Topic
6
Microscopic World II
Unit 23 Shapes of molecules
Unit 24 Bond polarity and intermolecular forces
Key
C o ncepts
Microscopic World II
Shapes of molecules
• Covalent molecules with non-octet
structures
• Shapes of molecules
• The VSEPR theory
• Molecular crystals of
buckminsterfullerene (C60)
• Carbon nanotubes
Bond polarity and
intermolecular forces
• Electronegativity and bond polarity
• Van der Waals’ forces
• Factors affecting the strength of van
der Waals’ forces
• Hydrogen bonding
• Surface tension and viscosity of liquids
Topic 6
Unit 23 Shapes of molecules
Microscopic World II
Unit 23
Shapes of molecules
23.1 – 23.7
Summary
23.1
Covalent molecules with non-octet structures
23.2
Shapes of molecules of methane, ammonia and water
23.3
Shapes of some other molecules
23.4
Influence of the nature of electron pairs on bond angles in
molecules of methane, ammonia and water
23.5
Shapes of some polyatomic ions
23.6
Shapes of molecules with multiple bonds
23.7
Molecular crystals of buckminsterfullerene (C60)
1 Examples of covalent molecules with non-octet structures are as follows:
Molecule
Electron diagram
Y Y
Y Y
Y
Y Y Y
YFY Y FY
Y r
Br Y
Yr
Y Y
YFY
YY
BF3
Y
Y YY
Y Cl Y
Y r
Y Y Yr
YCl Y
Y Y
PCl5
YY
Y
Y
Y Cl Y
Y r YY
Y
P r Cl YY
Y r YY
Y
Y
Y Cl Y
YY
Molecule
Electron diagram
SF4
YY YY YY Y Y
YF
FY
Y r Y Yr Y
Y Y Yr S rY Y Y
Y FY rr YF Y
YY
YY
SF6
YY
Y Y
Y
F
Y YY Y Y Y Y YY
Y
Y F Y r Y FY
Y r
Y
S Yr
Y
Y Y r Y r YY
Y FY Y Y r Y Y FYY
Y
Y
YFY
YY
2 The following table summarizes the relationship between the arrangement of electron
pairs around the central atom of a molecule or polyatomic ion and the shape of the
species.
Total number of electron
pairs* in the outermost
shell of the central
atom of a molecule or
polyatomic ion
Arrangement of
electron pairs*
Shape of the molecule
or polyatomic ion
B
2
A
linear
B
linear
Example(s)
BeCl2
CO2
CS2
B
A
3
B
trigonal planar
B
BF3
trigonal planar
B
B
CH4
A
B
B
tetrahedral
SiF4
NH4
+
PCl4+
NH3
4
B
tetrahedral
A
B
B
trigonal pyramidal
PH3
NCl3
H3O+
H2O
A
B
B
V-shaped
H2S
NH2–
Topic 6
Unit 23 Shapes of molecules
Microscopic World II
Total number of electron
pairs* in the outermost
shell of the central
atom of a molecule or
polyatomic ion
Arrangement of
electron pairs*
Shape of the molecule
or polyatomic ion
Example(s)
♦ Do NOT confuse the terms ‘tetrahedral’ and ‘trigonal planar’ in the
description of shapes of species.
B
B
B
5
A
B
B
trigonal
bipyramidal
♦ Students need to give good drawings of three-dimensional structures.
Use the conventions commonly used in the representation of threedimensional structures.
PCl5
♦ The shape of a PH3 / NCl3 molecule is trigonal pyramidal, NOT trigonal
planar. ✘
trigonal
bipyramidal
P
B
B
6
B
A
B
B
B
H
SF6
PCl6
Cl
H
–
octahedral
octahedral
N
H
Cl
Cl
♦ The shape of an NH4+ ion is tetrahedral, NOT square planar.
✘
* When using the VSEPR theory, multiple bonds can be counted as single bonds.
+
H
3 a) The C60 molecule is a spherical molecule with the carbon atoms arranged at the
60 vertices of 32 interconnecting pentagons and hexagons. Each carbon atom
forms three single covalent bonds with other carbon atoms.
N
H
H
H
b)C60 is as soft as graphite and is a semiconductor.
c) At room temperature, molecules in solid C60 are closely packed and bound by
weak van der Waals’ forces.
♦ Questions may ask students to draw the shapes of unfamiliar species.
d)Interest in the fullerenes has led to the discovery of a related group of carbon
structures referred to as nanotubes.
XeF2 e.g.
F
120°
Exam tips
Xe
♦ Examination questions may ask students to draw the electron diagrams
of common ions.
e.g.
F
CO32–
x
x
x
x
O
xx
key:
electron from an external source
Cxx
O x x O xx
x x
x
NO3– x
N
x x
Ox xO
SO
xx
F
–
O xx
xx
x x
2–
4
x
x
x
2–
xx
Ox
x
key:
electron from an external source
x
x
key:
electron from an external source
xx
O xx S x O xx
xx
x
x
2–
3
S2O
x
x
x
x
xx
XeF4 x
x
x
x
x
2–
x
x
xx
x
O xx
2–
O xx
xx
xx
O x S x O xx
x
x
S
key:
electron from an external source
F
Xe
F
F
Topic 6
Unit 23 Shapes of molecules
Microscopic World II
♦ Questions often ask about the shapes and bond angles of species
containing nitrogen.
Species
Electron diagram
xxx
O xx
+
NO
NO2+
NO3–
YY
Y
Y
YOY
YY
Y
YO
YY
+
N
Y
Y
YOY
YY
Y
+
YY
Y
Y
YOY
N
Shape
Bond angle
linear
180°
linear
180°
N O
Y
Y
Y
NOF
YY
Y
F N
YY
YY
Y
YO
YY
F
F
S
F
F
F
(1)
The SF6 has an octahedral shape.
(0.5)
The molecule has six bond pairs of electrons in the outermost shell of the central
sulphur atom.
(0.5)
The shape that puts the electron pairs furthest apart is octahedral.
b)Not agree
trigonal planar
120°
electron from an
external source
Y
Y
ii)
F
–
YY
(1)
Oxygen is in the second period of the periodic table. It cannot form compounds
with more than 8 electrons in the outermost shell of its atom.
(1)
Remarks*
Remarks
V-shaped
➤Questions may ask students to draw the three-dimensional structures of the
above two fluorides.
120°
♦ Students should be able to put in order the H–N–H bond angles in the
–
+
following species: NH2 (g), NH3(g) and NH4 (g) and explain the order in
terms of VSEPR theory.
OF2
O
F
F
♦ Buckminsterfullerene (C60) and graphite can be distinguished by
SF6
– physical method — solubility in an organic solvent (e.g. toluene).
– spectroscopic method — mass spectrometry: C60 gives a peak at m/e
= 720 for the molecular ion.
F
F
F
S
F
F
F
➤Questions may ask why phosphorus can form pentachloride (PCl5) while
nitrogen cannot.
Example
a) Draw the respective electronic structures of OF2 and SF6. Hence deduce the shape of
each species.
i) OF2 (2 marks)
ii)SF6
(2 marks)
b)Explain whether you agree with the following statement.
‘Both oxygen and sulphur can form hexafluoride, i.e. OF6 and SF6.
(2 marks)
Answer
a) i) F O
F
(0.5)
The OF2 molecule is V-shaped.
The molecule has two lone pairs and two bond pairs of electrons in the outermost
shell of the oxygen atom.
(0.5)
(0.5)
The four pairs of electrons will adopt a tetrahedral arrangement.
The shape of a molecule is determined only by the arrangement of atoms. Thus,
the OF2 molecule is V-shaped.
(0.5)
10
Topic 6
Unit 24 Bond polarity and intermolecular forces
Microscopic World II
Unit 24
Bond polarity and intermolecular forces
24.1 – 24.4
Summary
24.1
Electron sharing in a covalent bond
24.2
Electronegativity
24.3
How polar bonds and shape affect the polarity of a molecule
1 A polar covalent bond is formed when electrons are unequally shared between the
two bonded atoms.
Polar covalent bonding occurs because one bonded atom has a stronger attraction
for electrons than the other bonded atom does.
24.4
Effect of a charged rod on polar and non-polar liquids
24.5
Intermolecular forces
2 The electronegativity of an element represents the power of an atom of that element
to attract a bonding pair of electrons towards itself in a molecule.
24.6
Factors affecting the strength of van der Waals’ forces
3 For a molecule to be polar, the following conditions must apply:
24.7
Hydrogen bonding
• the molecule must contain one or more polar bonds;
• the shape of the molecule must not be completely symmetrical.
24.8
The density of water and ice
24.9
Surface tension and viscosity of liquids
4 The following table shows some polar molecules.
24.10 Intermolecular attractions in alcohols
24.11 Intermolecular attractions and properties of liquids
Molecule
HCl
H2O
δ–
Cl δ–
δ+ H
Shape
dipole
moment
NH3
CHCl3
H
δ–
O
N
H
H
δ+
δ+
net
dipole
H
moment δ+
H
net
dipole δ–
H
δ+ moment
Cl
δ+
C
Cl
δ–
Cl
δ+
net
dipole
moment
δ–
5 The following table shows three symmetrical molecules with polar bonds. They are
non-polar molecules because the identical bond dipole moments cancel one another
out exactly. The molecules have no net dipole moment.
Molecule
CO2
BF3
CCl4
δ–
δ–
F
Shape
δ–
O
C
δ+
O
Cl
δ+
δ–
B
δ– F
δ+
F δ–
δ–
Cl
Cl
δ–
Cl
δ–
11
12
Topic 6
Unit 24 Bond polarity and intermolecular forces
Microscopic World II
Exam tips
♦ Students should be able to explain whether a molecule has a net
dipole moment in terms of bond dipole moments and shape of the
molecule.
e.g.
Explain why sulphur dioxide possesses a net dipole moment while carbon
dioxide does not.
Each S–O bond is polar.
13
Example
a) Define ‘dipole moment’ in the case of a diatomic molecule consisting of two different
atoms. (1 mark)
b)Explain why the dipole moment of the H–F bond is greater than that of the H–I
bond. (2 marks)
Answer
a) Dipole moment of a diatomic molecule is the product of the charge and the distance
between the charges.
(1)
A sulphur dioxide molecule is V-shaped.
The individual S–O bond dipole moments reinforce each other.
Hence the molecule has a net dipole moment.
b)The electronegativity of fluroine is much higher than that of iodine.
Each C=O bond is polar.
(1)
For HF and HI, the effect of electronegativity difference outweighs the effect of
difference in the distance between the charges.
(1)
A carbon dioxide molecule is linear in shape.
The individual C=O bond dipole moments cancel each other out
exactly.
Remarks*
Remarks
➤Students should be able to give a precise definition of the ‘dipole moment’
of a diatomic molecule.
Hence the molecule has no net dipole moment.
♦ Questions may ask students to predict the effect of a charged rod on
a stream of a polar / non-polar liquid from a burette.
e.g.
Explain why the water stream is deflected as shown below.
24.5 – 24.11
Summary
water
water
1 In general, the relative strength of the three types of van der Waals’ forces decreases
in the following order:
m
Because of its V-shape, individual bond dipole moments in a water
molecule reinforce each other.
The water molecule has a net dipole moment.
permanent dipole-permanent dipole attractions > permanent dipole-induced dipole
attractions > instantaneous dipole-induced dipole attractions
2 Van der Waals’ forces are stronger between molecules
a) with greater number of electrons;
b)with more spread-out shapes (i.e. greater molecular surface areas).
3 A hydrogen bond is the particularly strong attractive force between the hydrogen atom
attached to an electronegative atom and the lone pair on another electronegative
atom. Usually the electronegative atom is oxygen, nitrogen or fluorine.
4 Intermolecular forces give rise to a number of structural features and physical properties
of liquids.
The jet of water is deflected by the positively charged rod.
a) Liquids having strong intermolecular forces tend to have high surface tension.
Negative ends of the molecules are attracted towards the rod.
b)The viscosity of a liquid is a measure of a liquid’s resistance to flow.
The jet of water is also deflected by the negatively charged rod.
Positive ends of the molecules are attracted towards the rod.
The viscosity of a liquid depends on:
• the strength of attractive forces between molecules; and
• the tendency of molecules to become entangled with each other.
Topic 6
14
Unit 24 Bond polarity and intermolecular forces
Microscopic World II
Exam tips
b)Compound X forms intramolecular hydrogen bonds.
♦ Hydrogen bonds form between trichloromethane and ethyl ethanoate
molecules. So, energy is released when trichloromethane and ethyl
ethanoate are mixed.
δ–
δ+
H
Cl
Cl
C
H
C
H
C
C
C
H
O
O
(1)
Compound Y does not form intramolecular hydrogen bonds. It can form more
intermolecular hydrogen bonds per molecule.
(1)
Thus, the melting point of compound X is lower than that of compound Y.
Remarks*
Remarks
➤Questions may ask students to show by drawing the hydrogen bonds
formed between molecules, such as those between ammonia and water
molecules.
C
H
cis isomer
boiling point
C
(1)
hydrogen bond
CH3
♦ Questions often ask students to compare the boiling points of cis and
trans isomers of unfamiliar compounds.
H
O
CH3CH2O
C
O
C
Cl
H
O
δ–
hydrogen bond
Cl3C
Cl
δ+
trans isomer
60 °C
lone pair
H
δ–
48 °C
H
(Refer to the example in Unit 30 of Topic 8 Chemistry of Carbon
Compounds.)
δ+
N
δ+
δ+
H
δ–
O
hydrogen bond H
H
δ+
➤Questions often ask students to predict the volatility and melting points
of compounds that form intramolecular hydrogen bonding in addition to
intermolecular hydrogen bonding.
Example
e.g.
Account for each of the following:
a) NH3(g) is very soluble in water, but PH3(g) is almost insoluble.
(3 marks)
b)The melting point of compound X shown below is lower than that of compound
Y. H
H
C
C
HOOC
melting point
HOOC
COOH
Which of the following compounds is more volatile?
OH
OH
NO2
and
H
C
C
H
NO2
2-nitrophenol
COOH
compound X
compound Y
130 °C
302 °C
4-nitrophenol
Only 2-nitrophenol forms intramolecular hydrogen bonds.
(3 marks)
O
N
Answer
a) NH3 molecules can form hydrogen bonds with water molecules extensively.
(1)
The electronegativity of P is very close to that of H. PH3 molecule has a very small
dipole moment.
(1)
Its interaction with water molecules is much weaker than the hydrogen bonds
between water molecules.
(1)
15
O
O
H
key:
hydrogen bond
It forms less intermolecular hydrogen bonds than 4-nitrophenol does. The
weaker intermolecular attractions make 2-nitrophenol more volatile.