l
"9N&O
/PUBUJPO
4IBQFPG.PMFDVMF
EJDIMPSPNFUIBOF
-FXJT4USVDUVSF
/BNFBOE%JBHSBN
CHEMISTRY 11 AP – INTERMOLECULAR FORCES WORKSHEET
B
5Z
"
C
B
1) Identify the main intermolecular force acting between the following molecules:
Licensed to Walnut Grove Secondary for school use only from September 1, 2015 to June
(a)
CH4 Valence E-
4+4=8
$PNQMFUFUIFGPMMPXJOHUBCMF
l
*l
Licensed to Walnut Grove Secondary
(g)
l
l
$PNQMFUFUIFGPMMPXJOHUBCMF
QIPTHFO
F
4IBQFPG.PMFDVMF
H
"9N&O
-FXJT4USVDUVSF
/PUBUJPOD
tetrahedral
C-
5ZQFPG*
"9N&O "DUJOH#
/BNFBOE%JBHSBN
EJDIMPSPNFUIBOF
't
non polar
-FXJT4USVDUVSF
/PUBUJPO
'PSFBDIQBJSPGDPMVNOTESBXMJOFTUPDPOOFDUUIF"9
B
N&OOPUBUJPOPOUIFMFGUUPUIFDPSSFDUT
C
POFJTEPOFGPSZPV
B
London (dispersion) forces
Dipole-Induced Dipole
.
(b)
"9N&O/PUBUJPO
l
"9
"9&
"9
polar
(c)
BOHVMBS
"9N&O/PUBUJPO
l
"9polar
&
l
TVMQIVSIFYBGMVPSJEF
Dipole-Dipole
Dipole-Dipole "9 &
USJHPOBMCJQZSBNJEBM
E
AlCl3
(i)
USJHPOBMQZSBNJEBM
C
EJDIMPSPNFUIBOF
"9&
5+21=26-6
ftp.j#3tCIII
l
l USJHPOBMQMBOBS
"9&
"9
polar
UFUSBIFESBM
"9&l
QIPTHFO
F.SN/r?pjr:trigonalbipyramidal..is..rthpjr
t.ci:5
l
"9&
"9
.PMFDVMBS4IBQF
l
l
EJDIMPSPNFUIBOF
QIPTHFOF
C
D
(h)
l
D
Ionic Bonding
MJOFBS
Dipole-Dipole "9&
JPEJOFQFOUBGMVPSJEF
QIPTHFOF
(d) TVMQIVSIFYBGMVPSJEF
(j)
N2
D
Valence
=
polar
$POTJEFSUIF-FXJTTUSVDUVSFTGPSNFUIBOFBOEBNNPOJB
:N=N
octahedral
B
8IJDINPMFDVMFXJMMIBWFUIFTNBMMFS9o"o9CPOEBOHMFBOEXIZ
linear
E
non
.
:
TVMQIVSIFYBGMVPSJEF
London (dispersion) forces
(e)
E
non
JPEJOFQFOUBGMVPSJEF
polar
Valence
Nitrogen tribromide
.
e
=
TVMQIVSIFYBGMVPSJEF
:
E
Hydrogen bonding
.
London (dispersion) forces
(k)
JPEJOFQFOUBGMVPSJEF
polar
-2=8
5+5=10
.
Dipole-Dipole
polar
(f)
:i÷i"
sit
F
÷
SiF4 valence E-
(l)
4+28=32-8--24
tetrahedral
.
F
.
.
:
non
:
Ion-Dipole force
.
polar
London (dispersion) forces
2) Predict which compound in each pair will have the higher melting point/boiling point. Explain your
answer.
§=C÷§
:p
:c
.ci
'q;&l
CS2
(a) CS2 or CCl4
valence
CCH
4+12=16 -4=12
e=
Valence e-
=
4+28--32-8=24
:
:
linear
non
.
:
polar
I
,
Cl
ntetrahedral
on
polar
.
CCl4 – Since both are non-polar, the compound that has the stronger London (dispersion) forces
will have the higher MP/BP. London (dispersion) forces get stronger as the molecular mass
increases.
IF
valence E
(b) HI or KI
=
1+7=8-2--6
.ie?ipolarKTC:IICl2&F2va1encee-=
KI – KI has ionic bonds while HI possesses weaker dipole-dipole forces.
(c) Cl2 or F2
0
14-2=12
:D
-
Cl
6
,
non
°
+
0
G
6
x
:
•
e
F- F
:
a
°
-
:
0
•
.
polar
non
.
polar
Cl2 – Since both are non-polar, the compound that has the stronger London (dispersion) forces
will have the higher MP/BP. London (dispersion) forces get stronger as the molecular mass
increases.
Hsovalenceei
(d) Na2O or H2O
-4=4
2+6=8
=
%
Hst
s
°
.
Nail
:D
:]
[
2-
.
[ NAI
st
.
bent
polar
H
Na2O – Na2O has ionic bonds while H2O possesses weaker hydrogen bonding.
e=
Cckvalencee
'
=
linear
g
0
g
non
:O
6
polar
.
4+12=16 -4=12
4=12
4+12=16
0=si=0
valence
9.02
(e) SiO2 or CO2
.=C=
:O
non
-
-
.
polar
SiO2 – SiO2 is a network covalent solid, so covalent bonds would need to be broken while CO2
has weak London (dispersion) forces.
Ctkvalencee
(f) CH4 or NH3
Ntbvalencee
µ
-
C
t
-
=
.
=
g+H
stt
ql
#
-6=2
5+3=8
H
non
.
g.
pyramidal
siingtrigonal
tetrahedral
1
H
4+4=8
.
polar
'
polar
,
NH3 – NH3 has hydrogen bonding while CH4 possesses weaker London (dispersion) forces.
(g) CHCl3 or CF4
£
CHA
⇐
:& ?
}uaknceE=
name
"
4+1+21=26
4+28
-8=18
'
?jEg÷¥
.ro#fpi.t:tftfaahredra
µ
*
'
0.5
of
1
a
0.5
,
tetrahedral
non
CHCl3 – CHCl3 has dipole-dipole forces while CF4 possesses weaker London (dispersion)
forces.
.
polar
÷
:p
F.
Btzvalencee
(h) BF3 or P4
,
,
B
/
;
Pyvalenoee
trigonalplanar
non
'
Polar
\
3+21=24 -6=18
20-6=14
'
=
'
=
.
p.=p
think
ftp.tshfnoture
infaotit
when
isthis
Parenon
:P
:P them
"
's
,
this
:
you may
-
~
P
,
both
of
.
polar
.
P4 – Since both are non-polar, the compound that has the stronger London (dispersion) forces
will have the higher MP/BP. London (dispersion) forces get stronger as the molecular mass
increases.
(i) CaF2 or HF
Hfualencee
EEICCATTEEI
1+7=8-2--6
.
=
SIFFF
polar
CaF2 – CaF2 has ionic bonds while HF possesses weaker hydrogen bonding
3) The industrial production of ammonia, NH3, from H2 and N2 is called the Haber process, named for
Fritz Haber, the German chemist who developed it just before World War I. During the process, in a
gaseous mixture of all three substances, NH3 must be separated from H2 and N2. This is done by
cooling the gaseous mixture so as to condense only the NH3. This leaves the elemental nitrogen and
hydrogen as gases to be recycled and produce more ammonia. Why does only the ammonia liquefy
upon cooling, but not the H2 or N2?
Since hydrogen bonds exist between ammonia molecules but only London (dispersion) forces exist
between diatomic molecules of nitrogen and hydrogen, the ammonia has a higher boiling point and
therefore liquefies (condenses) at a higher temperature than nitrogen and hydrogen that remain in the
gaseous phase upon cooling.
4) Rank the following compounds from weakest intermolecular forces to strongest. Justify your
answers.
'benFH
8TH
say
.si
Hasvalencee
Hst
sty
Haovalencee
'
H2S, I2, N2, H2O
'
-2+6--8-4=4
polar
.MY?f..et=H.2=t2N2valenceej=
-2+6=8-4--4
sioiqbent
10-2=8
:N±N
?!
.
:
a
non
polar
.
non
.
polar
polar
N2 and I2 are non-polar, so they only have London (dispersion) forces – I2 has stronger forces since it
is larger. H2S has dipole-dipole forces therefore it is stronger than I2. H2O has hydrogen bonding
therefore it is stronger than H2S.
N2 < I2 < H2S < H2O
5) Rank the following from weakest intermolecular forces to strongest. Justify your answers.
Pos
8TH
.si
Htbent
'H£v5H'
set
'Hs
0h
'benfH£
8TH o.io#.o.pdar....
Paige
polar
"qa3
all valence
^
5
E=
H2Se, H2S, H2Po, H2Te
2+6=8-4--4polar
.
polar
Te
.
'
bent
←
bent
'H
These compounds are all the same bent shape. Although H2S is slightly more polar than the others,
it is not very polar so it has very weak dipole-dipole forces. Therefore, the differences in London
(dispersion) forces are more important for these compounds. H2Po is the largest and therefore has
the strongest London (dispersion) forces.
6) Iodine is a non-polar diatomic molecule, yet its molecules have enough attraction for each other that
the element exists as a solid at room temperature. Identify the attractive force and explain why it is
strong enough to keep the molecules of I2 attached to each other even at room temperature.
I2 is a relatively large and massive diatomic molecule and the strength of London dispersion forces
increases as the size of the molecules involved increases. This is because large electron clouds are
more loosely held than smaller clouds and therefore more easily deformed or polarized by a nearby
dipole than compact tightly held clouds. In addition, large molecules with more surface area have
electron clouds that are spread out and so are more easily distorted by neighbouring dipoles. As a
result, the dispersion forces are strong enough to keep the molecules of I2 attached to each other
even at room temperature.
7) Surface tension is the ability of a fluid to act as a thin elastic membrane at its surface. Explain why
non-polar molecules usually have much lower surface tension than polar ones.
Surface tension results when the molecules at a surface attract each other to be able to form a thin,
elastic membrane. For example, H2O (due to their hydrogen bonding) has surface tension since the
positive hydrogen pole of one water molecule is attracted to the negative oxygen pole of a
neighbouring water molecule.
Since non-polar molecules are not attracted to each other as much as in polar molecules, these
molecules are much less likely to have high surface tension.
8)
TPUIFDSZTUBMTUSVDUVSFCFHJOTUPCSFBLEPXO"TUIFJP
JNNFEJBUFMZCFDPNFTVSSPVOEFEPSFODMPTFEJOXIBUD
Ionic compounds
such as NaCl have very high melting points because a great deal of energy is
required to overcome the many attractive forces between the oppositely charged ions in an ionic
crystal lattice. NaCl melts "UUIFDFOUSFPGPOFUZQFPGIZESBUJPOTIFMMUIFO
at 801°C, yet its ions will readily separate from each other at room
temperature when the solid is added to water. Explain this by discussing the predominant force that
allows an ionicPSJFOUUIFNTFMWFTOFYUUPBOETVSSPVOEBDBUJPO"UUIF
compound to dissolve in water.
CZXBUFSNPMFDVMFTPSJFOUFEXJUIUIFJSQPTJUJWFIZESPH
Although the ionic
bonds holding a crystal lattice together are strong, when the surface of that lattice
is in contact with water, each ion on that surface will attract the oppositely charged end of polar
water molecules
near them. That attraction between an ion and a polar molecule is called an ionTIPXTUIFJPOEJQPMFGPSDFTBDUJOHCFUXFFOXBU
dipole force. These attractive forces soon overcome those existing between the ions themselves and
so the crystal structure
begins to break down and the ionic compound dissolves. As the ions move
*POEJQPMFGPSDFTBSFUIFQSJNBSZGPSDFSFTQPOTJCMFGPS
away from the lattice surface, they immediately become surrounded or enclosed in what chemists
call a hydrationBRVFPVTTPMVUJPOTPGTPNFJPOJDDPNQPVOETBSFBMNPT
shell. Ion-dipole forces are the primary force responsible for the solubility of ionic
compounds in water.
OE5IFPSZ
SJEJ[BUJPO
0SCJUBMT
ODFQUPG
SHF
l
a
a
ater molecule
ith dipole
l
ion dipole attraction
'JHVSF *POEJQPMFGPSDFTCFUXFFOXBUFSNPMFDVMFTB
(/-FXJTTEPUTUSVDUVSFTQSPWJEFVTFGVMNPEFMTPGUIF
#ZFWBMVBUJOHUIFUPUBMOVNCFSPGWBMFODFFMFDUSPOTJO
BUPNT-FXJTTUSVDUVSFTQSPWJEFBXBZUPSFQSFTFOUXIF
PSNVMUJQMFCPOET5IFSFQVMTJPOUIBUPDDVSTCFUXFFOC
FMFDUSPOTDBOCFVTFEUPQSFEJDUNPMFDVMBSTIBQF5IJT
FMFDUSPOFHBUJWJUZPGBUPNTXJUIJOBNPMFDVMFT-FXJTTU