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UNIT V A-2021
INTERMOLECULAR FORCES, LIQUIDS, and SOLIDS
Sections 11.2; 11.3; 11.6

Non Polar Molecular Substances: Examples: F2, Cl2, Br2, I2, O2, N2, CH4, C2H6.
Consider the boiling of a non polar molecular substance such as liquid bromine (Br2), molar
mass = 160 g/mol.
b.p. 59oC
Br2(l) Br2(g)

From the periodic table we see that Z(Br) = 35, thus there are 70 electrons in one molecule of
Br2.

What is the nature of the intermolecular attractive forces in liquid Br2?

 Draw an electron cloud containing all the electrons for a molecule of Br2.
 Draw an electron cloud containing all the electrons for a second molecule of Br2.
 Make a sketch of the resulting London dispersion forces.

Br2 molecule a cloud of 70 electrons

a cloud of 70 electrons

d+ Dd+

a cloud of 70 electrons
Br2 molecule

Problem: Why does liquid fluorine, F2, have such a low boiling point? b.p = - 188oC.
Molar mass of F2 is 38 g/mol.
Ans: Cl2 (38 g/mol)has a much smaller molar mass than Cl2(71 g/mol) and Br2(160 g/mol).

Problem: Would you predict the boiling point of molecular iodine, I2, to be greater than the
boiling point of F2, Br2? Molar mass of I2 is 254 g/mol.
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Polar Molecular Substances: Examples: CO; HCl; HBr; ICl; ClF, H2S; CHCl3.
Consider the boiling (b.p.) of a polar molecular substance such as liquid ICl, molar mass of 162
g/mol.

ICl(l) ICl(g) b.p. 97oC

What is the nature of the intermolecular attractive forces in liquid ICl?

 Notice that the molar mass is almost the same as that of Br2.
 Thus the electron cloud containing all the electrons for a molecule of ICl will be about
the same size as that for a molecule of Br2. See the four sketches.
 Consider the difference in electronegativities of I and Cl. Which element is more
electronegative? Assign partial positive and negative charges to the atoms concerned. Do
the same for a second molecule of ICl, and a third molecule of ICl.
 Make a sketch of the permanent dipole-dipole attractive forces.
 Now make a sketch of the resulting London dispersion forces.
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Polar Molecular Substances having O – H, or N – H, or F – H bonds. Section 11.2,Chang

Molecule 1 Molecule 2 Molecule 1 Molecule 2 Molecule 1 Molecule 2

O – H -------- :N N – H --------- :N F – H ---------- :N

O – H -------- :O N – H ---------:O F – H ----------:O

O – H -------- :F N – H --------- :F F – H ---------- :F

Observation:
H2O, molar mass of 18 g/mol, b.p. of 100oC.
H2S, larger molar mass of 34 g/mol, b.p. of –58oC is lower than that of water.
Why does a compound with a larger molar mass have a lower b.p.?

Rationale:
 Oxygen is a smaller atom than S with a high nuclear charge. (Z = +8).
 Oxygen has a large electronegativity value of 3.5, the value for S is only 2.5.
 Hydrogen has a small electronegativity value of 2.1.
 The electrons in the covalent O – H bonds are pulled close to the oxygen atom. The O –
H bonds are thus very polar. The oxygen atoms have a considerable net negative charge,
-, and the hydrogen atoms are net positively charged, +.
 Relatively strong dipole-dipole intermolecular attractive forces result between water
molecules. These are named hydrogen bonds.

Sketches:

Intermolecular attractive forces (continued)

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Problem 1. Hydrogen bonding is thought to account for the differences in boiling points for:
NH3, b.p. –30oC, (molar mass, 17 g/mol) and PH3, b.p. –85oC, (molar mass, 34 g/mol). Make
sketches of the intermolecular forces involved. Electronegativity values: N, 3.1; P, 2.1; H, 2.1.

Problem 2. The boiling point of HF is 20oC; the boiling point of HCl is –75oC. Explain.

Problem 3.
Which of the following compounds would be expected to form intermolecular hydrogen bonds in
the liquid state?
a) CH3OCH3
b) CH4
c) HF
d) CH3CO2H
e) Br2
f) CH3OH
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MELTING
N.B. When a solid melts only a portion of the intermolecular forces are broken. The
intermolecular forces that remain are responsible for the fact that the substance is now in
the liquid state. Recall that covalent bonds (~ 400 kJ/mol) are not broken during melting or
boiling.
SUMMARY OF INTERMOLECULAR ATTRACTIVE FORCES

Increasing Strength

Hydrogen bonding > permanent dipole - dipole attractive forces > dispersion forces

Attractive Force Strength (kJ/mol)

Hydrogen bonding 10 – 40

Dipole-dipole 3–4

London Dispersion Forces 1 – 10

Problem 4. Water, H2O(s) , has a higher melting point than H2S(s)?

What is the reason for this difference? Consider the array of water molecules in ice.
Intermolecular hydrogen bonds lock water molecules into a rigid 3-D structure.

Sketch
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cnx.org/contents/0c34addb-482e-4cf4-bf20-153f2f1ec049@4/The_Hydrogen_Bond
Holes exist in the middle of the ice structure. This is why H2O(ice) is less denser than H2O(l),
Problem 5. The m.p. of Br2(s) with a molar mass of 160 g/mol is -7.2oC; ICl(s) with a molar
mass of 162 g/mol has a considerably higher m.p. of 27.2oC
What is the reason for the higher m.p. of ICl(s).
Ans: Both compounds have about the same molar mass and thus have similar dispersion forces.
ICl(s) -7.2oC ICl(l)
m.p.

Problem 6. Match the melting points with the correct compound. Briefly explain.
Compounds: C3H8, C4H10, C5H12. Melting points oC: -129 oC; -138 oC; -189 oC.

Intermolecular Forces: Review questions

7. What are the intermolecular atractive forces that must be broken in order to:
a) melt Cl2(s)

b) boil ICl(l)

c) boil CHCl3(l)

d) melt I2(s)

e) melt CH4(s)

f) boil SF4(l)
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g) Hexane (C6H14) and water are immiscible (not soluble in each other). Explain why.
COVALENT NETWORK SOLIDS (COVALENT CRYSTALS)
 These are solids where no simple molecules are present.
 The substance is “held together” by an extensive network of covalent bonds.
 The solids are characterized by a high melting point and an insolubility in
any solvent.
Rationale: The network of strong covalent bonds can only be broken at a very high
temperature. Also, there are no small individual molecules or ions that can be
surrounded and solvated by solvent molecules.

Consider two allotropes of carbon. (See Figure 11.29 in Chang).

chemistry.tutorvista.com/organic-chemistry/carbon-compounds.html

Diamond Graphite
Hardness = 10 Hardness = 1
Insoluble in all solvents Insoluble in all solvents
m.p. 3500oC m.p. sublimes at ~3652oC
density = 3.51 g/cm3 density = 2.22 g/cm3
non conductor conductor
(no mobile electrons are present) (due to the mobile valence electrons present
in the coplanar p orbitals of the pi bonds)

Since diamond is denser it is possible to convert graphite into diamond.

P>150,000 atm
Graphite Synthetic diamonds
T>2800 C o
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Mike’s Memo: Other allotropes of carbon, fullerenes and carbon nanotubes.

 Fullerenes are soccer ball shaped molecules discovered in 1985.*

Nobel prize in Chemistry to Kroto (University of Sussex, England) and to
Curl and Smalley (Rice University, Texas).
 Fullerenes are a natural component of soot (the solid residue following
combustion).
 C60 fullerene is called a buckyball and it has 12 pentagonal and 20 hexagonal
faces.
 Fullerenes have been used in AIDS research to develop drugs that prevent
the HIV virus from reproducing.
 Buckyballs also have a potential use as a high temperature superconductor
and lubricant.

Lab synthesis Graphite C60 Buckyball

Laser


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nanotechnologyuniverse.com/wp-content/uploads/2011/10/buckyball.jpg
 Carbon nanotubes are made of graphite sheets that are rolled into tubes. The
tubes have a diameter of 2 – 30 nm. On a mass basis they are about ten times
stronger than steel.

www.azonano.com/article.aspx?ArticleID=983

Another covalent crystal is silicon dioxide (SiO2) or simply named sand.

 See sketch p.501 (Figure 11.32) of Chang; each Si atom is bonded in a
tetrahedral fashion to four O atoms.

 Pure quartz glass is 100% SiO2.

 Pyrex glass has ~10 – 25% B2O3 added.

 Green glass has Fe2O3 added; red glass has Cu added; blue glass has CuO or
CoO added.

 No individual molecules or ions exist. It is insoluble in any solvent and

melts at a very high temperature.
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METALLIC CRYSTALS AND METALLIC BONDING

Properties of metals:

 Conduct heat and electricity.

 Many metals have a shiny luster.
 Low ionization energies (easily lose their valence electrons) and are good
reducing agents. Metallic character is said to increase in going down the
periodic table and in going from right to left across the table.
 Thus the most metallic elements are found on the lower left hand side of the
periodic table.
 Some metals such as Na, K, Cu and Au are malleable and ductile.
 Group 1A and 2A metals are soft while most transition metals are hard.
 Insoluble in most solvents.
The Electron-Sea Model of Bonding in Metals

Fixed, positively charged metal nuclei from group 1 (a) or group 2 (b) are
surrounded by a “sea” of mobile valence electrons. Because a group 2 metal has
twice the number of valence electrons as a group 1 metal, it should have a higher
melting point.
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Source: 2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/s16-05-correlation-between-
bonding-an.html

Problem: Arrange the following in order of increasing metallic character: Pb, Sn, C.

SOLUBILITY
Consider the dissolving of an ionic substance such as NaCl(s) in water:
 Ionic bonds in NaCl break.
 ion-dipole bonds result.

What are the intermolecular bonds or attractive forces that form when:
Benzoic acid dissolves in hot water?

Paraffin, CH3 ___ (CH2)n ___ CH3(s), dissolves in hexane, CH3CH2CH2CH2CH2CH3?

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INTERMOLECULAR FORCES LAB #9. Name: ________________________________

Date due: Friday, Nov 27, 5 p.m.

Chapter 11: Read Sections 11.2, 11.3 (The structure and properties of water); Section 11.6
(Covalent Crystals, Molecular crystals, Metallic crystals).
Attempt problems: 11.1, 11.3, 11.7 – 11.19.
Do Practice Exercise problems: pp. 470, 472. See answers on p. 517

Complete the following chart:

In column 2 put the classification of the substance: covalent crystal, ionic, metallic,
polar molecular (p.mol), non polar molecular (n.p. mol).
In column 3 put the most important intermolecular force: covalent bonds, ionic
bonds, metallic bonds, permanent dipole(dipole-dipole), dispersion.

Compound Column 2 Column 3 Compound Column 2 Column 3

CCl4(l) H2O(l)

C5H12(l) XeF4(l)

RbF(s) Al(s)

C(diamond) Cu(s)

SF4(l) HF(l)

ZnCl2(s) KBr(s)

CO2(s) Sulfur(s)

SiO2(s) SiBr4(l)

HBr(l) KBr(s)
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SO2(l) C2H2(l)

XeF4(s) NaNO2(s)

BrF5(l) C6H14(l)

CH3NH2(l) SiO2(s)