Why is aluminium used for transmission cables?

Per kilogram, aluminium conducts about twice as well as copper because it is much less dense. Lighter cables need cheaper towers. Aluminium is also cheaper and corrosion-resistant.

Detailed Discussions On Aluminium

Q: What are the main steps in the extraction of aluminium?

Aluminium is extracted from bauxite in two stages. Stage I Bayer's process purifies bauxite using hot concentrated NaOH and gives pure alumina by seeding and calcination. Stage II Hall-Heroult process electrolyses molten cryolite containing dissolved alumina to yield aluminium at the cathode.

Q: Why is molten cryolite added to alumina in the Hall-Heroult process?

Pure alumina melts at 2345 K which is too high for practical electrolysis. Molten cryolite dissolves alumina and gives a highly conducting melt at about 1225 K. Small amounts of CaF2 lower the temperature further and AlF3 improves conductivity.

Q: Why does aluminium show amphoteric behaviour?

Aluminium sits at the borderline between metal and non-metal. Its small highly charged Al3+ ion has strong polarising power. Al2O3 and Al(OH)3 react both with acids (as bases) and with alkalis (as acids), giving the metal amphoteric character.

Q: Why does aluminium not react with concentrated nitric acid?

Concentrated HNO3 is a strong oxidising acid. It very quickly thickens the natural surface oxide layer on aluminium into an impervious coating that prevents further attack. This is called passivation.

Q: Why is AlCl3 a dimer while BCl3 is a monomer?

In BCl3 the empty 2p orbital of B is stabilised by back-donation from Cl and BCl3 remains monomeric. In AlCl3 no such back-bonding is possible, so Al is electron-deficient and two units dimerise into Al2Cl6 with two bridging Cl atoms.

Q: What is an alum and what are alums used for?

An alum is a double sulphate of a monovalent and a trivalent metal with 24 water molecules. Potash alum K2SO4 Al2(SO4)3 24H2O is common. Uses include water purification, paper sizing, mordanting in dyeing and as a styptic.

Q: Which impurities give ruby and sapphire their colours?

Both ruby and sapphire are corundum alpha-Al2O3 with trace impurity ions. Ruby gets its red colour from Cr3+. Sapphire's blue arises from a Fe2+ to Ti4+ charge transfer that absorbs red-orange light.

ChemistryThe p-Block Elements: Group 13 & 14For NEET aspirants

Aluminium is the most abundant metal in the earth's crust ( $\sim 8.3%$ by weight) and the third most abundant element overall, after oxygen and silicon. Its silvery-white metallic look, low density ( $2.70$ g cm $^{- 3}$ ), excellent electrical conductivity and remarkable resistance to corrosion - thanks to a self-repairing oxide film - make it central to modern industry. This concept covers aluminium's occurrence, its two-stage extraction from bauxite (Bayer's process followed by Hall-Heroult electrolysis), its physical and chemical properties including the important amphoteric behaviour, and its key compounds such as alumina, aluminium chloride and alums.

Key Formulas & Facts - Quick Reference

Chief ore: bauxite, $A l_{2} O_{3} \cdot 2 H_{2} O$ ; abundance: $\sim 8.3%$ ; density: $2.70$ g cm $^{- 3}$ ; m.p.: $933$ K
Bayer's process: $A l_{2} O_{3} (s) + 2 N a O H (a q) + 3 H_{2} O (ℓ) \to 2 N a [A l (O H)_{4}] (a q)$ ; then $N a [A l (O H)_{4}] \to A l (O H)_{3} ↓ + N a O H$
Hall-Heroult (in molten cryolite $N a_{3} A l F_{6}$ ): cathode $A l^{3 +} + 3 e^{-} \to A l$ ; anode $C + 2 O^{2 -} \to C O_{2} + 4 e^{-}$
Amphoteric nature: $A l (O H)_{3} + 3 H C l \to A l C l_{3} + 3 H_{2} O$ and $A l (O H)_{3} + N a O H \to N a [A l (O H)_{4}]$
Thermite reaction: $2 A l + F e_{2} O_{3} \to A l_{2} O_{3} + 2 F e$ ( $Δ H = - 852$ kJ mol $^{- 1}$ )
$A l C l_{3}$ dimer: $A l_{2} C l_{6}$ , with two bridging Cl atoms
Potash alum: $K_{2} S O_{4} \cdot A l_{2} (S O_{4})_{3} \cdot 24 H_{2} O$ or $K A l (S O_{4})_{2} \cdot 12 H_{2} O$

1. Occurrence and Ores

Aluminium is far too reactive to occur free in nature. It is always found combined with oxygen or silicon. The commercially important ores are:

Bauxite: $A l_{2} O_{3} \cdot 2 H_{2} O$ - the principal ore worldwide.
Cryolite: $N a_{3} A l F_{6}$ - used both as an ore and as the electrolyte solvent in Hall-Heroult.
Corundum: $A l_{2} O_{3}$ (anhydrous crystalline) - gemstones (ruby with $C r^{3 +}$ , sapphire with $F e^{2 +} / T i^{4 +}$ ).
Feldspar: $K A l S i_{3} O_{8}$
Mica: $K_{2} O \cdot 3 A l_{2} O_{3} \cdot 6 S i O_{2} \cdot 2 H_{2} O$
Kaolinite (china clay): $A l_{2} (O H)_{4} S i_{2} O_{5}$

2. Extraction: Stage I - Purification of Bauxite (Bayer's Process)

Bauxite typically contains iron oxide ( $F e_{2} O_{3}$ ), silica ( $S i O_{2}$ ) and titanium dioxide ( $T i O_{2}$ ) as impurities. The Bayer's process exploits the amphoteric nature of $A l_{2} O_{3}$ to separate it from the basic $F e_{2} O_{3}$ and the acidic $T i O_{2}$ .

Digestion: Crushed bauxite is treated with concentrated $N a O H$ (about $45%$ ) at $473 - 523$ K and $35 - 36$ bar in an autoclave. $A l_{2} O_{3}$ and $S i O_{2}$ dissolve as sodium aluminate and sodium silicate; $F e_{2} O_{3}$ and $T i O_{2}$ remain insoluble and are filtered off as red mud.
$A l_{2} O_{3} (s) + 2 N a O H (a q) + 3 H_{2} O (ℓ) \to 2 N a [A l (O H)_{4}] (a q)$
$S i O_{2} + 2 N a O H \to N a_{2} S i O_{3} + H_{2} O$
Precipitation: The clear filtrate is cooled, diluted and seeded with a small quantity of freshly precipitated $A l (O H)_{3}$ . This induces $A l (O H)_{3}$ to crystallise out slowly, while sodium silicate stays in solution.
$N a [A l (O H)_{4}] (a q) seeding A l (O H)_{3} (s) ↓ + N a O H (a q)$
Calcination: The washed $A l (O H)_{3}$ is heated to $1473$ K to give pure alumina.
$2 A l (O H)_{3} 1473 K A l_{2} O_{3} + 3 H_{2} O$

Figure 1: Flow diagram of Bayer's process. Amphoteric

A l_{2} O_{3}

dissolves in hot concentrated

N a O H

while

F e_{2} O_{3}

(basic) and

T i O_{2}

(acidic) do not, so the impurities are filtered out as red mud.

JEE Advanced

Why seeding works: the sodium aluminate solution from the digester is supersaturated in $A l (O H)_{3}$ after cooling and dilution - but nucleation is very slow. Adding a small amount of pre-formed $A l (O H)_{3}$ crystals provides growth sites, so the excess ion crystallises out on them.

Meanwhile the sodium silicate (also present in solution) stays in solution because it is not supersaturated at these conditions. This is the key separation trick of the Bayer's process: seed-selective precipitation of aluminate ahead of silicate.

3. Extraction: Stage II - Hall-Heroult Electrolysis

Pure alumina has a very high melting point ( $2345$ K), which makes direct electrolysis impractical. Charles Hall and Paul Heroult independently discovered in $1886$ that alumina dissolves readily in molten cryolite ( $N a_{3} A l F_{6}$ ) at only $\sim 1225$ K, giving a highly conducting melt suitable for electrolysis.

Figure 2: An Ellingham-style comparison. The

A l / A l_{2} O_{3}

line lies below the

C / C O

line until about

2300

K, so carbon cannot reduce alumina at any workable furnace temperature. This is exactly why aluminium is won electrolytically in the Hall-Heroult cell rather than in a blast furnace.

Cell and electrolyte

Cell: a steel tank lined internally with graphite (which acts as the cathode).
Anodes: a set of graphite rods dipped into the melt from above.
Electrolyte: molten cryolite ( $80 - 85%$ ) + fluorite $C a F_{2}$ ( $5 - 7%$ ) + $A l F_{3}$ ( $5 - 7%$ ) + alumina ( $2 - 8%$ ). $C a F_{2}$ further lowers the melting point of the mixture and $A l F_{3}$ improves conductivity.
Voltage: $5 - 6$ V; current density $\sim 1$ A cm $^{- 2}$ ; temperature $\sim 1225$ K.

Figure 3: The Hall-Heroult cell. Alumina dissolved in molten cryolite is electrolysed between graphite anodes and a carbon-lined cathode. Molten Al (

d = 2.7

g cm

^{- 3}

) is denser than the melt (

d \approx 2.1

) so it collects at the bottom and is tapped off; the anodes are slowly burnt away as

C O

and

C O_{2}

Electrode reactions

In the melt, alumina ionises approximately as: $A l_{2} O_{3} \to 2 A l^{3 +} + 3 O^{2 -}$

Cathode: $A l^{3 +} (m e l t) + 3 e^{-} \to A l (ℓ)$
Anode: $2 O^{2 -} \to O_{2} + 4 e^{-}$ , but the liberated oxygen immediately attacks the hot graphite anode: $C + O_{2} \to C O_{2}$ and $2 C + O_{2} \to 2 C O$ .

Because the anode is consumed as

C O

and

C O_{2}

, it must be replaced every few weeks. Modern plants use pre-baked or Soderberg-type carbon anodes.

4. Physical and Chemical Properties

Physical properties

Silvery-white, malleable and ductile metal; can be beaten into very thin foils.
Density $2.70$ g cm $^{- 3}$ (light metal); m.p. $933$ K; b.p. $2740$ K.
Excellent conductor of heat and electricity ( $\sim 60%$ of copper on volume basis but $\sim 200%$ on mass basis).
Face-centred cubic (FCC) lattice, giving it high ductility.

Chemical properties

(i) Reaction with air: Pure Al is not attacked by dry air, but in moist air a thin, impervious film of $A l_{2} O_{3}$ ( $\sim 1 0^{- 5}$ cm) forms and protects the underlying metal from further corrosion. On strong heating in air, aluminium powder burns with a dazzling flame:

$4 A l + 3 O_{2} Δ 2 A l_{2} O_{3}$ $Δ H = - 1670$ kJ mol $^{- 1}$

(ii) Reaction with water: No reaction with cold or hot water because of the oxide film. If the film is removed (by amalgamation with Hg), Al reacts even with cold water:

$2 A l + 6 H_{2} O \to 2 A l (O H)_{3} + 3 H_{2}$

(iii) Reaction with dilute acids: Al liberates $H_{2}$ from dilute $H C l$ and dilute $H_{2} S O_{4}$ :

$2 A l + 6 H C l \to 2 A l C l_{3} + 3 H_{2}$

$2 A l + 3 H_{2} S O_{4} (d i l) \to A l_{2} (S O_{4})_{3} + 3 H_{2}$

(iv) Reaction with concentrated $H N O_{3}$ : Al becomes passive due to formation of a very thick oxide layer. Hence concentrated $H N O_{3}$ is transported in aluminium containers.

Figure 4: Aluminium owes its corrosion resistance to a thin, tightly adherent

A l_{2} O_{3}

film (about

1 0^{- 5}

cm) that re-forms the instant it is scratched. Concentrated

H N O_{3}

thickens this film into an impervious layer, so the metal becomes passive instead of dissolving.

(v) Reaction with alkalis (amphoteric behaviour): Al reacts with hot $N a O H$ solution to form sodium aluminate and $H_{2}$ :

$2 A l + 2 N a O H + 6 H_{2} O \to 2 N a [A l (O H)_{4}] + 3 H_{2}$

(older way of writing: $2 A l + 2 N a O H + 2 H_{2} O \to 2 N a A l O_{2} + 3 H_{2}$ )

Figure 5:

A l (O H)_{3}

is amphoteric. It acts as a base towards acids (giving

A l C l_{3}

) and as an acid towards strong alkalis (giving sodium aluminate). The same dual character in

A l_{2} O_{3}

is what lets Bayer's process strip it away from basic

F e_{2} O_{3}

and acidic

T i O_{2}

Why amphoteric?

A l

is on the borderline between metal and non-metal. Its

+ 3

ion is small and highly polarising, so

A l_{2} O_{3}

and

A l (O H)_{3}

can react with both acids (as bases) and alkalis (as acids). This is exactly what makes Bayer's process possible.

(vi) Reaction with halogens:

$2 A l + 3 X_{2} \to 2 A l X_{3}$ ( $X = F, C l, B r, I$ )

(vii) Reduction of metal oxides (thermite reaction): Al is a strong reducing agent because $A l_{2} O_{3}$ is very stable ( $Δ_{f} H^{\circ} = - 1676$ kJ mol $^{- 1}$ ). Aluminothermic reduction is used to produce metals like Cr, Mn from their oxides:

$2 A l + F e_{2} O_{3} Δ A l_{2} O_{3} + 2 F e$ ( $Δ H = - 852$ kJ mol $^{- 1}$ )

$8 A l + 3 M n_{3} O_{4} \to 4 A l_{2} O_{3} + 9 M n$

$2 A l + C r_{2} O_{3} \to A l_{2} O_{3} + 2 C r$

Figure 6: The thermite (aluminothermic) process for on-site welding of broken rails. A

M g

ribbon ignites the

A l + F e_{2} O_{3}

mixture; the reaction is so exothermic (

Δ H = - 852

kJ mol

^{- 1}

) that the iron is produced molten and runs straight into the gap. The same reduction gives Cr, Mn and V from their oxides.

Solved Example 1

Aluminium vessels should not be washed with strong washing soda solution. Explain.

Solution:

Strong washing soda ( $N a_{2} C O_{3}$ ) is a strong alkali in solution because it hydrolyses to give $N a O H$ . Aluminium is amphoteric and reacts with $N a O H$ to give sodium aluminate and hydrogen gas: $2 A l + 2 N a O H + 6 H_{2} O \to 2 N a [A l (O H)_{4}] + 3 H_{2}$ . This dissolves the protective oxide film and eats into the vessel, releasing $H_{2}$ and causing pitting. That is why aluminium cookware carries a warning against alkaline detergents.

5. Uses of Aluminium and its Alloys

Household utensils, cans and cooking foil (light, non-toxic, corrosion-resistant).
Transmission cables for electricity - lighter and cheaper than copper for the same current-carrying capacity.
Aircraft and automobile parts (mostly as duralumin: $A l 95%, C u 4%, M n 0.5%, M g 0.5%$ ).
Window frames, doors, roofing (via magnalium: $A l 90%, M g 10%$ ).
Aluminothermic process for on-site welding and for extraction of Cr, Mn, V, Mo.
Silvery paint (aluminium powder mixed with linseed oil) for radiators and boilers.
Wrappers for chocolates, medicines and cigarettes (Al foil).

6. Important Compounds of Aluminium

6.1 Aluminium oxide (alumina), $A l_{2} O_{3}$

Alumina exists in several crystalline forms; the most common are $α$ - $A l_{2} O_{3}$ (corundum, very hard, used as an abrasive) and $γ$ - $A l_{2} O_{3}$ (a porous form used as a catalyst support and drying agent).

JEE Advanced

Corundum coloured by trace impurities gives gemstones:

Gemstone	Colour	Impurity ion
Ruby	Red	$C r^{3 +}$
Sapphire	Blue	$F e^{2 +}$ + $T i^{4 +}$ (charge-transfer)
Topaz	Yellow	$F e^{3 +}$
Emerald	Green	$C r^{3 +}$ (in beryl matrix, not corundum)
Amethyst	Violet	$M n^{2 +}$ (in quartz)

$A l_{2} O_{3}$ is amphoteric:

$A l_{2} O_{3} + 6 H C l \to 2 A l C l_{3} + 3 H_{2} O$

$A l_{2} O_{3} + 2 N a O H + 3 H_{2} O \to 2 N a [A l (O H)_{4}]$

6.2 Aluminium chloride, $A l C l_{3}$

Anhydrous $A l C l_{3}$ is a white sublimable solid ( $m . p . 465$ K). In the gas phase and in non-polar solvents it exists as the dimer $A l_{2} C l_{6}$ , in which two Cl atoms bridge between two Al centres by donating lone pairs, completing the octet on both Al atoms.

Figure 7:

A l_{2} C l_{6}

has a four-membered

A l_{2} C l_{2}

ring. Each bridging Cl donates a lone pair coordinately (dashed) so both Al atoms complete their octet and become

s p^{3}

. Bond angles:

C l_{t}

A l

C l_{t} \approx 11 8^{\circ}

and

A l

C l_{b}

A l \approx 7 9^{\circ}

Preparation: anhydrous $A l C l_{3}$ is made by passing dry $C l_{2}$ over red-hot Al or over a hot mixture of $A l_{2} O_{3}$ and coke:

$2 A l + 3 C l_{2} \to 2 A l C l_{3}$

JEE Advanced

Industrial route from bauxite (without pure Al metal): a mixture of alumina and coke is heated in a stream of chlorine gas at $\sim 1100$ K. Carbon acts as a reductant, removing the oxygen as $C O$ :

$A l_{2} O_{3} + 3 C + 3 C l_{2} 1100 K 2 A l C l_{3} + 3 C O$

This is the cheaper industrial route for large-scale anhydrous $A l C l_{3}$ production. The gaseous $A l C l_{3}$ is condensed to a white sublimable solid.

Hydrolysis of hydrated salt: the hydrated salt $A l C l_{3} \cdot 6 H_{2} O$ cannot be dehydrated by direct heating - it hydrolyses instead:

$A l C l_{3} \cdot 6 H_{2} O Δ A l (O H)_{3} + 3 H C l + 3 H_{2} O$ ; then $2 A l (O H)_{3} \to A l_{2} O_{3} + 3 H_{2} O$

Uses: $A l C l_{3}$ is a strong Lewis acid and the classical catalyst for the Friedel-Crafts alkylation and acylation of aromatic compounds.

6.3 Aluminium hydroxide, $A l (O H)_{3}$

Formed as a gelatinous white precipitate when $N H_{4} O H$ is added to an aluminium salt solution:

$A l C l_{3} + 3 N H_{4} O H \to A l (O H)_{3} ↓ + 3 N H_{4} C l$

It is amphoteric:

$A l (O H)_{3} + 3 H C l \to A l C l_{3} + 3 H_{2} O$

$A l (O H)_{3} + N a O H \to N a [A l (O H)_{4}]$

Important: $N H_{4} O H$ (weak base) does not dissolve $A l (O H)_{3}$ , so ammonia is used to precipitate Al selectively in the presence of other cations that dissolve in $N a O H$ .

Figure 8: Both

A l (O H)_{3}

and

Z n (O H)_{2}

dissolve in excess

N a O H

, so a strong alkali cannot tell them apart. With excess

N H_{4} O H

only zinc dissolves (as the ammine complex

[Z n (N H_{3})_{4}]^{2 +}

) while

A l (O H)_{3}

stays behind as a precipitate, which is the standard way of separating the two in qualitative analysis.

6.4 Alums

An alum is a double sulphate of the general formula $M_{2}^{I} S O_{4} \cdot M_{2}^{I I I} (S O_{4})_{3} \cdot 24 H_{2} O$ , where $M^{I}$ is a monovalent cation (Na $^{+}$ , K $^{+}$ , NH $_{4}^{+}$ , Tl $^{+}$ ) and $M^{I I I}$ is a trivalent cation (Al $^{3 +}$ , Cr $^{3 +}$ , Fe $^{3 +}$ ).

The most familiar is potash alum: $K_{2} S O_{4} \cdot A l_{2} (S O_{4})_{3} \cdot 24 H_{2} O$ , or equivalently $K A l (S O_{4})_{2} \cdot 12 H_{2} O$ .

Figure 9: Anatomy of the alum formula. Change

M^{I}

(K to Na,

N H_{4}

, Tl) or

M^{I I I}

(Al to Cr, Fe) and the compound is still an alum. Common examples are sodium alum, ammonium alum and chrome alum (

K_{2} S O_{4} \cdot C r_{2} (S O_{4})_{3} \cdot 24 H_{2} O

Alums are isomorphous (they crystallise with the same octahedral shape). Uses include: purification of water (as a coagulant of colloidal impurities), sizing of paper, mordanting in dyeing, and as a styptic (to stop bleeding from minor cuts).

Solved Example 2

Anhydrous

A l C l_{3}

is covalent while

A l F_{3}

is ionic. Justify.

Solution:

Apply Fajans' rules. $A l^{3 +}$ is small and carries a $+ 3$ charge, so its polarising power is very high. The $F^{-}$ ion is small and hard to polarise, so $A l^{3 +}$ - $F^{-}$ interactions retain a strongly ionic character and $A l F_{3}$ is a high-melting ionic solid ( $m . p . 1563$ K). The $C l^{-}$ ion is much larger and more polarisable, so $A l^{3 +}$ heavily distorts its electron cloud, producing significant covalent character. This is why $A l C l_{3}$ is a low-melting, sublimable solid that exists as covalent $A l_{2} C l_{6}$ dimers in the vapour phase.

Figure 10: Fajans' rules in picture form.

A l^{3 +}

is small and triply charged, so it polarises strongly.

F^{-}

is small and hard, so

A l F_{3}

stays ionic and melts at

1563

C l^{-}

is large and soft, so its cloud is pulled towards

A l^{3 +}

, giving covalent

A l C l_{3}

that sublimes at

465

K and exists as

A l_{2} C l_{6}

in the vapour.

Solved Example 3

Why is cryolite added to alumina in the Hall-Heroult process even though it is not the source of aluminium?

Solution:

Pure $A l_{2} O_{3}$ melts at $2345$ K, which is far too high for practical electrolysis (energy cost, corrosion, container problems). Molten cryolite ( $N a_{3} A l F_{6}$ ) dissolves $A l_{2} O_{3}$ and forms a conducting melt at around $1225$ K, roughly $1100$ K lower than pure alumina's melting point. Cryolite also improves electrical conductivity of the melt (alumina alone is a poor conductor) and lowers the density of the electrolyte enough that molten Al ( $d = 2.7$ ) sinks to the bottom and can be tapped off. The added $C a F_{2}$ further lowers the operating temperature.

Common Mistakes to Avoid

Watch out

Do not confuse Bayer's process (purification of bauxite by NaOH) with the Hall-Heroult process (electrolysis of pure alumina in molten cryolite). Both are needed - Bayer's gives pure $A l_{2} O_{3}$ , Hall-Heroult reduces it to metal.
Cryolite is not decomposed by the current in the Hall-Heroult cell; it acts only as a solvent for alumina.
Anhydrous $A l C l_{3}$ cannot be obtained by heating hydrated $A l C l_{3} \cdot 6 H_{2} O$ - hydrolysis occurs instead giving $A l (O H)_{3}$ and $H C l$ . Use passage of dry $C l_{2}$ over Al metal.
$A l (O H)_{3}$ dissolves in $N a O H$ (a strong base) but not in $N H_{4} O H$ (weak base). Use this to separate $A l^{3 +}$ from $Z n^{2 +}$ - both hydroxides dissolve in $N a O H$ but only $Z n (O H)_{2}$ dissolves in excess $N H_{4} O H$ .
In the aluminate anion, the formula is $N a [A l (O H)_{4}]$ or $N a A l O_{2}$ (older). Never write $N a [A l (O H)_{2}]$ - that is unbalanced.
Concentrated $H N O_{3}$ passivates Al by thickening the oxide film. Do not write a "no reaction" - write "passivation".

Frequently Asked Questions

Q1. What are the main steps in the extraction of aluminium?

Aluminium is extracted from bauxite in two stages. Stage I - Bayer's process: bauxite is digested with hot concentrated $N a O H$ at $473 - 523$ K, filtered to remove $F e_{2} O_{3}$ and $T i O_{2}$ , and the aluminate solution is seeded with $A l (O H)_{3}$ to precipitate pure aluminium hydroxide, which is calcined at $1473$ K to give pure alumina. Stage II - Hall-Heroult process: pure $A l_{2} O_{3}$ is dissolved in molten cryolite ( $N a_{3} A l F_{6}$ ) at $\sim 1225$ K and electrolysed with graphite electrodes to yield molten aluminium at the cathode.

Q2. Why is molten cryolite added to alumina in the Hall-Heroult process?

Pure alumina melts at $2345$ K, which is impractically high. Molten cryolite ( $N a_{3} A l F_{6}$ ) dissolves $A l_{2} O_{3}$ and forms a highly conducting melt at only $\sim 1225$ K. Small amounts of $C a F_{2}$ further lower the melting point and $A l F_{3}$ improves conductivity. Cryolite itself is not reduced - it is just the solvent.

Q3. Why does aluminium show amphoteric behaviour?

Aluminium sits at the borderline between metals and non-metals. Its small, highly-charged $A l^{3 +}$ ion has strong polarising power, giving its oxide and hydroxide dual character: they behave as bases towards strong acids ( $A l (O H)_{3} + 3 H C l \to A l C l_{3} + 3 H_{2} O$ ) and as acids towards strong alkalis ( $A l (O H)_{3} + N a O H \to N a [A l (O H)_{4}]$ ). This amphoteric nature is exploited in Bayer's process to separate $A l_{2} O_{3}$ from basic $F e_{2} O_{3}$ and acidic $T i O_{2}$ .

Q4. Why does aluminium not react with concentrated nitric acid?

Concentrated $H N O_{3}$ is a strong oxidising acid. When it contacts Al, it very quickly thickens the natural surface oxide layer into an impervious coating that prevents any further attack. This state is called passivation. Because of passivation, concentrated $H N O_{3}$ is safely transported in aluminium tankers.

Q5. What is the thermite reaction and where is it used?

The thermite reaction is the reduction of a metal oxide by aluminium powder: $2 A l + F e_{2} O_{3} \to A l_{2} O_{3} + 2 F e$ ( $Δ H = - 852$ kJ mol $^{- 1}$ ). The heat released melts the iron produced. The classic application is on-site welding of broken railway tracks - the molten iron flows into the gap and solidifies as a strong joint. The same aluminothermic reduction is used industrially to extract Cr, Mn and V from their oxides.

Q6. Why is $A l C l_{3}$ a dimer while $B C l_{3}$ is a monomer?

In $B C l_{3}$ , boron's empty $2 p$ orbital is stabilised by $p π$ - $p π$ back-donation from Cl lone pairs, and BCl3 is trigonal planar and monomeric. In $A l C l_{3}$ , no effective back-bonding to $3 p$ -Al is possible from Cl, so Al is genuinely electron-deficient. Two $A l C l_{3}$ units therefore dimerise: two Cl atoms bridge across, donating lone pairs coordinately to both Al centres and completing the octet.

Q7. What is an alum and what are alums used for?

An alum is a double sulphate of a monovalent metal ( $M^{I}$ : Na, K, NH $_{4}$ , Tl) and a trivalent metal ( $M^{I I I}$ : Al, Cr, Fe) with $24$ molecules of water: $M_{2}^{I} S O_{4} \cdot M_{2}^{I I I} (S O_{4})_{3} \cdot 24 H_{2} O$ . Potash alum $K_{2} S O_{4} \cdot A l_{2} (S O_{4})_{3} \cdot 24 H_{2} O$ is the most common. Uses include water purification (coagulating colloidal impurities), sizing of paper, mordanting in dyeing and as a styptic to stop bleeding from small cuts.

Q8. Why is aluminium used to make electricity transmission cables even though copper is a better conductor?

On a mass basis (per kilogram), aluminium actually conducts electricity about twice as well as copper because it is much less dense. For transmission lines, weight matters - heavier cables need stronger, more expensive towers. Aluminium is also much cheaper and is corrosion-resistant. So even though copper wins on conductivity per unit volume, aluminium wins on cost and weight for long-distance transmission.

Q9. Which impurities give ruby and sapphire their colours?

Both ruby and sapphire are corundum ( $α$ - $A l_{2} O_{3}$ ) with trace impurity ions. Ruby gets its red colour from $C r^{3 +}$ substituting for $A l^{3 +}$ in the lattice. Sapphire's characteristic blue arises from a metal-to-metal charge transfer between $F e^{2 +}$ and $T i^{4 +}$ impurity ions, which absorbs red-orange light and leaves the transmitted light blue.

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1. Occurrence and Ores

2. Extraction: Stage I - Purification of Bauxite (Bayer's Process)

3. Extraction: Stage II - Hall-Heroult Electrolysis

Cell and electrolyte

Electrode reactions

4. Physical and Chemical Properties

Physical properties

Chemical properties

5. Uses of Aluminium and its Alloys

6. Important Compounds of Aluminium

6.1 Aluminium oxide (alumina), Al2​O3​

6.2 Aluminium chloride, AlCl3​

6.3 Aluminium hydroxide, Al(OH)3​

6.4 Alums

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Frequently Asked Questions

Q1. What are the main steps in the extraction of aluminium?

Q2. Why is molten cryolite added to alumina in the Hall-Heroult process?

Q3. Why does aluminium show amphoteric behaviour?

Q4. Why does aluminium not react with concentrated nitric acid?

Q5. What is the thermite reaction and where is it used?

Q6. Why is AlCl3​ a dimer while BCl3​ is a monomer?

Q7. What is an alum and what are alums used for?

Q8. Why is aluminium used to make electricity transmission cables even though copper is a better conductor?

Q9. Which impurities give ruby and sapphire their colours?

Ready to master The p-Block Elements: Group 13 & 14?

6.1 Aluminium oxide (alumina), $A l_{2} O_{3}$

6.2 Aluminium chloride, $A l C l_{3}$

6.3 Aluminium hydroxide, $A l (O H)_{3}$

Q6. Why is $A l C l_{3}$ a dimer while $B C l_{3}$ is a monomer?