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Unit 1 Mod 3 Group IV
page 1 of 8
a) explain the variation in physical properties of the elements in terms of
structure and bonding ( refer to variations in metallic character and
electrical conductivity)
b) describe the bonding of the tetrachlorides
c) explain the reactions of the tetrachlorides with water
d) discuss the trends in:- (i) bonding (ii) acid/base character (iii) thermal
stability of the oxides of oxidation states II and IV (make reference to Eө
values of the elements)
e) discuss the uses of ceramics based on silicon(IV) oxide
GROUP IV ELEMENTS
Group IV elements:C
Si
Ge
Sn
Pb
Variation in melting points and electrical conductivities of the elements
Property
m.p (K)
C
4003
Si
1683
Ge
1210
Sn
505
Pb
600
The structure of the elements range from macro-molecular non-metals
through metalloids to metallic lattices of close-packed ions.
Carbon exists either as a diamond formation or as graphite, in either
form, there are strong covalent bonds between the carbon atoms which
must be broken, resulting in a extremely high melting point. Only in
graphite is electrical conductivity observed but as expected it is very low.
Both silicon and germanium exist in distorted diamond-like lattices, with
germanium’s structure more distorted than silicon’s. The larger the
atom, the greater the deviation of its structure from that of diamond.
Their melting points are high, however not as high as carbon, but they
decrease because of increasing size of the atom, which results in weaker
degree of orbital overlap (i.e. weaker bond). Both of them are semiconductors.
Unit 1 Mod 3 Group IV
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Tin has 2 allotropes, one stable below 286.2K and another stable above
286.2K, the allotrope stable below 286.2K exists as a structure similar to
that of germanium while the allotrope above 286.2K has a metallic
structure. It is this transition between states which results in the lower
than expected melting point. In its metallic form, it conducts electricity
as do all metals, while in its germanium form, it may conduct only in
certain conditions like a semi-conductor.
Lead exists as a metallic lattice and thus would have a moderately high
melting point, and it would conduct electric current.
Checkpoint A
1. Using this circle
as the size of the silicon atom, draw circles
to represent the atomic radii of the elements carbon, germanium, tin and
lead.
C
Ge
Sn
Pb
2. Using the graph below, sketch the electrical conductivities of the
group IV elements
3. Does metallic character of the group IV elements INCREASE or
DECREASE down the group? …………………
Unit 1 Mod 3 Group IV
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4. Why does graphite conduct electricity but diamond does not, even
though they are both made from carbon atoms?
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The tetrachlorides
The tetrachlorides of group IV elements are covalent, tetrahedral
molecules. These exist as liquids at rtp. The molecules exist as discrete
units with weak van der Waals’ forces between them and this results in
the liquids being quite volatile at room temperature. Within the molecule
are strong covalent bonds between the atoms involved.
Note that the larger the molecules, the LESS volatile they are
Shape of a tetrachloride of a group IV element (tetrahedral)
Group IV tetrachlorides are hydrolysed by water (except CCl4) to form
the hydroxide (or hydrated oxide) and hydrochloride gas.
MCl4(l) + 2H2O(l)  M(OH)4(s) or MO2.xH2O(s) + HCl(g)
In the case of silicon and all other group IV elements except carbon,
hydrolysis of the tetrachloride is accomplished by attack of the water
molecule to the electron-deficient metal atom. The group IV metal atom
can use a vacant low lying 3d orbital to accommodate the lone pair from
the oxygen atom, forming a short- lived 5 co-ordinate intermediate
which dissociates to form the hydrogen chloride and the metal trichloride
hydroxide, MCl3OH. Repetition of these steps gives the hydrated metal
oxide or the metal hydroxide.
Unit 1 Mod 3 Group IV
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Carbon cannot accommodate the water molecule as it does not have
any empty low-lying 3d orbitals, remember carbon has electronic
configuration 1s2 2s2 2p2 therefore the 3d orbitals are TOO HIGH in
energy to be utilized. Therefore the lone pair of electrons from the
water molecule cannot be accepted and thus no reaction i.e. no
hydrolysis of CCl4 can occur.
Checkpoint B
1. Would the rate of hydrolysis of the tetrachlorides increase or
decrease down the group? …………………
2. Suggest a reason for your answer in question 1 above
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Acid-base nature of the oxides in oxidation state +2 and +4
CO is neutral, CO2 is acidic
SiO & GeO do not exist
SiO2 is weakly acidic
GeO2, SnO, SnO2, PbO and PbO2 are amphoteric
Unit 1 Mod 3 Group IV
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Reactions of the oxides in the +2 and +4 oxidation states to show
their acidic or basic nature
Since CO is neutral, it reacts with neither acids nor bases.
CO2 (g) + H2O (l)  H2CO3 (aq)
CO2 forms an acid when reacted with water
SiO2 (s) + OH- (aq)  SiO32- (aq) + H2O(l) It react with alkali to form
a salt and water
General form of group IV monoxide MO
General form of group IV dioxide MO2
Reaction of MO with HCl is:
Note: Reaction of PbO with HCl forms solid PbCl2 and not (aq)
Reaction of MO with NaOH is:
Reaction of MO2 with HCl is:
Reaction of MO2 with NaOH is:
Unit 1 Mod 3 Group IV
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Thermal stability of oxides in +2 and +4 oxidation states
Note that +4 oxidation state is more thermally stable than +2 for
ALL the group IV elements EXCEPT lead. At lead, the situation
reverses and the +2 is MORE thermally stable than the +4 oxidation
state.
This stability occurs for the aqueous ions as well.
Note: There is no +2 oxidation state for Si and Ge
Checkpoint C
1. CO can be easily oxidised to CO2. True or False?
2. SnO2 is easily reduced to SnO. True or False?
3. PbO2 is easily reduced to PbO. True or False?
4. Would SnO or SnO2 be a powerful reducing agent? Give a reason for
your answer.
5. Would PbO or PbO2 be a powerful reducing agent? Give a reason for
your answer.
Unit 1 Mod 3 Group IV
Inert pair effect
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The reason for the reversal in difference in stabilities of the +II and +IV
oxidation state for lead is the inert pair effect.
On descending group IV, the atoms get larger. The shielding of the
nuclear charge from the valence electrons by the core electrons gets
more and more ineffective. At lead, the shielding is so poor that the 2
electrons of the s orbital in the valence shell feel the nuclear charge
more so than the p electrons of the valence shell and are pulled towards
the core electrons.
In essence the two electrons are now part of the core electrons and
do not readily take part in bonding and thus are “inert”. Hence lead
acts as if it only has 2 valence electrons instead of 4. Thus the term
“inert pair effect”. This makes lead to be more stable utilizing two
electrons i.e. oxidation state +II as opposed to utilizing 4 electrons i.e.
oxidation state +IV.
Ceramics
These are materials which are fashioned at room temperature from clays
or other natural earths and then permanently hardened by heat.
Feldspar + carbon dioxide + water  clay + silica + potassium carbonate
When silica is reacted with other oxides in the process of glazing (the
final stage in making ceramics), the important silicate ion (SiO32-) is
formed.
Properties of ceramics
1. Hard and brittle (unidirectional bonds)
2. Non-corrosive and insulating to heat and electricity (no mobile
electrons)
3. Resistant to high and low voltages and high melting point (because
of the high strength of the Si—O)
Unit 1 Mod 3 Group IV
Practice Questions
1.
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