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CHAPTER 11
Intermolecular Forces
11.1 – 11.3, (11.5 – 11.9)
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electrostatic in nature
Intramolecular forces bonding forces
Intermolecular forces nonbonding forces
ATTRACTIVE FORCES
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Phase Changes
solid liquid gas
melting
freezing
vaporizing
condensing
sublimination
endothermic
exothermic
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Changes of State
At the phase-change temperature, all the energy goes into changing the state of the substance and, therefore, the temperature remains constant during a phase change. For melting, energy goes into the system; for freezing, energy leaves the system.
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Table 12.1
A Macroscopic Comparison of Gases, Liquids, and Solids
State Shape and Volume Compressibility Ability to Flow
Gas Conforms to shape and volume of container
high high
Liquid Conforms to shape of container; volume limited by surface
very low moderate
Solid Maintains its own shape and volume
almost none almost none
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bond length
van der Waal’s distance
Figure 12.10 Covalent and van der Waals radii.
covalent radius
100 pm
van der Waal’s radius 180 pm
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Periodic trends in covalent and van der Waals radii (in pm).
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Figure 12.12 Polar molecules and dipole-dipole forces.
solid
liquid
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HH
water dipole••
••
O-δ
+δ
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OH
Hδ+
δ-• • • O
H
Hδ+
δ-• • • O
H
Hδ+
δ-• • •
Na+Mg2+
Cs+
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Dipole–Dipole Attractions • Polar molecules have a permanent dipole
– because of bond polarity and shape – dipole moment – as well as the always present induced dipole
• The permanent dipole adds to the attractive forces between the molecules – raising the boiling and melting points relative to nonpolar
molecules of similar size and shape
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Tro: Chemistry: A Molecular Approach, 2/e
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replace with the figure 11.8
Tro: Chemistry: A Molecular Approach, 2/e
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Example 11.1b: Determine if dipole–dipole attractions occur between CH2Cl2 molecules
molecules that have dipole–dipole attractions must be polar
CH2Cl2, EN C = 2.5, H = 2.1, Cl = 3.0 If there are dipole–dipole attractions
Solution:
Conceptual Plan:
Relationships:
Given: Find:
EN Difference Shape
Lewis Structure
Bond Polarity
Molecule Polarity Formula
Cl—C 3.0−2.5 = 0.5 polar C—H 2.5−2.1 = 0.4 nonpolar
polar molecule; therefore dipole–dipole attractions
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polar bonds and tetrahedral shape = polar molecule
4 bonding areas no lone pairs = tetrahedral shape
Tro: Chemistry: A Molecular Approach, 2/e
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THE HYDROGEN BOND
a dipole-dipole intermolecular force
The elements which are so electronegative are N, O, and F.
A hydrogen bond may occur when an H atom in a molecule, bound to small highly electronegative atom with lone pairs of electrons, is attracted to the lone pairs in another molecule.
.. F ..
..
.. H O ..
N
.. F H ..
..
..
O.. ..
.. N H
hydrogen bond donor
hydrogen bond acceptor
hydrogen bond acceptor
hydrogen bond donor
hydrogen bond donor
hydrogen bond acceptor
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The Hydrogen Bond
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H-Bonding in Water
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H-Bonds
• Hydrogen bonds are very strong intermolecular attractive forces – stronger than dipole–dipole or dispersion forces
• Substances that can hydrogen bond will have higher boiling points and melting points than similar substances that cannot
• But hydrogen bonds are not nearly as strong as chemical bonds – 2 to 5% the strength of covalent bonds
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H-Bonding in Water
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Hydrogen bonding and boiling point.
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H-Bonding in Biology
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H-Bonding in DNA
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Base Pairing through H-Bonding
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Base Pairing through H-Bonding
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Base Pairing through H-Bonding
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Dipole - Induced-Dipole
The water dipole INDUCES a dipole in the O2 electric cloud.
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Dipole - Induced-Dipole
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Dipole - Induced-Dipole
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distortion of an electron cloud
• Polarizability increases down a group
size increases and the larger electron clouds are further from the nucleus
• Polarizability decreases left to right across a period
increasing Zeff shrinks atomic size and holds the electrons more tightly
• Cations are less polarizable than their parent atom because they are smaller.
• Anions are more polarizable than their parent atom because they are larger.
Polarizability and Charged-Induced Dipole Forces
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Dispersion Force
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The Noble gases are all nonpolar atomic elements
As the molar mass increases, the number of electrons increases. Therefore the strength of the dispersion forces increases.
Effect of Molecular Size���on Size of Dispersion Force
The stronger the attractive forces between the molecules, the higher the boiling point will be.
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Boiling Points of n-Alkanes
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Figure 12.17
more points for dispersion
forces to act
fewer points for dispersion
forces to act
Molecular shape and boiling point.
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Alkane Boiling Points • Branched chains
have lower BPs than straight chains
• The straight chain isomers have more surface-to-surface contact
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Tro: Chemistry: A Molecular Approach, 2/e
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Practice – Choose the Substance in Each Pair with the Higher Boiling Point
a) CH4 CH3CH2CH2CH3
b) CH3CH2CH=CHCH2CH3 cyclohexane
Both molecules are nonpolar larger molar mass
Both molecules are nonpolar, but the flatter ring molecule has larger surface-to-surface contact
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Intermolecular Forces
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Figure 12.18
Summary diagram for analyzing the intermolecular forces in a sample.
INTERACTING PARTICLES (atoms, molecules, ions)
ions only IONIC BONDING (Section 9.2)
ion + polar molecule ION-DIPOLE FORCES
ions present ions not present
polar molecules only DIPOLE-DIPOLE
FORCES
HYDROGEN BONDING
polar + nonpolar molecules DIPOLE- INDUCED DIPOLE FORCES
nonpolar molecules only DISPERSION FORCES only
DISPERSION FORCES ALSO PRESENT
H bonded to N, O, or F
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Step 3. Evaluate the strengths of the total intermolecular attractive forces. The substance with the strongest will be the liquid.
Formaldehyde: dispersion forces: MM 30.03, trigonal planar dipole–dipole: very polar C=O bond uncancelled H-bonding: no O–H, N–H, or F–H therefore no H-bond
Fluoromethane: dispersion forces: MM 34.03, tetrahedral dipole–dipole: very polar C–F bond uncancelled H-bonding: no O–H, N–H, or F–H therefore no H-bond
Hydrogen peroxide: dispersion forces: MM 34.02, tetrahedral bent dipole–dipole: polar O–H bonds uncancelled H-bonding: O–H, therefore H-bond
Because the molar masses are similar, the size of the dispersion force attractions should be similar
Because only hydrogen peroxide has the additional very strong H-bond additional attractions, its intermolecular attractions will be the strongest. We therefore expect hydrogen peroxide to be the liquid.
Example 11.2: One of these compounds is a liquid at room temperature (the others are gases). Which
one and why?
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Step 1. Determine the kinds of intermolecular attractive forces
MM = 30.03 Polar No H-Bonds
MM = 34.03 Polar No H-Bonds
MM = 34.02 Polar H-Bonds
Step 2. Compare intermolecular attractive forces
Because all the molecules are polar, the size of the dipole–dipole attractions should be similar
Only the hydrogen peroxide also has additional hydrogen bond attractions
Tro: Chemistry: A Molecular Approach, 2/e
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a) CH3OH CH3CHF2
b) CH3-O-CH2CH3 CH3CH2CH2NH2
Practice – Choose the substance in each pair that is a liquid at room temperature (the other is a gas)
can H-bond
can H-bond
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break IM bonds
make IM bonds
Add energy
Remove energy
LIQUID VAPOR
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Figure 13.14
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Liquid boils when its vapor pressure equals atmospheric pressure.
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C 2 H 5 H 5 C 2 H H 5 C 2 H H
water alcohol ether
increasing strength of IM interactions
extensive H-bonds H-bonds
dipole- dipole
O O O
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11-60 l nP2
P1 =
ΔHvapR 1
T1 - 1
T2
⎡
⎣ ⎢
⎤
⎦ ⎥
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concave
meniscus
H 2 O in
glass
tube
ADHESIVE FORCES
between water
and glass
COHESIVE FORCES
between water
molecules
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Separate phases
Increasing pressure
More pressure
Supercritcal CO2
Figure 13.41
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O
N
CH3
N
H3C
NNH3C
O
CO2 Decaffeination Process 1.SOAKING green coffee beans in water doubles their size, allowing the caffeine to dissolve into water inside the bean.
2. CAFFEINE REMOVAL occurs in an extraction vessel, which may be 70 feet high and 10 feet in diameter, suffused with carbon dioxide at roughly 200 degrees Fahrenheit and 250 atmospheres. Caffeine diffuses into this supercritical carbon dioxide, along with some water. Beans enter at the top of the chamber and move toward the bottom over five hours. To extract the caffeine continuously, the beans lower in the column are exposed to fresher carbon dioxide, which ensures that the caffeine concentration inside beans is always higher than in the surrounding solvent. Caffeine therefore always diffuses out of the beans.
3. DECAFFEINATED BEANS at the bottom of the vessel are removed, dried and roasted.
4. RECOVERY of dissolved caffeine occurs in an absorption chamber.
www.sweetmarias.com/decaf.co-2method.html
