Chapter 15.2 - Some important Organic compounds

In the previous section, we completed a discussion on combustion  and thermal cracking. In this section we will see some important organic compounds.

Alcohols

• We have learned the nomenclature details about alcohols in the previous chapter. 
• They contain the OH functional group.  
• CH₃ ㅡOH (Methanol) and  CH₃ ㅡCH₂ ㅡOH (Ethanol) are two important alcohols.
• Methanol is commonly known as wood spirit
• Ethanol is commonly known as grape spirit
■ First we will see methanol in detail:
Uses of methanol
1. Methanol is used as a solvent in the manufacture of paint
2. It is also used as a reactant in the manufacture of varnish and formalin
• So it is a very valuable industrial substance. It is very poisonous too.

Industrial preparation of methanol
• Methanol is industrially prepared by the reaction between carbon monoxide with hydrogen in the presence of catalysts. The equation is shown below:

Note that 3 items are required for the reaction:
(i) Catalyst
(ii) Temperature of 573 k. (Details about the units of temperature can be seen here)
(iii) Pressure of 200 atm (Details about pressure can be seen here)

■ Next we will see ethanol:
Uses of ethanol
1. Ethanol is used as an organic solvent 
2. It is used in the manufacture of various organic compounds and paints. 
3. It can also be used as a fuel by itself or in combination with other compounds.

Industrial preparation of ethanol

1. During the manufacture of sugar, we get a liquid mixture of sugar crystals. 
• That is., a liquid containing crystals of sugar
2. When the solid sugar crystals are removed, a liquid is left behind. 
• This liquid is a concentrated solution of sugar. 
• This concentrated solution of sugar is called molasses. 
3. This molasses is diluted and then fermented. 
• Fermentation is done by adding yeast. The process of fermentation lasts for a few days. 
• During the fermentation, some reactions occur due to the presence of enzymes in the yeast. 
• The enzymes are: Invertase and Zymase. 
The reactions are shown below:

4. The reaction takes place in stages:
• First the sugar is converted into glucose and fructose. 
    ♦ This is accomplished with the help of the enzyme invertase
    ♦ Note that glucose and fructose have the same molecular formula. 
    ♦ But they have different structural formulae
• Then the gucose and fructose are converted into ethanol. 
    ♦ This is accomplished with the help of the enzyme zymase
5. The ethanol thus obtained will be about 8-10% strong. 
• That is., if there are a total of 100 molecules in a sample, only 8 to 10 of them will be ethanol molecules. 
• This ethanol is called wash. 
6. Wash is subjected to fractional distillation. Some basics about fractional distillation can be seen here.
• By this process, we get 95.6% strong ethanol solution. It is known as rectified spirit.
7. Rectified spirit is a very important industrial substance. So it would be produced on a large scale.
• This rectified spirit meant for industrial purpose should not be used as a beverage. 
• To prevent it's use as a beverage, poisonous substances are added to those ethanol which are meant for industrial purposes. 
• This ethanol is called denatured spirit.
8. 99.5% ethanol is known as absolute alcohol. 
• A mixture of absolute alcohol and petrol is known as power alcohol. It is used as fuel in automobiles.
9. What we saw above is the production of ethanol from molasses. 
• Ethanol can be manufactured from starchy substances like barley, rice, tapioca etc., also.

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Carboxylic acids

These are compounds containing carboxylic acid group (ㅡCOOH). We have seen their nomenclature details in the previous chapter. Some examples are:
1. IUPAC name: Methanoic acid
• Common name: Formic acid
• Formula:  H ㅡCOOH 
• Structural formula: See fig.15.20(a) below 
2. IUPAC name: Ethanoic acid 
• Common name: Acetic acid
• Formula:  CH₃ ㅡCOOH 
• Structural formula: See fig.15.20(b) below
3. IUPAC name: Propanoic acid
• Common name: Propionic acid
• Formula:  CH₃ ㅡCH₂ ㅡCOOH 
• Structural formula: See fig.15.20(c) below

Fig.15.20

• Many substances obtained from nature contains carboxylic acids. For example:
• The product obtained as a result of the fermentation of fruits will contain Ethanoic acid (Acetic acid)
• About 5-8% strong ethanoic acid is called vinegar.
• Carboxylic acids are usually prepared by the oxidation of alcohols. 
    ♦ Vinegar can be prepared from ethanol in this way
• Carboxylic acids containing 12 or more carbon atoms are called fatty acids.

Industrial preparation of Ethanoic acid

• Ethanoic acid can be manufactured by treating methanol with carbon monoxide in the presence of catalyst. The reaction is shown below:

• Note that when the number of carbon atoms become 2, the 'meth' becomes 'eth'
• Vinegar can be obtained by the fermentation of ethanol in the presence of air using the bacteria acetobactor

Uses of ethanoic acid:
• In the manufacture of rayon
• In the rubber and silk industry
• Gives flavour to food items
• As a preservative

Esters

• Esters are obtained by the reaction between alcohols and carboxylic acids. 
• We have seen the details about both alcohols and carboxylic acids above. 
■ So let us see an example of a reaction between them:
• Consider a carboxylic acid: Ethanoic acid
• Consider a alcohol: Ethanol
• The equation of the reaction between them is shown below:

• Note how the two oxygen atoms come together
• From the structural formula of the product Ethyl ethanoate, it is clear that, the functional group in esters is:  ㅡCOOㅡ
• We have seen that fatty acids are carboxylic acids having 12 or more carbon atoms
    ♦ Palmitic acid and stearic acid are fatty acids
    ♦ Because they are carboxylic acids having 12 or more carbon atoms
• Consider an alcohol: Glycerol
    ♦ It reacts with palmitic acid and stearic acid. 
    ♦ The products are oils and fats. 
    ♦ Oils and fats are esters
Solved example 15.6:
Examine the structural formulae given below and identify the esters. Write the chemicals required for the manufacture of each of those esters.
1.  CH₃ ㅡCH₂ㅡCOOㅡCH₃  
2.  CH₃ ㅡCH₂ㅡCOOH
3.  CH₃ ㅡCH₂ㅡCOㅡCH₃    
4.  CH₃ ㅡOH
5.  CH₃ ㅡCH₂ㅡCH₂ㅡOH.
6.  CH₃ ㅡCOOH
7.  CH₃ ㅡCH₂ㅡCH₂ㅡCOOㅡCH₃  
Solution:
Esters contain the 'ㅡCOOㅡ' functional group. So (1) and (7) are esters
1.  CH₃ ㅡCH₂ㅡCOOㅡCH₃
• In the 'ㅡCOOㅡ', the CO comes from a carboxylic acid
• The second O comes from an alcohol
• So we can split it as: CH₃ ㅡCH₂ㅡCO│OㅡCH₃  . 
• On the left side, we have 3 carbon atoms. So it indicates Propanoic acid
• On the right side, we have 1 carbon atom. So it indicates Methanol
• So the chemicals required are: Propanoic acid and methanol
7.  CH₃ ㅡCH₂ㅡCH₂ㅡCOOㅡCH₃
• In the 'ㅡCOOㅡ', the CO comes from a carboxylic acid
• The second O comes from an alcohol
• So we can split it as: CH₃ ㅡCH₂ㅡCH₂ㅡCO│OㅡCH₃  . 
• On the left side, we have 4 carbon atoms. So it indicates Butanoic acid
• On the right side, we have 1 carbon atom. So it indicates Methanol
• So the chemicals required are: Butanoic acid and methanol

Aromatic compounds

Consider the aromatic compound shown in fig.15.21(a) below:

Fig.15.21

• It is the simplest aromatic compoundstructure. It's name is benzene. 
• It has 6 carbon atoms. These carbon atoms are arranged at the corners of a regular hexagon
• But each of those carbon atoms have only one hydrogen atom.
• So to satisfy the valency requirements, there should be 3 double bonds. 
• These double bonds will be arranged alternately. 
    ♦ That is., no two double bonds in the structure will come near each other.
■ So there are two possible ways to draw the structure
• In fig.a, the first double bond starts at top left   
• In fig.b, the first double bond starts at top right
■ A simple way for drawing the structures is shown in fig.15.22 below

Fig.15.22

Compounds of Benzene
• All aromatic compounds have a ring structure. 
• There are double bonds between alternate carbon atoms
• In benzene, each carbon atom has one hydrogen atom. 
• Any one of those hydrogen can be replaced by a functional group. 
• Thus we get different aromatic compounds. Some examples are shown in fig.15.23 below:

Fig.15.23

• These compounds have immense chemical and industrial significance. 
• Many useful substances can be prepared from them. 
• Coal tar obtained by the distillation of coal in the absence of air is the source of aromatic compounds.

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In the next section, we will see some important applications of Chemistry

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Chapter 13.3 - Production of Aluminium and Copper

In the previous section, we saw the industrial production of steel. In this section, we will see the industrial production of aluminium and copper.

Industrial production of Aluminium

Extraction of Aluminium
• Aluminium is the most abundant metal on the earth's crust. 
• But it's reactivity is more than gold and silver. 
• So it is found in a combined state in nature. 
• A 'cost effective method to separate pure aluminium from it's ore' was discovered only in the nineteenth century. Until then aluminium was costlier than gold.
• The method is known as the Hall-Heroult process. It was discovered by two scientists Charles Martin Hall and Paul Heroult.
• Let us see the details of this process. We will write it in steps:
1. Bauxite (Al₂O₃.2H₂O) is the ore of aluminium. 
• The chief impurity in it, is the silicon dioxide (SiO₂) 
2. We know that any ore is to be powdered before it enters the first stage. It is true for bauxite also. So the bauxite is finely powdered.
• Stage 1 is the concentration of the powdered ore. 
• We have seen the different methods used for concentration. In the case of bauxite, leaching is used.
3. For that, the powdered bauxite is treated with hot concentrated NaOH solution. 
• Then the Al₂O₃ in the bauxite dissolves in the solution. 
• The impurities will not dissolve. So they are filtered off. 
4. But the aluminium that we want, is now in a dissolved form inside the solution. We want it back.
• Inside the solution, the aluminium exists as sodium aluminate (NaAlO₂). 
5. So, to this solution, a little aluminium hydroxide (Al(OH)₃) is added. 
• After that, the solution is diluted with water. 
• Then the Al gets separated from NaAlO₂ and will precipitate as Al(OH)₃. 
• This precipitate is filtered and washed. 
• Once we get this Al(OH)₃, leaching is complete. 
6. Next step is to convert this hydroxide into the oxide. 
• For that, calcination is used. 
• The hydroxide is heated strongly. It decomposes to give the oxide. 
• The equation is: 2Al(OH)₃ (s) ➙ Al₂O₃ (s) + H₂O (l)
• Al₂O₃ is called alumina
7. So the second stage is also over. The third and final stage is refining. 
• We have to reduce the oxide Al₂O₃ into pure aluminium. 
8. The Al₂O₃ is so stable that carbon or carbon monoxide is not sufficient to reduce it. 
• We will need the 'most powerful reducing agent', which is electricity. 
• That is., we will need to conduct electrolysis to obtain pure aluminium from Al₂O₃.
9. We will have to melt Al₂O₃ into a liquid form. 
• This molten Al₂O₃ will be the electrolyte. 
10. But Al₂O₃ will not melt easily. Because it's melting point is very high. 
• But when heated after adding cryolite, the cryolite melts and Al₂O₃ dissolves in it. 
• Cryolite helps to achieve two things:
    ♦ Bring down the melting point of Al₂O₃ 
    ♦ Increase the electrical conductivity of the molten mixture.
11. Consider the fig.13.8 below:

Fig.13.8

• The molten mixture is inside a steel tank 
    ♦ The steel tank has a carbon lining on the inner sides. 
    ♦ This carbon lining is the negative electrode. 
• Carbon rods act as positive electrodes. 
12. The Al₂O₃ exists as ions. Al³⁺ ions and O^2- ions. 
• When electricity is passed through the electrodes, the positive Al³⁺ ions gets separated from the O^2- ions and move towards the negative electrode. 
• There they accept electrons and get reduced to Al atoms. 
13. The negative O^2- atoms move towards the positive electrodes. 
• There they donate the electrons and get oxidised to Oxygen atoms. 
• Thus oxygen is produced at the positive electrodes. 
• The positive electrodes are carbon rods. The newly produced oxygen reacts with the carbon rods. So these carbon rods will have to be replaced from time to time. 
14. The newly produced pure aluminium is in a molten state. 
• It is denser than the electrolyte. So it is collected at the bottom of the tank. 
• It is taken out through the plug.
• Thus the refining process is complete

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Aluminium is a metal with many special properties:
• It is a light metal
• It has good thermal conductivity. So it is used for making cooking vessels  
• It has good electrical conductivity. So it is used for making electrical wires
• It has good metallic lustre. So it is used for making reflectors. See images here.
• It has great malleability. So it can be made into thin sheets.

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Refining of copper

We have seen that there are 3 stages in the production of metals:
1. Concentration of ore 
2. Extraction of metals from the concentrated ore
3. Refining of the metals
• But in this section we are discussing only the refining of copper. Why is that? What about the first two stages?
Ans: We have already mentioned the first two stages related to copper.
• When we discussed froth floatation, we mentioned that copper pyrites is concentrated by that method
• When we discussed roasting, we mentioned that Cu₂S is extracted by that method.
• So now we will see refining of copper
• Copper is a metal that is widely used in electrical appliances. 
• For such purposes, copper used must be very pure. Electrolytic method is used to obtain pure copper. Consider the fig.13.9 below:

Fig.13.9

1. 'Copper obtained during extraction', which contains many impurities is used as the positive electrode. We have seen the reason here. 
2. Pure copper is used as the negative electrode. 
3. Copper sulphate solution to which a little H₂SO₄ added is used as the electrolyte. 
4.When the switch is turned on:
• At the positive electrode, the Cu²⁺ ions separates themselves from the impurities and go into the solution. 
• These positive ions are attracted towards the negative electrode. There they receive two electrons and become pure copper atoms. 
• The reaction taking place is: Cu²⁺ + 2e^- ➙ Cu
5. This is a reduction reaction. 
• In any type of cell (galvanic cell or electrolytic cell), cathode is where reduction takes place.  So in our present case:
• The original pure metal piece used as the negative electrode is the cathode. 
• And the impure metal piece is the anode.
6. As more and more Cu²⁺ ions begin to leave the anode, the impurities cannot be bound together. 
• So The impurities disintegrate into smaller pieces and gets deposited below the anode. 
• This is called anode mud. Anode mud may contain valuable metals like gold or silver. 
7. The pure copper is removed from the cathode after a few days

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Now we will see some solved examples:
Solved example 13.2
During the extraction of iron, a mixture of coke, limestone and haematite is added from the top into the blast furnace. What is the significance of adding limestone in the mixture?
Solution:
1. The calcium carbonate (in the form of limestone) which is supplied along with the haematite, will decompose into calcium oxide and carbon dioxide. 
The equation is: CaCO₃ (s) ➙ CaO (s) + CO₂ (g)
2. The calcium oxide (CaO) thus formed will combine with silicon dioxide (SiO₂). 
• This SiO₂ is the main impurity. It is sand. 
• The reaction between sand and CaO converts the sand into calcium silicate.
• The equation is: CaO (s) + SiO₂ (s) ➙ CaSiO₃ (s)
• Thus the sand is removed.
3. Sand (SiO₂) is the gangue. It is acidic
• Calcium oxide (CaO) is the flux. It is basic
• The above two react to form the slag calcium silicate (CaSiO₃).
4. We can remove the gangue only by converting it into slag
• To convert the gangue into slag, we need the flux CaO
• To obtain CaO, we need CaCO₃. That is why we add limestone

Solved example 13.3
Which one does not belong to the group? Why?
(a) Leaching, Distillation, Froth floatation, Hydraulic washing
(b) Bauxite, Haematite, Clay, Cryolite 
Solution:
(a) Distillation does not belong to the group because it is a refining process
All others are concentration processes
(b) Haematite does not belong to the group because it is an ore of iron
All others are minerals of aluminium

Solved example 13.4
Match the following

	A	B	C
1	Sulphide ore	Bauxite	Electrolysis
2	Ore of iron	Zinc blende	Calcination
3	Aluminium ore	Calamine	Blast furnace
4	Carbonate ore	Haematite	Froth floatation process

Solution:

	A	B	C
1	Sulphide ore	Zinc blende	Froth floatation process
2	Ore of iron	Haematite	Blast furnace
3	Aluminium ore	Bauxite	Electrolysis
4	Carbonate ore	Calamine	Calcination

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In the next chapter, we will see Nomenclature of organic compounds

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Chapter 13.2 - Industrial Production of Iron

In the previous section, we saw the various stages in the production of metals. In this section, we will see how these stages are applied in the case of iron.

Industrial production of iron

We will write it in steps:
1. Haematite is the principal ore of iron.
• As we have seen in the previous sections, which ever be the metal, the ore has to be powdered.
• The 'concentration of the powdered ore' is the first stage.
2. In the case of haematite, we use the levigation method for the concentration. 
• As iron is heavier than the impurities, it will settle to the bottom of the tank. 
• The impurities will be washed away.
• So first stage is over
3. The second stage is the extraction of the metal from the concentrated ore. 
• We have seen that there are two steps to be completed for this. They are:
(a) Conversion of the ore into it's oxide.
(b) Reduction of the oxide
4. Let us see how they are accomplished in the case of iron:
(a) First we have to convert the iron into iron oxide. There are two methods available.
(i) Calcination and (ii) roasting. 
• In the case of iron, roasting is used. 
• Haematite is Fe₂O₃. So it already contains the oxide of iron. Even then, it is subjected to roasting.
• During roasting, impurities like sulphur, arsenic, phosphorus etc., are removed as their gaseous oxides. Water is also expelled. 
5. Next, we have to reduce the oxide.
• The haematite obtained after roasting will contain large quantities of silicon dioxide (sand). So we have to remove it to get the pure iron. The following procedure is used:
6. A mixture of roasted haematite, coke and lime stone (CaCO₃) is fed into a blast furnace. See fig.13.7 below:

Fig.13.7

• Blast furnace is a huge steel furnace. 
• Steel cannot withstand the high temperature produced inside the furnace. 
• So the inner side of the furnace is lined with a refractory material.
7. A blast of hot air is sent into the furnace from the bottom. 
• At the same time, the mixture containing haematite falls from the top.
• Because of the heat, chemical reactions will take place. Let us see each of those reactions:
• The calcium carbonate (in the form of limestone) which is supplied along with the haematite, will decompose into calcium oxide and carbon dioxide. 
The equation is: CaCO₃ (s) ➙ CaO (s) + CO₂ (g)
8. The calcium oxide (CaO) thus formed will combine with silicon dioxide (SiO₂). 
• This SiO₂ is the main impurity. It is sand. 
• The reaction between sand and CaO converts the sand into calcium silicate.
• The equation is: CaO (s) + SiO₂ (s) ➙ CaSiO₃ (s)
• Thus the sand is removed. 
9. Now we are left with the iron oxide Fe₂O₃
• It will soon be reduced to iron. Let us see:
10. At the bottom of the furnace, coke combines with the oxygen. This oxygen is obtained from the hot blast of air. 
• The equation of the reaction is: C (s) + O₂ (g) ➙ CO₂ (g) + heat
• From the equation, we can see that it is an exothermic reaction. So the temperature rises upto 1800^o C. 
• The CO₂ thus produced rises to the upper parts of the furnace. It will get  reduced by coke. 
• The equation is: CO₂ (g) + C (s) + heat ➙ 2CO (g)
11. The temperature in these upper parts is lower than the temperature at the bottom parts. This is because, the reaction here is endothermic. 
• Note that, the 'heat' is on the left side of the arrow. So there is an 'absorption of heat'
• The fig.13.7 above indicates that the lower parts are 'white hot'. While upper parts are 'red hot'. 
    ♦ 'White hot' is much hotter than 'red hot'.
    ♦ There is a gradation from white to red as we go upwards  
• The carbon monoxide reaches the middle portion of the furnace and reacts with the iron oxide.
• The iron oxide gets reduced to iron. 
• The equation is:  Fe₂O₃  (s) +3CO (g) ➙ 2Fe (s) + 3CO₂ (g)
12. The iron thus formed moves downwards. 
• Due to the high temperatures in the lower parts, it will melt into liquid form. 
• This liquid gets collected at the bottom of the furnace. From there it can be taken out. This is shown in fig.13.7 above

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Thus we successfully produce iron. Now we have to learn some technical terms involved in the above process.
■ The impurity in the ore is called gangue.
• See step (5) above. In our present case, sand (SiO₂) is the gangue.
■ The chemical that is used to remove the gangue is called the flux.
• In step (8), we see that the calcium oxide (CaO) combines with sand to form calcium silicate.
• So in our present case, CaO is the flux
■ The product formed as a result of the reaction between gangue and flux is called the slag. 
• The only way to remove the sand is by converting it to 'another material'.   
• In step (8) we see that this 'another material' is calcium silicate (CaSiO₃). It is formed as a result of the reaction between the gangue and the flux.
• So in our present case, CaSiO₃ is the slag.
■ CaO is chosen as the flux because it is basic in nature. It will react with the gangue (SiO₂) which is acidic in nature

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13. The slag formed in our present case will melt due to the high temperature. So like the molten iron, it also becomes a liquid and gets collected at the bottom of the furnace
• But the two liquids do not mix together. This is because, the slag is lighter. 
• So it floats above the molten iron. So it can be removed easily. This is shown in the fig.13.7 above
• The formation of the separate 'slag layer' above the 'molten iron layer' has the following advantages:
    ♦ The two can be removed from the furnace easily
    ♦  Contact between molten iron and oxygen is avoided, thus preventing the formation of iron oxide.
14. The iron produced in this way has 4% carbon.
• That is., if we take 100 grams of this iron, 4 grams will be carbon
• This much carbon is not allowable. It will prevent the iron from acquiring it's unique properties. 
• This iron has other impurities also like manganese, silicon, phosphorus etc., 
• This iron is called pig iron. Images can be seen here. It has only limited use and is an intermediate product of the iron industry.
15. The pig iron is taken to another special furnace (The electric arc furnace is used in modern days).
• There, it is mixed with scrap iron and coke and again melted.
• The product obtained is called cast iron. It contains 3% carbon
• When molten cast iron cools and solidifies, there will be an increase in volume. 
• So the molten cast iron is poured into moulds to make various shapes. Some products can be seen here.
• The shapes thus obtained are very hard. But they will break if subjected to bending.
16. The cast iron can be purified further. 
• The product obtained will contain only 0.2 - 0.5% carbon. 
• It will also contain some traces of phosphorus and silicon. This is called wrought iron.
• Some products using wrought iron can be seen here.
17. So we can write:
Pig iron ➙ Cast iron ➙ Wrought iron

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• We often hear the word steel. Words 'iron' and 'steel' are sometimes used interchangeably. 

• But they are not the same. Let us see the difference:

■ Iron is an element in the periodic table. It's symbol is Fe

• We try to make pure iron in the blast furnace. 

• But the product obtained contain some carbon and other impurities. 

• It is not possible to completely avoid the 'presence of some impurities'.

• However pure iron can be prepared in specially equipped labs

■ Steel is an alloy. 

• We know that alloy is:

    ♦ a combination of two or more metals

    ♦ or a combination of a metal and another element

• Steel is an alloy of 'the metal iron' and 'the element carbon'. 

• It should not contain any other element

• Also, the amount of carbon should not exceed 2%

• By varying the content of carbon within the 2%, steels having various properties can be obtained.

    ♦ Low carbon steels have carbon less than 0.3%

    ♦ Medium carbon steels have carbon from 0.3% to 0.6%

    ♦ High carbon steels have carbon more than 0.6%

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Following are some of the processes done on steel:
■ Annealing:
• Heat the steel to high temperature
• And then allow it to cool slowly. 
• This will make the steel ready to be shaped into different forms
■ Hardening:
• Heat the steel to high temperature
• And then cool it rapidly by dipping in water or oil
• This will make the steel very hard. But it will be brittle
■ Tempering:
• Heat the hardened steel until it becomes blue in colour
• And then allow it to cool slowly.
• This will make the hardened steel less brittle, while maintaining the hardness 

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We will now see 3 alloys in which iron is a component:
1. Name: Stainless steel
Components: Iron (Fe), Chromium (Cr), Nickel (Ni) and Carbon (C)
Properties: Strong, will not rust.
Uses: For the manufacture of utensils, parts of vehicles
2. Name: Alnico
Components: Iron (Fe), Nickel (Ni), Aluminium (Al) and Cobalt (Co)
Properties: Magnetic in nature
Uses: For the manufacture of permanent magnets
3. Name: Nichrome
Components: Iron (Fe), Nickel (Ni),  Chromium (Cr), and Carbon (C)
Properties: High resistance
Uses: For the manufacture of heating coils
■ If we examine the above alloys, we can note an interesting point:
• Stainless steel and nichrome have the same components. But their properties are different. 
• Why is that so?
Ans: The difference in properties is achieved by increasing or decreasing the quantity of each component of the alloys. For example:
• Take 100 grams of stainless steel
Let the amount of chromium contained in it be x₁ grams
• Take 100 grams of nichrome
Let the amount of chromium contained in it be x₂ grams
• Then x₁ will not be equal to x₂
• In other words, the 'percentage of chromium' in stainless steel and nichrome are different
• Same is applicable for the other components iron, nickel and carbon
■ We can make different alloys by:
• Combining appropriate metals and also by 
• Changing the percentage content of each metal or element

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In the next section, we will see the industrial production of aluminium and copper. 

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Chapter 13.1 - Extraction and Refining

In the previous section, we completed a discussion on 'concentration of ore'. In this section, we will see the next stage.
The second stage after concentration of ore is:
■ Extraction of metal from concentrated ore
There are two stages in it:
(i) Conversion of the concentrated ore into it's oxide
(ii) Reduction of the oxide
• We will now see them in detail:
For the first stage, any one of the following two methods can be used. 
(a) Calcination
(b) Roasting
The choice depends on the type of the ore
(a) Calcination:
1. The concentrated ore is heated to a high temperature in the absence of air
• But the temperature should not exceed the melting point of the ore
2. So the impurities which have lower melting points will melt and then evaporate away.
• The most common impurities which are removed in this way are water and organic matter
3. When the impurities are removed, the useful ore remains. 
• These are the carbonates and hydroxides of the metal. 
4. Due to the heat, these carbonates and hydroxides decompose to form the oxides of the metal. 
• When oxides of the metal are formed, they are ready to be passed on to the next stage
5. Since there is no air, oxygen will not take part in this decomposition reaction.
Example: Zinc carbonate is converted into Zinc oxide
(b) Roasting:
1. The concentrated ore is heated to a high temperature in a current of air
• But the temperature should not exceed the melting point of the ore
• Because of the current of air, oxygen will also take part in the reaction
2. The impurities which have lower melting points will melt and then evaporate away.
• The most common impurities which are removed in this way are water and organic matter
3. When the impurities are removed, the useful ore remains. 
• These remaining ore is converted into oxides because of the presence of oxygen
• When oxides of the metal are formed, they are ready to be passed on to the next stage

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■ At this stage we will draw a flow chart to show various steps that we have seen so far:

Fig.13.4

• As we learn more details, we will fill up the missing spaces in the above chart.
• Now we move to the next step in 'extraction of metals', which is: 

Reduction of the oxide

• We have seen the details about reducing agents and reduction reactions here. 
• We have seen some redox reactions in the previous chapter on 'Reactivity' also. Details here. 
• When we 'reduce' the oxide of a metal, we will get the original metal. 
• Suitable reducing agents can be used for this purpose. 

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Before we proceed further, let us discuss the reason for saying:
'reduce' the oxide of a metal.
The following steps will help us to understand the reason:
1. Let 'X' be a metal. So we can denote it's oxide as 'XO', where 'O' stands for oxygen.
2. Thus 'XO' is a compound. The individual elements in that compound are:
• The metal 'X'
• The oxygen 'O'
3. Now, X is a metal. Metals are electron donors. 
• So in the compound XO, the metal X will be having a positive oxidation state. Let it be +m
• And O will be having a negative oxidation state. Let it be -n
4. So we can write: X^+m O^-n.
• Mg⁺² O^-2 is an example
5. Now we allow this X^+m O^-n to react with 'another compound'.
This 'another compound' must be able to donate electrons
6. The electrons thus donated will be received by X^+m .
• Then X^+m will become X⁰ or simply X
• The oxidation number is reduced from +m to zero
7. This X will no longer attach itself to O
• It becomes independent and we get the pure metal X
8. So we see that some 'reduction' is indeed involved
• The oxidation number decreases from +m to zero
• That is why we say:
'reduce' the oxide of a metal to get the pure metal  
9. The 'another compound' that we saw in step (5) above is the reducing agent.
10. Electricity is the strongest reducing agent. It has an unlimited supply of electrons. Also, it's energy level can be increased by increasing the voltage.

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Now we continue our original discussion:
Some examples of the application of reducing agents are:
• Carbon monoxide is used as the reducing agent to extract iron from haematite

    ♦ The equation is: Fe₂O₃ (s) + 3CO (s) → 2Fe (s) + 3CO₂ (g)

    ♦ In Fe₂O₃, The Fe exists as positive ion. It receive electrons from the reducing agent CO. 

    ♦ The Fe ion thus becomes Fe atom

• Carbon is used as reducing agent to extract zinc from zinc oxide.

    ♦ The equation is: ZnO (s) + C (s) → Zn (s) + CO (g)

    ♦ In ZnO, The Zn exists as positive ion. It receive electrons from the reducing agent C. 

    ♦ The Zn ion thus becomes Zn atom

• Electricity is the strongest reducing agent. This electricity is required to extract highly reactive  metals like sodium, potassium and calcium from their oxides.
■ So we completed the second stage also. The next stage is:

Refining of metals

• The metal obtained by the reduction process that we discussed above, will still contain some impurities. 
• Such impurities include small quantities of other metals and may be some non metals. 
• We must remove them also to get the pure metal. 
■ Following are some of the methods that can be used. The choice of the method will depend on the nature of the metals and the impurities present in them.
(a) Liquation
(b) Distillation
(c) Electrolytic refining
• Before seeing each of them in detail, we will complete the chart. 
• This will give us a better understanding about our exact position in the whole discussion. 
• The completed chart is shown in fig.13.5 below:

Fig.13.5

• Now we can continue the discussion. We were discussing the last step, which is 'Refining of the metal'.
• So we are in the bottom most green box. We can use any one of the three. The choice depends on the type of metal and the impurities.
(a) Liquation
(b) Distillation
(c) Electrolytic refining

Liquation

• This method is used when both the following condition are satisfied:
    ♦ The pure metal to be extracted, has a low melting point
    ♦ The impurities are metals having high melting points 
• Consider fig.13.6 below. We can write the steps:

Fig.13.6

1. The impure metal is supplied from the top of a specially shaped funnel
• This funnel is kept in an inclined position
2. The funnel is heated from it's bottom side
3. The pure metal which has low melting point will melt and will flow down along the inclined surface.
• It are collected at the bottom
4. The impurities having high melting points will be left behind
• Tin and lead, which have low melting points can be refined using this method
5. Note that, the structure in which heat is produced, is made of bricks
• But they are not ordinary bricks. They are 'refractory bricks'. They can withstand very high temperatures.

Distillation

• This method is used when both the following condition are satified:
    ♦ The pure metal to be extracted, has a low boiling point
    ♦ The impurities are metals having high boiling points 
• We can write the steps:
1. The impure metal is heated in a retort. 
2. The pure metal alone will boil and turn into vapours
3. These vapours are collected and condensed to get the pure metal.
4. The impurities will be left behind
• Zinc, cadmium and mercury have low boiling points. So they can be refined by this method

Electrolytic refining

• In this method, we use electrolysis to obtain the metal. 
• We have seen the details about electrolysis in the previous chapter. (Details here). 
• For electrolysis, we need an electrolyte. For our present case, we use a solution of the 'salt of the metal' as the electrolyte. 
■ A small piece of the pure metal is used as the negative electrode. 
■ A piece of the impure metal is used as the positive electrode.
• We want to know why pure metal is used as negative and the impure metal is used as positive electrodes.
• We will see the reason when we analyse the reaction that takes place when we turn on the switch. 
We will write it in steps:
1. We want to refine the impure piece. 
• For that, the pure metal should separate from the impurities and dissolve in the solution. 
2. How does such a dissolution occur?
Ans: The positive metal ions get separated from the impure piece and goes into the solution.
3. Now we want these positive ions to be converted into pure metal. 
• For that, the positive ions move to the negative electrode. 
• There they receive electrons and will be converted into pure metal atoms. 
4. So the formation of the 'new pure metal' occurs at the negative electrode. 
• This newly formed pure metal should stick to the already available pure metal. Then only we can use it. 
■ So the already available pure metal piece is used as the negative electrode. 
5. The reaction taking place at the negative electrode is reduction. 
• So the negative electrode is the cathode. 
• The impure piece is the positive electrode and it is the anode

The chemistry of metallurgy

Consider the table given below. The metals are arranged in the decreasing order of their reactivity.
Also they are arranged in four groups. 

Let us write the details:
1. The five members of the top most group are very reactive. 
• We will need electricity to extract them from their compounds. These compounds are first heated and melted into the liquid state. 
Then electrolysis is carried out
2. The five members of the second group are extracted by reducing their oxides. 
• Depending on the metal, carbon or carbon monoxide can be used as the reducing agent.
3. There is only one member in the third group. It is copper. 
• It has a lower reactivity than the first and second groups. It can be extracted more easily than the metals above it. 
• First it's sulphide is converted to oxide and then the oxide is reduced.
4. The fourth group has two members. 
• Their reactivity is very low. So they do not form any compounds. 
• Thus they are found in free state in nature.

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We can write a summary:
1. Metals have a tendency to lose electrons.
• So they are seen as positive ions in their compounds
2. The positive ions must be converted to ordinary atoms to obtain the pure metal
• So we must supply electrons to the positive ions
3. The positive ions receive these electrons and get reduced to the original atoms
■ So in short, if we are asked to choose between oxidation and reduction:
We must choose reduction to extract metal from it's compounds.

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So we have seen the basics of metallurgy and the various stages involved in the extraction of metals.  In the next section, we will see how these stages are applied in the case of iron. 

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