Group IV Elements
42.1
42.2
42.3
42.4
1
Introduction
Characteristic Properties of the Group IV
Elements
Composition of Chlorides and Oxides of
the Group IV Elements
Silicon and Silicates
Introduction
The Group IV elements
2

carbon

silicon

germanium

tin

lead

exhibit a marked change (dissimilarity)
among the elements in the same group
Introduction
Carbon
 dull black in the form of graphite
3
Introduction
Carbon
 hard and transparent in the form of
diamond
4
Introduction
Silicon and germanium
 dull grey or black
Si
5
Ge
Introduction
Tin and lead  shiny grey
Sn
6
Pb
Introduction
The Group IV elements
 outermost shell electronic configuration
of ns2np2
Element
7
Electronic configuration
Carbon
[He] 2s22p2
2,4
Silicon
[Ne] 3s23p2
2,8,4
Germanium
[Ar] 3d104s24p2
2,8,18,4
Tin
[Kr] 4d 105s25p2
2,8,18,18,4
[Xe] 4f 145d 106s26p2
2,8,18,32,18,4
Lead
Structure and Bonding
Moving down the group
carbon  non-metal
silicon
germanium
tin
lead
8
 metalloids
 metals
Structure and Bonding
Most common structure : giant covalent structure
Examples
 carbon
 silicon
 germanium
 grey tin (an allotrope of tin)
9
Allotropes are different structures of the
same element
1. Carbon
10

two important allotropic forms

diamond and graphite
Structures of diamond
11
each carbon atom is bonded to four other C
atoms
12
extremely hard and chemically inert
13
All electrons are localized  non-conductor
14
Structure of graphite
15
Graphite

layered structure
Covalent bonds
van der Waals’
forces
16
The layers slide over each other easily

17
brittle and soft
Electrons between layers are delocalized
 conducts electricity along the layers
18
2. Silicon and Germanium
19

network lattice

the atoms are covalently bonded to
one another
3. Tin and Lead
Tin

two allotropes

white tin and grey tin
White tin
20
cold
heat
Grey tin
3. Tin and Lead
White tin
21

stable form

metallic lattice structure

atoms are held together by metallic
bonding

conducts electricity

shows the properties of a typical metal
3. Tin and Lead
Grey tin

network lattice structure

similar to that of diamond
White tin
more dense
cold
heat
Grey tin
less dense
White tin expands and crumbles on cooling
Napoleon’s retreat from Russia
22
3. Tin and Lead
Lead
23

typical metallic lattice

atoms are held together by metallic
bonding
Some physical properties of the Group IV elements
C
Si
Ge
Electronegativity value
2.5
1.74
2.0
Electronic configuration
1s22s22p2
[Ne]
3s23p2
[Ar]
3d104s24p2
0.077
0.117
0.122
347
226
188
Melting point (C)
3527
1414
1211
Boiling point (C)
4027
3265
2833
716
456
376
Atomic radius (nm)
Bond enthalpy (kJ mol–1)
Enthalpy change of
atomization (kJ mol–1)
24
Some physical properties of the Group IV elements
Tin (Sn)
Lead (Pb)
Electronegativity value
1.7
1.55
Electronic configuration
[Kr]4d10
5s25p2
[Xe] 4f145d10
6s26p2
0.140
0.154
Bond enthalpy (kJ mol–1)
150
–
Melting point (C)
232
327
Boiling point (C)
2602
1749
302
195
Atomic radius (nm)
Enthalpy change of
atomization (kJ mol–1)
25
Variation in Melting Point
 on going down the group
26
C
3527C
Si
1414C
Ge
1211C
Sn
232C
Pb
327C
Variation in Melting Point
The very high m.p. of diamond
is due to the strong C – C bonds
& the giant structure
C
3527C
Si
1414C
Going from C to Ge
Ge
1211C

bond length 
Sn
232C

bond strength 
Pb
327C

melting point 
27
Variation in Melting Point
28
Sn and Pb have exceptionally
low m.p. because
C
3527C
Si
1414C
1. metallic structures
Ge
1211C
 extent of bond breaking
on melting is small
Sn
232C
Pb
327C
Variation in Melting Point
Sn and Pb have exceptionally
low m.p. because
2. only two (ns2) of the four
valence electrons are
involved in the sea of
electrons
Tin (Sn)
Lead (Pb)
[Kr]4d10 5s25p2 [Xe] 4f145d10 6s26p2
29
C
3527C
Si
1414C
Ge
1211C
Sn
232C
Pb
327C
Variation in Boiling Point
The general trend and explanation
 similar to those for m.p.
C
4027C
Si
3265C
Ge 2833C
30
Sn
2602C
Pb
1749C
Chlorides
Two series of chlorides formed by the
Group IV elements
31

the dichlorides (MCl2)

the tetrachlorides (MCl4)
Chlorides
All Group IV elements

form tetrachlorides

liquids at room temperature
and pressure
GeCl4
all are simple covalent molecules
with a tetrahedral shape
PbCl4

32
CCl4
SiCl4
SnCl4
-
+
-
-
M – Cl bonds are polar with ionic character
Molecules as a whole are non-polar
33
Reactions with water
CCl4 + H2O  no reaction
SiCl4 + H2O  Si(OH)Cl3 + HCl
34
H
Cl
+ Si
Cl
Cl
O
Cl
Cl
Cl
H
Si
Cl
H
O
Cl
H
Si in SiCl4 is more positively charged than C in CCl4
 More susceptible to nucleophilic attack
Si, unlike C, can expand its octet to accept an
additional electron pair
35
H
Cl
+ Si
Cl
Cl
O
Cl
Cl
Cl
Si
Cl
H
O
H
Cl
H
Cl
Si
Cl
Cl
OH
H4SiO4, silicic acid
36
+ HCl
Reactions with water
CCl4 + H2O  no reaction
SiCl4 + H2O  Si(OH)Cl3 + HCl
Si(OH)Cl3 + H2O  Si(OH)2Cl2 + HCl
Si(OH)2Cl2 + H2O  Si(OH)3Cl + HCl
Si(OH)3Cl + H2O  Si(OH)4 + HCl
H4SiO4, silicic acid
37
Chlorides
 tendency to form dichlorides, MCl2
down the group

38
all possess covalent character
though they exist as crystalline
solids at room temperature and
pressure
GeCl2
SnCl2
PbCl2
Chlorides
On moving down the group,
Metallic character of elements 
Ionic character of MCl2 
mainly covalent
mainly ionic
39
GeCl2
SnCl2
PbCl2
Chlorides
On moving down the group,
 the relative stability of +4 oxidation state 
 the relative stability of +2 oxidation state 
40
Tin (Sn)
Lead (Pb)
[Kr]4d10 5s25p2
[Xe] 4f145d10 6s26p2
The outermost ns2 electrons are less
shielded by the more diffused inner d and/or
f electrons.
 They are attracted more by the positive
nucleus
 Less available for forming bonds
 Form only two bonds using np2
41
Oxides
Two series of oxides formed by the Group IV
elements
42

the monoxides (MO)

the dioxides (MO2)
Oxides
All Group IV elements

form the dioxides
Carbon dioxide
43

the only Group IV dioxide which
consists of simple molecules

exists as a gas at room temperature
and pressure
Oxides
The dioxides of other Group IV elements
44

crystalline solids of high melting points

either giant covalent or giant ionic
structures
Oxides
All Group IV elements (except silicon)
 form the monoxides at normal
conditions
 Stability of MO  down the
group
CO
GeO
SnO
PbO
45
Oxides
CO2
SiO2
GeO2
SnO2
PbO2
46
Decreasing
stability of
dioxide
Tin (Sn)
Lead (Pb)
[Kr]4d10 5s25p2
[Xe] 4f145d10 6s26p2
The outermost ns2 electrons are less
shielded by the more diffused inner d and/or
f electrons.
 They are attracted more by the positive
nucleus
 Less available for forming bonds
 Form only two bonds using np2
47
The bond type and the relative stabilitiy of the monoxides
and dioxides formed by the Group IV elements
Group IV
element
Carbon
Silicon
Oxides
formed
Bond type of the
oxide
Relative
stability
CO
Covalent
Unstable
(reducing)
CO2
Covalent
Stable
(SiO)
–
Very unstable
SiO2
Covalent
Stable
GeO
Predominantly ionic
Unstable in the
presence of O2
GeO2
Partly ionic,
partly covalent
Stable
Germanium
48
The bond type and the relative stabilitiy of the monoxides
and dioxides formed by the Group IV elements
Group IV
element
Oxides
formed
SnO
Tin
SnO2
PbO
Lead
49
PbO2
Bond type of the oxide
Relative stability
Predominantly ionic
Unstable
(reducing)
Partly ionic,
partly covalent
Unstable
(oxidizing)
Ionic
Stable
Predominantly ionic
Unstable
(oxidizing)
Silicon and Silicates
Silicon
50

the second most abundant element in
the Earth’s crust

about 28% by mass

commonly found as silicon oxide
(also known as silica)
oxygen
46.6%
silicon
27.7%
aluminium
iron 8.1%
5.0%
all other elements
1.5%
magnesium 2.1%
51
calcium 3.6%
sodium 2.8%
potassium 2.6%
Percentage abundance by weight of major
elements in the Earth’s crust.
Silicon
Example:
52

in a variety of forms such as sand,
quartz and flint

also found as silicates in rocks and clay
Preparation of Silicon
1.
by reduction of silica with carbon in
an electric furnace
SiO2(s) + 2C(s)  Si(s) + 2CO(g)
2.
Extremely pure silicon can be obtained
by the reaction of silicon(IV) chloride
with hydrogen
SiCl4(s) + 2H2(g)  Si(s) + 4HCl(l)
followed by zone refining of the
resultant silicon
53
Applications of Silicon
Silicon is the basic material
54

for making semi-conductors used in the
construction of transistors and rectifiers

for making steel and aluminium alloys
Chemistry of Silicon

dominated by its strong tendency to
form Si–O single bond

reflected by its formation of silica
and a variety of silicates
Silicates consist of Si, O and
one or more metals
55
Structures and Bonding of Silicates
1. SiO44– as the Basic Chemical Unit of Silicates
Covalent bond
one electron (symbol ‘O’)
of each oxygen atom is
gained from another atom
(usually metal atom)
56
4-
O
109.5º
Si
O
O
O
SiO44
57
For simplicity, a SiO44– tetrahedron can
be represented by a pyramid as follows:
or
Si > O not drawn to scale
58
SiO44- tetrahedron
apex towards us
Types of silicates
•
•
•
•
•
Nesosilicates (lone tetrahedron) - [SiO4]4−, eg olivine.
Sorosilicates (double tetrahedra) - [Si2O7]6−, eg epidote.
Cyclosilicates (rings) - [SinO3n]2n−, eg tourmaline group.
Inosilicates (single chain) - [SinO3n]2n−, eg pyroxene group.
Inosilicates(double chain) - [Si4nO11n]6n−, eg amphibole
group. (not required in A-Level)
• Phyllosilicates (sheets) - [Si2nO5n]2n−, eg micas and clays.
• Tectosilicates (3D framework) - [AlxSiyO2(x+y)]x−,
eg quartz, feldspars.
http://www.windows.ucar.edu/tour/link=/earth/geology/silicates2.html
59
Different structures of silicates
1. Isolated silicates
Contain isolated SiO44– tetrahedra.
or
60
1. Isolated silicates
are not polymerized.
are bonded to the metal ions (e.g. Mg2+ or
Fe2+) by ionic bonds.
tend to have high densities and are not
easy to cleave.
61
Olivine ((Mg,Fe)2SiO4) is an example of
isolated silicates.
It is the most abundant mineral in the
Earth’s mantle.
62
Deducing the chemical formula of an isolated
silicate, (Fe,Mg)2SiO4
The diagram below shows an incomplete part of an
isolated silicate. This part, when complete, can
repeat infinitely to give a three-dimensional
structure of the silicate.
silicate tetrahedron
magnesium ion
63
(a) Add an appropriate number of
(representing
iron(II) ion) in the figure above to make up a
complete part of the silicate.
Charge on each Mg2+ ion is +2
Charge on each SiO4 4– ion is –4,
the net charge of the
incomplete part
= (+2)(8) + (–4)(8) = –16
To balance the charge, the number of Fe2+ ions
needed = 8
64
For regular packing of particles,
one of the feasible way of
putting Fe2+ ions is shown below:
(b) Hence, deduce the chemical
formula of the silicate.
From the picture, the formula
of the part is Fe8Mg8(SiO4)8
 FeMgSiO4.
65
M1 = Mg2+; M2 = Fe2+
No. of Fe2+ = 4
Fe4Mg4(SiO4)4
 FeMgSiO4.
No. of Mg2+ = 4¼ + 4½ + 1 = 4
66
1. SiO44– as the Basic Chemical Unit of Silicates
•
Zircon (ZrSiO4)
 an example of isolated silicate mineral
 the principal ore of zirconium metal
67
1. SiO44– as the Basic Chemical Unit of Silicates
•
Zircon (ZrSiO4)
 brilliant appearance
 high refractive index
 used as a diamondlike gem
68
Double tetrahedra – Sorosilicates
The Si2O76– anion is formed by joining two SiO44–
tetrahedra together through a common oxygen atom
The negative charges are present only on the oxygen
atoms that are not shared by the two silicon atoms
69
Double tetrahedra – Sorosilicates
70
Double tetrahedra – Sorosilicates
71
2. Structures of Silicates
•
The SiO44– tetrahedra can be joined up
 by sharing oxygen atoms
 form ring, chain, sheet or network
silicates
72
Cyclosilicates (rings) - [SiO3]n2n−
O shared by two Si
[SiO3]626
[Si6O18]12
Carry one –ve charge
73
Cyclosilicates (rings) - [SiO3]n2n−
[SiO3]626
[Si6O18]12
74
Single chain silicates
Each SiO44– tetrahedron possesses 1 Si
atom and 3 (i.e. 1 + 1 + 0.5 + 0.5) O atoms.
The ratio of Si to O is 1 : 3.
75
Stoichiometry : (SiO3)n2n–
Single chain silicates
76
(a) Draw a diagram to show that
when two oxygen atoms of each
SiO44– tetrahedron are used for
sharing, an infinite polymer
chain can be formed.
Your diagram should have four
SiO4 4– tetrahedra.
77
O
Si
O
O
O
SiO44– tetrahedron
(b) Hence, show that the general formula of the
silicate is (SiO3)n2n–.
Key:
oxygen
silicon
Number of Si atoms = 4
Number of O atoms
= 11 (inside the chain)  0.5 × 2 (ends of chain) = 12
Charge on oxide = 8 (bridging O has no charge)
 the formula is Si4O128
 the general formula is (SiO3)48 or (SiO3)n2n
78
Single chain silicates - pyroxenes
Generally dark-coloured
Commonly found in igneous rocks
79
Pyroxene minerals
2+
(SiO3)24
2+
• Schefferite, Ca(Mg,Fe,Mn)Si2O6
2+
2+
• Zinc schefferite, Ca(Mg,Mn,Zn)Si2O6
2+
2+
• Jeffersonite, Ca(Mg,Fe,Mn,Zn)Si2O6
3+
• Leucaugite, Ca(Mg,Fe,Al)(Al,Si)2O6
2+3+
• Calcium-Tschermak's molecule, CaAlAlSiO6
Half of Si atoms replaced by Al
80
(AlO3)3(SiO3)2
Double chain silicates
Every other tetrahedron in a single chain
shares a third oxygen atom with a
neighbouring chain.
81
Double chain silicates
In every two SiO44– tetrahedra,
No. of Si = 2 Si : O = 2 : 5.5 = 4 : 11
Stoichiometry : Si4O116–
No. of O
= (1 + 1 + 0.5 + 0.5) + (1 + 0.5 + 0.5 + 0.5) = 5.5
82
1st
tetrahedron
2nd
tetrahedron
+27
-62
-12
Amphibole (Ca2Mg5(Si4O11)2(OH)2) is an
example of double chain silicates.
83
It is more difficult to cleave across the chains
For both single chain silicates and double
chain silicates, the Si–O covalent bonds
within the chain silicate crystals are strong.
84
It is easier to cleave along the chain direction
The ionic bonds between the metal ions and
polymeric silicate chains are comparatively
weaker than the Si–O covalent bonds.
85
Also, cleave between the front and the back
chains
Two planes of cleavage are perpendicular to
each other
86
Chain silicate anions are found in fibrous asbestos
87
Single chain silicates
Glass : CaSiO3
long chain anions
 can be drawn into long wire or glass wool.
Limestone : CaCO3
 Individual anions  hard and brittle
88
Carbonate ions do not exist as long chains.
Why?
Carbon can form strong  bond with oxygen.
Moreover, carbonate ions can be stabilized by
delocalization of  electrons.
-
89
O
O
O-
O-
C
C
C
O-
O
O-
-
O
O
Also, polymerization of CO32 ions gives
polymeric (CO2)2 which is electrically neutral
O
O
O
C
C
C
O
O
O
Stability : CO2 > (CO2)n
O
90
C
O
O
Sheet silicates
Each SiO44– tetrahedron shares three oxygen
atoms with neighbouring SiO44– tetrahedra
91
Each tetrahedron
has 1 Si atom and 2.5 O atoms (1+0.5+0.5+0.5)
92 carries one negative charge
(SiO2.5)nn or (Si2O5)n2n or (Si4O10)n4n
93
94
Structure of sheet silicates
4
1
Talc (Mg3(Si4O10)(OH)2) is an example
of sheet silicates.
It is an important filler material for
paints, rubber and insecticides.
Talc(滑石)
95
van der Waals’
forces
a sandwichlike layer
with Mg2+
and OH
trapped in
the middle
Only weak van der Waals’
forces exist between the
sheets of the SiO44– anions
 only one plane of cleavage
96
Si
O OH Mg2+
Since talc is a relatively
soft silicate, it can be
pulverized to make talcum
powder. It is a soft and
fine powder used after
bath to make our body feel
smooth and dry.
It is slippery because the
layers can slide over one
another
97
Sheet silicates
Mica(雲母)
Clay(黏土)
Mica and clay readily cleave into thin slices
(only one plane of cleavage)
98
Sheet silicates
Mica(雲母)
Clay(黏土)
Mica is used as window glass and waveguide
cover in microwave oven
99
4. Network silicates
Consist of SiO44– tetrahedra in which ALL
oxygen atoms are shared with adjacent
SiO44– tetrahedra
 Electrically neutral
Each tetrahedron has one Si atom and
2 O atoms (40.5)
 Si : O = 1 : 2
 SiO2 (Quartz)
100
4. Network silicates
each Si atom is
attached to 4 O atoms
each O atom
is attached to
2 Si atoms
Si
SiO2
O
 No plane of cleavage
 much harder than other types of silicates
101  very high melting points
4. Network silicates
Examples:-
102
Quartz
Feldspar(長石)
4. Network silicates
Feldspar
- the most abundant group of minerals
in the Earth’s crust
- every oxygen atom is shared between
SiO44– tetrahedra
- some Si atoms are replaced by Al atoms
Formula of anions :
[(AlO2)x(SiO2)y]x− or [AlxSiyO2(x+y)]x
103
4. Network silicates
Feldspar
Formula of anions :
[(AlO2)x(SiO2)y]x− or [AlxSiyO2(x+y)]x
when x=0  SiO2
104
Types of Feldspar
K-feldspar endmember
KAlSi3O8  K+[(AlO2)(SiO2)3]
Albite endmember
NaAlSi3O8  Na+[(AlO2)(SiO2)3]
Anorthite endmember
CaAl2Si2O8  Ca2+[(AlO2)2(SiO2)2]2
105
Comparison of different types
of silicates
Structure
Number of O atoms
Formula of
shared by each
Si : O
anion
tetrahedron
Isolated silicate
0
SiO44–
1:4
Single chain silicate
2
SiO32– or
1:3
Si2O64–
Double chain
silicate
106
2.5
Si4O116–
4:11
Number of O atoms
Structure
Sheet silicate
shared by each
tetrahedron
3
Formula of
anion
Si2O52- or
Si : O
2:5
Si4O104-
Network silicate
4
SiO2 or
[(AlO2)x(SiO2)y]x
107
1:2
Structure
Isolated silicate
Properties
• high density
• not easy to cleave in a
particular direction
Single chain silicate
• not easy to cleave the
crystal across the chains
Example
Olivine,
(Mg,Fe)2SiO4
Pyroxene,
CaMg(Si2O6)
• Two planes of cleavage
Double chain silicate
• not easy to cleave the
crystal across the chains
• Two planes of cleavage
108
Amphibole,
Ca2Mg5(Si4O11)2(OH)2
Structure
Sheet silicate
Properties
• soft
• easy to be pulverized to make
powder
• cleavage usually occurs
between sheets (only one
plane of cleavage)
Network silicate
Example
•very hard
•high melting point
Talc,
Mg3(Si4O10)(OH)2
Quartz,
SiO2
Feldspar,
Ca(Al2Si2O8)
109
Predict the type of each silicate from its
chemical formula.
(a) CaMg(SiO3)2
single chain
(b) Fe3Al2Si3O12
isolated
(c) Si12Mg8O30(OH)4
sheet
(d) Na2Mg3Al2Si8O22(OH)2 double chain
(e) Mg(Al2Si2O8)
network (feldspar)
(f) MgAl(AlSiO6)
single chain
110