(a) H3PO4 has two resonance forms and formal charges indicate the more important form.
O
PO O
O
H
H
H
O
PO O
O
H
H
H
-1
0
0
0
00
0
+10
0
0
0
0
0
0
0
lower formal charges
more weight
(b) BFCl2 has only 1 Lewis structure.
F
BCl Cl
Dot Structures - Octet Expansion & Contraction
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.
VSEPR - Valence Shell Electron Pair Repulsion Theory
linear trigonal planar tetrahedral
trigonal bipyramidal octahedral
Electron-pair repulsions and the five basic molecular shapes.
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 electron-group arrangement.
Class
Shape
Examples:
SO3, BF3, NO3-, CO3
2-
Examples:
SO2, O3, PbCl2, SnBr2
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.
C O
H
Hideal
1200
1200
larger EN
greater electron density
Effect of Double Bonds
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.
C O
H
Hideal
1200
1200
larger EN
greater electron density
1220
1160
real
Lone pairs take more space than bonding pairs.. Sn
Cl Cl
950
Effect of Double Bonds
Effect of Nonbonding(Lone) Pairs
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.
C O
H
Hideal
1200
1200
larger EN
greater electron density
1220
1160
real
Effect of Double Bonds
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.
C O
H
Hideal
1200
1200
larger EN
greater electron density
1220
1160
real
Lone pairs repel bonding pairs more strongly than bonding pairs repel each other.
Sn
Cl Cl
950
Effect of Double Bonds
Effect of Nonbonding(Lone) Pairs
Figure 10.5 The three molecular shapes of the tetrahedral electron-group arrangement.
Examples:
CH4, SiCl4, SO4
2-, ClO4-
NH3
PF3
ClO3
H3O+
H2O
OF2
SCl2
Figure 10.6 Lewis structures and molecular shapes.
Figure 10.7 The four molecular shapes of the trigonal bipyramidal electron-group arrangement.
SF4
XeO2F2
IF4+
IO2F2-
ClF3
BrF3
XeF2
I3-
IF2-
PF5
AsF5
SOF4
Figure 10.8 The three molecular shapes of the octahedral electron-group arrangement.
SF6
IOF5
BrF5
TeF5-
XeOF4
XeF4
ICl4-
A summary of common molecular shapes with two to six electron groups.
SAMPLE PROBLEM Predicting Molecular Shapes with Five or Six Electron Groups
PROBLEM: 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 Predicting Molecular Shapes with Five or Six Electron Groups
PROBLEM: Determine the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) SbF5 and (b) BrF5.
SOLUTION: (a) SbF5 - 40 valence e-; all electrons around central atom will be in bonding pairs; shape is AX5 - trigonal bipyramidal.
F
SbF
F F
FF Sb
F
F
F
F
(b) BrF5 - 42 valence e-; 5 bonding pairs and 1 nonbonding pair on central atom. Shape is AX5E, square pyramidal.
BrF
F F
F
F
SAMPLE PROBLEM Predicting Molecular Shapes with More Than One Central Atom
SOLUTION:
PROBLEM: Determine the shape around each of the central atoms in acetone, (CH3)2C=O.
PLAN: Find the shape of one atom at a time after writing the Lewis structure.
C C C
OH
H
H
HH
H
tetrahedral tetrahedral
trigonal planar
C
O
HC
HHH
CH
H>1200
<1200
Figure 10.11 The tetrahedral centers of ethane and ethanol.
ethane
CH3CH3
ethanol
CH3CH2OH
Figure 10.12 The orientation of polar molecules in an electric field.
Electric field OFF Electric field ON
04/19/23Zumdahl Chapter 13 62
Dipole moment = μ = QR
04/19/23Zumdahl Chapter 13 63
Non-Polar covalent bonding
SAMPLE PROBLEM 10.9 Predicting the Polarity of Molecules
(a) Ammonia, NH3 (b) Boron trifluoride, BF3
(c) Carbonyl sulfide, COS (atom sequence SCO)
PROBLEM: 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:
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
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.
1200
(c) COS is linear. C and S have the same EN (2.0) but the C=O bond is quite polar(EN) 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
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.
1200
(c) COS is linear. C and S have the same EN (2.0) but the C=O bond is quite polar(EN) so the molecule is polar overall.
