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Chemical Bonding I:Basic Concepts
Chapter 9
Intermolecular Forces
11.2
Intermolecular forces are attractive forces between molecules.
Intramolecular forces hold atoms together in a molecule.
Intermolecular vs Intramolecular
• 41 kJ to vaporize 1 mole of water (inter)
• 930 kJ to break all O-H bonds in 1 mole of water (intra)
Generally, intermolecular forces are much weaker than intramolecular forces.
“Measure” of intermolecular force
boiling point
melting point
Hvap
Hfus
Hsub
9.1
Valence electrons are the outer shell electrons of an atom. The valence electrons are the electrons thatparticpate in chemical bonding.
1A 1ns1
2A 2ns2
3A 3ns2np1
4A 4ns2np2
5A 5ns2np3
6A 6ns2np4
7A 7ns2np5
Group # of valence e-e- configuration
9.2
Li + F Li+ F -
The Ionic Bond
1s22s11s22s22p5 1s21s22s22p6[He][Ne]
Li Li+ + e-
e- + F F -
F -Li+ + Li+ F -
9.1
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 F+
7e- 7e-
F F
8e- 8e-
F F
F F
Lewis structure of F2
lone pairslone pairs
lone pairslone pairs
single covalent bond
single covalent bond
9.4
8e-
H HO+ + OH H O HHor
2e- 2e-
Lewis structure of water
Double bond – two atoms share two pairs of electrons
single covalent bonds
O C O or O C O
8e- 8e-8e-double bonds double bonds
Triple bond – two atoms share three pairs of electrons
N N8e-8e-
N N
triple bondtriple bond
or
9.4
Bond Type
Bond Length
(pm)
C-C 154
CC 133
CC 120
C-N 143
CN 138
CN 116
Lengths of Covalent Bonds
Bond Lengths
Triple bond < Double Bond < Single Bond 9.4
9.4
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.
Writing Lewis Structures
9.6
Write the Lewis structure of nitrogen trifluoride (NF3).
Step 1 – N is less electronegative than F, put N in center
F N F
F
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
9.6
Write the Lewis structure of the carbonate ion (CO32-).
Step 1 – C is less electronegative than O, put C in center
O C O
O
Step 2 – Count valence electrons C - 4 (2s22p2) and O - 6 (2s22p4) -2 charge – 2e-
4 + (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
9.6
Step 5 - Too many electrons, form double bond and re-check # of e-
2 single bonds (2x2) = 41 double bond = 4
8 lone pairs (8x2) = 16Total = 24
Valence shell electron pair repulsion (VSEPR) model:
Predict the geometry of the molecule from the electrostatic repulsions between the electron (bonding and nonbonding) pairs.
AB2 2 0
Class
# of atomsbonded to
central atom
# lonepairs on
central atomArrangement of electron pairs
MolecularGeometry
10.1
linear linear
B B
Cl ClBe
2 atoms bonded to central atom0 lone pairs on central atom 10.1
AB2 2 0 linear linear
Class
# of atomsbonded to
central atom
# lonepairs on
central atomArrangement of electron pairs
MolecularGeometry
VSEPR
AB3 3 0trigonal planar
trigonal planar
10.1
10.1
AB2 2 0 linear linear
Class
# of atomsbonded to
central atom
# lonepairs on
central atomArrangement of electron pairs
MolecularGeometry
VSEPR
AB3 3 0trigonal planar
trigonal planar
10.1
AB4 4 0 tetrahedral tetrahedral
10.1
AB2 2 0 linear linear
Class
# of atomsbonded to
central atom
# lonepairs on
central atomArrangement of electron pairs
MolecularGeometry
VSEPR
AB3 3 0trigonal planar
trigonal planar
10.1
AB4 4 0 tetrahedral tetrahedral
AB5 5 0trigonal
bipyramidaltrigonal
bipyramidal
10.1
AB2 2 0 linear linear
Class
# of atomsbonded to
central atom
# lonepairs on
central atomArrangement of electron pairs
MolecularGeometry
VSEPR
AB3 3 0trigonal planar
trigonal planar
10.1
AB4 4 0 tetrahedral tetrahedral
AB5 5 0trigonal
bipyramidaltrigonal
bipyramidal
AB6 6 0 octahedraloctahedral
10.1
10.1
bonding-pair vs. bonding
pair repulsionlone-pair vs. lone pair
repulsionlone-pair vs. bonding
pair repulsion> >
Class
# of atomsbonded to
central atom
# lonepairs on
central atomArrangement of electron pairs
MolecularGeometry
VSEPR
AB3 3 0trigonal planar
trigonal planar
AB2E 2 1trigonal planar
bent
10.1
Class
# of atomsbonded to
central atom
# lonepairs on
central atomArrangement of electron pairs
MolecularGeometry
VSEPR
AB3E 3 1
AB4 4 0 tetrahedral tetrahedral
tetrahedraltrigonal
pyramidal
10.1
Class
# of atomsbonded to
central atom
# lonepairs on
central atomArrangement of electron pairs
MolecularGeometry
VSEPR
AB4 4 0 tetrahedral tetrahedral
10.1
AB3E 3 1 tetrahedraltrigonal
pyramidal
AB2E2 2 2 tetrahedral bent
H
O
H
Class
# of atomsbonded to
central atom
# lonepairs on
central atomArrangement of electron pairs
MolecularGeometry
VSEPR
10.1
AB5 5 0trigonal
bipyramidaltrigonal
bipyramidal
AB4E 4 1trigonal
bipyramidaldistorted
tetrahedron
Class
# of atomsbonded to
central atom
# lonepairs on
central atomArrangement of electron pairs
MolecularGeometry
VSEPR
10.1
AB5 5 0trigonal
bipyramidaltrigonal
bipyramidal
AB4E 4 1trigonal
bipyramidaldistorted
tetrahedron
AB3E2 3 2trigonal
bipyramidalT-shaped
ClF
F
F
Class
# of atomsbonded to
central atom
# lonepairs on
central atomArrangement of electron pairs
MolecularGeometry
VSEPR
10.1
AB5 5 0trigonal
bipyramidaltrigonal
bipyramidal
AB4E 4 1trigonal
bipyramidaldistorted
tetrahedron
AB3E2 3 2trigonal
bipyramidalT-shaped
AB2E3 2 3trigonal
bipyramidallinear
I
I
I
Class
# of atomsbonded to
central atom
# lonepairs on
central atomArrangement of electron pairs
MolecularGeometry
VSEPR
10.1
AB6 6 0 octahedraloctahedral
AB5E 5 1 octahedral square pyramidal
Br
F F
FF
F
Class
# of atomsbonded to
central atom
# lonepairs on
central atomArrangement of electron pairs
MolecularGeometry
VSEPR
10.1
AB6 6 0 octahedraloctahedral
AB5E 5 1 octahedral square pyramidal
AB4E2 4 2 octahedral square planar
Xe
F F
FF
10.1
Predicting Molecular Geometry1. 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?
SO O
AB2E
bent
S
F
F
F F
AB4E
distortedtetrahedron
10.1
9.7
Two possible skeletal structures of formaldehyde (CH2O)
H C O HH
C OH
An atom’s formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure.
formal charge on an atom in a Lewis structure
=1
2
total number of bonding electrons( )
total number of valence electrons in the free atom
-total number of nonbonding electrons
-
The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion.
H C O HC – 4 e-
O – 6 e-
2H – 2x1 e-
12 e-
2 single bonds (2x2) = 41 double bond = 4
2 lone pairs (2x2) = 4Total = 12
formal charge on C = 4 -2 - ½ x 6 = -1
formal charge on O = 6 -2 - ½ x 6 = +1
formal charge on an atom in a Lewis structure
=1
2
total number of bonding electrons( )
total number of valence electrons in the free atom
-total number of nonbonding electrons
-
-1 +1
9.7
C – 4 e-
O – 6 e-
2H – 2x1 e-
12 e-
2 single bonds (2x2) = 41 double bond = 4
2 lone pairs (2x2) = 4Total = 12
HC O
H
formal charge on C = 4 -0 - ½ x 8 = 0
formal charge on O = 6 -4 - ½ x 4 = 0
formal charge on an atom in a Lewis structure
=1
2
total number of bonding electrons( )
total number of valence electrons in the free atom
-total number of nonbonding electrons
-
0 0
9.7
Formal Charge and Lewis Structures
9.7
1. For neutral molecules, a Lewis structure in which there are no formal charges is preferable to one in which formal charges are present.
2. Lewis structures with large formal charges are less plausible than those with small formal charges.
3. Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms.
Which is the most likely Lewis structure for CH2O?
H C O H
-1 +1 HC O
H
0 0
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+ -
OOO+-
O C O
O
- -O C O
O
-
-
OCO
O
-
- 9.8
What are the resonance structures of the carbonate (CO3
2-) ion?
Exceptions to the Octet Rule
The Incomplete Octet
H HBeBe – 2e-
2H – 2x1e-
4e-
BeH2
BF3
B – 3e-
3F – 3x7e-
24e-
F B F
F
3 single bonds (3x2) = 69 lone pairs (9x2) = 18
Total = 24
9.9
Exceptions to the Octet Rule
Odd-Electron Molecules
N – 5e-
O – 6e-
11e-
NO N O
The Expanded Octet (central atom with principal quantum number n > 2)
SF6
S – 6e-
6F – 42e-
48e-
S
F
F
F
FF
F
6 single bonds (6x2) = 1218 lone pairs (18x2) = 36
Total = 48
9.9
Covalent
share e-
Polar Covalent
partial transfer of e-
Ionic
transfer e-
Increasing difference in electronegativity
Classification of bonds by difference in electronegativity
Difference Bond Type
0 Covalent
2 Ionic
0 < and <2 Polar Covalent
9.5
Classify the following bonds as ionic, polar covalent, or covalent: The bond in CsCl; the bond in H2S; andthe 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
9.5
H F FH
Polar covalent bond or polar bond is a covalent bond with greater electron density around one of the two atoms
electron richregion
electron poorregion e- riche- poor
+ -
9.5
Electronegativity is the ability of an atom to attract toward itself the electrons in a chemical bond.
Electron Affinity - measurable, Cl is highest
Electronegativity - relative, F is highest
X (g) + e- X-(g)
9.5
9.5
Dipole Moments and Polar Molecules
10.2
H F
electron richregion
electron poorregion
= Q x rQ is the charge
r is the distance between charges
1 D = 3.36 x 10-30 C m
10.2
10.2
10.2
Which of the following molecules have a dipole moment?H2O, CO2, SO2, and CH4
O HH
dipole momentpolar molecule
SO
O
CO O
no dipole momentnonpolar molecule
dipole momentpolar molecule
C
H
H
HH
no dipole momentnonpolar molecule
Does CH2Cl2 have a dipole moment?
10.2
Intermolecular Forces
Dipole-Dipole Forces
Attractive forces between polar molecules
Orientation of Polar Molecules in a Solid
11.2
Intermolecular Forces
Ion-Dipole Forces
Attractive forces between an ion and a polar molecule
11.2
Ion-Dipole Interaction
11.2
Intermolecular ForcesDispersion Forces
Attractive forces that arise as a result of temporary dipoles induced in atoms or molecules
11.2
ion-induced dipole interaction
dipole-induced dipole interaction
Intermolecular ForcesDispersion Forces Continued
11.2
Polarizability is the ease with which the electron distribution in the atom or molecule can be distorted.
Polarizability increases with:
• greater number of electrons
• more diffuse electron cloud
Dispersion forces usually increase with molar mass.
SO
O
What type(s) of intermolecular forces exist between each of the following molecules?
HBrHBr is a polar molecule: dipole-dipole forces. There are also dispersion forces between HBr molecules.
CH4
CH4 is nonpolar: dispersion forces.
SO2
SO2 is a polar molecule: dipole-dipole forces. There are also dispersion forces between SO2 molecules.
11.2
Intermolecular ForcesHydrogen Bond
11.2
The hydrogen bond is a special dipole-dipole interaction between they hydrogen atom in a polar N-H, O-H, or F-H bond and an electronegative O, N, or F atom.
A H…B A H…Aor
A & B are N, O, or F
Hydrogen Bond
11.2
Why is the hydrogen bond considered a “special” dipole-dipole interaction?
Decreasing molar massDecreasing boiling point
11.2
Types of Crystals
Molecular Crystals• Lattice points occupied by molecules• Held together by intermolecular forces• Soft, low melting point• Poor conductor of heat and electricity
11.6
Types of Crystals
Metallic Crystals• Lattice points occupied by metal atoms• Held together by metallic bonds• Soft to hard, low to high melting point• Good conductors of heat and electricity
11.6
Cross Section of a Metallic Crystal
nucleus &inner shell e-
mobile “sea”of e-
Types of Crystals
Ionic Crystals• Lattice points occupied by cations and anions• Held together by electrostatic attraction• Hard, brittle, high melting point• Poor conductor of heat and electricity
CsCl ZnS CaF2
11.6
9.3
Lattice energy (E) increases as Q increases and/or
as r decreases.
cmpd lattice energyMgF2
MgO
LiF
LiCl
2957
3938
1036
853
Q= +2,-1
Q= +2,-2
r F < r Cl
Electrostatic (Lattice) Energy
E = kQ+Q-r
Q+ is the charge on the cation
Q- is the charge on the anionr is the distance between the ions
Lattice energy (E) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions.
Types of Crystals
Covalent Crystals• Lattice points occupied by atoms• Held together by covalent bonds• Hard, high melting point• Poor conductor of heat and electricity
11.6diamond graphite
carbonatoms
Types of Crystals
11.6
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
10.4
10.4
10.4
10.4
Predict correctbond angle
Formation of sp Hybrid Orbitals
10.4
Formation of sp2 Hybrid Orbitals
10.4
# of Lone Pairs+
# of Bonded Atoms Hybridization Examples
2
3
4
5
6
sp
sp2
sp3
sp3d
sp3d2
BeCl2
BF3
CH4, NH3, H2O
PCl5
SF6
How do I predict the hybridization of the central atom?
Count the number of lone pairs AND the numberof atoms bonded to the central atom
10.4
Valence Bond Theory and NH3
N – 1s22s22p3
3 H – 1s1
If the bonds form from overlap of 3 2p orbitals on nitrogenwith the 1s orbital on each hydrogen atom, what would the molecular geometry of NH3 be?
If use the3 2p orbitalspredict 900
Actual H-N-Hbond angle is
107.30
10.4
10.5
10.5
Sigma () and Pi Bonds ()
Single bond 1 sigma bond
Double bond 1 sigma bond and 1 pi bond
Triple bond 1 sigma bond and 2 pi bonds
How many and bonds are in the acetic acid(vinegar) molecule CH3COOH?
C
H
H
CH
O
O H bonds = 6 + 1 = 7
bonds = 1
10.5
The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy.
H2 (g) H (g) + H (g) H0 = 436.4 kJ
Cl2 (g) Cl (g)+ Cl (g) H0 = 242.7 kJ
HCl (g) H (g) + Cl (g) H0 = 431.9 kJ
O2 (g) O (g) + O (g) H0 = 498.7 kJ O O
N2 (g) N (g) + N (g) H0 = 941.4 kJ N N
Bond Energy
Bond Energies
Single bond < Double bond < Triple bond
9.10
Average bond energy in polyatomic molecules
H2O (g) H (g) + OH (g) H0 = 502 kJ
OH (g) H (g) + O (g) H0 = 427 kJ
Average OH bond energy = 502 + 427
2= 464 kJ
9.10
Bond Energies (BE) and Enthalpy changes in reactions
H0 = total energy input – total energy released= BE(reactants) – BE(products)
Imagine reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products.
9.10
9.10
H2 (g) + Cl2 (g) 2HCl (g) 2H2 (g) + O2 (g) 2H2O (g)
Use bond energies to calculate the enthalpy change for:H2 (g) + F2 (g) 2HF (g)
H0 = BE(reactants) – BE(products)
Type of bonds broken
Number of bonds broken
Bond energy (kJ/mol)
Energy change (kJ)
H H 1 436.4 436.4
F F 1 156.9 156.9
Type of bonds formed
Number of bonds formed
Bond energy (kJ/mol)
Energy change (kJ)
H F 2 568.2 1136.4
H0 = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ
9.10