MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements

MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements

General Principles and Processes of Isolation of Elements NCERT Intext Exercises

Question 1.
Which of the ores mentioned in Table 6-1 ? Can be concentrated by mag-netic separation method ?
Answer:
Magnetic ores are separated from non-magnetic impurities by magnetic separa¬tion method. For example, magnetic (anion ore) is separated from non-magnetic silica and other impurities by this method.

Question 2.
What is the significance of leaching in the extraction of aluminium ?
Answer:
Main ore of aluminium is bauxite (Al2O3 xH2O). It contains the impurities of SiO2, FeO and Titanium oxide (TiO2). These impurities are removed by Leaching. The sig-nificance of leaching is to prepare pure alumina from bauxite. The powdered bauxite ore is heated with NaOH solution at 473 – 523K. Alumina dissolves as sodium meta aluminate whereas impurities of iron and titanium remains behind.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 1
The impurities are filtered off. The filtrate is neutralized by passing CO2. Aluminium hydroxide separates out while sodium silicate remains in the solution. Aluminium hydroxide on heating gives pure alumina.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 2

Question 3.
The reaction,
Cr2O3 + 3Al → Al2O3 + 2Cr; (∆°G = -421 kJ)
is thermodynamically feasible as is apparent from the Gibb’s energy value. Why does it not take place at room temperature ?
Answer:
All the reactants and products are solid at room temperature. Therefore, reaction does not occur at room temperature. At high temperature reactants melt and react.

Question 4.
Is it true that under certain conditions, Mg can reduce SiO2 and Si can reduce MgO ? What are those conditions ?
Answer:
Mg can reduce SiO2 below or at 1693K. Silicon can reduce MgO above 1693K.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 3

MP Board Solutions

General Principles and Processes of Isolation of Elements NCERT TextBook Exercises

Question 1.
Copper can be extracted by hydrometallurgy but not zinc. Explain.
Answer:
Copper is comparatively less acitve metal as its reduction potential i.e. E° (Cu2+/Cu) is high (+0-34V). It can be displaced from solution of Cu2+ ions by more active metals which have E° value lower than copper. For example, E° of zinc (Zn2+/Zn) is -0.76V and thus, zinc can displace copper from solution of Cu2+ ions. In contrast to displace zinc from solution of Zn2+ ions a more reactive metal than zinc is required like, Na, K, Mg, Ca, etc. But, the more active metals readily react with water forming their corresponding ions and evolve hydrogen gas.
[2Na + 2H2O → 2NaOH + H2],
Thus, it is difficult to displace zinc from solution of Zn2+ ions. Elence, copper can be extracted by hydrometallurgy but not zinc.

Question 2.
What is the role of depressant in froath floatation process ?
Answer:
The role of depressant is to prevent one type of sulphide ore particles from forming the froth with air bubbles. NaCN is used as a depressant to separate lead sulphide (PbS) ore from zinc sulphide (ZnS) ore. NaCN forms a zinc complex, Na[Zn(CN)4] on the surface of ZnS thereby preventing it from the formation of the froath.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 4

Question 3.
Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction ?
Answer:
The standard free energy (∆fG°) of formation of Cu2 S is more -ve than those of H2S and CS2. Therefore, Cu2S cannot be reduced by carbon or H2. The following two reactions do not occur. ∆rG° for these reactions is positive.
Cu2S + H2 → 2Cu + H2S;
2CU2S + C → 4Cu + CS2
In contrast ∆fG° of Cu2O is less -ve than that of CO and hence carbon can easily reduced Cu2O to Cu.
Cu2O(s) + C(s) → 2Cu + CO(g)
It is because of this reason that the extraction of copper from pyrite is difficult than from its oxide ore.

Question 4.
Explain: (i) Zone refining, (ii) Column chromatography.
Answer:
(i) Zone refining (Fractional crystallization): This method is used to prepare metals in extremely pure state. Boron, gallium, indium, silicon, germanium etc. are purified by this method. This method is based on the fact that when molten impure metal is allowed to cool, the pure metal crystallizes out whereas the impurities remain behind in the melt.

Method: A thin bar of impure metal is taken. One zone of bar is heated with a moving circular heater in the atmosphere of an inert gas. A narrow zone of the metal is melted. The pure metal crystallizes out of the melt while the impurities move into the adjacent molten zone.

The process is repeated several times till the impurities are completely driven to one end and the pure metal towards the other end. Then end of the rod where the impurities have collected is cut off. The metal obtained is extremely pure.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 6

(ii) Column chromatography: In column chromatography, a suitable adsorbent such as alumina (Al2O3) is packed in glass tube having a stop-cock near the bottom. This constitutes the stationary phase. The mixture to be separated is dissolved in some suitable solvents and added to the column. The different components of the mixture get adsorbed to different extent. Some suitable solvent (called eluent) is then added to the column. The eluent constitute the mobile phase. The weakly adsorbed component reaches the bottom of the column first.

It is followed by more strongly adsorbed components. Thus, different components of the mixture reach the bottom one by one and in this way get separated. This method is suitable for purification of those elements which are available in small quantities and the impurities are not much different in chemical behaviour from the element to be purified. Lanthanoides (rare earth elements) are purified by this technique.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 7

Question 5.
Out of C and CO, which is a better reducing agent at 673 K ?
Answer:
When carbon reacts with dioxygen, two reactions are possible
C(s) + O2(g) → CO2(g) ……(i)
2C(s) + O2(g) → 2CO(g) …..(ii)
When CO is used as a reducing agent, it gets oxidized to CO2
2CO + O2 → 2CO2 …(iii)
It is clear from the Ellingham diagram that at 673K the ∆G° for the oxidation of CO to CO2 is more negative than the reaction (i) and reaction (ii). Therefore, CO is better reducing agent than C. It is supported by the fact that the curve for the reaction (iii) lies below the curve for the reaction (i) and reaction (ii) at 673K. An element below in Ellingham diagram reduces the oxide of other metal which lies above it.

Question 6.
Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present ?
Answer:
The common elements present in the anode mud are Ag, Au, Pt, Sb, Se etc. These elements are less reactive and not affected by CuSO4 and H2SO4 solution and hence, settle down under anode as anode mud.

Question 7.
Write down the reactions taking place in different zones in the blast furnace during the extraction of iron.
Answer:
Reactions in blast furnace :
(i) Coke burns in presence of air to form CO2 in excess of CO is formed.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 8
(ii) CO reduces haematite ore (Fe2O3) to ferric oxide (FeO) at 600°C.
CO + Fe2O3 → 2 FeO + CO2
(iii) About 750°C CO reduces FeO into Fe.
FeO + CO → Fe + CO2
(iv) At 110° CaCO3 decomposes into CaO.
CaCO3 → CaO + CO2
This way the iron sobtained is called spongy iron.
(v) CaO reacts with silica present in ore and form slag.
CaO + SiO2 → CaSiO3
calcium silicate
Slag floats over molten iron from where it is removed through the tapping hole.
Labelled diagram of blast furnace :
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 9

Question 8.
Write chemical reactions taking place in the extraction of zinc from zinc blende.
Answer:
Zinc oxide is obtained from zinc blende, then zinc is extracted from zinc oxide. The whole process is described below :
The zinc blende (ZnS) ore is concentrated by froth floatation process, then roasted to give oxide.
2ZnS + 2O3 → 2ZnO + 2SO2
The zinc oxide is mixed with coke and clay to make bricketters. These bricketters are heated at 1673 K to give zinc. The temperature in this case is higher than that in case of copper.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 10
The metal is distilled off and collected by rapid chilling.
(Note : For detail, refer to Other Important Long Ans. Type Q. No. 7.)

Question 9.
State the role of silica in the metallurgy of copper.
Answer:
Silica acts as acidic flux in order to remove basic impurity of iron oxide.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 11

Question 10.
What is meant by the term “Chromatography” ?
Answer:
Chromatography is a technique for the separation and purification based on the difference in the adsorbing tendencies of the metal and its impurities on a suitable adsorbent. It is based on the principle that different compounds of a mixture are differently adsorbed on an adsorbent. The terms is originally derived from the Greek word ‘Chroma’ meaning colour and ‘graphy’ for writing because it is first used to the colour plant pigments.

Question 11.
What criterion is followed for the selection of the stationary phase in chro-matography ?
Answer:
The stationary phase is selected in such a way that it is capable of adsorbing the impurities more strongly than the elements to be purified. Under this condition, the impurities are retained by stationary phase i. e„ these cannot be eluted easily while the pure component, which is weakly adsorbed is easily eluted.

Question 12.
Describe a method for refining nickel.
Answer:
Mond process : Nickel is refined by this method. Impure nickel is heated in a current of CO at 330-350 K. Volatile nickel carbonyl Ni(CO)4 is formed while the impurities remain behind. Nickel carbonyl is now heated to 450-470 K. It decomposes to give pure nickel.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 12

Question 13.
How can you separate alumina from silica in bauxite ore associated with silica ? Give equations, if any.
Answer:
Main ore of aluminium is bauxite (Al2O3 xH2O). It contains the impurities of SiO2, FeO and Titanium oxide (TiO2). These impurities are removed by Leaching. The sig-nificance of leaching is to prepare pure alumina from bauxite. The powdered bauxite ore is heated with NaOH solution at 473 – 523K. Alumina dissolves as sodium meta aluminate whereas impurities of iron and titanium remains behind.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 13
The impurities are filtered off. The filtrate is neutralized by passing CO2. Aluminium hydroxide separates out while sodium silicate remains in the solution. Aluminium hydroxide on heating gives pure alumina.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 14

Question 14.
Giving examples, differentiate between ‘Roasting’ and ‘Calcination’.
Answer:
Difference between Roasting and Calcination :
Roasting
Heating ore in the presence of air below its melting point. This is generally done for sulphide ore.
Calcination
Heating ore either in the presence or limited supply of air below its melting point. This is generally done for carbonate ore.

Question 15.
How is ‘cast iron’ different from ‘pig iron’ ?
Answer:
Iron obtained from blast furnace is called pig iron. It contains about 4% carbon and other impurities of S, P, Si, Mn etc. When pig iron is mixed with scrap iron and coke and then heated in a blast of hot air, some impurities are removed. Cast iron is obtained. It contains 3% carbon and some other impurities. It is hard and brittle.

Question 16.
Differentiate between “Minerals” and “Ores”.
Answer:
Differences between Minerals and Ores :
Minerals

  1. Compounds of metals which contain less amount of metal and are found in in the earth’s crust with impurities are called mineral.
  2. 0It is difficult to obtain metals in pure form from minerals.
    Example : Mica, Feldspar, Limonite etc.

Ores

  1. Compounds of metals which contain sufficient amount of metal are known as ores.
  2. Metals can be obtained easily in pure form from the ores.
    Example: Ores of Aluminium, Cryolite, Bauxite, Alunite etc.

Question 17.
Why copper matte is put in silica lined converter ?
Answer:
Copper matte contains Cu2S and FeS. Copper matte is heated with silica to re-move the impurities of FeS as FeSiO3 (slag).
2FeS + 3O2 → 2FeO + 2SO2
FeO + SiO2 → FeSiO3
Slag

Question 18.
What is the role of cryolite in the metallurgy of aluminium ?
Answer:
Cryolite serve the following two purposes :

  1. It makes alumina a good conductor of electricity.
  2. It lowers fusion temperature of electrolyte.

Question 19.
How is leaching carried out in case of low grade copper ores ?
Answer:
The leaching of the low grade copper ores is carried out with acids in the presence of air when copper goes into solution as Cu2+ ions. Thus,
2Cu + 2H2SO4 + O2 → 2CUSO4 + 2H2O

Question 20.
Why is zinc not extracted from zinc oxide through reduction using CO ?
Answer:
In Ellingham diagram, the curve for the oxidation of CO to CO2 lies above the curve for the oxidation of Zn. Therefore, CO cannot reduce ZnO to Zn. On the other hand, the curve for the oxidation of C to CO, lies below the curve for the oxidation of Zn at temperature 1120K or above. Therefore, ‘C’ can be used to reduce ZnO at 1120K or above this temperature.

Question 21.
The value of ∆fG° for formation of Cr2O3 is -540 kJ mol-1 and that of Al2O3 is -827 kj mol-1. Is the reduction of Cr2O3 possible with A1 ?
Answer:
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 15
rG° of the above reaction is -ve. Therefore Cr2O3 can be reduced by A1 to free chromium metal.

Question 22.
Out of C and CO, which is a better reducing agent for ZnO ?
Answer:
Carbon is better reducing agent. It has been explained in Q. No. 20.

Question 23.
The choice of a reducing agent in a particular case depends on thermody-namic factor. How far do you agree with this statement ? Support your opinion with two examples.
Answer:
The thermodynamic factor helps us in choosing a suitable reducing agent for the reduction of a particular metal oxide to metal. The feasibility of thermal reduction can be predicted on the basis of ∆fG° vs T plots for the formation of oxides, known as Ellingham diagram. From the diagram, it can be predicted that metals for which ∆fG° oxides is more negative can reduce those metal oxides for which the ∆fG°, oxides is less negative. In other words, one will reduce the oxides of other metals which lie above in Ellingham diagram, because ∆rG° of the combined redox reaction will be -ve by an amount equal to the difference in ∆fG° of the two metal oxides. For example, Al can reduce Cr2O3 but not MgO. Similarly C can reduce ZnO to Zn but not CO. Thus, the choice of a particular reducing agent depends upon thermodynamic factor.

Question 24.
Name the processes from which chlorine is obtained as a by-product. What will happen if an aqueous solution of NaCl is subjected to electrolysis ?
Answer:
Sodium metal is prepared by Down’s process. It involves the electrolysis of a fused mixture of NaCl and CaCl2 at 873 k
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 16
If however, an aqueous solution of NaCl is electrolysed, H2 is evolved at the cathode while Cl2 is obtained at anode. The reason being \(\mathrm{E}_{\mathrm{Na}^{+} / \mathrm{Na}}^{\circ}\), (-2.71) is much lower than \(\mathrm{E}_{\mathrm{H}_{2} \mathrm{O} / \mathrm{H}_{2}}^{\circ}\) (-0.832) and hence water is reduced to H2 in preference to Na+ ions.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 17

Question 25.
What is the role of graphite rod in the electrometallurgy of aluminium ?
Answer:
In the electrolytic reduction of alumina, graphite anode facilitates reduction of Al2O3 to aluminium by electrolysis. Carbon reacts with oxygen liberated at anode producing CO and CO2.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 18

Question 26.
Outline the principles of refining of metals by the following methods:
(i) Zone refining
(ii) Electrolytic refining
(iii) Vapour phase refining.
Answer:
(i) Zone refining (Fractional crystallization): This method is used to prepare metals in extremely pure state. Boron, gallium, indium, silicon, germanium etc. are purified by this method. This method is based on the fact that when molten impure metal is allowed to cool, the pure metal crystallizes out whereas the impurities remain behind in the melt.

Method: A thin bar of impure metal is taken. One zone of bar is heated with a moving circular heater in the atmosphere of an inert gas. A narrow zone of the metal is melted. The pure metal crystallizes out of the melt while the impurities move into the adjacent molten zone.

The process is repeated several times till the impurities are completely driven to one end and the pure metal towards the other end. Then end of the rod where the impurities have collected is cut off. The metal obtained is extremely pure.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 19

(ii) Electrolytic refining : This method is based upon the phenomenon of electroly-sis. The crude metal is made anode whereas the thin sheet of pure metal is made cathode.

Electrolyte is dissolved in the solution of some salt of the metal. On passing electricity the metal from the anode goes into the solution as ions due to oxidation while pure metal gets deposited at the cathode due to reduction of metal ions. The less electropositive impurities settle down below the anode as anode mud.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 20

(iii) Vapour phase refining : The crude metal is heated with a specific reagent to obtain a volatile compound at lower temperature. The volatile compound is then heated at a higher temperature. The volatile compound decomposes at the higher temperature to give the metal Ni, Ti, Zr etc. are purified by this method.

(a) Mond process : Nickel is refined by this method. Impure nickel is heated in a current of CO at 330-350 K. Volatile nickel carbonyl Ni(CO)4 is formed while the impurities remain behind. Nickel carbonyl is now heated to 450-470 K. It decomposes to give pure nickel.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 21
(b) Van Arkel method: By this method ultrapure metals are obtained. In this method, the metal is converted into a volatile unstable compound (e.g., iodide). The impurities re main behind. The unstable volatile compound is collected and it is then decomposed to get the pure metal. Titanium and Zirconium are purified by this method. These metals are used in space technology.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 22

Question 27.
Predict conditions under which Al might be expected to reduce MgO.
Answer:
Above 1623K, Al can reduce MgO in Mg.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 23
Al2O3 is more stable than MgO, hence Al reduces MgO above 1623K.

MP Board Solutions

General Principles and Processes of Isolation of Elements Other Important Questions and Answers

General Principles and Processes of Isolation of Elements Objective Type Questions

Question 1.
Choose the correct answer :

Question 1.
Cryolite is mixed during the electrolysis of alumina because it:
(a) Decreases the m.p. of alumina
(b) Decreases electrical conductance
(c) Separates the impurity of alumina
(d) Decreases anode effect.

Question 2.
Which of the following metal is preserve by the film of its own oxide :
(a) Ag
(b) Fe
(c)Cu
(d) Al.

Question 3.
Which method is applicable for the extraction of Al from bauxite :
(a) Reduction by Mg
(b) Reduction by coke
(c) Electrolytic reduction
(d) Reduction by Fe.

Question 4.
The bauxite containing impurity of iron oxide is purified by :
(a) Hoop’s method
(b) Serpeck’s process
(c) Bayer’s process
(d) Electrolytic process.

Question 5.
Blister copper is:
(a) Ore of copper
(b) Alloy of copper
(c) Pure copper
(d) Copper containing 1% impurities.

Question 6.
In blast furnace iron oxide is reduced by :
(a) SiO2
(b) CO
(c) C
(d) CaCO3.

Question 7.
In the extraction of Fe from haematite, lime stone acts as :
(a) Reducing agent
(b) Slag
(c) Gangue
(d) Flux.

Question 8.
Cupellation is used in the metallurgy of:
(a) Cu
(b) Ag
(c) Al
(d) Fe

Question 9.
Which is the essential part of photographic plates and films :
(a) AgNO3
(b)Ag2S2O3
(c) AgBr
(d) Ag2CO3.

Question 10.
The compound obtained by the reaction of Zn with the excess of caustic soda :
(a) Zn(OH)2
(b) ZnO
(c) Na2ZnO2
(d) ZnH2.

Question 11.
Calomel is:
(a) Hg2Cl2
(b) HgCl2
(c) Hg2Cl2 + Hg
(d) Hg + HgCl2.

Question 12.
Is formed on adding potassium iodide solution in excess amount of mercuric iodide:
(a) Hg2Cl2
(b) K2HgI4
(c) Hg
(d) Hg + KI3.

Question 13.
The colour obtained by the mixing of excess of KI in to HgCl2 solution:
(a) Orange
(b) Brown
(c) Red
(d) Colourless.

Question 14.
Which metal does not form amalgum :
(a) Zn
(b) Cu
(c) Mg
(d) Fe.

Question 15.
Malachite is:
(a) Cu2S
(b) CUCO3.CU(OH)2
(c) Cu2O
(d) CuCO3.

Question 16.
AgBr is soluble in hypo because its forms :
(a) Ag2SO3
(b) Ag2S2O3
(c) [Ag(S2O3)]
(d) [Ag(S2O3)2]3-

Question 17.
AgCl is soluble in ammonia due to the formation of:
(a) [Ag(NH3)4]+
(b) [Ag(NH3)2]2+
(c) [Ag(NH3)4]3+
(d) [Ag(NH3)2]+.

Question 18.
When KI mixed with solution of CuSO4, it form :
(a) Cul2
(b) Cul-22
(c) K2[CUI4]
(d) Cu2I2 + I2.

Question 19.
When KCN reacts with solution of CuSO4, it forms :
(a) CU(CN)2
(b) CuCN
(c) K2[Cu(CN)4]
(d) K3[Cu(CN)4]

Question 20.
Na2S2O3 used in photography as a :
(a) Reducing agent
(b) Developer
(c) Fixer
(d) Tonning agent.

Answers:
1. (a), 2. (d), 3. (c), 4. (c), 5. (d), 6. (b), 7. (d), 8. (b), 9. (c), 10. (c), 11. (a), 12. (b), 13. (c), 14. (d), 15. (b), 16. (c), 17. (d), 18. (d), 19. (d), 20. (c).

MP Board Solutions

Question 2.
Fill in the blanks :

  1. Malachite is an ore of ……………..
  2. In stainless steel, along with iron …………….. and …………….. metals form alloys.
  3. …………….. is used as a purgative.
  4. Colloidal solution of …………….. is used as a medicine of eyes.
  5. AgNO3 is known as ……………..
  6. Chemical formula of corrosive sublimate is ……………..
  7. Froath floatation process is generally employed for …………….. ores.
  8. The process in which metal oxide is reduced by A1 is known as ……………..
  9. Is used for drying ammonia ……………..
  10. …………….. is called lunar caustic.
  11. The chemical formula of fluorspar is ……………..
  12. Alkaline solution of HgCl2 and KI is known as ……………..
  13. Red hot steel is slowly cooled when it gets converted to soft steel, this is known as ……………..

Answers:

  1. Cu
  2. Cr, Ni
  3. Calomel
  4. Ag
  5. Lunar caustic
  6. HgCl2
  7. Sulphide
  8. Alum- inothermic
  9. CuO
  10. Silver Nitrate
  11. CaF2
  12. Nesseler’s reagent
  13. Annealing.

Question 3.
Match the following :

I.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 24
Answers:

  1. (g)
  2. (d)
  3. (c)
  4. (e)
  5. (b)
  6. (a)
  7. (f).

II.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 25
Answers:

  1. (d)
  2. (f)
  3. (b)
  4. (e)
  5. (c)
  6. (a).

III.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 26
Answers:

  1. (d)
  2. (e)
  3. (f)
  4. (c)
  5. (b)
  6. (a).

Question 4.
Answer in one word / sentence :

  1. The mixture of Cu2S and FeS obtained from blast furnace is known as.
  2. Which metal is purified by polling?
  3. After developing which solution is used for the fixing of photographic films?
  4. Chemical formula of philosophers’s wool is.
  5. Chemical formula of horn silver.
  6. Which compound is normally used for toning in photography?
  7. Which are known as coinage metals?
  8. Give the name of ore used for extraction of Cu.
  9. Give the name of ore used for extraction of Iron.
  10. What is the name of graph which is drawn between the absolute temperature and standard free energy change for formation of metal oxide?
  11. What is the name of iron obtained after the removal of impurities from cast iron?
  12. What is the method of slow cooling of hot hard steel known as?
  13. What is the method of heating hard steel known as?
  14. What iš the method of heating steel in pressure of ammonia known as?
  15. What is Lunar Caustic?

Answers:

  1. Matte
  2. Copper
  3. Hypo solution (Na2S2O3)
  4. ZnO
  5. AgCl
  6. Aurric chloride
  7. Cu, Ag and Au
  8. Copper pyrite
  9. Haematite
  10. Ellinghum diagram
  11. Wrought iron
  12. Annealing
  13. Softening
  14. Nitriding
  15. AgNO3.

MP Board Solutions

General Principles and Processes of Isolation of Elements Very Short Answer Type Questions

Question 1.
Cryolite is added during the electrolysis of alumina to obtain aluminium. Give reason.
Answer:
Melting point of Alumina is very high (2050°C). By adding cryolite, melting point of alumina is lowered as well as electrolysis becomes easier.

Question 2.
When Al is placed in contact with cone. HNO3, then no reaction seems to take place. Why ?
Answer:
Since, Al reacts with HNO3 to form Al2O3 which forms a protective layer on the metal by which reaction do not proceed further.

Question 3.
Which compound of A1 is a good reducing agent ?
Answer:
Complex hydride Li[AlH4] of Al is mainly a good reducing agent for organic compounds.

Question 4.
Write the chemical formula of the following :

  1. Calomel
  2. Lunar caustic
  3. Philosopher’s wool
  4. Corrosive sublimate.

Answer:

  1. Calomel: Mercurous chloride (Hg2Cl2)
  2. Lunar caustic : Silver nitrate (AgNO3)
  3. Philosopher’s wool: Zinc oxide (ZnO)
  4. Corrosive sublimate : Mercuric chloride (HgCl2).

Question 5.
Write the main ores of copper.
Answer:
Main ores of copper are as follows :
1. Oxide: Cuprite or Ruby copper (Cu2O)
2. Carbonate:Azurite – 2CuCO3.Cu(OH)2
Malachite – CuCO3.Cu(OH)2 or CuCO3.CuO.H2O
3. Sulphide: Copper pyrites or Calcopyrite – Cu2S + Fe2S3 or CuFeS2
Copper glance or Calcocite – Cu2S.

Question 6.
Give the chemical name, formula and two application of lunar caustic.
Answer:
Chemical name of lunar caustic : Silver nitrate (AgNO3)
Applications : It is used for 1. Silvering of mirror. 2. Making marking inks and hair dyes.

MP Board Solutions

General Principles and Processes of Isolation of Elements Short Answer Type Questions

Question 1.
Explain following process of steel:
(a) Annealing
(b) Hardening
(c) Tempering
(d) Nitriding.
Answer:
(a) Annealing: It is process of heating steel to redness followed by slow cooling due to this steel becomes soft and pliable.
(b) Hardening or Quenching: It is a process of heating steel to redness followed by sudden cooling by plunging the red hot steel into water or oil. Steel becomes brittle and hard.
(c) Tempering : It is a process of heating the hardened or quenched steel to a temperature much below redness (200-350°C) followed by slow cooling steel loses its brittleness completely or partially depending upon time and temperature.
(d) Nitriding : Formation of a layer of iron nitride by heating steel at 500-600°C in chamber of ammonia for 3-4 days is called nitriding. Steel becomes harder by this treatment.

Question 2.
Write method of preparation of copper sulphate (Blue vitriol) and its use.
Answer:
Method of preparation of copper sulphate :
1. By heating copper with dilute H2SO4 in the presence of air.
2Cu + 2H2SO4 + O2 → 2CUSO4 + 2H2O

2. By the action of H2SO4 on copper oxide, hydroxide or carbonate :
CuO + H2SO4 → CuSO4 + H20
CU(OH)2 + H2SO4 → CuSO4 + 2H2O
CuCO3 + H2SO4 → CuSO4 + H2O + CO2
Uses : (1) In dying and printing textile, (2) As a bacteriocidal, (3) As an insectiside, (4) In electrical cells.

Question 3.
Give the chemical reaction when CuSO4 solution reacts with :
(i) NaOH solution
(ii) NH4OH
(iii) KI solution
(iv) KCN.
Answer:
(i) It forms light blue precipitate of cuprichydroxide.
CuSO4 + 2NaOH → Cu(OH)2 + Na2SO4
(ii) It forms deep blue colour solution which is due to the formation of cupric ammonium sulphate complex.
CuSO4 + NH4OH → [CU(NH3 )4 ]SO4 + 4H2O
(iii) It liberates iodine.
CuSO4 + 2 KI → K2SO4 + Cul2
2CUI2 → I2 + Cu2I2
(iv) It forms potassium cupracyanide.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 27

Question 4.
What is blue vitriol ? What is the effect of heat on it?
Answer:
Commercial name of blue vitriol is blue vitriol which contain five molecules of water of crystallization. ¡t is a blue coloured crystalline solid soluble in water.

Effect of heat : On heating 100°C it loses 4 molecules of water of crystallization. At 230°C, it loses 5th molecules of water and becomes white amorphous powder. The white powder again becomes blue in presence of water. It undergoes dissociation on strong heating.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 28

Question 5.
What are the main ores of Al ? How can impure aluminium ¡s purified?
Answer:
Main ores of Al:
I. Bauxite – Al2O3.2H2O
2. Felspar – K2O.Al2O3.6H2O
3. Cryolite – Na3AIF6
Electrolytic refining of aluminium (Hoope’s electrolytic process): Aluminium is further purified by Hoope’s process. Process is carried out in an iron cell lined inside with carbon. The cell consist of three layers which differ in specific gravities. The lowest layer acting as anode consists of impure aluminium. A mixture of fluorides of Al, Ba and Na forms middle layer and acts as electrolyte. The uppermost layer is of pure aluminium and it acts as anode.

Following changes occur on passing electric current:
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 29
On electrolysis, aluminium ions of the electrolyte are discharged at cathode and pure aluminium collects as the uppermost layer. An equivalent amount of aluminium is taken up by the middle layer from the bottom layer. Thus, aluminium is transferred from bottom layer to the upper layer through the middle layer while impurities are left behind. Crude aluminium is added from time to time to continue the process. Thus 99-98% pure aluminium is obtained.

Question 6.
Differentiate between cast iron, wrought iron and steel.
Answer:
Differences between Cast Iron, Wrought Iron and Steel:
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 30

Question 7.
Give names and formula of any two ores of following metals :
(i) Al
(ii) Zn
(iii) Fe
(iv) Cu
(v) Ag.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 31

Question 8.
Write four different chemical reactions of copper with nitric acid. Give equations also.
Answer:
Chemical reaction of Cu with HNO3:
(a) With cold and very dilute nitric acid, nitrogen gas is evolved.
5Cu + 12HNO3 → 5CU(NO3)2 + 6H2O + N2

(b) With hot and dilute Nitric acid, nitrous oxide is obtained.
4CU + 10HNO3 → 4CU(NO3)2 + N2O + 5H2O

(c) With 50% HN03, Nitric oxide is formed.
3Cu + 8HNO3 → 3CU(NO3)2 + 2NO + 4H2O

(d) With cone. HNO3, Nitrogen peroxide is obtained.
CU + 4HNO3 → CU(NO3)2 + 2NO2 + 2H2O.

Question 9.
Write the name of three alloys of copper and give the composition and one use of each.
Answer:
Alloys of copper, their composition and uses are given ahead :
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 32

Question 10.
Explain why a bulb placed in parallel series to the electrical circuit for extraction of Al from Al2O3 ?
Answer:
In the extraction of Al form Al2O3 by electrolysis. Al2O3 converted into Al with decrease in concentration of electrolyte bulb placed in parallel glass indicating that cell resistance has increased. Therefore, more alumina is added at this stage and carbon rods are also replaced.

Question 11.
Explain the purification of bauxite with chemical equations.
Answer:
Baeyer’s process applied for the purification of bauxite ore containing ferric oxide as chief impurity. This ore is usually red in colour and hence called red bauxite.

The powdered ore is roasted first to convert ferrous oxide. If any, to ferric oxide and then digested with a concentrated solution of sodium hydroxide under pressure at 150°C for several hours. Alumina dissolves in sodium hydroxide forming soluble sodium meta aluminate (NaAlO2) while ferric oxide and silica being insoluble settles down and are removed by filtration.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 33
Sodium meta aluminate is then agitated with freshly precipitated Al(OH)3 for 36 hours. As a result sodium meta aluminate gets hydrolysed forming precipitate of aluminium hydroxide.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 34
The precipitate is washed, dried and ignited to get pure alumina while solution of NaOH is concentrated and used again.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 35

Question 12.
Give equation for the Auto-reduction process of copper pyrite to obtain pure copper.
Answer:
Molten matte obtained by smelting is mixed with sand and taken in a Bessemer converter and air is blown through tuyers when FeS gets converted to FeO which reacts with SiO2 to form slag (FeSiO3) and some part of Cu2S gets oxidized to Cu2O.
2FeS + 3O2 → 2FeO + 2SO2
2FeO + 2SiO2 → 2FeSiO3
2Cu2S + 3O2 → 2Cu2O + 2SO2
Which reacts with the remaining Cu2S and gets reduced to metallic copper. This is known as Auto-reduction.
2CU2O + Cu2S → 6Cu + SO2

Question 13.
How many type of iron is found ? Give the name of each and two characteristics of each.
Answer:
Iron is of three types : 1. Cast Iron, 2. Wrought Iron, 3. Steel.
Characteristics 1. Cast Iron : (i) It consists of 2-5 to 4-5% carbon.
(ii) It is very hard and its melting point is 1200°C.

2. Wrought Iron : (i) It consists of 0-12 to 0-25% carbon.
(ii) It is soft and its melting point is 1500°C.

3. Steel: (i) It consists of 0-5 to 1 -5% carbon.
(ii) It is normally hard and melting point is nearly 1300°C.

Question 14.
Describe the purification method of bauxite having silica as impurity.
Answer:
Serpeck’s process : This process is used when bauxite ore contains consider-able amount of silica as impurity. In this process ore mixed with coke is heated to 1800°C in presence of nitrogen. Alumina gets converted into aluminium nitride, AIN.
Al2O3 + 3C + N2 → 2 AIN + 3CO
Silica gets reduced to silicon which volatilises at this temperature.
SiO2 + 2C → Si + 2CO
AIN is hydrolysed to aluminium hydroxide by water.
AIN + 3H2O → Al(OH)3 + NH3
As usual aluminium hydroxide is converted into alumina.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 36

MP Board Solutions

General Principles and Processes of Isolation of Elements Long Answer Type Questions

Question 1.
Write note on photography.
Answer:
The process of producing an exact impression of an object on an art paper by using light is called photography.
Modern photography involves following steps :

(i) Preparation of sensitive plate or film : A paste of colloidal silver bromide or silver iodide in gelatin is the sensitive emulsion. It is prepared by adding ammonium silver nitrate solution to ammonium bromide solution containing gelatin.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 37
This is allowed to stand in a warm room for sometime when silver bromide particle grow in size and at the sametime becomes more sensitive to light.

(ii) Exposure : The sensitive plate or film is mounted in a camera and exposed for a few second to the image of properly focussed object. The light reduces silver bromide to extremely small particles of silver.
2 AgBr → 2Ag + Br2
Inverted image of the object which is formed on the plate is not visible and is therefore called latent image.

(iii) Developing : The exposed film or plate is immersed in a solution of developer. Developer is a weak reducing agent such as potassium ferrous oxalate or an alkaline solu-tion of organic reducing agents like pyrogallol or quinol.

When the plate dipped in developing bath, the parts affected by light are reduced to maximum extent while part not affected by light remain unattached by the developer.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 38
The plate is therefore called negative plate.

(iv) Fixing : Sodium thiosulphate is used in fixing negative plates. Negative plate is washed and dipped in a hypo, it dissolves unaffected silver bromide and leaves metallic silver unchanged.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 39
(v) Printing (Printing out paper or P.O.P.): Coated with silver chloride and silver citrate mixture. The process is slow and can be watched and controlled when the image of required shade is obtained, it is fixed and toned.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 40

Question 2.
What is Blister copper ? Describe with labelled diagram of its purification by electrolysis.
Answer:
Copper is removed by inverting Bessemer converter. The metal thus obtained contains sufficient quantity of SO2. On cooling, SO2 comes out from the metal in the form of bubbles producing small blisters on the metal surface. The metal is 98% pure and is called Blister copper.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 41
Purification of copper by electrolysis : Mixture of 15% CUSO4 and 5% H2SO4 is taken in a big tank which is fitted with thick plates of copper which act as anode. Plates of pure copper act as cathode. On passing electric current the impure metal dissolves at anode and goes into solution and pure metal from solution starts depositing at cathode. The impurities of Pt, Au and Ag settle down as anode mud. The impurities of Ni, Fe, Zn remain in the solution. Metal of 99-99% purity is obtained by this method.

Question 3.
Explain Siemen-Martin open hearth process for the manufacture of steel.
Answer:
Siemen-Martin’s open hearth process : The mixture of cast iron (30 – 50%), scrap iron haematite (Fe2O3) and lime is melted in the open hearth furnace which has a basic or acidic lining depending upon impurities. Fe2O3 acts as oxidizing agent in the formation of steel and carbon gets oxidized to CO and escapes
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 42
Fe2O3 + 3C → 3CO + 2Fe
2Fe2O3 +3S → 4Fe + 3SO2
10F22O3 + 3P4 → 20Fe + 6P2O5
2Fe2O3 + 3Si → 4Fe + 3SiO2
Fe2O3 + 3Mn → 2Fe + 3MnO
Now steel containing required amount of carbon is obtained by addition of spiegel.

Advantages :

  1. As the process proceeds slowly therefore, its control is easy.
  2. Better quality of steel can be manufactured.
  3. The composition and temperature can be easily controlled.
  4. As no blast of air is passed therefore, very little steel is lost.
  5. The steel can be directly obtained by the use of ore and scrap iron.

Question 4.
Write the names and formula of two ores of silver and describe the method of extraction of Ag from any ore.
Or
Write the names and formula of the two ores of silver and describe the cyanide method for the extraction of Ag.
Answer:
Two ores of silver: (i) Argentite or Silver glance – Ag2S.
(ii) Horn silver or Silver chloride – AgCl.

Extraction of silver by cyanide process (Mac Arthur Forest Process):
Silver metal is extracted from the Argentite ore (Ag2S) by cyanide process.
It is a modern process and based on the following principle :

  1. Silver reacts with the cyanides of alkal metals and forms a soluble complex.
  2. Impurities do not react with alkali cyanides.
  3. More electropositive metals (like zinc and aluminium) displace silver from its complex cyanides.

The process involves the following steps :
1. Crushing and concentration: The ore is crushed and then finely powdered in ball mills. It is then subjected to the froath floatation process for concentration.

2. Treatment of the ore with sodium cyanide : The concentrated ore is treated with a dilute solution (about 0-5%) of sodium cyanide for several hours. The solution is continuously agitated by passing a current of air. Silver sulphide goes into the solution in the form of soluble complex.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 43
The reaction is reversible. Hence, Na2S must be removed so that the reaction may proceed in the forward direction.
The air blown in oxidizes the sulphide formed to thiosulphate and sulphate and thus enables the reaction to proceed in the forward direction.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 44

3. Precipitation of silver : The above solution is filtered to remove insoluble impurities. It is then treated with zinc dust. Silver being less electropositive, gets displaced by more electropositive zinc and is precipitated.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 45
Uses of silver glance :
(i) In silver plating on metallic articles e.g., tableware and silvering of mirrors
(ii) Used in medicine.

Question 5.
Explain the composition, uses of alloys of :
(i) Aluminium
(ii) Copper
(iii) Iron.
Answer:
Composition and uses of alloys of given elements :
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 46
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 47

MP Board Solutions

Question 6.
Write the principle of extraction of copper from copper pyrite.
Or, Explain extraction of copper under the following heads :
(a) Smelting
(b) Bessemerization.
Answer:
Principle : Copper pyrite is mixed ore of copper and iron. As sulphur has got more affinity for copper. Therefore on heating the ore first iron oxide is formed which is removed completely and then sulphur is extracted.
Process involves following steps :
(i) Concentration : By froath floatation process, powdered ore is taken in tank containing water and palm or eucalyptus oil. Air stream is passed, sulphide ore form froth which can be removed and impurity settles down.

(ii) Roasting : Concentrated ore is heated in the presence of air in a reverberatory furnace, sulphide ore is converted into oxides.
2CuFeS2 + O2 → 2FeS + Cu2S + SO2
2Cu2S + 3O2 → 2CU2O + 2SO2
2FeS + 3O2 → 2FeO + 2SO2

MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 48

(iii) Smelting : Roasted ore is mixed with coke sand and acidic flux and smelted in blast furnace of 15-20ft height following reaction takes place:
Cu2O + FeS → Cu2S + FeO
FeO + SiO2 → FeSiO3 (Slag)
flux slag
At last matte is obtained which is a mixture of cuprous sulphide and small quantity of ferrous sulphide.

(iv) Bessemerization : Matte is mixed with little sand and melted in oval shape Bessemer converter lined with CaO or MgO. Here cuprous sulphide first converted into cuprous oxide which on reduction (autoreduction) gives copper.
2FeS + 3O2 → 2FeO + 2SO2
FeO + SiO2 → FeSiO3(slag)
2Cu2S + 3O2 → 2Cu2O + 2SO2
Cu2S + 2Cu2O → 6Cu + SO2
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 49
Metal obtained contains SO2 on cooling SO2 comes out from the metal in the form of bubbles producing blisters on the metal surface called blister copper.

(v) Purification (By electrolytic method) : Thin strip of pure copper is taken as cathode and thick strip of impure copper is taken as anode. CuSO4 as electrolyte when electric current is passed pure copper get deposited at cathode.

Question 7.
What are the main ores of Zn ? Describe the vertical retort method for the extraction of Zn.
Or,
Describe the modern method for the extraction of Zn.
Answer:
Main ores of Zn:
(i) Zinc blende – ZnS
(ii) Calamine – ZnCO3
(iii) Willemite – ZnSiO4.
Extraction : Zinc is extracted from its ore by two methods :
1. Reduction process
2. Electrolytic process.

1. Reduction Process : It involves the following steps :
(i) Concentration : Zinc blende is concentrated by froath floatation process while calamine ore is concentrated by gravity separation process. Iron oxide if present in ore is removed by magnetic separation.
(ii) Roasting : Concentrated ore is heated at about 900°C in excess of oxygen. Zinc sulphide is oxidised to zinc oxide.
2ZnS + 3O2 → 2ZnO + 2SO2
A portion of ZnS is also oxidised to zinc sulphate but it decomposes at 900°C into ZnO
ZnS + 2O2 → ZnSO4
2ZnSO4 → 2ZnO + 2SO2 + O2
In case of calamine ore, calcination is done instead of roasting.
ZnCO3 → ZnO + CO2
(iii) Reduction : Roasted or calcined ore is mixed with coke and heated to 1400°C. Ore gets reduced into the metal.
ZnO + C → Zn + CO.

Reduction can be done by the following method :
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 50
Vertical retort process : Process carried out in vertical retort is continuous. Vertical retorts are made up of highly refractory silicon carbide bricks capable of with standing high temperature. Heating is done externally using producer gas. A screw conveyer is provided in the extension at the bottom for the removal of ash. Just near the top an outlet is there. It is connected with condenser. Roasted ore and coke is fed into the retort through the charging door. Vapour of zinc along with CO passes into the condenser where zinc condenses while CO escaping out is used as fuel for heating the furnace. Zinc is removed periodically from the condenser.

Question 8.
What happens when (Give only equations):
(i) Blue vitriol is heated.
(ii) Ammonium hydroxide is added to silver nitrate solution.
(iii) Mercuric chloride is added to stannous chloride solution.
(iv) Copper reacts with hot and concentrated sulphuric acid.
(v) Mercuric chloride reacts with potassium iodide.
(vi) Silver nitrate reacts with hydrochloric acid.
(vii) AgNO3 is heated.
Answers:
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 51

Question 9.
Explain with labelled diagram the method of electrolysis to obtain pure aluminium metal.
Answer:
Electrolysis of alumina (Hall-Heroult process) : Electrolysis of alumina is done in an iron tank fitted with a lining of gas carbon which acts as cathode. Graphite rods act as anode and a bulb is fitted in parallel as indicator for alumina content. The electrolyte is a mixture of alumina and cryolite (Na3AlF6). The melting point of alumina is very high (2050°C). But, in presence of cryolite (Na3AlF6) and fluorspar (CaF2), it melts at 870°C. Then pure alumina is added to the tank which dissolves in cryolite mixture.
MP Board Class 12th Chemistry Solutions Chapter 6 General Principles and Processes of Isolation of Elements 52
Reaction : Reactions take place as below :
Na3AlF6 → 3NaF + AlF3
AlF3 → Al+3 +3F
3F → 3e + 3F (oxidation) at anode
Al3+ + 3e → Al (reduction) at cathode
2Al2O3 +12F → 4AlF3 + 3O2
The reaction proceeds continuously. When amount of Al2O3 decreases resistance of electrolyte increases and the bulb in the parallel glows up. Therefore at this junction more alumina is added. The Al liberated at cathode collects at the bottom of the tank in molten state which is taken out through the exit. Coke is added to the mixture of cryolite and alumina which does not allow it to cool and further protects the eyes from light. 99% pure aluminium metal is obtained by this process.

MP Board Class 12th Chemistry Solutions