Chemistry 1000 (Fall 2013)
Problem Set #6: Intermolecular Forces and Chromatography
Answers to Questions in Silberberg (only those w/out answers at the back of the book)
11.3 Phase changes involve intermolecular forces (IMF).
(a)
Intermolecular forces (IMF) prevent ice cubes from adopting the shape of their container.
(b)
Intermolecular forces (IMF) are overcome when ice melts.
(c)
Intermolecular forces (IMF) are overcome when liquid water is vapourized.
(d)
Intramolecular forces (in this case, covalent bonds) are overcome when gaseous water is
converted to hydrogen gas and oxygen gas.
11.30 Covalent bonds involve the sharing of electrons between two atoms.
Intermolecular forces do not involve sharing of electrons.
11.31
(a)
A shows dipole-dipole
B shows dipole-dipole
C shows ion-dipole
D shows hydrogen bonding
(b)
Dipole-dipole (A and B) is weakest
Hydrogen bonding (D) is in the middle
Ion-dipole (C) is strongest
11.33 A covalent bond involves sharing of electrons between two atoms and is what holds the
two atoms in a single I2 molecule together. It is a stronger attractive interaction than the
induced dipole-induced dipole forces active between I2 molecules. As such, the distance
between atoms within a single I2 molecule is shorter than the distance between atoms of
different I2 molecules.
11.35 From the glossary:
Polarizability is “the ease with which a particle’s electron cloud can be distorted”.
Polarity (adapted from polar molecule definition) is an unequal distribution of charge
across a molecule as a result of polar bonds and shape. i.e. the presence of a molecular
dipole
Polarity is required for a molecule to experience dipole-dipole forces. It is also required
for a molecule to experience ion-dipole forces when interacting with an ion. It is also
required that at least one molecule in a dipole-induced dipole attraction be polar.
Polarizability affects the strength of induced dipoles. Highly polarizable molecules can
develop strong induced dipoles and will experience strong induced dipole-induced dipole
attractions. Note that table 11.2 shows that the maximum strength of induced dipoleinduced dipole forces is higher than that of all other IMF except ion-dipole forces (and is
the same as the maximum strength of a hydrogen bond).
11.36 see lecture notes and/or explanation on p.412
11.54 The molecules in motor oil are large and therefore very polarizable. As such, the
dispersion forces (aka induced dipole-induced dipole forces) within motor oil are strong.
22.75 see lecture notes and/or explanation on pp. 990-992
Additional Practice Problems
1.
List all the intermolecular forces, in decreasing order of strength, which operate in an
aqueous sodium chloride solution. Between which solution components, do each of these
forces operate.
STRONGEST
 ion-dipole forces between ions (Na+ or Cl−) and water;
 hydrogen bonding between water molecules;
 ion-ion forces between Na+ ions, between Cl− ions and between Na+ and Cl− ions
(these forces are weaker than might be anticipated because the ions are solvated by
water molecules so the distance between them is large);*
 ion-induced dipole forces between ions and water (not discussed in class, could be
left out of your answer);*
 dipole-dipole forces between water molecules;*
 dipole-induced dipole forces between water molecules;
 induced dipole-induced dipole forces (London dispersion forces) between water
molecules
WEAKEST
* These three types of IMF are comparable in strength in this situation. It is not
important which order you had them in relative to each other.
2.
For each of the compounds below, indicate whether it would be volatile* (to an
appreciable extent) at room temperature.
a. potassium iodide
b. tartaric acid (structure on question sheet)
c. bromine
*volatile = “evaporates readily”; you should be familiar with this term. If a compound
experiences only weak intermolecular forces, it will evaporate readily.
Only bromine (Br2) is volatile under these conditions. The intermolecular forces in
potassium iodide (KI) and tartaric acid are too strong.
3.
Compare the intermolecular forces present in pure CCl4 and in pure CI4. One of these
compound is a liquid under standard conditions; the other is a solid. Which is which?
Justify your answer.
 Both tetrahalides are tetrahedral, non- polar compounds, so the only intermolecular
forces present are induced dipole-induced dipole forces
 CCl4 is smaller than CI4. Therefore, CI4 has a much higher polarizibility and has
much stronger induced dipole-induced dipole forces. As a consequence, CI4 has a
lower vapour pressure and a higher boiling point than CCl4.
 The difference in strengths of the intermolecular forces results in different states of
matter for these two compounds. Carbon tetrachloride is a liquid and carbon
tetraiodide is a solid.
4.
For each pair of compounds, identify the compound with the higher boiling point. Justify
each answer.
Boiling point increases as the strength of intermolecular forces increases.
(a)
octane (CH3CH2CH2CH2CH2CH2CH3) or hexane (CH3CH2CH2CH2CH3)
Octane
Both molecules are nonpolar. Since octane is a larger molecule, it has stronger induced
dipole-induced dipole forces. Hence, octane has a higher boiling point.
(b)
helium or argon
Argon
Both atoms are non-polar. Since argon is a larger atom, it has stronger induced dipoleinduced dipole forces. Hence, argon has a higher boiling point.
(c)
SO2 or CO2
SO2
Sulfur dioxide is a polar molecule (bent geometry). On the other hand, carbon dioxide is
a nonpolar molecule (linear geometry). The dipole-dipole forces between SO2 molecules
are stronger than the induced dipole-induced dipole forces between CO2 molecules.
Hence, sulfur dioxide has a higher boiling point.
(d)
water or ethanol (CH3CH2OH)
Water and ethanol can form hydrogen bonds. Water can form two hydrogen bonds per
molecule, while ethanol has a non-polar carbon group attached to the O. The
intermolecular forces in water are stronger and, hence, the boiling point of water is higher
than that of ethanol.
5.
List the intermolecular forces present in pure samples of each of the following:
In order to answer this question you have to draw the Lewis structures and determine the
molecular geometries.
(a)
NaF
This compound consists of Na+ cations and F- anions. Neither ion has a dipole moment.
Intermolecular forces:
Ion-Ion forces
Ion-induced dipole forces
Induced dipole-induced dipole forces
(b)
H2S
This molecule has a bent molecular geometry and is polar.
Intermolecular forces:
Dipole-dipole forces
Dipole-induced dipole forces
Induced dipole-induced dipole forces
(c)
SF6
This molecule has an octahedral molecular geometry and is non-polar.
Intermolecular forces:
Induced dipole-induced dipole forces
(d)
KClO4
This compound consists of K+ cations and ClO4- anions. Neither ion has a dipole moment
(perchlorate is tetrahedral).
Intermolecular forces:
Ion-Ion forces
Ion-induced dipole forces
Induced dipole-induced dipole forces
(e)
NO2F
This molecule has a trigonal pyramidal molecular geometry and is polar.
Intermolecular forces:
Dipole-dipole interactions
Dipole-induced dipole interactions
Induced dipole-induced dipole forces
(f)
SeF4
This molecule has a seesaw molecular geometry and is polar.
Intermolecular forces:
Dipole-dipole interactions
Dipole-induced dipole interactions
Induced dipole-induced dipole forces
(g)
OF2
This molecule has a bent molecular geometry and is polar.
Intermolecular forces:
Dipole-dipole interactions
Dipole-induced dipole interactions
Induced dipole-induced dipole forces
6.
Which of the following substances are capable of hydrogen bonding with water?
Hydrogen bonding is an intermolecular attraction between atoms of different molecules.
It is not the same as covalent bonding!!!
(a)
NaF
F- anions can hydrogen bond with H in water
(b)
H2S
No atoms in H2S can hydrogen bond with H or O in water.
(c)
SF6
F atoms in SF6 can hydrogen bond with H in water.
(d)
KClO4
O atoms in ClO4- anion can hydrogen bond with H in water.
(e)
NO2F
O and F atoms in NO2F can hydrogen bond with H in water.
N in NO2F cannot hydrogen bond because it has no available lone pairs
(see Lewis structure).
7.
(a)
In what ways are GC and HPLC similar techniques?
Since both are forms of chromatography, they both allow separation of mixtures based on
how strongly each component is attracted to a stationary phase compared to a mobile
phase. Usually, these attractions are determined by intermolecular forces. Components
with higher attractions to the stationary phase will spend a higher proportion of their time
adsorbed* to it and will therefore take longer to travel through the instrument.
* adsorbed = “stuck to”; it is *not* the same as “absorbed” (which implies penetration)
** “They are both chromatography” would be too vague an answer to be worth any marks.
An answer to a question like this should show understanding of what chromatography is.
(b)
In what ways do GC and HPLC differ?
In GC, the stationary phase is a liquid (coated onto a solid column) and the mobile phase
is a gas (traveling through the column).
In HPLC, the stationary phase is a solid (packed into a column) and the mobile phase is a
liquid (traveling through the column; it is pushed through using pressure).
(c)
What sort of factors might you take into account when choosing whether to analyze a
mixture using GC or using HPLC?
For a mixture to be separable using GC, the components have to be somewhat volatile
(evaporate relatively easily – though the column is usually heated to facilitate
vaporization). Otherwise, they will not travel through the column with the gaseous
mobile phase.
For a mixture to be separable using HPLC, the components have to be soluble in the
mobile phase. Otherwise, they will not dissolve in it at all and will not be able to travel
through the column with it.