Group Quiz: Intermolecular Forces
1
Give a reason for the fact that, of the elements of Group 17, under standard conditions
fluorine and chlorine are gases, while bromine is a liquid and iodine is a solid.
2
Give a reason for the fact that water, the hydride of oxygen, is a liquid at room
temperature while the hydride of sulfur (the next member of Group 16) is a gas?
3
(1.) Hydrogen fluoride is a liquid at room temperature and a weak acid but hydrogen
chloride is a gas and a strong acid. Explain these facts.
(2.) For similar reasons water and ammonia have unexpected properties. What are they?
4
Explain, using diagrams and a brief statement, how hydrogen bonding produces each of
the following results:
(1.) Ammonia gas is readily liquefied despite its low density.
(2.) Acetic (ethanoic) acid is strongly associated in the liquid state.
(3.) The boiling point of ethanol is much higher than that of dimethyl ether of identical
molecular weight.
5
List an example of each of the following types of species:
(1.) a molecule containing no polar bonds.
(2.) a non-polar molecule containing polar bonds.
(3.) a polar molecule.
(4.) a molecule subject to hydrogen bonding.
(5.) a compound containing the H- ion.
6
Explain the following observations:
(1.) The boiling point of CCl4 (350 K) is greater than the boiling point of CH4 (111 K).
(2.) the boiling point of ammonia (NH3, 240 K) is greater than the boiling point of
phosphine (PH3, 185 K).
7
List the following substances in order of increasing normal boiling point:
HBr, HCl, HF, HI.
8
List by formula the following substances in order of increasing boiling point:
methoxymethane (dimethyl ether), CH3-O-CH3
1,2-ethanediol, OH-CH2-CH2-OH
ethanol, CH3-CH2-OH
ethane, CH3-CH3.
Intermolecular Forces - Answers
1
All group 17 elements (the halogens) have the same valence electron configuration and
exhibit the same type of bonding. They all exist as covalent, diatomic molecules (F2, Cl2,
Br2, I2). Between individual molecules there exists dispersion forces, which arise from
the randomness of electron distribution within the individual molecules. Dispersion forces
are intermolecular forces which are relatively weak when compared with covalent or ionic
bonds, so the melting points of the halogens are low. The increase in melting point down
the group is due to the increase in intermolecular dispersion forces experienced as a
result of the increased number of electrons. The number of electrons in F2 is 18, Cl2 has
34, Br2 has 70, and I2 has 106.
2
The small size of the hydrogen atom and high electronegativity of oxygen result in highly
polar O-H bonds in H2O. These highly polar bonds lead to extensive hydrogen bonding
between water molecules. Sulfur is larger and less electronegative than oxygen, so the SH bonds in H2S are much less polar and no hydrogen bonding between molecules occurs.
These stronger intermolecular forces present between H2O molecules requires the supply
of considerably more energy to break individual molecules from each other than is the
case for H2S molecules - sufficient to give water a boiling point of 100 °C, while the
weaker intermolecular forces present between H2S molecules results in a boiling point of
only -60.3 °C (at 1 atm pressure).
3
(1.) Hydrogen fluoride exhibits hydrogen bonding between HF molecules. This results in a
boiling point much higher than might be expected from consideration of molecular mass
alone and thus hydrogen fluoride is a liquid at room temperature and pressure whereas
the other hydrogen halides are all gases at those conditions.
Ionisation in water is incomplete (unlike the other hydrogen halides),
ie the dissociation equation HF
H+ + F- lies to the left.
Strong hydrogen bonding between HF molecules and also between HF and H2O
molecules leads to extensive association of HF molecules in water solution, and results in
relatively few free H+ ions. Therefore HF is a weak acid. Conversely, HCl molecules do not
hydrogen bond to each other or to water molecules, so it exists as a gas at room
temperature and ionises completely in water solution (thus acting as a strong acid).
(2.) Water and ammonia both have much higher boiling points than might be expected by
considering molecular mass alone. This is due to hydrogen bonding between the highly
polar molecules.
4
5
(1.) Molecules containing no polar bonds include H2, Br2, O3.
(2.) Non-polar molecules containing polar bonds include CO2, CCl4, Br-CC-Br.
(3.) Polar molecules include HCl, H2O, H2S.
(4.) Molecules subject to hydrogen bonding include H2O, HF, NH3.
(5.) For hydrogen to exist as a hydride (ionic H-), it must be bonded to a substantially
more electropositive atom in an ionic solid. Such atoms include the Group 1 and 2
elements (except Be, which forms the covalent BeH2 molecule). Examples include NaH,
KH and SrH2.
6
(1.) CCl4 would be expected to have a higher boiling point than CH4 since it posesses
more electrons than CH4. Thus the magnitude of the dispersion forces present between
CCl4 molecules is higher than that between CH4 molecules, and this is the main reason for
the higher boiling point. Note that the increased molecular mass of CCl4 contributes only
very slightly to the boiling point. The dominant factor is the increased dispersion force.
(2.) NH3 exhibits hydrogen bonding in addition to dispersion forces. This significantly
increases the intermolecular force, and raises the boiling point. PH3 does not exhibit
hydrogen bonding and the dominant intermolecular force holding these molecules
together is dispersion forces.
7
In order of increasing boiling point: HCl, HBr, HI, HF.
The trend is determined by strength of dispersion force which is related to the number of
electrons, except for HF, which exhibits hydrogen bonding sufficiently strong to more than
compensate for the smaller number of electrons in the HF molecule. This results in a
boiling point higher than even the most electron rich hydrogen halide, HI.
8
The C-O bonds of methoxymethane (dimethyl ether) (CH3-O-CH3) are polar. The
geometry of the molecule is angular, resulting in an overall molecular dipole. Hence the
molecule will be subject to dipole-dipole and dipole/induced dipole interactions as well as
the stronger dispersion forces.
1,2-ethanediol (OH-CH2-CH2-OH), due to the presence of the O-H, bonds is capable of
hydrogen bonding which is usually the strongest intermolecular interaction. There are two
sites for hydrogen bonding in this molecule, so this will enhance the possible hydrogen
bonding interactions. This compound will of course also experience dispersion forces and
dipole/dipole and dipole/induced dipole forces between molecules but the hydrogen
bonding interaction would be most significant.
Ethanol (CH3CH2OH) experiences the same types of intermolecular forces as 1,2ethanediol but the hydrogen bonding can only occur at one site per molecule rather than
two. This results in reduced interactions between molecules compared with 1,2-ethanediol
but still more than in ethane and dimethyl ether which lack hydrogen bonding between
their molecules.
Ethane (CH3-CH3) is non-polar, and subject only to dispersion forces.
As hydrogen bonding is usually the strongest of the intermolecular forces, one would
expect the boiling points of these compounds to correlate with hydrogen bonding
interactions present. Hence ethanol would have a lower boiling point than 1,2-ethanediol
but ethane and dimethyl ether would both have lower boiling points. Of the latter two
compounds, ethane is smaller than dimethyl ether so has less dispersion forces and also
it is non-polar so lacks the dipole/dipole and dipole/induced dipole interactions which are
present in the ether. Thus the order of increasing boiling point of all four compounds would
be:
CH3-CH3, CH3-O-CH3, CH3CH2OH, OH-CH2-CH3-OH.