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Class 12 Chemistry Chapter 6 NCERT Solutions in English
Adda247 provides NCERT Solutions for Class 12 Chemistry Chapter 6. These solutions will not only help students to boost their board exams and score brilliant marks but for competitive exams. The Solutions are according to the NCERT guidelines.
Complete 16 chapters wise solutions were provided for the benefit of students.12th class sets the base for higher education for every student. This makes it the most crucial class for any student who is aiming for his/her dream of quality education. Scoring good marks in the 12th class is equivalent to a quality high education. So, it becomes extremely important for students to give a boost to their class 12 Chemistry preparation with Adda247 NCERT solutions.
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Advantages of NCERT Solutions for class 12 Chemistry
- In – depth explanation of all the questions with logical reason.
- Numericals are solved with step by step process.
- Free PDF download option.
- All the answers are given to the point of theoretical questions.
The detailed, explanative NCERT solutions are widely known for being extremely student-friendly and easy to digest. Solutions to relative complex questions are always broken down into simpler one in order to help students focus on the ‘method to solve’ rather than the solution itself. This concept centered approach is what enables students that use the Adda247 NCERT Solutions to solve all similar questions with ease.
NCERT Solution class 12 Chemistry Chapter 6: General Principles and Processes of Isolation of Elements
The most abundant metal found in the earth’s crust is Aluminium. Approx. 8.3% by weight. Cleaning of ore or removal of particles like sand, clay etc. from the ore is called dressing of ore or concentration of ore. This is done in the following steps: Hydraulic washing, Magnetic separation, froth floatation method, Leaching. This was a brief on Principles Isolation of Elements.
The chapter of General Principles and Processes of Isolation of Elements Class 12 NCERT PDF discusses the extraction of elements from their mineral ores.
The extraction and isolation of metals from ores involve the following major steps: Concentration of the ore, Isolation of the metal from its concentrated ore and Purification of the metal. The entire scientific and technological process used for isolation of the metal from its ores is known as metallurgy.
Principles involved in the extraction of element:
- Pyrometallurgy extraction of metals takes place at very high temperature.
- Bydrometallurgical process. In this method, metals are extracted by the use of their aqueous solution.
- Electrometallurgical process Na, K, Li, Ca, etc. are extracted from their molten salt solution through electrolytic method.
Carbon can be used to extract metals from some metal oxides. A metal can either be reduced or oxidized in a reaction. Reduction occurs when a metal in a compound loses oxygen, to form the elemental metal. Oxidation occurs when a metal gains oxygen, to form an oxide compound.
The chapter concludes with the refining of these metals and the uses of aluminium, copper, zinc, and iron. Chapter 6 Chemistry Class 12 is a detailed and complicated chapter, which can prove challenging for students to perform well in.
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Key features of NCERT Solution for class 12 Chemistry chapter 6
- The NCERT Solution provide clear and precise answer.
- The columns are used wherever necessary.
- The solutions are answered in a clear and straightforward style. Where needed it is responded to many equations.
- Diagrams are provided wherever necessary.
- A step by step approach makes it easier to apprehend.
- Long answers are broken into points.
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Important Question of NCERT Solution of Chemistry Chapter 6
Question:1 Out of C and CO, which is a better reducing agent at
673 K?
Answer:
The value of G° for change of C to CO2 is less than the value of G° for change of CO to CO2. Therefore, coke (C) is a better reducing agent than CO at 983 K or above temperature. However below this temperature (e.g., at 673 K), CO is more effective reducing agent than C.
Question:2 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 antimony, tellurium, silver, gold and platinum. These elements settle down under anode as anode mud because they are less reactive and are not effected by CuSO4 – H2SO4 solution.
Question:3 Write down the reactions taking place in different zones in the blast furnace during the extraction of iron.
Answer:
In blast furnace, iron oxides are reduced at different temperature ranges. In the lower part of the blast furnace, the temperature is as high as 2200 K. It is called combustion zone. At the top, the temperature is as low as 500−800 K. It is called reduction zone. In the lower temperature range, carbon is the reducing agent and in the higher temperature range, CO is the reducing agent. In the reduction zone (500−800 K), following reactions occur.
3Fe2O3 + CO → 2Fe3O4 + CO2
Fe3O4 + 4CO → 3Fe + 4CO2
Fe2O3 + CO → 2FeO + CO2
In the temperature range 900-1500 K, following reactions occur.
C + CO2 → 2CO
FeO + CO → Fe + CO2
Around 1270 K (middle portion), decomposition of limestone gives lime (CaO) and CO2. Lime is a flux and combines with silicate impurity to form slag of calcium silicate.
CaCO3 → CaO + CO2
CaO + SiO2 → CaSIO3
Question:4 Write chemical reactions taking place in the extraction of zinc from zinc blende.
Answer:
During extraction of Zn from zinc blende (ZnS), following reactions occur:
(i) ZnS is roasted in excess air at 1200 K to form ZnO.
2ZnS + 2O2 Δ 2ZnO + 2SO2
(ii) ZnO is heated with crushed coke at 1673 K, where it is reduced to Zn.
ZnO + C 1676K Zn + CO
(iii) Electrorefining is carried out for refining of impure zinc. Anode is impure zinc and cathode is pure zinc. Electrolyte is a mixture of zinc sulphate and dilute sulphuric acid (small amount). When current is passed, zinc is deposited on cathode and is collected.
At anode: Zn → Zn2+ + 2e–
At cathode: Zn2+ + 2e– → Zn
Question:5 State the role of silica in metallurgy of copper.
Answer:
During roasting, copper pyrites are converted into a mixture of FeO and Cu2O. Thus, acidic flux silica is added during smelting to remove FeO (basic). FeO combines with SiO2 to form famous silicate slag which floats over molten matte.
Question:6 What is meant by the term “chromatography”.
Answer:
It is the technique of separating the components of a mixture in which separation is achieved by the differential movement of individual components through a stationary phase under the influence of a mobile phase. The stationary is made up of aluminium oxide or silica gel.
Question:7 What criterion is followed for the selection of the stationary phase in chromatography?
Answer:
The stationary phase is selected in such a way that the impurities are more strongly absorbed or are more soluble in the stationary phase than component to be purified. Thus, when the column is extracted, the impurities will be retained by stationary phase while the pure component is easily eluted.
Question:8 Describe the method of refining nickel.
Answer:
When impure nickel is heated in presence of CO at 330 – 350 K, it forms volatile nickel tetra carbonyl leaving behind the impurities. The nickel tetra carbonyl thus obtained is than heated to high temperature (450 – 470 K), then it undergoes thermal decomposition to give pure nickel.
Ni + 4CO 330 – 350 K Ni(CO)4
(Impure) Nickel teracarbonyl
Ni(CO)4 450 – 470 K Ni + 4CO
Question:9 How can you separate alumina from silica in a bauxite ore associated with silica? Give equations, if any.
Answer:
Bauxite ore containing silica is heated with conc. NaOH solution at 473−523 K and 35−36 bar pressure. Alumina forms sodium aluminate and silica forms sodium silicate. Impurities are left behind.
Al2O3 + 2NaOH + 3H2O → 2Na[Al(OH)4]
Carbon dioxide neutralizes the aluminate to precipitate hydrated alumina.
2Na[Al(OH)4] + 2CO2 → Al2O3.xH2O + 2NaHCO3
Sodium silicate remains in the solution and hydrated alumina is filtered and dried. On heating, hydrated alumina gives pure alumina.
Al2O3.xH2O 1473 Al2O3(s) + xH2O
Question:10 Giving examples, differentiate b/w ‘roasting’ and ‘calcination’.
Answer:
| Calcination | Roasting
|
| 1.) Calcination is a process in which ore is heated in the absence of air or air might be supplied in limited quantity. | 1.) Roasting involves heating of ore lower than its melting point in the presence of air or oxygen.
|
| 2.) Calcination involves thermal decomposition of carbonate ores. | 2.)Roasting is carried out mostly for sulphide minerals.
|
| 3.) During calcination, moisture is driven out from an ore. | 3.)Roasting does not involve dehydrating an ore.
|
| 4.) Carbon dioxide is given out during calcination.
|
4.)During roasting large amount of toxic, metallic and acidic compounds are released. |
| For example: Decomposition of limestone to lime.
CaCO3 à CaO + CO2 |
For example: Zinc sulphide to zinc oxide.
2ZnS + 3O2 à 2ZnO + 2 SO2 |
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