INORGANIC CHEMISTRY ASSIGNMENT

GROUP THREE CHEMICAL ENGINEERING

EVENING

NAME REG.NO
ASIO DAISY MERCY 24/U/CHE/
03696/PE
NTENDA SHANICE 24/U/CHE/
10469/PE
MUWANGUZI ISAAC 24/U/CHE/
07982/PE
ATAMBA BRIGHT 24/U/CHE/
03718/PE
KISEMBO IGNATIUS 24/U/CHE/
06268/PE
OMOLLO DENY 24/U/CHE/
NASANAIRI 11134/PE
SSEKAGGO JOHN 24/U/CHE/
11690/PE
ALEMA HILLARY 24/U/CHE/
17466/PE
OKETTAYOT RONALD 24/U/CHD/585/GV
MAKANGA HAZAHAL 24/U/CHE/
06946/PE
QUESTION
1(a) . Explain the different chemical
properties of group III elements.
Group III elements are categorized into two, 3A and 3B
Group 3A elements, the boron group.
I. Boron (B)
II. Aluminium (Al)
III. Gallium (Ga)
IV. Indium (In)
V. Thallium (Tl) exhibits unique chemical characteristics.
Their general electronic configuration is ns2np1

Electronic Configuration and Periodic Trends

- Atomic radii increase down the group (B < Al < Ga < In < Tl)
- Electronegativity decreases down the group (B > Al > Ga > In
> Tl)
- Ionization energy decreases down the group (B > Al > Ga >
In > Tl)

Valency and Oxidation States

- +3 valency common among group members
- +1 and +3 oxidation states exhibited by Gallium, Indium, and
Thallium
Element-Specific Properties
 Reaction with oxygen.
They all burn in oxygen to form the corresponding
oxide
 4X(s) + 3O 2(g) 2X2O3(s)
X=(B, Al, Ga, In, Tl)
 Reaction with chlorine gas
They all react to form chlorides
2X(s) +3 Cl2(g)  2XCl3(s)
 Reaction with acids
Conc. acids
 Boron does react with nitric acid and sulphuric acids
when concentrated
Reaction:
B + 6HNO3 → H3BO3 + 3NO2 + 3H2O
The reactions of Group III elements with acids can vary, but
here are some general reactions:

1. Boron (B):
 Boron does not react with non-oxidizing acids like
hydrochloric acid (HCl) under normal conditions.
However, it can react with strong oxidizing acids
like nitric acid (HNO₃)
2. Aluminium (Al):
 Aluminium reacts with hydrochloric acid to produce
aluminium chloride and hydrogen gas
 It also reacts with sulphuric acid (H₂SO₄) to form
aluminium sulphate and hydrogen gas
3. Gallium (Ga), Indium (In), and Thallium (Tl):
 These elements generally react with acids to form
the corresponding metal salts and hydrogen gas.
For example, gallium with hydrochloric acid

These reactions illustrate the typical behaviour of Group III

elements with acids, forming salts and releasing hydrogen gas

 Reaction with bases and alkalis

Group III elements react with alkalis and bases in various ways:
 1. Boron (B):
 Boron reacts with strong bases like sodium
hydroxide (NaOH) to form borates
 2B +6NaOH+ 3H2O2Na3BO3 + 3H2
2. Aluminium (Al):
 Aluminium reacts with sodium hydroxide to produce
sodium aluminate and hydrogen gas:
 2Al+ 2NaOH +6H2O2NaAl(OH)4 + 3H2
3. Gallium (Ga), Indium (In), and Thallium (Tl):
 These elements can also react with strong bases,
forming corresponding metal hydroxides and
hydrogen gas. For example, gallium with sodium
hydroxide:
 2Ga+2NaOH + 6H2O  2NaGa(OH)4 +3H2

These reactions typically involve the formation of metal

hydroxides or aluminates and the release of hydrogen gas.

 Reaction with water

The reactions of Group III elements with water vary depending
on the element:

1. Boron (B):
 Boron does not react with water under normal
conditions.
2. Aluminium (Al):
 Aluminium reacts with water in the presence of air
or when finely powdered, forming aluminium
hydroxide and hydrogen gas:
 2Al +6H2O 2Al(OH)3 + 3H2
3. Gallium (Ga):
 Gallium reacts slowly with water, forming gallium
hydroxide and hydrogen gas:
 2Ga +6H2O 2Ga(OH)3 +3H2
4. Indium (In):
 Indium reacts with water at elevated temperatures
to form indium hydroxide and hydrogen gas:
  2In + 6H2O2In(OH)3 + 3H2
5. Thallium (Tl):
 Thallium does not react significantly with water
under normal conditions
 Reaction with metals
Group III elements can form alloys with other metals, but they
do not typically react with metals in the same way they do with
nonmetals or acids. Here are some general interactions:

1. Boron (B):
 Boron can form borides with metals, which are
typically hard and have high melting points. For
example, with titanium:
 Ti+B TiB2
2. Aluminium (Al):
 Aluminium forms alloys with many metals, such as
copper, magnesium, and zinc, enhancing properties
like strength and corrosion resistance. An example
is the formation of aluminium-copper alloys
 Al +Cu AlCu
3. Gallium (Ga), Indium (In), and Thallium (Tl):
 These elements can form alloys with metals like tin
and lead. For instance, gallium can form an alloy
with tin: Ga+ SnGaSn

These interactions are primarily alloy formations rather than

chemical reactions, enhancing material properties for various
applications

 Reaction with hydrogen

Group III elements react with hydrogen to form hydrides, but
the reactivity varies among the elements:

1. Boron (B):
 Boron forms boranes, which are a group of
compounds with varying hydrogen content. The
simplest is diborane:
 2B +3H2  B2H6
 2. Aluminium (Al):
 Aluminium can form aluminium hydride, although it
is not formed directly from the elements. It is
typically synthesized using complex chemical
reactions:
 AlCl3 +LiAlH4 AlH3 + LiCl
3. Gallium (Ga), Indium (In), and Thallium (Tl):
 These elements do not readily form stable hydrides
under normal conditions. Their hydrides are
generally unstable and not well-characterized.

These reactions illustrate the varying ability of Group III

elements to form hydrides, with boron being the most reactive
in this context.

 Reaction with Nitrogen

Group III elements react with nitrogen to form nitrides, with
varying reactivity:

1. Boron (B):
 Boron reacts with nitrogen at high temperatures to
form boron nitride:
 2B +N2 2BN
2. Aluminium (Al):
 Aluminium reacts with nitrogen when heated to
form aluminium nitride:
 2Al + N2  2AlN
3. Gallium (Ga), Indium (In), and Thallium (Tl):
 These elements also form nitrides when reacted
with nitrogen at elevated temperatures:
 2Ga + N2  2GaN
 2In + N2  2InN
 2Tl+ N2 2TlN

These nitrides are important in various applications, such as

semiconductors and ceramics.
Reactivity Trends
- Reactivity increases down the group (B < Al < Ga < In < Tl)
- Thallium's +1 oxidation state is more stable than +3
Key Takeaways
- Atomic radius and electronegativity trends influence chemical
behaviour
- Valency and oxidation states determine reactivity

b. Give three ores of aluminium.

Bauxite, Al2O3.2H2O
Corundum, Al2O3
Cryolite, Na3AlF6

c. Explain how aluminium can be purified

using any ore of your choice.
Bauxite
Concentration of the ore
The bauxite is first roasted at a low temperature to convert all
iron oxide to a +3 oxidation state. The finely powdered product
is then heated at about 430K in autoclaves with sodium
hydroxide under pressure. Iron (III) oxide and titanium (IV)
oxide remain not dissolved, as does most of the silica, though
some may dissolve as sodium silicate.
SiO2(s)+2OH (aq) → SiO3-2(aq)+H2O (1)
Aluminium oxide reacts to form sodium aluminate (III) in
solution
Al2O3(s)+2OH (aq)+3H2O (l) → 2Al (OH)4(aq)
The undissolved impurities are filtered off. The dilute liquid is
then seeded with freshly precipitated aluminum hydroxide and
agitated. This induces hydrolysis of the dissolved
aluminates (III) and aluminum hydroxide precipitates out. Any
silicates present are unaffected and remain in solution.
Al (OH)4-(aq)  Al(OH)3(s)+ OH-(aq)
Alternatively, carbon dioxide may be bubbled through the
solution
2Al(OH)4(aq) + CO2(g)→2Al (OH)3(s)+CO3-2(aq)+H2O(l)-
The hydroxide is filtered off, washed, dried, and heated
strongly to leave pure aluminum oxide which is then
electrolyzed
2Al(OH)3(s)  Al2O3(s)+ 3H2O(g)
Electrolysis of aluminum oxide
The aluminum oxide is electrolyzed in a molten mixture of
cryolite, and fluorspar. The steel container is lined with sheets
of carbon acting as the cathode and the anode is also made up
of carbon rods. Aluminum collects at the cell's floor and is
tapped off at intervals. Oxygen is evolved at the anodes which
are slowly burnt away as carbon dioxide, so the anodes require
regular renewal.
Figure 1: Diagram for electrolysis of aluminum

Reaction equations
At the cathode
A+3(aq)+3e→Al(s)
At the anode
202(ag)→O2(g)+4e
Note
(i)A low voltage is used to avoid the decomposing of the
solvent cryolite
(ii)The disadvantage involved is that it uses a lot of energy
hence not economical, and the anode must be replaced from
time to time because it burns off in the oxygen produced
(iii) Cryolite is added to lower the melting point of aluminum
oxide
(iv). Fluorspar is calcium fluoride

d. Define diagonal relationships and explain

the relation between beryllium and
Aluminium on the periodic table
Diagonal relationships is the similarity in chemical properties
between elements in the second period of the periodic table
and those in the next higher group in the third period.
Diagonal relationships arise because the variations of
properties across the period and down the group are in
opposing tendencies which causes the elements along the
diagonal to have similar properties. For example, across the
period atomic radius of the elements decreases while down the
group it increases. This and other properties approximately
balance across the diagonal.
Reasons why elements are similar.
1. Form cations with similar polarizing power
2. Similar standard electrode potentials
3. Their atoms have similar values of electronegativity
4. Their atoms have similar values of electropositive (Boron
and Silicon only)
Diagonal relationship between beryllium and aluminum
1. Both metals are rendered passive by concentrated nitric
acid. This is because concentrated nitric acid is an oxidizing
agent, and it forms a very thin layer of oxide on the metal
surface which prevents it from further attack
2. Both beryllium and aluminum react with a hot concentrated
solution of alkalis to produce hydrogen gas and a soluble
complex
Be(s) + 2OH(aq)+ 2H2O(l)→Be(OH)4-2(ag)+H2(g)
2Al(s)+2OH(aq) + 6H2O(l)→2Al(OH)4(aq)+ 3H2(g)
3. Beryllium and aluminum oxides and hydroxides are
amphoteric. Dissolve in acids and hot concentrated alkalis.
BeO(s)+2OH(aq)+ H2O(l)→Be(OH)42-(aq)
Al2O3(s)+ 2OH(aq)+ 3H2O(l)2Al(OH)4 - (aq)

Be(OH)2(s) + 2OH- (aq)Be(OH)42-(ag)

Al(OH)3(s) + OH-(aq)-→Al(OH)4 -(aq)
With acids
BeO(s) + 2H+(aq)  Be2+(aq)+ H2O(l)
Al2O3(s) + 6H+(aq)  2Al3+(aq)+3H2O(l)

Al(OH)3(s)+6H+(aq) →2Al3+(aq)+6H2O(l)
Be(OH)2(s) +2H+(aq)  Be2+(aq)+2H2O(l)
4. Their carbides are hydrolyzed to form methane
Be2C(s) +4H2O(l) →CH4(g)+2Be(OH)2(s)
Al4C3(s) + 12H2O(l) →3CH4(g)+4Al(OH)3(s)
5. Both beryllium and aluminum form covalent: chlorides.
Their chlorides are hydrolyzed by water to form acidic
solutions. They have lower melting and boiling points and
dimerize when heated.
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