Class 12 Chemistry Chapter 6 Ncert Solutions | Download Free PDF

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.12^th 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 12^th 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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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:

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