Section 1.1 Section 1.2 Section 1.3 - The Student Room

SECTION 5
a Pentane has the strongest intermolecular forces and
hence the highest boiling point. Dimethylpropane has
the weakest intermolecular forces and hence the
lowest boiling point.
b The molecules of pentane, because it is a straight
chain alkane, can approach closely to each other
which increases the opportunities for
instantaneous–induced dipole interactions and hence
stronger intermolecular attractions. Methylbutane has
one methyl side chain and so it is more difficult for
these molecules to approach each other and
instantaneous–induced dipole interactions are weaker.
Dimethylpropane has two methyl side-chains and so it
is even more difficult for molecules of this compound
to approach each other.
4 The strength of instantaneous dipole–induced dipole
forces between molecules increases as the relative
molecular masses of the molecules increase. To be sure
that the higher boiling point of the polar substance is
due only to the increased strength of dipole–dipole
attractions it will be necessary to ensure that the
instantaneous dipole–induced dipole forces in both
polar and non-polar substances are of similar strength.
This can be done by comparing substances of similar
molecular mass.
5 a A and D; C and G.
b A and G have the stronger intermolecular forces
compared to D and C, respectively.
6 Polarity will occur as follows:
C-F; H-Cl; H-N; C-O with charges in the order d+ d– in
each case.
7 CHCl 3
, CH 3
OH, (CH 3
) 2
CO, cis-1,2-difluoroethene and
1,2-dichlorobenzene possess dipoles.
8 a i Eighteen
ii The attractions will be similar.
iii H 2
S has a permanent dipole. It is a bent molecule
with two lone pairs. SiH 4
does not have an overall
permanent dipole as it is a symmetrical molecule.
b Both compounds have similar instantaneous
dipole–induced dipole forces. However, H 2
S also has
permanent dipole–permanent dipole attractions so its
boiling point is higher than that of SiH 4
.
9 a Instantaneous dipole–induced dipole.
b Instantaneous dipole–induced dipole.
c Instantaneous dipole–induced dipole, permanent
dipole–permanent dipole.
d Instantaneous dipole–induced dipole.
e Instantaneous dipole–induced dipole, permanent
dipole–permanent dipole.
f Instantaneous dipole–induced dipole.
g Instantaneous dipole–induced dipole.
h Instantaneous dipole–induced dipole, permanent
dipole–permanent dipole.
Section 5.4
1 a As the temperature rises, solids and liquids expand.
The temperature increase raises the kinetic energy of
the particles present. In solids, the rotational and
vibrational energy increases. In liquids, rotational,
vibrational and translational energy increases. The
increases in vibrational and translational energy
increase the volume occupied by the particles. As they
occupy an increasing volume, the density of the solid
or liquid decreases.
b i When ice melts, much of the open, hydrogenbonded
structure collapses. This enables the
molecules to occupy less space so the density
increases on melting.
ii The boiling point of water is higher than
expected, as more energy is needed to break the
hydrogen bonding.
iii The specific heating capacity of water is higher
than expected, as more energy is absorbed by the
water to break hydrogen bonds in the liquid.
2 a i H 2
O instantaneous dipole–induced dipole
permanent dipole–permanent dipole
hydrogen bonding.
H 2
S instantaneous dipole–induced dipole
permanent dipole–permanent dipole.
H 2
Se instantaneous dipole–induced dipole
permanent dipole–permanent dipole.
H 2
Te instantaneous dipole–induced dipole
permanent dipole–permanent dipole.
ii Intermolecular forces must be overcome when a
liquid boils.
Hydrogen bonding present between molecules of
H 2
O – but not between those of H 2
S, H 2
Se or H 2
Te.
Hydrogen bonding forces are much stronger than
other intermolecular forces and so the boiling point of
water is higher than that of the other hydrides.
b The strength of instantaneous dipole–induced dipole
and permanent dipole–permanent dipole attractions
in a substance gets weaker as its relative molecular
mass gets smaller. This produces a lower boiling point.
The boiling point of H 2
O should be lower than that of
H 2
S but it is in fact much higher. This suggests that,
compared to H 2
S, a different and much stronger type
of intermolecular bonding exists in H 2
O.
c ii All have instantaneous dipole–induced dipole
forces.
iii The shape of the graph is nearly a straight line
with positive slope. There is no hydrogen
bonding between the hydride molecules of
Group 4. The increase in boiling points down the
group is a result of their regularly increasing
molecular masses.
3 A, D, E and F
4 a Hydrogen bonding will be present in NH 3
, CH 3
OH,
and HF.
b
N H CH 3
F
H H O
H
H
H
H
H CH 3 F
F
N
O
H H
H
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