Monthly Archives: January 2022

In This Post we are  providing Chapter- 4 CHEMICAL KINETICS NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON CHEMICAL KINETICS

Question 1.
What is meant by rate of a reaction? Differentiate between average rate and instantaneous rate of a reaction.
Answer:
Rate of reaction: It is the change in concentration of the reactants or products in a unit time. Average rate : Average rate depends upon the change in concentration of reactants or products and the time taken for the change to occur. R → P

Instantaneous rate: It is defined as the rate of change in concentration of any one of the reactant or product at a particular moment of time.

Question 2.
(a) What is the physical significance of energy of activation ? Explain with diagram.
(b) In general, it is observed that the rate of a chemical reaction doubles with every 10 degree rise in temperature. If the generalization holds good for the reaction in the temperature range of 295 K to 305 K, what would be the value of activation energy for this reaction ?
[R = 8.314 J mol^-1 K^-1]
Answer:
(a) The minimum extra amount of energy absorbed by the reactant molecules so that their energy becomes equal to threshold value is called activation energy. Less is the activation energy, faster is the reaction or greater is the activation energy, slower is the reaction

Question 3.
(a) A reaction is second order in A and first order in B.
(i) Write the differential rate equation,
(ii) How is the rate affected on increasing the concentration of A three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?
(b) A first order reaction takes 40 minutes for 30% decomposition. Calculate t_1/2 for this reaction. (Given log 1.428 = 0.1548)
Answer:
(a) (i) Differential rate equation :
dxdt = K [A]²[B]
(ii) When concentration of A is increased to three times, the rate of reaction becomes 9 times
r = K[3A]²B ∴ r = 9KA²B i.e. = 9 times
(iii) r = K[2A]²[2B] ∴ r = 8KA²B i.e. = 8 times

(b) Given : Time, t = 40 minutes, t =?
Let a = 100, ∴ x = 30% of 100 = 30
Using the formula :

Question 4.
(a) For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction. .
(b) Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the equation:
log k = log A – Ea2.303R(1T)
where E_a is the activation energy. When a graph is plotted for log k vs. 1T¯¯¯¯,a straight line with a slope of – 4250 K is obtained. Calculate ‘E_a‘ for the reaction. (R = 8.314 JK^-1 mol^-1)
Answer:

∴ E_a = 2.303 × 8.314 (JK^-1 mol^-1) × 4250 K
= 81.375 J mol^-1 or 81.375 kJ mol^-1

Question 5.
(a) The decomposition of A into products has a value of K as 4.5 × 10³ s^-1 at 10°C and energy of activation 60 kj mol-1. At what temperature would K be 1.5 × 10⁴ s^-1?
(b) (i) If half life period of a first order reaction is x and 3/4,^th life period of the same reaction is y, how are x and y related to each other?
(ii) In some cases it is found that a large number of colliding molecules have energy more than threshold energy, yet the reaction is slow. Why?
Answer:
(a) Given : K₁ = 4.5 × 10³ s^-1,
T₁ = 10K + 273K = 283K
K₂ = 1.5 × 10⁴ s^-1, T₂ = ?
E_a = 60 KJ mol^-1
Using formula :

∴ Temperature, T₂ will be = 297° – 273° = 24° C

(b) (i) t_1/2 = 0.693K (For first order reaction)
t_3/4 = K ⇒ t_3/4 = 1.3864K
According to condition
(The value 1.3864 is double of 0.693)
From the above equation it is clear that
t_3/4 = 2t_1/2 ∴ y = 2X
(ii) It is due to improper orientation of the colliding molecules at the time of collision.

Question 6.
(a) A first order reaction takes 100 minutes for completion of 60% of the reaction. Find the time when 90% of the reaction will be completed.
(b) With the help of diagram explain the role of activated complex in a reaction. (Comptt. Delhi 2013)
Answer:
(a) For the first order reaction,

(b) In order that the reactants may change into products, they have to cross an energy barrier as shown in the diagram

This diagram is obained by plotting potential energy vs. reaction coordinate. It is believed that when the reactant molecules absorb energy, their bonds are loosened and new bonds are formed between them. The intermediate complex thus formed is called activated complex. It is unstable and immediately dissociates to form the stable products.

Question 7.
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained :

t/s	0	30	60
[CH3COOCH3]/mol L-1	0.60	0.30	0.15

(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds. (Given log 2 = 0.3010, log 4 = 0.6021)
Answer:

As k is constant in both the readings, hence it is a pseudo first order reaction.
(ii) Rate = – Δ[R] /Δt, Average rate between 30 to 60 seconds
= −(0.15−0.30)60−30=0.1530
= 0.5 × 10^-2 mol L^-1 sec^-1

Question 8.
(a) For a reaction A + B → P, the rate is given by Rate = k[A] [B]²
(i) How is the rate of reaction affected if the concentration of B is doubled?
(ii) What is the overall order of reaction if A is present in large excess?
(b) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction.
(log 2 = 0.3010)
Answer:
(a) For the reaction A + B → P rate is given
by Rate = k[A]¹[B]²
(i) r₁ = k[A]1 [B]2
r₂ = k[ A]¹[2B]² =
r₂ = k[A]¹ [2B]²=4k[A]¹ [B]²
r₁ = 4r₂, rate will increase four times of actual rate.

(ii) When A is present in large amount, order w.r.t. A is zero.
Hence overall order = 0 + 2 = 2, second order reaction.

Question 9.
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained:

t/s	0	10	20
[CH3COOCH3]/mol L-1	0.10	0.05	0.025

(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 10 to 20 seconds. (Given : log 2 = 0.3010, log 4 = 0.6021)
Answer:

As k₁ and k₂ are equal, hence pseudo rate constant is same.
It follows the pseudo first order reaction.
(ii) Average rate of reaction between 10 to 20 seconds
= −Δ[R]Δt=−(0.025−0.05)(20−10)=0.02510
= 0.0025 mol lit^-1 sec^-1

Question 10.
(a) For a reaction A + B → P, the rate is given by Rate = k[A] [B]²
(i) How is the rate of reaction affected if the concentration of B is doubled?
(ii) What is the overall order of reaction if A is present in large excess?
(b) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction.
Answer:
(a) For the reaction A + B → P
rate is given by Rate = k[A]¹[B]²
(i) r₁ = k[A]¹ [B]²
r₂= k[A]¹ [2B]² = 4k[A]¹ [B]²
r₁ = 4r₂ (rate of reaction becomes 4 times)

(ii) When A is present in large amounts, order w.r.t. A is zero.
Hence overall order = 0 + 2 = 2

In This Post we are  providing Chapter- 6 GENERAL PRINCIPLES AND PROCESSESS OF ISOLATION OF ELEMENTS NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON GENERAL PRINCIPLES AND PROCESSESS OF ISOLATION OF ELEMENTS

1. How are gold and silver extracted?
Ans. Gold and silver are extracted by leaching the metal with CN^–. The metal is later recovered by displacement method in which zinc acts as reducing agent.


2. Give two examples of metal refined by
 a) Distillation
 b) Liquation
 c) Electrolytic refining
Ans. a) Distillation – Zinc and Mercury
b) Liquation – Tin and Antimony
c) Electrolytic refining – Copper and Zinc
3. Explain electrolytic refining of copper.   
Ans. Electrolytic refining of copper-
In this method impure copper acts as anode and a strip of the same metal in the pure form is used as cathode. The electrolyte is acidified solution of copper sulphate. The net result is the transfer of copper in pure form from the anode to cathode.


Impurities from the blister copper like antimony, selenium, tellurium, silver, gold and platinum deposit as anode mud.
4. Write a short note on Mond’s process.
Ans. Mond’s Process- In this process, nickel is heated in a stream of carbon monoxide to give a volatile complex, nickel tetra carbonyl.

The carbonyl is heated to higher temperature

5. Which method is used for refining of zirconium? Explain.
Ans. Zirconium and Titanium are refined by van Ankle process, Here the crude metal is heated in an evacuated vessel with iodine.

The metal iodide is decomposed on a tungsten filament, electrically heated to about 1800K and pure metal is deposited on the filament.

6. What is the principle behind chromatography? Name some types of chromatographic techniques.
Ans. The principle behind chromatography is that different components of a mixture are differently adsorbed on an adsorbent. Some of the chromatographic techniques are paper chromatography, column chromatography, gas chromatography etc.
7. Explain the extraction of copper?   
Ans. The sulphide ores of copper are roasted to give oxides:

The oxide can then be easily reduced to metallic copper using coke.

The impurities like iron oxide are removed as slag by reacting with, added as flux.

8. What is the role of depressant in froth floatation process?
Ans. In the froth floatation process, the role of the depressants is to separate two sulphide ores by selectively preventing one ore from forming froth. For example, to separate two sulphide ores (ZnS and Pbs), NaCN is used as a depressant which selectively allows PbS to come with froth, but prevents ZnS from coming to froth. This happens because NaCN reacts with ZnS to form .

9.  Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?
Ans. The Gibbs free energy of formation of is less than that of and. Therefore, and C cannot reduce to Cu.
On the other hand, the Gibbs free energy of formation of is greater than that of CO. Hence, C can reduce  to Cu.

Hence, the extraction of copper from its pyrite ore is difficult than from its oxide ore through reduction.
10. State the role of silica in the metallurgy of copper.
Ans. During the roasting of pyrite ore, a mixture of FeO and is obtained.



The role of silica in the metallurgy of copper is to remove the iron oxide obtained during the process of roasting as ‘slag’. If the sulphide ore of copper contains iron, then silica is added as flux before roasting. Then, FeO combines with silica to form iron silicate, (slag).

In This Post we are  providing Chapter- 5 SURFACE CHEMISTRY NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON SURFACE CHEMISTRY

1. Explain modern theory of heterogeneous catalysis.
Ans. According to modern theory of catalysis, the mechanism of heterogeneous catalysis involves following steps –
(i) Diffusion of reactants on the surface of catalyst.
(ii) Adsorption of reactant molecules on the surface.
(iii) Occurrence of reaction on the catalysts surface through formation of an intermediate.
(iv) Desorption of products from surface.
(v) Diffusion of products away from surface.
2.  Differentiate between lyophobic and lyophillic sol?
Ans.

Lyophobic sol	Lyophillic sol.
1. It is relatively unstable due to Repulsion between dispersion ium and dispersed  phase.	1. It is relatively more stable due to med- attraction between dispersion medium and dispersed Phase.
2. It is irreversible.	2.  It is reversible.
3. It cannot be easily peptised.	3.  It can be easily peptised.
4. Small quantities of electrolyte cause precipitation.	4.  Small quantities of electrolyte has no  effect larger concentration causes precipitation.

3.  Distinguish between the meaning of the terms adsorption and absorption.
Give one example of each.
Ans. Adsorption is a surface phenomenon of accumulation of molecules of a substance at the surface rather than in the bulk of a solid or liquid. The substance that gets adsorbed is called the ‘adsorbate’ and the substance on whose surface the adsorption takes place is called the ‘adsorbent’. Here, the concentration of the adsorbate on the surface of the adsorbent increases. In adsorption, the substance gets concentrated at the surface only. It does not penetrate through the surface to the bulk of the solid or liquid. For example, when we dip a chalk stick into an ink solution, only its surface becomes coloured. If we break the chalk stick, it will be found to be white from inside.
On the other hand, the process of absorption is a bulk phenomenon. In absorption, the substance gets uniformly distributed throughout the bulk of the solid or liquid.
4. Why is adsorption always exothermic?
Ans. Adsorption is always exothermic. This statement can be explained in two ways.
(i) Adsorption leads to a decrease in the residual forces on the surface of the adsorbent. This causes a decrease in the surface energy of the adsorbent. Therefore, adsorption is always exothermic.
(ii) of adsorption is always negative. When a gas is adsorbed on a solid surface, its movement is restricted leading to a decrease in the entropy of the gas i.e., is negative. Now for a process to be spontaneous,should be negative.
Therefore, 
Since is negative, has to be negative to make negative. Hence, adsorption is always exothermic.
5. How are colloids classified on the basis of
(i) Physical states of components
(ii) Nature of dispersion medium and
(iii) Interaction between dispersed phase and dispersion medium?
Ans. Colloids can be classified on various bases:
(i) On the basis of the physical state of the components (by components we mean the dispersed phase and dispersion medium). Depending on whether the components are solids, liquids, or gases, we can have eight types of colloids.
(ii) On the basis of the dispersion medium, sols can be divided as:

Dispersion medium	Name of sol
Water	Aquasol or hydrosol
Alcohol	Alcosol
Benzene	Benzosol
Gases	Aerosol

(iii) On the basis of the nature of the interaction between the dispersed phase and dispersion medium, the colloids can be classified as lyophilic (solvent attracting) and lyophobic (solvent repelling).
6.  Explain what is observed
(i) When a beam of light is passed through a colloidal sol.
(ii) An electrolyte, NaCl is added to hydrated ferric oxide sol.
(iii) Electric current is passed through a colloidal sol?
Ans. (i) When a beam of light is passed through a colloidal solution, then scattering of light is observed. This is known as the Tyndall effect. This scattering of light illuminates the path of the beam in the colloidal solution.
(ii) When NaCl is added to ferric oxide sol, it dissociates to give and ­ ions. Particles of ferric oxide sol are positively charged. Thus, they get coagulated in the presence of negatively charged ions.
(iii) The colloidal particles are charged and carry either a positive or negative charge. The dispersion medium carries an equal and opposite charge. This makes the whole system neutral. Under the influence of an electric current, the colloidal particles move towards the oppositely charged electrode. When they come in contact with the electrode, they lose their charge and coagulate.
7. Action of soap is due to emulsification and micelle formation. Comment.
Ans. The cleansing action of soap is due to emulsification and micelle formation. Soaps are basically sodium and potassium salts of long chain fatty acids, . The end of the molecule to which the sodium is attached is polar in nature, while the alkyl-end is non-polar. Thus, a soap molecule contains a hydrophilic (polar) and a hydrophobic (non-polar) part.
When soap is added to water containing dirt, the soap molecules surround the dirt particles in such a manner that their hydrophobic parts get attached to the dirt molecule and the hydrophilic parts point away from the dirt molecule. This is known as micelle formation. Thus, we can say that the polar group dissolves in water while the non-polar group dissolves in the dirt particle. Now, as these micelles are negatively charged, they do not coalesce and a stable emulsion is formed.
8. What do you mean by activity and selectivity of catalysts?
Ans. (a) Activity of a catalyst:
The activity of a catalyst is its ability to increase the rate of a particular reaction. Chemisorption is the main factor in deciding the activity of a catalyst. The adsorption of reactants on the catalyst surface should be neither too strong nor too weak. It should just be strong enough to make the catalyst active.
(b) Selectivity of the catalyst:
The ability of the catalyst to direct a reaction to yield a particular product is referred to as the selectivity of the catalyst.
9. Explain the terms with suitable examples: (i) Aerosol (ii) Aerosol (iii) Hydrosol
Ans. (i) Aerosol:
A colloidal solution having alcohol as the dispersion medium and a solid substance as the dispersed phase is called an aeroosol.
For example: colloidal sol of cellulose nitrate in ethyl alcohol is an aerosol.
(ii) Aerosol:
A colloidal solution having a gas as the dispersion medium and a solid as the dispersed phase is called an aerosol.
For example: fog
(iii) Hydrosol
A colloidal solution having water as the dispersion medium and a solid as the dispersed phase is called a hydrosol. For example: starch sol or gold sol.

10- Explain:

(a) The same substance can act both as colloids & crystalloids.

(b) Artificial rain is done by spraying salt over clouds. 

Answer:

(a) Sodium chloride acts as a colloid when dissolved in benzene and acts as a crystalloid when dissolved in water. 

(b) Artificial rain is done by spraying common salt over the clouds because it is an electrolyte and thus results in coagulation of water particles.

In This Post we are  providing Chapter- 2 SOLUTIONS NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON SOLUTIONS

• Question 1-  Differentiate between molality and molarity of a solution. What is the effect of change in temperature solution on its molality and molarity molality and molarity

Molality (m) = Moles of solute / Mass of solvent in kg 

For example, 1.00 mol kg^–1 (or 1.00 m) solution of KCl means that 1 mol (74.5 g) of KCl is dissolved in 1 kg of water. 

Molarity (M) is defined as the number of moles of solute dissolved in one litre (or one cubic decimetre) of solution. It is a function of temperature (the volume depends on temperature and mass does not).

Molarity (M) = Moles of solute Molarity / Volume of solution in litre 

For example, 0.25 mol L^–1 (or 0.25 M) solution of NaOH means that 0.25 mol of NaOH has been dissolved in one litre (or one cubic decimetre).

Question 2-  State Raoult’s law. Non-ideal solutions exhibit either positive or negative deviations from Raoult’s law. What are these deviations and why are they caused? Explain with one example for each type. Derive an expression for Raoult’s law when the solute is non-volatile. 

Answer: Raoult’s Law: For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction. 

Thus, 

For component 1 

p₁ ∝ x₁ and p₁ = p^o ₁ x₁ (where p^o ₁ is the vapour pressure of pure component 1 at the same temperature). 

Similarly, 

For component 2 

p₂ = p₂ ⁰ x₂ (where p₂ ⁰ represents the vapour pressure of the pure component 2). 

According to Dalton’s law of partial pressures, the total pressure (p_total ) over the solution phase in the container will be the sum of the partial pressures of the components of the solution and is given as: 

p_total = p₁ + p₂ Substituting the values of p₁ and  p_2, 

we get p_total = p^o ₁ + (p₂ ⁰ –p^o ₁ ) x₂

Non-ideal solutions exhibit Negative deviation from Raoult’s law: For any composition of the non-ideal solution, the partial vapour pressure of each component and total vapour pressure of the solution is less than expected from Raoult’s law. Such solutions show a negative deviation.

Example: Mixture of CHCl₃ and acetone.

Non-ideal solutions show positive deviations from Raoult’s law on mixing of two volatile components of the solution.

Example: Mixture of acetone and benzene solutions show positive deviation

Question 3- Define ‘osmosis’ and ‘osmotic pressure’. What is the advantage of using osmotic pressure as compared to other colligative properties for the determination of molar masses of solutes in solutions?

Answer: Osmosis: The flow of the solvent molecules from the solvent to the solution or from a less concentrated solution to a more concentrated solution through a semipermeable membrane.

Osmotic pressure: The minimum excess pressure that has to be applied to the solution to prevent the entry of the solvent into the solution through the semipermeable membrane.

The advantage of using osmotic pressure as compared to other colligative properties for the determination of molar masses of solutes in solutions is that it uses molarities instead of molalities and it can be measured at room temperature.

Question 4- 

(a) What is van’t Hoff factor? 

(b) A 1.00 molal aqueous solution of trichloroacetic acid (CCl₃COOH) is heated to its boiling point. The solution has a boiling point of 100.18°C. Determine the van’t Hoff factor for trichloroacetic acid. (K_b for water = 0.512 K kg mol^-1)

Answer: (a) In 1880 van’t Hoff introduced factor i, known as the van’t Hoff factor, to account for the extent of dissociation or association. This factor i is defined as: 

i= Normal molar mass / Abnormal molar mass

i = Observed colligative property / Calculated colligative property 

i= (Total number of moles of particles after association/dissociation) / (Number of moles of particles before association/dissociation)

(b) i = 0.3 (Hint-  ΔT_b= iK_bm)

Question 5- Define the following: 

1. Mole fraction
2. Isotonic solutions
3. Hypertonic Solution

Answer: 1. Mole fraction: It is denoted by x and subscript used on the right-hand side of x denotes the component. It is defined as:

 Mole fraction of a component = Number of moles of the component / Total number of moles of all the components

Mole fraction unit is very useful in relating some physical properties of solutions, say vapour pressure with the concentration of the solution and quite useful in describing the calculations involving gas mixtures.

2. Isotonic Solutions: Two solutions having the same osmotic pressure at a given temperature. When such solutions are separated by semipermeable membrane no osmosis occurs between them. 

For example, the osmotic pressure associated with the fluid inside the blood cell is equivalent to that of 0.9% (mass/ volume) sodium chloride solution, called normal saline solution and it is safe to inject intravenously. 

3. Hypertonic Solution: A solution with higher osmotic pressure than another solution. If we place the cells in a solution containing more than 0.9% (mass/volume) sodium chloride, water will flow out of the cells and they would shrink. 

Question 6- State Henry’s law and mention two of its important applications. What is the effect of temperature on the solubility of a gas in a liquid?

Answer: Henry was the first to give a quantitative relation between pressure and solubility of a gas in a solvent which is known as Henry’s law. The law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas. 

Two important applications are: 

1. To increase the solubility of CO₂ in soft drinks and soda water, the bottle is sealed under high pressure. 
2.  Scuba divers must cope with high concentrations of dissolved gases while breathing air at high pressure underwater. Increased pressure increases the solubility of atmospheric gases in the blood. When the divers come towards the surface, the pressure gradually decreases. This releases the dissolved gases and leads to the formation of bubbles of nitrogen in the blood. This blocks capillaries and creates a medical condition known as bends, which are painful and dangerous to life. To avoid bends, as well as, the toxic effects of high concentrations of nitrogen in the blood, the tanks used by scuba divers are filled with air diluted with helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen).

Question 7-18 g of glucose, C₆H₁₂O₆ (Molar mass – 180 g mol^-1) is dissolved in 1 kg of water in a saucepan. At what temperature will this solution boil? (K_b for water = 0.52 K kg mol^-1, boiling point of pure water = 373.15 K)

Answer: The boiling point of the solution is 373.202 K 

Hint- ∆T_b= W_B/ M_B X (100x K_b/ wt. of solvent)

Question 8- 100 mg of a protein is dissolved in just enough water to make 10.0 mL of solution. If this solution has an osmotic pressure of 13.3 mm Hg at 25°C, what is the molar mass of the protein?

(R = 0.0821 L atm mol^-1 K^-1 and 760 mm Hg = 1 atm.)

Answer: Molar mass of protein = 13980.4 g mol^-1

Hint- Use the formula: M= wRT/ π V

Question 9- What mass of NaCl (molar mass = 58.5 g mol^-1) must be dissolved in 65 g of water to lower the freezing point by 7.5°C? The freezing point depression constant, K_f, for water is 1.86 K kg mol^-1. Assume van’t Hoff factor for NaCl is 1.87.

Answer: Mass of NaCl to be dissolved, w₂ = 8.199 g

Hint- Use formula- ΔT_f = iK_fm

Question 10- Define: 

1. Hypotonic Solution
2. Ideal solution 
3. Colligative properties 

Answer:

1. Hypotonic Solution: A solution with lower osmotic pressure than another solution.  If the salt concentration is less than 0.9% (mass/volume), the solution is said to be hypotonic. In this case, water will flow into the cells if placed in this solution and they would swell.

2. Ideal Solutions: The solutions which obey Raoult’s law over the entire range of concentration. The ideal solutions have two other important properties. The enthalpy of mixing of the pure components to form the solution is zero and the volume of mixing is also zero, i.e.,

Δ_mixH = 0,                   Δ_mixV = 0 

It means that no heat is absorbed or evolved when the components are mixed. Also, the volume of solution would be equal to the sum of volumes of the two components. At the molecular level, ideal behaviour of the solutions can be explained by considering two components A and B. In pure components, the intermolecular attractive interactions will be of types A-A and B-B, whereas in the binary solutions in addition to these two interactions, A-B type of interactions will also be present. If the intermolecular attractive forces between the A-A and B-B are nearly equal to those between A-B, this leads to the formation of an ideal solution. A perfectly ideal solution is rare but some solutions are nearly ideal in behaviour. The solution of n-hexane and n-heptane, bromoethane and chloroethane, benzene and toluene, etc. fall into this category.

3. Colligative properties: The properties which depend on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution. Such properties are called colligative properties (colligative: from Latin: co means together, ligare means to bind). 

The above-mentioned questions for CBSE Class 12th Chemistry Chapter 2- Solutions are based on the previous year question papers, NCERT Textbook and sample papers. The students appearing for CBSE Class 12th Chemistry Examination 2020 can go through the links below:

In This Post we are  providing Chapter- 3 ELECTROCHEMISTRY NCERT MOST IMPORTANT QUESTIONS for Class 12 BIOLOGY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON ELECTROCHEMISTRY

Question 1.
What is meant by ‘limiting molar conductivity’?
Answer:
The molar conductivity of a solution at infinite dilution is called limiting molar conductivity and is represented by the symbol Λ_m.

Question 2.
Express the relation between conductivity and molar conductivity of a solution held in a cell. (Delhi 2011)
Answer:
Λ_m = KC= Conductivity  Concentration 

Question 3.
What is the effect of catalyst on:
(i) Gibbs energy (ΔG) and
(ii) activation energy of a reaction? (Delhi 2017)
Answer:
(i) There will be no effect of catalyst on Gibbs .energy.
(ii) The catalyst provides an alternative pathway by decreasing the activation energy of a reaction.

Question 4.
What is the effect of adding a catalyst on
(a) Activation energy (Ea), and
(b) Gibbs energy (AG) of a reaction? (All India 2017)
Answer:
(a) On adding catalyst in a reaction, the activation energy reduces and rate of reaction is fastened.
(b) A catalyst does not alter Gibbs energy (AG) of a reaction.

Question 5.
Two half cell reactions of an electrochemical cell are given below :
MnO^–₄(aq) + 8H⁺ (aq) + 5e^– → Mn²⁺ (aq) + 4H₂O (I), E° = + 1.51 V
Sn²⁺ (aq) → 4 Sn⁴⁺ (aq) + 2e^–, E° = + 0.15 V
Construct the redox equation from the two half cell reactions and predict if this reaction favours formation of reactants or product shown in the equation.
Answer:
The reactions can be represented at anode and at cathode in the following ways :
At anode (oxidation) :
Sn²⁺ → = Sn⁴⁺ (aq) + 2e^– ] × 5 E° = + 0.15 V
At cathode (reduction) :
MnO^–₄(aq) + 8H⁺ (aq) + 5e^– → Mn²⁺ (aq) + 4H₂O (I)] × 2 E° = + 1.51 V
The Net R × M = 2MnO^–₄(aq) + 16H⁺ + 5Sn²⁺ → 2Mn²⁺ + 5Sn⁴⁺ + 8H₂O
Now E°_cell = E°_cathode – E°_anode
= 1.51 – 0.15 = + 1.36 V
∴ Positive value of E°_cell favours formation of product.

Question 6.
Express the relation among the cell constant, the resistance of the solution in the cell and the conductivity of the solution. How is the conductivity of a solution related to its molar conductivity? (All India 2010)
Answer:
1R×1a = Conductance (C) × Cell constant
Molar conductance : (Λ_m) = K×1000c.

Question 7.
Given that the standard electrode potentials (E°) of metals are :
K⁺/K = -2.93 V, Ag⁺/Ag = 0.80 V, Cu²⁺/Cu = 0.34 V,
Mg²⁺/Mg = -2.37 V, Cr³⁺/Cr = -0.74 V, Fe²⁺/Fe = -0.44 V.
Arrange these metals in increasing order of their reducing power. (All India 2010)
Answer:
Ag⁺/Ag < Cu²⁺/Cu < Fe²⁺/Fe < Cr³⁺/Cr < Mg²⁺/ Mg < K⁺/K
More negative the value of standard electrode potentials of metals is, more will be the reducing power.

Question 8.
Two half-reactions of an electrochemical cell are given below :
MnO^–₄ (aq) + 8H⁺ (aq) + 5e^– → Mn²⁺ (aq) + 4H₂O (I), E° = 1.51 V
Sn²⁺ (aq) → Sn⁴⁺ (aq) + 2e^–, E° = + 0.15 V.
Construct the redox reaction equation from the two half-reactions and calculate the cell potential from the standard potentials and predict if the reaction is reactant or product favoured. (All India 2010)
Answer:
The reactions can be represented at anode and at cathode in the following ways :
At anode (oxidation) :
Sn²⁺ → Sn⁴⁺ (aq) + 2e^– ] × 5 E° = + 0.15 V
Af cathode (reduction) :
MnO^–₄(aq) + 8H⁺ (aq) + 5e^– → Mn²⁺ (aq) + 4H₂O (I)] × 2 E° = + 1.51 V
The Net R × M = 2MnO^–₄(aq) + 16H⁺ + 5Sn²⁺ → 2Mn²⁺ + 5Sn⁴⁺ + 8H₂O
Now E°_cell = E°_cathode – E°_anode
= 1.51 – 0.15 = + 1.36 V
∴ Positive value of E°_cell favours formation of product.

Question 9.
The chemistry of corrosion of iron is essentially an electrochemical phenomenon. Explain the reactions occurring during the corrosion of iron in the atmosphere.
Answer:
The mechanism of corrosion is explained on the basis of electrochemical theory. By taking example of rusting of iron, we Refer tothe formation of small electrochemical cells on the surface of iron.
The redox reaction involves
At anode : Fe(S) → Fe²⁺ (aq) + 2e^–
At cathode : H₂O + CO₂ ⇌ H₂CO₃ (Carbonic acid)
H₂CO₃ ⇌2H⁺ + CO₂^2-
H₂O ⇌ H⁺ + OH^–
H⁺ + e^– → H
4H + O₂ → 2H₂O
Then net resultant Redox reaction is
2Fe(s) + O₂ (g) + 4H⁺ → 2Fe²⁺ + 2H₂O

Question 10.
Determine the values of equilibrium constant (K_c) and ΔG° for the following reaction :
Ni(s) + 2Ag⁺ (aq) → Ni²⁺ (aq) + 2Ag(s),
E° = 1.05 V
(1F = 96500 C mol^-1) (Delhi 2011)
Answer:
According to the formula
ΔG° = -nFE° = – 2 × 96500 ×1.05
or ΔG° = -202650 J mol^-1 = -202.65 KJ mol^-1
Now ΔG° ⇒ -202650 J Mol^-1
R = 8.314 J/Mol/K, T = 298 K

In This Post we are  providing Chapter- 1 THE SOLID STATE NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON THE SOLID STATE

Question 1.
Give reason:
(a) Why is Frenkel defect found in AgCI?
(b) What is the difference between phosphorus doped and gallium doped silicon semiconductors? (CBSE Sample Paper 2011)
Answer:
(a) Due to small size of Ag⁺ ion, it can fit into interstitial sites.
(b) Phosphorus doped silicon is n-type semiconductor while gallium doped silicon is p-type semiconductor.

Question 2.
Why does LiCl acquire pink colour when heated in Li vapours? (CBSE Sample Paper 2011)
Answer:
On heating LiCl in Li vapours, the excess of Li atoms deposit on the surface of the crystal. The CT ions diffuse to the surface of the crystal and combine with Li atoms to form LiCl. The electrons produced by ionisation of Li atoms diffuse into the crystal and get trapped at anion vacancies called F-centres. These absorb energy from visible light and radiate pink colour.

Question 3.
Account for the following:
(i) Schottky defects lower the density of related solids.
(ii) Conductivity of silicon increases on doping it with phosphorus. (C.B.S.E. 2013)
Answer:
(i) In Schottky defect, there are holes due to missing cations and anions. Due to the presence of holes in solid, the density decreases.

(ii) Pure silicon has a network lattice in which all the four valence electrons are bonded to four other atoms. Therefore, it is an insulator. However, when silicon is doped with phosphorus having five valence electrons, the impurity leads to excess of electrons after forming four covalent bonds like silicon. The extra electrons serve to conductivity and therefore, the conductivity of silicon doped with phosphorus increases.

Question 4.
(i) What type of non-stoichiometric point defect is responsible for the pink colour of LiCl?
Answer:
Metal excess defect due to anion vacancies filled by free electrons i.e. F-centres.

(ii) What type of stoichiometric defect is shown by NaCI?
Answer:
Schottky defect.

OR

How will you distinguish between the following pairs of terms:
(i) Tetrahedral and octahedral voids
Answer:
Tetrahedral void is surrounded by 4 constituent particles (atoms, molecules or ions). Octahedral void is surrounded by 6 constituent particles (atoms, molecules or ions).

(ii) Crystal lattice and unit cell? (C.B.S.E. 2014)
Answer:
A regular three dimensional arrangement of points in space is called crystal lattice.
The smallest repeating pattern in crystal lattice which when repeated in three dimensional space gives the entire lattice is called the unit cell.

Question 5.
An element crystallises in a fcc lattice with cell edge of 250 pm. Calculate its density if 300 g of this element contain 2 × 10²⁴ atoms.
Answer:
Length of edge, a = 250 pm = 250 × 10^-12 m
= 250 × 10^-10 cm
Volume of unit cell = (250 × 10^-10 cm)³
= 15.625 × 10^-24 cm³
Mass of unit cell = No. of atoms in unit cell × Mass of each atom
Since the element has fcc arrangement, the number of atoms per unit cell, Z = 4
Mass of an atom = \(\frac{300}{2 \times 10^{24}}\) g
∴ Mass of unit cell = \(\frac{300}{2 \times 10^{24}}\) x 4
= 6.0 × 10^-22 g

Question 6.
How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.
Answer:
Consider a unit cell of edge a cm
Volume of unit cell = a³ cm³

Mass of unit cell = No. of atoms in a unit cell × Mass of each atom = Z × m …… (ii)
Mass of an atom present in a unit cell

Knowing density (d), edge length (a), number of atoms per unit cell (Z) and Avogadro’s number (6.02 × 10²³), atomic mass can be calculated.

Question 7.
ZnO turns yellow on heating. Why?
Answer:
When ZnO is heated, it loses oxygen as:

The Zn²⁺ ions are entrapped in the interstitial sites and electrons are entrapped in the neighbouring interstitial sites to maintain electrical neutrality. This results in metal excess defect. Due to the presence of free electrons in the interstitial sites the colour is yellow.

Question 8.
(a) What type of semiconductor is obtained when silicon is doped with boron?
Answer:
(a) When silicon is doped with boron having three valence electrons, the bonds formed create electron deficient sites called holes. Under the influence of applied electric field, one electron from neighbouring atom moves to fill the hole but creates another hole at its own place. Therefore, the electrical conductance is due to movement of positive holes. Therefore, this type of semi-conductor is called p-type semi-conductor.

(b) What type of magnetism is shown in the following alignment of magnetic moments?
↑    ↑    ↑    ↑   ↑   ↑    ↑
Answer:
Ferromagnetic.

(c) What type of point defect is produced when AgCI is doped with CdCl₂?
Answer:
CdCl₂ on adding to AgCl introduces impurity defect. The addition of one Cd²⁺ ion will replace two Ag⁺ ions to maintain electrical neutrality. One of the positions of Ag⁺ will be occupied by Cd²⁺ ion and the other will be left as a hole. Thus, a hole is created similar to Schottky defect.

Question 9.
Account for the following:
(i) Schottky defects lower the density of related solids.
Answer:
In Schottky defect, there are holes due to missing cations and anions. Due to the presence of holes in solid, the density decreases.

(ii) Conductivity of silicon increases on doping it with phosphorus.
Answer:
Pure silicon has a network lattice in which all the four valence electrons are bonded to four other atoms. Therefore, it is an insulator. However, when silicon is doped with phosphorus (s²p³) having five valence electrons, the impurity leads to excess of electrons after forming four covalent bonds like silicon. The extra electrons serve to conductivity and therefore, the conductivity of silicon doped with phosphorus increases.

Question 10.
Examine the given defective crystal

Answer the following questions:
(i) What type of stoichiometric defect is shown by the crystal?
Answer:
Schottky defect

(ii) How is the density of the crystal affected by this defect?
Answer:
Density of the crystal decreases.

(iii) What type of ionic substances show such defect?
Answer:
Crystals having
(a) high coordination number and
(b) ions (cations and anions) of almost similar sizes. For example: KCl, KBr.

SHORT ANSWER TYPE QUESTION:

Q1.When can a hazard become a disaster?

Answer
A hazard can become a disaster when it becomes active. A disaster causes losses to property and life on a large scale.
Q2. Why are there more earthquakes in the Himalayas and in the north-eastern region of India?
Answer

The Indian plate is moving at a speed of one centimetre per year towards the north and northeastern direction and the movement of plates is being constantly obstructed by the Eurasian plate from the north. This result in accumulation of energy at different points of time. Excessive accumulation of energy results in building up of stress, which ultimately leads to the breaking up of the lock and the sudden release of energy causes earthquakes in the Himalayas and in the north-eastern region of India

Q3. What are the basic requirements for the formation of a cyclone?

Answer

The basic requirements for the formation of a cyclone are:
• Large and continuous supply of warm and moist air that can release enormous latent heat.
• Strong Coriolis force that can prevent filling of low pressure at the centre.
• Unstable condition through the troposphere that creates local disturbances around which a cyclone develops.
• Absence of strong vertical wind wedge, which disturbs the vertical transport of latent heat.

Q4. How are the floods in Eastern India different from the ones in Western India?

Answer

The flood occur more in Eastern India as compared to Western India because of high rainfall in eastern India than western India. Also, the floods of eastern India are more disastrous in comparison to western India.

Q5. Why are there more droughts in Central and Western India?
Answer

Central and Western India experiences less rainfall because the intensity of monsoon winds becomes weak as they reach in these areas.

Q6. Identify the Landslide-prone regions of India and suggest some measures to mitigate the disasters caused by these.
Answer

The Landslide-prone regions of India are:
• Young mountainous areas in the Himalayas and Andaman and Nicobar.
• High rainfall regions with steep slopes in the Western Ghats and Nilgiris.
• The north-eastern regions.
• Trans-Himalayan areas of Ladakh and Spiti (Himachal Pradesh)
• Undulated yet stable relief and low precipitation areas in the Araval
• Rain shadow areas in the Western and Eastern Ghats and Deccan plateau.• Landslides due to the mining and subsidence are most common in the states like Jharkhand, Orissa, Chhattisgarh, Madhya Pradesh, Maharashtra, Andhra Pradesh, Karnataka etc.

Some measures to mitigate the disasters are:
• Restriction on the construction and other developmental activities such as roads and dams, limiting
agriculture to valleys.
• Areas with moderate slopes, and control on the development of large settlements in the high vulnerability zones, should be enforced.
• Terrace farming should be encouraged in the northeastern hill states where Slash and Burn/Shifting
Cultivationis still prevalent.
Q7. What is vulnerability? Divide India into natural disaster vulnerability zones based on droughts and suggest some mitigation measures.
Answer

Vulnerability means the risk becoming victim to the disaster.

On the basis of severity of droughts, India is divided into following regions:

• Extreme Drought Affected Areas: Most parts of Rajasthan, particularly areas to the west of the Aravali hills, i.e. Marusthali and Kachchh regions of Gujarat. Included here are also the districts like Jaisalmer and Barmer from the Indian desert that receive less that 90 mm average annual rainfall.

• Severe Drought Prone Area: Parts of eastern Rajasthan, most parts of Madhya Pradesh, eastern parts of Maharashtra, interior parts of Andhra Pradesh and Karnataka Plateau, northern parts of interior Tamil Nadu and southern parts of Jharkhand and interior Odisha are included in this category.

• Moderate Drought Affected Area: Northern parts of Rajasthan, Haryana, southern districts of Uttar Pradesh, the remaining parts of Gujarat, Maharashtra except Konkan, Jharkhand and Coimbatore plateau of Tamil Nadu and interior Karnataka are included in this category. The remaining parts of India can be considered either free or less prone to the drought.

Some mitigation measures are:

Identification of ground water potential in the form of aquifers.• Transfer of river water from the surplus to the deficit areas.• Planning for inter-linking of rivers and construction of reservoirs and dams.• Remote sensing and satellite imageries can be useful in identifying the possible river-basins that can be inter-linked and in identifying the ground water potential.

Q8.When can developmental activities become the cause of disasters?

Answer

There are many times when developmental activities carried by human beings that are directly responsible for disasters.

Bhopal Gas tragedy, Chernobyl nuclear disaster, wars, release of CFCs (Chlorofluorocarbons) and increase of green house gases, environmental pollutions like noise, air, water and soil are some of the disasters which are created due to developmental activities done by humans. 
There are some other activities of human beings that accelerate or intensify disasters indirectly. Landslides and floods due to deforestation, unscientific land use and construction activities in fragile areas are some of the disasters that are the results of indirect human actions.
The human-made disasters have increased both in their numbers and magnitudes over the years and serious efforts are necessary are on at various levels to prevent and minimise their occurrences.

Long Answer Type Questions:

Q1.Explain about different types of drought.
Answer:
Different types of droughts are as follows:

• Meteorological Drought: It is a situation when there is a prolonged period of inadequate rainfall marked with mal-distribution of the same over time and space.
• Agricultural Drought: It is also known as soil moisture drought, characterised by low soil moisture that is necessary to support the crops, thereby resulting in crop failures. Moreover, if an area has more than 30 percent of its gross cropped area under irrigation, the area is excluded from the drought-prone category.
• Hydrological Drought: It results when the availability of water in different storages and reservoirs like aquifers, lakes, reservoirs, etc. falls below what the precipitation can replenish.
• Ecological Drought: When the productivity of a natural ecosystem fails due to shortage of water and as a consequence of ecological distress, damages are induced in the ecosystem. Various parts of India experience these droughts recurrently which result in some serious socio-economic and ecological problems.

Q2.On the basis of past experiences, frequency and certain causal relationships with the controlling factors like geology, geomorphic agents, slope, land-use, vegetation cover and human activities, India has been divided into how many zones?
Answer:
On the basis of past experiences, frequency and certain causal relationships with the controlling factors like geology, geomorphic agents, slope, land-use, vegetation cover and human activities, India has been divided into a number of zones.

1. Very High Vulnerability Zone: Highly unstable, relatively young mountainous areas in the Himalayas and Andaman and Nicobar, high rainfall regions with steep slopes in the Western Ghats and Nilgiris, the north-eastern regions, along with areas that experience frequent ground-shaking due to earthquakes, etc. and areas of intense human activities, particularly those related to construction of roads, dams, etc. are very highly vulnerable.

2. High Vulnerability Zone: Areas that have almost similar conditions to those included in the very high vulnerability zone are also included in this category. All the Himalayan states and the states from the north-eastern regions except the plains of Assam are included in the high vulnerability zones.

3. Moderate to Low Vulnerability Zone: Areas that receive less precipitation such as Trans-Himalayan areas of Ladakh and Spiti, undulated yet stable relief and low precipitation areas in the Aravali, rain shadow areas in the Western and Eastern Ghats and Deccan plateau also experience occasional landslides. Landslides due to mining and subsidence are most common in states like Jharkhand, Orissa, Chhattisgarh, Madhya Pradesh, Maharashtra, Andhra Pradesh, Karnataka, Tamil Nadu, Goa and Kerala.

4. Other Areas: The remaining parts of India, particularly states like Rajasthan, Haryana, Uttar Pradesh, Bihar, West Bengal, Assam and Coastal regions of the southern States are safe as far as landslides are concerned.

SHORT ANSWER TYPE QUESTION:

Q1. What is soil?

Answer
Soil is the mixture of rock debris and organic materials which develop on the earth’s surface. Components of the soil are mineral particles, humus, water and air.

Q2.What are the main factors responsible for the formation of soil?
Answer
The major factors affecting the formation of soil are relief, parent material, climate, vegetation and other life-forms and time.

Q3.Mention the three horizons of a soil profile.
Answer
There are three horizons of a soil profile:
• ‘Horizon A’ is the topmost zone, where organic materials have got incorporated with the mineral matter, nutrients and water, which are necessary for the growth of plants.
• ‘Horizon B’ is a transition zone between the ‘horizon A’ and
• ‘Horizon C’ is composed of the loose parent material.
Q4. What is soil degradation?
Answer
Soil degradation can be defined as the decline in soil fertility, when the nutritional status declines and depth of the soil goes down due to erosion and misuse. 
Q5. What is the difference between Khadar and Bhangar?
Answer
Khadar is the new alluvium and is deposited by floods annually, which enriches the soil by depositing fine silts while Bhangar represents a system of older alluvium, deposited away from the flood plains.

Q6.What are black soils? Describe their formation and characteristics.
Answer
Black soils are also known as the ‘Regur Soil’ or the ‘Black Cotton Soil’. It covers most of the Deccan Plateau which includes parts of Maharashtra, Madhya Pradesh, Gujarat, Andhra Pradesh and some parts of Tamil Nadu.
The black soil is formed by the weathering of igneous rocks and cooling of lava after a volcanic eruption.
Characteristics of black soils are:
• The black soils are generally clayey, deep and impermeable.
• They swell and become sticky when wet and shrink when dried.
• The black soils are rich in lime, iron, magnesia and alumina.
• They also contain potash.
• The colour of the soil ranges from deep black to grey.
Q7. What is soil conservation? Suggest some measures to conserve soil.
Answer
Soil conservation is a methodology to maintain soil fertility, prevent soil erosion and exhaustion, and improve the degraded condition of the soil.
Some measures to conserve soil are:
• Afforestation should be encouraged.
• Over-grazing and shifting cultivation should be regulated and controlled by educating villagers about the consequences.
• In arid and semi-arid areas, efforts should be made to protect cultivable lands from encroachment by sand dunes through developing shelter belts of trees and agro-forestry.
• Contour terracing should be done in sloppy areas to reduce surface flow.
• Efforts should be made to prevent gully erosion and control their formation. Finger gullies can be eliminated by terracing. In bigger gullies, the erosive velocity of water may be reduced by constructing a series of check dams.
• Lands not suitable for cultivation should be converted into pastures for grazing.
Q8. How do you know that a particular type of soil is fertile or not? Differentiate between naturally determined fertility and culturally induced fertility.
Answer
We can know about the fertility of a particular type of soil by having a test of the soil sample. To check fertility of the soil we need to know about their inherent characteristics and external features such as texture, colour, slope of land and moisture content.
The fertility which is present in a particular type of soil naturally, it is called naturally determined fertility. Some soils have phosphorous, potassium, calcium, humus content and nitrogen naturally.
Some soils are made fertile by adding fertilizers and manures, it is called culturally induced fertility. They are deficient and minerals and humus content.

Long Answer Type Questions:

Q1.Explain about the regions affected by soil erosion.
Answer:
West Bengal, Uttar Pradesh, Maharashtra, Tamil Nadu, Karnataka, Delhi, Rajasthan and in many parts of the country soil erosion has been a big problem. In mountain regions, there is erosion due to over grazing. In Meghalaya and Nilgiri hills due to potato cultivation and in Himalaya region due to deforestation, soil erosion is increasing.
Wind erosion is significant in arid and semi-arid regions.

In regions with heavy rainfall and steep slopes, erosion by running water is more significant. Water erosion which is more serious and occurs extensively in different’ parts of India, takes place mainly in the form of sheet and gully erosion. Gully erosion is common on steep slopes. Ravines are widespread, in the Chambal basin. Besides this, they are also found in Tamil Nadu and West Bengal. The country is losing about 8,000 hectares of land to ravines every year.

Q2.Explain about causes responsible for soil erosion.
Answer:
Following causes are responsible for soil erosion:

1. Human Activities: Human activities too are responsible for soil erosion to a great extent. As the human population increases, the demand on the land also increases. Forest and other natural vegetation is removed for human settlement, for cultivation, for grazing animals and for various other needs. Wind and water are powerful agents of soil erosion because of their ability to remove soil and transport it.

2. Deforestation: Deforestation is one of the major causes of soil erosion. Plants keep soils bound in locks of roots, and thus, prevent erosion. They also add humus to the soil by shedding leaves and twigs. Forests have been denuded practically in most parts of India but their effect on soil erosion are more in hilly parts of the country.

3. Wind and water: Wind erosion is significant in arid and semi-arid regions.
In regions with heavy rainfall and steep slopes, erosion by running water is more significant. Water erosion which is more serious and occurs extensively in different parts of India, takes place mainly in the form of sheet and gully erosion. Sheet erosion takes place on level lands after a heavy shower and the soil removal is not easily noticeable.

Q3.In India, fertility of soil is also destroyed by over irrigation. Explain.
Answer:
A fairly large area of arable land in the irrigated zones of India is becoming saline because of over-irrigation.

• The salt lodged in the lower profiles of the soil comes up to the surface and destroys its fertility.
• Chemical fertilisers in the absence of organic manures are also harmful to the soil.
• Unless the soil gets enough humus, chemicals harden it and reduce its fertility in the long run. This problem is common in all the command areas of the river valley projects, which were the first beneficiaries of the Green Revolution.
• According to estimates, about half of the total land of India is under some degree of degradation.

Every year, India loses millions of tonnes of soil and its nutrients to the agents of its degradation, which adversely affects our national productivity. So, it is imperative to initiate immediate steps to reclaim and conserve soils.

Q4.Wind and water are two important agents of soil erosion. Explain.
Answer:
Wind and water are powerful agents of soil erosion because of their ability to remove soil and transport it. Wind erosion is significant in arid and semi-arid regions. Erosion by running water is more significant in regions with heavy rainfall and steep slopes. Water erosion which is more serious and occurs extensively in different parts of India, takes place mainly in the form of sheet and gully erosion. Sheet erosion takes place on level lands after a heavy shower and the soil removal is not easily noticeable.

But it is harmful since it removes the finer and more fertile top soil. Gully erosion is common on steep slopes. Gullies deepen with rainfall, cut the agricultural lands into small fragments and make them unfit for cultivation. A region with a large number of deep gullies or ravines is called bad land topography. Ravines are widespread, in the Chambal basin. They are also found in Tamil Nadu and West Bengal. The country is losing about 8,000 hectares of land to ravines every year.

Q5.Explain about laterite soils.
Answer:
The word ‘laterite’ has been derived from a Latin word ‘later’ meaning ‘brick’. The laterite soil is widely spread in India and is mainly found on the summits of the Western Ghats, Eastern Ghats, Rajmahal Hills, Vindhyas, Satpuras and Malwa plateau. It’s well- developed in southern Maharashtra, and parts of Orissa, West Bengal, Karnataka, Andhra Pradesh, Kerala, Bihar, Assam and Meghalaya.

• The laterite soil is formed under conditions of high temperature and heavy rainfall with alternate wet and dry periods.
• Such climatic conditions promote leaching of soil. Leaching is a process in which heavy rains wash away the fertile part of the soil.
• The laterite soil is red in colour and composed of little clay and much gravel of red sandstones.
• Laterite soil generally is poor in lime and deficient in nitrogen. The phosphate contents are generally high.

Due to intensive leaching, the laterite j soil generally lacks fertility and is of low value for crop production. But when manured and timely irrigated, the soil is suitable for producing plantation crops like tea, coffee, rubber, coconut, areca nut, etc. It also provides valuable building materials.

SHORT ANSWER TYPE QUESTION:

Q1.What is natural vegetation? Under what climatic conditions are tropical evergreen forests develop?

Answer

Natural vegetation refers to a plant community that has been left undisturbed over a long time, so as to allow its individual species to adjust themselves to climate and soil conditions as fully as possible.
Tropical evergreen forests develop in warm and humid areas with an annual precipitation of over 200 cm and mean annual temperature above 22°C.

Q2. What do you understand by social forestry?

Answer

Social forestry means the management and protection of forests and afforestation on barren lands with the purpose of helping in the environmental, social and rural development.

Q3. Define Biosphere reserves?

Answer

A Biosphere Reserve is a unique and representative ecosystem of terrestrial and coastal areas which are internationally recognised within the framework of UNESCO’s Man and Biosphere (MAB) Programme.

Q4. What is the difference between forest area and forest cover?

Answer

The forest area and forest cover are not the same.

• The forest area is the area notified and recorded as the forest land irrespective of the existence of trees, while the actual forest cover is the area occupied by forests with canopy.

• The forest area is based on the records of the State Revenue Department, while the forest cover is based on aerial photographs and satellite imageries.

• According to state records, the forest area covers 23.28 per cent of the total land area of the country while the actual forest cover in India is only 21.05 percent.

Q5.What steps have been taken up to conserve forests?

Answer

There are various steps taken based on the forest conservation policy. These are:

• Social forestry: It means the management and protection of forests and afforestation on barren lands with the purpose of helping in the environmental, social and rural development. The National Commission on Agriculture (1976) has classified social forestry into three categories. These are Urban forestry, Rural forestry and Farm forestry.

→ Urban forestry: It pertains to the raising and management of trees on public and privately owned lands in and around urban centres such as green belts, parks, roadside avenues, industrial and commercial green belts, etc.–

→ Rural forestry: It lays emphasis on promotion of agro-forestry and community-forestry.

→ Agro-forestry: It is the raising of trees and agriculture crops on the same land inclusive of the waste patches.

• Community forestry: It involves the raising of trees on public or community land such as the village pasture and temple land, roadside, canal bank, strips along railway lines, and schools etc.

• Farm forestry is a term applied to the process under which farmers grow trees for commercial and non-commercial purposes on their farm lands.

Q6.How can people’s participation be effective in conserving forests and wildlife?

Answer

The government can make policies on the conservation of forests and wildlife but it is on the people who take part in this and make them successful. It is mostly the local people who take part in the illegal activities by damaging the environment for their benefits knowingly or unknowingly. The government need the cooperation of common people for ensuring the protection against the deforestation, poaching, hunting etc. This can effective in many ways:

• Holding a regular meeting of local people and make them aware of the advantages of conservation of forests and wildlife.
→ The government officials can inform them about the various policies of the government about the conservation and how they can be part of it.

• Incentives for the good work: The government can provide the incentives to the local people for their good work in protecting the local trees or wildlife which will encourage others to do the same work. Gradually, it will spread to mass scale.

• Various NGOs can also help in doing this activity by providing the skills to the people in conserving forests and wildlife through their experts.

Long Answer Type Questions:

Q1.“Natural vegetation is an outcome of climate.” Substantiate the statement by taking example of Indian vegetation.
Answer:
India is a land of great variety of natural vegetation. Himalayan heights are marked with temperate vegetation; the Western Ghats and the Andaman Nicobar Islands have tropical rain forests.

• Tropical evergreen forests are found in warm and humid areas with an annual precipitation of over 200 cm and mean annual temperature above 22°C. In these forests, trees reach at great heights up to 60 m or above. The semi¬evergreen forests are found in the less rainy parts of these regions.
• Tropical deciduous forests are spread over regions which receive rainfall between 70-200 cm.
• The moist deciduous forests are more pronounced in the regions which record rainfall between 100-200 cm. Dry deciduous forest covers vast areas of the country, where rainfall ranges between 70-100 cm.
• Tropical thorn forests occur in the areas which receive rainfall less than 50 cm. In mountainous areas, the decrease in temperature with increasing altitude leads to a corresponding change in natural vegetation.
• The Himalayan ranges show a succession of vegetation from the tropical to the tundra, with change in the altitude. Deciduous forests are found in the foothills of the Himalayas. It is succeeded by the wet temperate type of forests between an altitude of 1,000-2,000 m. In the higher hill ranges of north-eastern India, hilly areas of West Bengal and Uttarakhand, evergreen broad leaf trees such as oak and chestnut are predominant. Between 1,500-1,750 m, pine forests are also well-developed in this zone, with Chir Pine as a very useful commercial tree.

Q2.Mention the reasons for the decline of wildlife in India?
Answer:
Important reasons for the decline of Wildlife in India are-

• Industrial and technological advancement brought about a rapid increase in the exploitation of forest resources.
• More and more lands were closed for agriculture, human settlement, roads, mining, resources, etc.
• Pressure on forests maintained due to looping for fodder and fuel, wood and removal of small timber by the local people.
• Grazing by domestic cattle caused an adverse effect on wildlife and its habitat.
• Hunting was taken up as a sport by the elite and hundreds of wild animals were killed in a single hunt. Now commercial poaching is rampant.
• Incidence of forest fire.

Q3.According to the statistics received from state records, there are differences in forest area and actual forest cover. Explain.
Answer:
According to state records, the forest area covers 23.28 percent of the total land area of the country. It is important to note that the forest area and the actual forest cover are not the same. The forest area is the area notified and recorded as the forest land irrespective of the existence of trees, while the actual forest cover is the area occupied by forests with canopy.
Forest area is based on the records of the State Revenue Department, while the actual forest cover is based on aerial photographs and satellite imageries.

According to India State of Forest Report 2011, the actual forest cover in India is only 21.05 percent. Of the forest cover, the share of dense and open forests is 12.29 and 8.75 percent respectively. Both forest area and forest cover vary from state to state. Lakshadweep has zero percent forest area; Andaman and Nicobar Islands have 86.93 percent. Most of the states with less than 10 percent of the forest area lie in the north and northwestern part of the country. These are Rajasthan, Gujarat, Punjab, Haryana and Delhi.

States with 10-20 percent forest area are Tamil Nadu and West Bengal. In Peninsular India, excluding Tamil Nadu, Dadra and Nagar Haveli and Goa, the area under forest cover is 20-30 percent. The northeastern states have more than 30 percent of the land under forest. Hilly topography and heavy rainfall are good for forest growth. There is a lot of variation in actual forest cover, which ranges from 9.56 percent in Jammu and Kashmir to 84.01 percent in Andaman and Nicobar Islands.

Q4.On the basis of actual forest cover, in how many categories have Indian states been divided?
Answer:
On the basis of the percentage of the actual forest cover, the states have been grouped into four regions:

• The region of high concentration > 40: It includes Andaman and Nicobar islands, Mizoram, Nagaland, Arunachal Pradesh which have 80% of their total area under forests. Manipur, Tripura, Meghalaya, Sikkim and Dadar and Haveli have forest cover between 40-80%.
• The region of medium concentration 20-40: It includes Madhya Pradesh, Odisha, Goa, Kerala, Assam and Himachal Pradesh. In Goa, actual forest cover is 33.27% which is highest in this range. Thereafter, comes Assam and Orissa. In other states 30% of their area is covered with forests.
• The region of low concentration 10-20: It includes states of Maharashtra, Andhra Pradesh, Karnataka, Tamil Nadu, Bihar and Uttar Pradesh.
• The region of very low concentration < 10: It includes states of Rajasthan, Punjab, Haryana, and Gujarat. It also includes union territories of Delhi and Chandigarh. It also includes West Bengal.

Q5.Explain in short about four important Biospheres of India.
Answer:
Four Biosphere Reserves have been recognised by the UNESCO on World Network of Biosphere Reserves. These are as follows:

1. Nilgiri Biosphere Reserve: The Nilgiri Biosphere Reserve (NBR) is the first of the fourteen biosphere reserves of
India. It was established in September 1986. It embraces the sanctuary complex of Wyanad, Nagarhole, Bandipur and Mudumalai, the entire forested hill slopes of Nilambur, the Upper Nilgiri plateau, Silent Valley and the Siruvani hills. The total area of the biosphere reserve is around 5,520 sq. km. The largest south Indian population of elephant, tiger, gaur, sambar and chital as well as a good number of endemic and endangered plants are also found in this reserve. The topography of the NBR is extremely varied, ranging from an altitude of250 m to 2,650 m. About 80 percent of the flowering plants reported from the Western Ghats occur in the Nilgiri Biosphere Reserve.

2. Nanda Devi Biosphere Reserve: The Nanda Devi Biosphere Reserve is situated in Uttarakhand. It includes parts of Chamoli, Almora, Pithoragarh and Ba’geshwar districts. The major forest types of the reserve are temperate. A few important species are silver weed and orchids like latifolie and rhododendron. The biosphere reserve has a rich fauna like the snow leopard, black bear, brown bear, musk deer, snow- cock, golden eagle and black eagle.

3. Sunderbans Biosphere Reserve: It is located in the swampy delta of the river Ganga in West Bengal. It extends over a vast area of 9,630 sq. km and consists of mangrove forests, swamps and forested islands. Sunderbans is the home of nearly 200 Royal Bengal tigers. More than 170 birds species are known to inhabit these mangrove forests. In the Sunderbans, the mangrove forests are characterised by Heritiera fomes, a species valued for its timber.

4. Gulf of Mannar Biosphere Reserve: The Gulf of Mannar Biosphere Reserve covers an area of 105,000 hectares on the south-east coast of India. It is one of the world’s richest regions from a marine biodiversity perspective. The biosphere reserve comprises 21 islands with estuaries, beaches, forests of the nearshore environment, sea grasses, coral reefs, salt marshes and mangroves.

SHORT ANSWER TYPE QUESTION:

Q1. What are the three important factors which influence the mechanism of Indian weather?
Answer
The three important factors which influence the mechanism of Indian weather are:• Distribution of air pressure and winds on the surface of the earth.
• Upper air circulation caused by factors controlling global weather and the inflow of different air masses and jet streams.
• Inflow of western cyclones generally known as disturbances during the winterseason and tropical depressions during the south-west monsoon period into India, creating weather conditions favourable to rainfall.

Q2. What is the Inter-Tropical Convergene Zone?
Answer
The Inter Tropical Convergence Zone (ITCZ) is a low pressure zone located at the equatorwhere trade winds converge, and so, it is a zone where air tends to ascend.
Q3. What is meant by ‘bursting of monsoon’? Name the place of India which gets the highest rainfall.
Answer
The sudden onset of the moisture-laden winds associated with violent thunder and lightening, is often termed as the “break” or “burst” of the monsoons.Mawsynram in Meghalaya is the place of India which gets the highest rainfall.
Q4. Define ‘climatic region’? What are the bases of Koeppen’s classification?
Answer
 A climatic region has a homogeneous climatic condition which is the result of a combination of factors. The bases of Koeppen’s classification are:• Temperature• Precipitation

Q5.Which type(s) of cyclones cause rainfall in north-western India during winter? Where do they originate?
Answer
Western cyclonic disturbances cause rainfall in north-western India during winter. They originate over the Mediterranean Sea.

Q6.Notwithstanding the broad climatic unity, the climate of India has many regional variations. Elaborate this statement giving suitable examples.
Answer
Due to the influence of monsoon, India as a whole has broad climatic unity. However, the climate of India has many regional variations expressed in the pattern of winds, temperature and rainfall, rhythm of seasons and the degree of wetness or dryness. There are various examples of this:
• Temperature: In the summer the mercury occasionally touches 55°C in the western Rajasthan, it drops down to as low as minus 45°C in winter around Leh. Churu in Rajasthan may record a temperature of 50°C or more on a June day while the temperature in Tawang (Arunachal Pradesh) is 19°C on the same day.
• Rainfall: Cherrapunji and Mawsynram in the Khasi Hills of Meghalaya receive rain fallover 1,080 cm in a year while Jaisalmer in Rajasthan rarely gets more than 9 cm of rainfall during the same period.
• Monsoon: The Ganga delta and the coastal plains of Orissa are hit by strong rain-bearing stormsal most every third or fifth day in July and August while the Coromandal coast, a thousand km to the south, goes generally dry during these months. 
Q7.How many distinct seasons are found in India as per the Indian Meteorological Department? Discuss the weather conditions associated with any one season in detail.
Answer
The meteorological department recognise the following four seasons :(i) the cold weather season(ii) the hot weather season(iii) the southwest monsoon season(iv) the retreating monsoon season.
• The cold weather season: The cold weather season sets in by mid-November in northern India. December and January are the coldest months in the northern plain. The mean daily temperature remains below 21°C over most parts of northern India. The night temperature may be quite low.The Peninsular region of India, however, does not have any well-defined cold weather season. There is hardly any seasonal change in the distribution pattern of the temperature in coastal areas because of moderating influence of the sea and the proximity to equator.
During the winters, the weather in India is pleasant. The pleasant weather conditions, however, at intervals, get disturbed by shallow cyclonic depressions originating over the east Mediterranean Sea and travelling eastwards.
Winter monsoons do not cause rainfall as they move from land to the sea. It is because they have little humidity and due to anti cyclonic circulation on land. So, most parts of India do not have
rainfall in the winter season. However, there are some exceptions, Rainfall occurs in northwestern India due to western disturbances. Central parts of India and northern parts of southern peninsula also get winter rainfall occasionally.

Long Answer Type Questions:

Q1.Explain the important features of Winter Season of India.
Answer:
By October, the rainy season comes to an end all over the country and the days become short and the night become long. The rays of the sun are not overhead. The air turns cooler in the plains marking the coming of the winter season.

The winter season lasts from November to February every year. In the Northern Plains, very cold wind blow making the winter months severe. December and January are the coldest months in the northern plain. The mean daily temperature remains below 21°C over most parts of northern India.

Most of the hilly areas receive heavy snowfall. January is the coldest month in the Northern Plains. The winter season in Peninsular India is mild as a result of the influence of the surrounding water bodies. The coromandel Coast receives heavy rainfall during this season.

The desert is cool during the day but cold at night. The air starts warming up in the month of March, and the weather is neither cold nor hot. This time of early summer is also called the spring season. Every part of our country regularly experiences this cycle of seasons.

Q2.According to Koeppen, in how many groups- can you classify the climate of India?
Answer:
Koeppen identified a close relationship between the distribution of vegetation and climate. He selected certain values of temperature and precipitation and related them to the distribution of vegetation and used these values for classifying the climates. Koeppen introduced the use of capital and small letters to designate climatic groups and types. Koeppen recognised five major climatic groups, four of them are based on temperature and one on precipitation.

• Tropical climates,
• Dry climates,
• Warm temperate climates,
• Cool temperate climates,
• Ice climates

Classification of Indian climatic regions: Amw – Monsoon with short dry season.
Areas: West coast of India south of Goa
As – Monsoon with dry summer
Areas: Coromandel coast of Tamil Nadu
Aw – Tropical savannah.
Areas: Most of the Peninsular plateaus, south of the Tropic of Cancer
Bwhw – Semi-arid steppe climate.
Areas: North-western Gujarat, some parts of western Rajasthan and Punjab
Bwhw – Hot desert. Areas: Extreme western Rajasthan
Cwg – Monsoon with dry winter.
Areas: Ganga plain, eastern Rajasthan, northern Madhya Pradesh, most of North-east India
Dfc – Cold humid winter with short summer.
Areas: Arunachal Pradesh
E – Polar type. Areas: Jammu and Kashmir, Himachal Pradesh and Uttarakhand.

Q3.Explain the spatial variation in the rainfall throughout the country.
Answer:
There is great variation in rainfall throughout the country.

• While Cherrapunji and Mawsynram in the Khasi Hills of Meghalaya receive rainfall over 1,080 cm in a year, Jaisalmer in Rajasthan rarely gets more than 9 cm of rainfall during the same period.
• Tura situated in the Garo Hills of Meghalaya may receive an amount of rainfall in a single day which is equal to 10 years of rainfall at Jaisalmer. While the annual precipitation is less than 10 cm in the north-west Himalayas and the western deserts, it exceeds 400 cm in Meghalaya.
• The highest rainfall occurs along the west coast, on the western Ghats as well as in the sub-Himalayan areas in the north-west and the hills of Meghalaya, rainfall exceeding 200 cm. In some parts of Khasi and Jaintia hills, the rainfall exceeds 1,000 cm. In the Brahmaputra valley and the adjoining hills, the rainfall is less than 200 cm.
• Rainfall between 100-200 cm is received in southern parts of Gujarat, east Tamil Nadu, North-eastern Peninsular covering Orissa, Jharkhand, Bihar, eastern Madhya Pradesh, Northern Ganga Plain along the sub-Himalayas and the Cachar valley and Manipur.
• Western Uttar Pradesh, Delhi, Haryana, Punjab, Jammu and Kashmir, eastern Rajasthan, Gujarat and Deccan Plateau receive rainfall between 50-100 cm.

Q4.Explain the spatial variation in temperature in India.
Answer:
India has hot monsoonal climate which is the prevalent climate in south and south-east Asia.

• While in the summer the mercury occasionally touches 55°C in the western Rajasthan, it drops down to as low as minus 45°C in winter around Leh.
• Churu in Rajasthan may record a temperature of 50°C or more on a June day while the mercury hardly touches 19°C in Tawang (Arunachal Pradesh) on the same day.
• On a December night, temperature in Drass (Jammu and Kashmir) may drop down to minus 45°C while Thiruvananthapuram or Chennai on the same night records 20°C or 22°C.
• In Kerala and in the Andaman Islands, the difference between day and night temperatures may be hardly seven or eight degree Celsius. But in the Thar desert, if the day temperature is around 50°C, at night, it may drop down considerably upto 15°-20°C.
• While snowfall occurs in the Himalayas, it only rains over the rest of the country. Similarly, variations are noticeable not only in the type of precipitation but also in its amount.

Q5.How economic life in India is affected by monsoon?
Answer:
Economic life of India is extremely affected by the monsoon.

• Monsoon is that axis around which revolves the entire agricultural cycle of India. Around 64 % people of India depend on agriculture for their livelihood and agriculture itself is based on south-west monsoon.
• Except Himalayas all the parts of the country have temperature above the threshold level to grow the crops or plants throughout the year.
• Regional variations in monsoon climate help in growing various types of crops.
• Variability of rainfall brings droughts or floods every year in some parts of the country.
• Agricultural property of India depends very much on timely and adequately distributed rainfall. If it fails, agriculture is adversely affected particularly in those regions where means of irrigation are not developed.
• Sudden monsoon burst creates problems of soil erosion over large areas in India.
• Winter rainfall by temperate cyclones in north India is highly beneficial for Rabi crops.
• Regional climatic variation in India is reflected in the vast variety of food, clothes and house types.

Q6.What is Global Warming? What are the effects of Global Warming?
Answer:
Due to global warming the polar ice caps and mountain glaciers would melt and the amount of water in the ocean would increase. It leads to rise in the sea level and melting of glaciers and sea-ice due to warming.

Effects:

• Sea level will rise 48 cm by the end of twenty first century.
• Increase the incidence of annual flooding.
• Insect-borne diseases like malaria, and leads to shift in climatic boundaries, making some regions wetter and other dries.
• Agricultural pattern would shift and human population as well as the ecosystem would experience change.
• The peninsular India would be submerged.
• Global warming refers to the increase in average ground temperatures refers to the increase in average ground temperatures on earth. These higher temperatures across the planet are caused by an intensification of the greenhouse effect.

Q7.What is Break in Monsoon? What are its causes? When is the monsoon expected to break in Kerala and reach the plains of Punjab?
Answer:
During the south-west monsoon period after having rains for a few days, if rain fails to occur for one or more weeks, it is known as break in the monsoon. These dry spells are quite common during the rainy season. These breaks in the different regions are due to different reasons:

• In northern India rains are likely to fail if the rain-bearing storms are not very frequent along the monsoon trough or the ITCZ over this region.
• Over the west coast the dry spells are associated with days when winds blow parallel to the coast.

Breaking of Monsoon:

• Breaking of the Monsoon in Kerala: Beginning of June.
• Reaching Punjab: First week of July.