Isolation of Elements

In calcination, we convert a compound or ore into oxide.

In option (D) you can see there is ZnO and MgO, as they are already in oxide form so no calcination required for then.

In the Hall-Heroult process the cathode is made of carbon.

Siderite : FeCO₃

Kaolinite : Al₂(OH)₄Si₂O₅

Malachite : CuCO₃ .Cu(OH)₂

Calamine : ZnCO₃

Note : ( Remember all those ores names. Any one of those can be asked in the exam.)

Oxides Ores :

(1) ZnO - Zincite

(2) Fe₂O₃ - Haematite

(3) Fe₃O₄ - Magnetite (FeO + Fe₂O₃ mixture)

(4) Fe₂O₃ . 3H₂O - Limonite

(5) MnO₂ - Pyrolusite

(6) Cu₂O - Cuprite or Ruby Copper

(7) TiO₂ - Rutile

(8) FeCr₂O₄ - Chromite (FeO + Cr₂O₃)

(9) FeTiO₃ - Illmenite (FeO + TiO₂)

(10) Na₂B₄O₇ . 10H₂O - Borax or Tincal

(11) U₃O₈ - Pitch Blende

(12) SnO₂ - Tin Stone or Cassiterite

(13) Ca₂B₆O₁₁ . 5H₂O - Colemanite (2 Cao + 3 B₂O₃)

(14) Al₂O₃ . 2H₂O - Bauxite

(15) Al₂O₃ . H₂O - Diaspore

(16) Al₂O₃ - Corundum

Sulphides Ores :

(1) ZnS - Zinc Blende or Sphalerite

(2) PbS - Galena

(3) Ag₂S - Argentite or Silver Glance

(4) HgS - Cinnabar

(5) Cu₂S - Chalcocite or Copper glance

(6) CuFeS₂ - Copper pyrites or Chalco pyrites (Cu₂S + Fe₂S₃ mixture)

(7) FeS₂ - Iron pyrites or Fool's Gold

(8) 3Ag₂S . Sb₂S₂ - Pyrargyrite or ruby silver

Halides Ores :

(1) NaCl - Rock Salt

(2) KCl - Sylvine

(3) Na₃AlF₆ - Cryolite [3NaF + AlF₆]

(4) CaF₂ - Fluorspar

(5) KCl . MgCl₂ . 6H₂O - Carnalite

(6) AgCl - Horn Silver

Carbonates Ores :

(1) CaCO₃ - Limestone

(2) MgCO₃ - Magnesite

(3) CaCO₃ . MgCO₃ - Dolomite

(4) ZnCO₃ - Calamine

(5) PbCO₃ - Cerrusite

(6) FeCO₃ - Siderite

(7) CuCO₃ . Cu(OH)₂ or Cu₂CO₃(OH)₂ - Malachite or Basic Copper Carbonates

(8) 2 CuCO₃ . Cu(OH)₂ - Azurite

Sulphates Ores :

(1) CuSO₄ . 2H₂O - Gypsum

(2) MgSO₄ . 7H₂O - Epson Salt

(3) Na₂SO₄ . 10 H₂O - Glauber's Salt

(4) PbSO₄ - Anglesite

(5) ZnSO₄ . 7H₂O - White Vitriol

(6) FeSO₄ . 7H₂O - Green Vitriol

(7) CuSO₄ . 5H₂O - Blue Vitriol or Chalcanthite

Nitrate Ores :

(1) KNO₃ - Indian Saltpetre

(2) NaNO₃ - Chile Saltpetre

Arsenides Ores :

(1) NiAs - Kupfernickel

(2) NiAsS - Nickel glance

[Au(CN)₂]^– and [Ag(CN)₂]^– both are soluble complexes.

[Au(CN)₂]^– ⇌ Au⁺ + 2CN^-

[Ag(CN)₂]^– ⇌ Ag⁺ + 2CN^-

The concentration of produced Au⁺ ion and Ag⁺ ion due to ionisation is very less which ensure slow deposition and this slow deposition will generate uniform layer on the substance to be electroplated.

In Hall-Heroult's process is given by

At cathode :- $4Al_{\left( \ell \right)}^{3 + }{\mkern 1mu} {\mkern 1mu} + {\mkern 1mu} {\mkern 1mu} 12{e^ - }{\mkern 1mu} {\mkern 1mu} \,\mathrel{\mathop{\kern0pt\longrightarrow} \limits_\,} \,\,4Al\left( \ell \right)$

At Anode :- $6O_{\left( \ell \right)}^{2 - }\,\,\mathrel{\mathop{\kern0pt\longrightarrow} \limits_\,} \,\,3{O_2}(g) + 12{e^ - }$

$3C + 3{O_2}\,\,\mathrel{\mathop{\kern0pt\longrightarrow} \limits_\,} \,\,3C{O_2}\left( \uparrow \right)$

According to the given diagram A1 can reduce ZnO.

3ZnO + 2Al $ \to $ 3Zn + Al₂O₃

Copper pyrites	:	CuFeS2
Malachite	:	CuCO3 . Cu(OH)2
Azurite	:	2CuCO3 . Cu(OH)2
Dolomite	:	CaCO3 . MgCO3

Cyanide process of gold extraction involves leaching out gold from its ore with CN^- in the presence of O₂ (Q) in water to form [Au(CN)₂]^- (R). When [Au(CN)₂]^- reacts with Zn (T), it forms [Zn(CN)₄]^2- (Z) and Au.

The corresponding reactions are as follows :


Hence, options (a, b, c) are correct.

ZnCO₃ - Calamine (zinc ore)

CuCO₃.Cu(OH)₂ - Malachite (copper ore)

Fe₃O₄ - Magnetite (iron ore)

Na₃AlF₆ - Cryolite (aluminum ore)

Thus, option (d) is correct.

Note : ( Remember all those ores names. Any one of those can be asked in the exam.)

Oxides Ores :

(1) ZnO - Zincite

(2) Fe₂O₃ - Haematite

(3) Fe₃O₄ - Magnetite (FeO + Fe₂O₃ mixture)

(4) Fe₂O₃ . 3H₂O - Limonite

(5) MnO₂ - Pyrolusite

(6) Cu₂O - Cuprite or Ruby Copper

(7) TiO₂ - Rutile

(8) FeCr₂O₄ - Chromite (FeO + Cr₂O₃)

(9) FeTiO₃ - Illmenite (FeO + TiO₂)

(10) Na₂B₄O₇ . 10H₂O - Borax or Tincal

(11) U₃O₈ - Pitch Blende

(12) SnO₂ - Tin Stone or Cassiterite

(13) Ca₂B₆O₁₁ . 5H₂O - Colemanite (2 Cao + 3 B₂O₃)

(14) Al₂O₃ . 2H₂O - Bauxite

(15) Al₂O₃ . H₂O - Diaspore

(16) Al₂O₃ - Corundum

Sulphides Ores :

(1) ZnS - Zinc Blende or Sphalerite

(2) PbS - Galena

(3) Ag₂S - Argentite or Silver Glance

(4) HgS - Cinnabar

(5) Cu₂S - Chalcocite or Copper glance

(6) CuFeS₂ - Copper pyrites or Chalco pyrites (Cu₂S + Fe₂S₃ mixture)

(7) FeS₂ - Iron pyrites or Fool's Gold

(8) 3Ag₂S . Sb₂S₂ - Pyrargyrite or ruby silver

Halides Ores :

(1) NaCl - Rock Salt

(2) KCl - Sylvine

(3) Na₃AlF₆ - Cryolite [3NaF + AlF₆]

(4) CaF₂ - Fluorspar

(5) KCl . MgCl₂ . 6H₂O - Carnalite

(6) AgCl - Horn Silver

Carbonates Ores :

(1) CaCO₃ - Limestone

(2) MgCO₃ - Magnesite

(3) CaCO₃ . MgCO₃ - Dolomite

(4) ZnCO₃ - Calamine

(5) PbCO₃ - Cerrusite

(6) FeCO₃ - Siderite

(7) CuCO₃ . Cu(OH)₂ or Cu₂CO₃(OH)₂ - Malachite or Basic Copper Carbonates

(8) 2 CuCO₃ . Cu(OH)₂ - Azurite

Sulphates Ores :

(1) CuSO₄ . 2H₂O - Gypsum

(2) MgSO₄ . 7H₂O - Epson Salt

(3) Na₂SO₄ . 10 H₂O - Glauber's Salt

(4) PbSO₄ - Anglesite

(5) ZnSO₄ . 7H₂O - White Vitriol

(6) FeSO₄ . 7H₂O - Green Vitriol

(7) CuSO₄ . 5H₂O - Blue Vitriol or Chalcanthite

Nitrate Ores :

(1) KNO₃ - Indian Saltpetre

(2) NaNO₃ - Chile Saltpetre

Arsenides Ores :

(1) NiAs - Kupfernickel

(2) NiAsS - Nickel glance

2Cu₂S  +  3O₂   $ \to $   2Cu₂O  +   2SO₂

2Cu₂O  +  Cu₂S  $ \to $  6Cu  +  SO₂

It is a example of auto reduction.

First a part of sulphide ore is converted into oxide which then reacts with remaining sulphide to give the metal.

The gangue material in the ore is removed by adding another chemical substance called flux. If ore has basic impurities, such as FeO, CaO, MgO, etc. suitable acid flux such as SiO₂ , P₂O₅ , etc. are used. For example, in the extraction of copper, ferrous oxide (FeO), a basic impurity, is removed by adding silica. FeO (gangue) and SiO₂ (flux) together form slag FeSiO₃.

Froth floatation method is mainly applicable for sulphide ores.

$(1)$ Malachite ore : $Cu{\left( {OH} \right)_2},\,\,CuC{O_3}$

$(2)$ Magnetite ore : $F{e_3}{O_4}$

$(3)$ Siderite ore : $FeC{O_3}$

$(4)$ Galena ore : $PbS$ (Sulphide Ore)

(a) In the extraction of copper from copper pyrite (CuFeS₂), after crushing, concentration of ore is done by froth floatation process.

(b) Iron is removed as slag.

(c) Auto-reduction :

(d) Blister copper is finally purified by electrolytic refining.

In the metallurgy of aluminium purified $A{l_2}{O_3}$ is mixed with $N{a_3}Al{F_6}\,\,$ $\,\,\,$ or $\,\,\,$ $Ca{F_2}$ which lowers the melting point of the mix and brings conductivity.

(a) is wrong statement. Impure copper is set as anode where copper is oxidized to Cu²⁺ and goes into electrolytic solutions.

(b) CuSO₄ is used as an electrolyte in purification process.

(c) Pure copper is deposited at cathode as : Cu²⁺ + 2e^$-$ $\to$ Cu : (At cathode)

(d) Less active metals like Ag, Au etc. settle down as anode mud.

Carbonate ores are
(P) Siderite : FeCO₃
(Q) Malachite : CuCO₃.Cu(OH)₂
(S) Calamine : ZnCO₃

Sulphide ore is (T) Argentite : Ag₂S.

Hydroxide ion is present in
(Q) Malachite : CuCO₃.Cu(OH)₂
(R) Bauxite : Al₂O₃.2H₂O or AlO_x (OH)_3–2x where 0 < x < 1

Oxide ore is bauxite (R) only

Note : ( Remember all those ores names. Any one of those can be asked in the exam.)

Oxides Ores :

(1) ZnO - Zincite

(2) Fe₂O₃ - Haematite

(3) Fe₃O₄ - Magnetite (FeO + Fe₂O₃ mixture)

(4) Fe₂O₃ . 3H₂O - Limonite

(5) MnO₂ - Pyrolusite

(6) Cu₂O - Cuprite or Ruby Copper

(7) TiO₂ - Rutile

(8) FeCr₂O₄ - Chromite (FeO + Cr₂O₃)

(9) FeTiO₃ - Illmenite (FeO + TiO₂)

(10) Na₂B₄O₇ . 10H₂O - Borax or Tincal

(11) U₃O₈ - Pitch Blende

(12) SnO₂ - Tin Stone or Cassiterite

(13) Ca₂B₆O₁₁ . 5H₂O - Colemanite (2 Cao + 3 B₂O₃)

(14) Al₂O₃ . 2H₂O - Bauxite

(15) Al₂O₃ . H₂O - Diaspore

(16) Al₂O₃ - Corundum

Sulphides Ores :

(1) ZnS - Zinc Blende or Sphalerite

(2) PbS - Galena

(3) Ag₂S - Argentite or Silver Glance

(4) HgS - Cinnabar

(5) Cu₂S - Chalcocite or Copper glance

(6) CuFeS₂ - Copper pyrites or Chalco pyrites (Cu₂S + Fe₂S₃ mixture)

(7) FeS₂ - Iron pyrites or Fool's Gold

(8) 3Ag₂S . Sb₂S₂ - Pyrargyrite or ruby silver

Halides Ores :

(1) NaCl - Rock Salt

(2) KCl - Sylvine

(3) Na₃AlF₆ - Cryolite [3NaF + AlF₆]

(4) CaF₂ - Fluorspar

(5) KCl . MgCl₂ . 6H₂O - Carnalite

(6) AgCl - Horn Silver

Carbonates Ores :

(1) CaCO₃ - Limestone

(2) MgCO₃ - Magnesite

(3) CaCO₃ . MgCO₃ - Dolomite

(4) ZnCO₃ - Calamine

(5) PbCO₃ - Cerrusite

(6) FeCO₃ - Siderite

(7) CuCO₃ . Cu(OH)₂ or Cu₂CO₃(OH)₂ - Malachite or Basic Copper Carbonates

(8) 2 CuCO₃ . Cu(OH)₂ - Azurite

Sulphates Ores :

(1) CuSO₄ . 2H₂O - Gypsum

(2) MgSO₄ . 7H₂O - Epson Salt

(3) Na₂SO₄ . 10 H₂O - Glauber's Salt

(4) PbSO₄ - Anglesite

(5) ZnSO₄ . 7H₂O - White Vitriol

(6) FeSO₄ . 7H₂O - Green Vitriol

(7) CuSO₄ . 5H₂O - Blue Vitriol or Chalcanthite

Nitrate Ores :

(1) KNO₃ - Indian Saltpetre

(2) NaNO₃ - Chile Saltpetre

Arsenides Ores :

(1) NiAs - Kupfernickel

(2) NiAsS - Nickel glance

Extraction of various metals from their respective concentrated ores is carried out by different methods:

(i) Carbon Reduction of Hematite:

The process for the extraction of iron from hematite (Fe₂O₃) involves carbon reduction:

3 Fe2O3 + CO → 2 Fe3O4 + CO2(g)

Fe3O4 + 4 CO → 3 Fe(s) + 4 CO(g)

This reaction occurs at temperatures between 500-800 K.

FeO(s) + CO(g) → Fe(s) + CO2(g)

Carbon monoxide needed for the reduction is produced as follows:

C(s) + CO2(g) → 2 CO

The net reaction can be written as:

FeO(s) + C(s) → Fe(s) + CO(g)

Thus, carbon and its oxides are utilized for the reduction of iron ores such as Fe₂O₃.

(ii) Carbon Reduction of Cassiterite (SnO₂):

The reduction process involves carbon:

SnO2(s) + C(s) → Sn(s) + CO2(g)

Therefore, carbon is used for the reduction of cassiterite ore.

(iii) Electrolytic Refining of Aluminium (Al₂O₃):

Aluminium is extracted from bauxite using electrolysis. The reduction occurs at the cathode:

Al3+ + 3 e- → Al(s)

The reduction of Al³⁺ takes place at the cathode.

(iv) Dow's Process for Extraction of Magnesium from Dolomite (MgCO₃.CaCO₃):

Dolomite is treated with hydrochloric acid (HCl) to produce magnesium chloride (MgCl₂). Electrolysis of MgCl₂ reduces Mg²⁺ to magnesium metal at the cathode:

Mg2+(aq) + 2 e- → Mg

Magnesium metal is obtained by the reduction of Mg²⁺.

The correct answer is Option A: Ag, Cu and Pb. Here's why :

Sulphide ores are minerals where the metal is combined with sulfur. Many metals are found in this form due to the chemical reactivity of sulfur.

• Ag (Silver): Silver is often found as the sulfide mineral argentite (Ag₂S).
• Cu (Copper): Copper is commonly found as chalcopyrite (CuFeS₂) and chalcocite (Cu₂S).
• Pb (Lead): Lead is extracted from the sulfide mineral galena (PbS).

Let's look at why the other options are incorrect :

• Option B: Ag, Mg and Pb - Magnesium (Mg) is typically found in oxide ores, not sulfide ores.
• Option C: Ag, Cu and Sn - Tin (Sn) is primarily found in oxide ores, not sulfide ores.
• Option D: Al, Cu and Pb - Aluminum (Al) is mainly extracted from bauxite ore, an oxide mineral.

Van Arkel is a method in which heat treatment it used to purify metal in this process metals are converted into other metal compound for loosly coupled like as iodine to make metal iodide which are easily decomposed and give pure metal.

The process is known as Van Arkel method.

Silver ore is oxidized using oxygen from the air, as shown in the reaction below :

$ 4\text{Ag} + 8\text{NaCN} + 2\text{H}_2\text{O} + \text{O}_2(\text{air}) \longrightarrow 4\text{NaAg(CN)}_2 + 4\text{NaOH} $

Here, silver ($ \text{Ag}(0) $) is undergoing oxidation to form $ \text{Ag}(+1) $ :

$ \text{Ag}(0) \xrightarrow{\text{oxidation}} \text{Ag}(+1) $

Silver is then precipitated from the solution by adding zinc dust in a finely divided form :

$ 2\text{NaAg(CN)}_2 + \text{Zn} \longrightarrow \text{Na}_2\text{Zn(CN)}_4 + 2\text{Ag} $

In this reaction, $ \text{Ag}(+1) $ is reduced back to $ \text{Ag}(0) $ :

$ \text{Ag}(+1) \xrightarrow{\text{reduction}} \text{Ag}(0) $

In summary, oxygen from the air acts as the oxidizing agent, and zinc dust serves as the reducing agent.

Tin is obtained by reducing the ore cassiterite with coal in a reverberatory furnace. Limestone is added to produce a slag with the impurities, which can be separated.

$ \mathrm{SnO}_2+2 \mathrm{C} \longrightarrow \mathrm{Sn}+2 \mathrm{CO} $

Crude tin obtained from this process is contaminated with iron, copper, lead, and other metals. To purify it, the crude tin is remelted in an inclined furnace—a process known as liquation. During liquation, the easily fusible tin melts away, leaving the less fusible impurities behind.

Finally, the molten tin is refined using green poles of wood in contact with air. This step, known as poling, helps to oxidize any remaining metal impurities. These impurities form a scum on the surface, which can be removed.

This step-by-step process ensures the extraction and purification of tin from cassiterite.

The molecular formula of the mineral haematite is $\mathrm{Fe}_2 \mathrm{O}_3$ and that of magnetite is $\mathrm{Fe}_3 \mathrm{O}_4$ respective. Both contain iron ( Fe ) but in different oxidation states.

Oxidation state of iron ( Fe ) in :

(a) Haematite $\left(\mathrm{Fe}_2 \mathrm{O}_3\right)$

Let the oxidation state of iron ( Fe ) be $x$

$ \begin{aligned} 2 x+3 \times(-2) & =0 \\\\ 2 x & =6 \\\\ x & =+3 \end{aligned} $

Iron exist as Fe (III).

(b) Magnetite $\left(\mathrm{Fe}_3 \mathrm{O}_4\right)$

The mineral magnetite is made by two iron oxides $\mathrm{FeO} . \mathrm{Fe}_2 \mathrm{O}_3$.

The oxidation state of iron in $\mathrm{Fe}_2 \mathrm{O}_3$ is +3 and oxidation state of iron in FeO can be calculated as :

$ \begin{aligned} x+2 \times(-2) & =0 \\\\ x & =+2 \end{aligned} $

Iron exist as Fe (II).

Hence, the oxidation state of iron in haematite is III and that in magnetite is II, III respectively.

Partial roasting of chalcopyrite gives

2CuFeS$_2$ + O$_2$ $\to$ Cu$_2$S + 2FeS + SO$_2$

2CuFeS$_2$ + 4O$_2$ $\to$ Cu$_2$S + 2FeO + 3SO$_2$

Since iron is more electropositive than copper, so its sulphide is oxidised in preference. Cu$_2$S remains mostly unaffected and the small amount of Cu$_2$O formed on oxidation, reacts with FeS to give back Cu$_2$S.

2Cu$_2$S + 3O$_2$ $\to$ 2Cu$_2$O + 2SO$_2$

Cu$_2$O + FeS $\to$ Cu$_2$S + FeO

Iron obtained in form of its oxide is removed by slag formation with SiO$_2$.

FeO + SiO$_2$ $\to$ FeSiO$_ 3$ (slag)

In the self-reduction step,

Cu$_2$S + $\frac{3}{2}$O$_2$ $\to$ Cu$_2$O + SO$_2$

Cu$_2$S + 2Cu$_2$O $\to$ $\mathrm{\mathop {6Cu(l)}\limits_{Blister\,copper} }$ + SO$_2$

In the reaction, the oxidation number of sulphur (S$^{2-}$) changes to (S$^{4+}$) so, S$^{2-}$ is the reducing agent.

NOTE : The reduction of metal sulphides by carbon reduction process is not spontaneous because $\Delta G$ for such a process is positive. The reduction of metal oxide by carbon reduction process is spontaneous as $\Delta G$ for such a process is negative.

From this we find that on thermodynamic considerations $C{O_2}$ is more stable than $C{S_2}$ and the metal sulphides are more stable than corresponding oxides.

In view of above the factor listed in choice (c) is incorrect and so is of no significance.

Hence the correct answer is $(c)$

A. For PbS $\to$ O$_2$ PbO, roasting is done.

B. For CaCO$_3$ $\to$ CaO, calcinations is done in the absence of air.

C. For ZnS $\to$ Zn, there is roasting is followed by reduction

2ZnS + 3O$_2$ $\buildrel \Delta \over \longrightarrow $ 2ZnO + 2SO$_2$ $\uparrow$

2ZnSO$_4$ $\buildrel \Delta \over \longrightarrow $ ZnO + SO$_2$ + $\frac{1}{2}$O$_2$ $\uparrow$

ZnO $\buildrel \Delta \over \longrightarrow $ Zn + CO $\uparrow$

D. For Cu$_2$S $\to$ Cu, roasting is done followed by self-reduction.

2Cu$_2$S + 3O$_2$ $\to$ 2Cu$_2$O + 2SO$_2$

Cu$_2$S + 2Cu$_2$O $\to$ 6Cu + SO$_2$

Reaction involving leaching of silver ores is reversible add, so to get the maximum yield of the product the reaction is carried out in presence of oxygen. Also, oxygen oxidises silver which then produces soluble complex with CN$^-$ ions.

4Ag + 8NaCN + 2H$_2$O + O$_2$ $\to$ 4Na[Ag(CN)$_2$] + 4NaOH

Zinc from zinc blende is obtained by first roasting of zinc. After roasting, a reduction process is carried out in the presence of carbon which leads to the formation of free metal that is zinc. The molecular formula of zinc blende is ZnS. Zinc blende is an ore. Any metal can be extracted by following two major steps:

1. Conversion of the ore into metallic oxide

2. Reduction of the metallic oxide into free metal

In order to get zinc from ZnS, first convert the zinc blende (ore) to zinc oxide (metallic oxide). The conversion of ore into metallic oxide can be done by calcination and roasting.

Calcination is generally utilised to convert hydrated oxides or hydroxides and carbonates into respective oxides while roasting is utilised to convert sulphide ores into respective oxides. Zinc blende is sulphide ore. Thus, the roasting process will be carried out to first convert zinc blende to Zinc oxide (ZnO).

2ZnS + 3O$_2$ $\to$ 2ZnO + 2SO$_2$

Now, reduction of metallic oxide to free metal is carried out. The reduction of zinc oxide in presence of carbon leads to the formation of zinc metal.

ZnO + C $\to$ Zn + CO

(A) The self-reduction is the reduction in which the reduction of metal sulphide occur by other compound of that same metal without addition of any external reducing agent.

The ores of lead and copper undergoes self-reduction during extraction of metals from these ores.

The lead (II) sulphide undergoes oxidation, producing lead (II) oxide and sulphur dioxide. The lead (II) oxide formed reduces lead (II) sulphide to produce, metallic lead and sulphur dioxide.

$ \begin{aligned} 2 \mathrm{PbS}+3 \mathrm{O}_2 & \rightarrow 2 \mathrm{PbO}+2 \mathrm{SO}_2 \\ 2 \mathrm{PbO}+\mathrm{PbS} & \rightarrow 3 \mathrm{~Pb}+\mathrm{SO}_2 \end{aligned} $

Thus, lead (II) oxide acts as a reducing agent to reduce lead (II) sulphide to metallic lead.

The sulphide ore of copper are known as chalcocite containing copper (I) sulphide with chemical formula $\mathrm{Cu}_2 \mathrm{~S}$.

The copper (I) sulphide undergoes oxidation, producing cuprous oxide and sulphur dioxide. The cuprous oxide will then act as a reducing agent to reduce copper (I) sulphide to produce, metallic copper and sulphur dioxide.

$ \begin{aligned} 2 \mathrm{Cu}_2 \mathrm{~S}+2 \mathrm{O}_2 & \rightarrow 2 \mathrm{Cu}_2 \mathrm{O}+2 \mathrm{SO}_2 \\ \mathrm{Cu}_2 \mathrm{~S}+2 \mathrm{Cu}_2 \mathrm{O} & \rightarrow 6 \mathrm{Cu}+\mathrm{SO}_2 \end{aligned} $

Hence, option A in column I matches with options P and R in column II.

(B) The carbon reduction process uses coke or coal as a reducing agent and the reduction is carried out by strong heating. It is used for reducing oxides of electropositive metals.

The lead (II) oxide is reduced by coke to yield lead along with carbon monoxide. This carbon monoxide formed will be further reduced to lead (II) oxide to produce lead.

$ \begin{gathered} \mathrm{PbO}+\mathrm{C} \rightarrow \mathrm{~Pb}+\mathrm{CO} \\ \mathrm{PbO}+\mathrm{CO} \rightarrow \mathrm{~Pb}+\mathrm{CO}_2 \end{gathered} $

The copper oxide is reduced by coke to produce copper.

$ \mathrm{Cu}_2 \mathrm{O}+\mathrm{C} \rightarrow 2 \mathrm{Cu}+\mathrm{CO}_2 $

Hence, option $B$ in column I matches with options P and R in column II.

(C) The process of extracting silver by complex formation and displacement by metal is known as Macarthur-Forrest process or cyanide process. In this process, the silver sulphide ore is treated with dilute solution of potassium cyanide or sodium cyanide.

Silver on treatment with sodium cyanide in presence of air will form a complex first known as sodium argentocyanide complex. The complex when heated with zinc metal, results in displacement of silver by zinc and producing silver in metallic form.

The silver from sodium argentocyanide is displaced by more reactive zinc.

$ \begin{aligned} 4 \mathrm{Ag}+8 \mathrm{CN}^{-}+2 \mathrm{H}_2 \mathrm{O} \xrightarrow{\mathrm{O}_2} 4\left[\mathrm{Ag}(\mathrm{CN})_2\right]^{-} +4 \mathrm{OH}^{-} \end{aligned} $

$2 \mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_2\right]+\mathrm{Zn} \rightarrow \mathrm{Na}_2\left[\mathrm{Zn}(\mathrm{CN})_4\right]+2 \mathrm{Ag} \downarrow$

This method is useful for ores containing less concentration of metal that needs to be extracted.

Hence, option C in column I matches with option Q in column II.

(D) The boron is extracted by the method of decomposition of iodide.

The thermal decomposition of boron iodide at very high temperature (approximately $700^{\circ} \mathrm{C}$ or more) will yield boron and iodine. The method is useful for the production of crystalline boron.

The boron iodide is not stable above $700^{\circ} \mathrm{C}$. Hence, boron is isolated from its iodide by this method.

$ 2 \mathrm{BI}_3 \xrightarrow{700^{\circ} \mathrm{C}} 2 \mathrm{~B}+3 \mathrm{I}_2 $

Hence, option D in column I matches with option S in column II.

During the process of electrolytic refining Ag and Au are obtained as anode mud.

NOTE : Galena is PbS and thus purified by froth floatation method.

Froath floatation method is used to concentrate sulphide ores. This method is based on the preferential wetting properties with the froathing agent and water.

Silver ore forms a soluble complex with $NaCN$ from which silver is precipitated using scrap zinc.

$A{g_2}S + 2NaCN \to $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Na\left[ {Ag{{\left( {CN} \right)}_2}} \right]\buildrel {Zn} \over \longrightarrow $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,N{a_2}\left[ {Zn{{\left( {CN} \right)}_4}} \right] + Ag \downarrow $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$sodargento-cynanide(soluble)

Pure aluminium can be obtained by electrolysis of a mixture containing alumina, crayolite and fluorspar in the ratio $20:24:20.$ The fusion temperature of this mixture is ${900^ \circ }C$ and it is a good conductor of electricity.