PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry

Very Short Answer Type Questions

Question 1.
Define desorption.
Answer:
The process of removal of an adsorbed substance from a surface on which it is adsorbed is called desorption.

Question 2.
What is the effect of temperature on chemisorption?
Answer:
Chemisorption initially increases then decreases with rise in temperature. The initial increase is due to the fact that heat supplied acts as activation energy. The decrease afterwards is due to the exothermic nature of adsorption equilibrium.

Question 3.
What is the role of diffusion in heterogeneous catalysis?
Answer:
The gaseous molecules diffuses on to the surface of the solid catalyst and get adsorbed. After the required chemical changes, the products diffuse away from the surface of the catalyst leaving the surface free for more reactant molecules to get adsorbed and undergo reaction.

PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry

Question 4.
What is the type of charge on Agl colloidal sol formed when AgNO3 solution is added to KI solution?
Answer:
Negatively charged sol, Agl/I is formed when AgNO3 solution is added to KI solution.

Question 5.
What causes Brownian movement in a colloidal solution?
Answer:
Unbalanced bombardment of the particles of dispersed phase by molecules of dispersion medium causes Brownian motion. This stabilises the sol.

Question 6.
Based on the type of dispersed phase, what type of colloid is micelles?
Answer:
Associated colloids

PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry

Question 7.
Name the temperature above which the formation of micelles takes place.
Answer:
Kraft temperature.

Question 8.
How do emulsifying agents stabilise the emulsion?
Answer:
The emulsifying agent forms an interfacial layer between suspended particles and the dispersion medium thereby stabilising the emulsion.

Question 9.
Write the dispersed phase and dispersion medium of butter.
Answer:
Dispersed phase — Liquid
Dispersion medium — Solid.

PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry

Question 10.
Write the main reason for the stability of colloidal sols.
Answer:
All the particles of colloidal sol carry the same charge so they keep on repelling each other and do not aggregate together to form bigger particles.

Question 11.
How is Brownian movement responsible for the stability of sols?
Answer:
The Brownian movement has a stirring effect, which does not allow the particles to settle down.

Short Answer Type Questions

Question 1.
Differentiate among a homogeneous solution, a suspension and a colloidal solution, giving a suitable example of each.
Answer:

Property

Homogeneous solution

Colloidal solution

Suspension
(i) Particle size Less than 1 nm Between 1 nm to 1000 nm More than 1000 nm
(ii) Separation by
ordinary filtration Not possible Not possible Not possible
ultra filtration Not possible Possible Possible
(iii) Settling of particles Do not settle Settle only on coagulation Settle under gravity
(iv) Appearance Transparent Opaque Translucent
(v) Example Glucose dissolved in water Smoke, milk, gold sol Sand in water

PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry

Question 2.
Classify colloids where the dispersion medium is water. State their characteristics and write an example of each of these classes.
Answer:
These are of two types
(i) Hydrophilic
Stability: More stable as the stability is due to charge and water envelope surrounding the sol particles.
Nature: Reversible
Examples: Starch, gum etc.

(ii) Hydrophobic
Stability: Less stable as the stability is due to charge only.
Nature: Irreversible
Examples: Metal hydroxide like Fe(OH)3 and metal sulphide like As2S3.

Question 3.
Explain the cleansing action of soap. Why do soaps not work in hard water?
Answer:
The cleansing action of soap such as sodium stearate is due to the fact that soap molecules form micelle around the oil droplet in such a way that hydrophobic part of the stearate ions is in the oil droplet and hydrophilic part projects out of the grease droplet like the bristles. Since the polar groups can interact with water, the oil droplet surrounded by stearate ions is now pulled in water and removed from the dirty surface. Thus, soap helps in emulsification and washing away of oils and fats.

Hard water contains calcium and magnesium salts. In hard water, soap gets precipitated as calcium and magnesium soap which being insoluble stick to the clothes as gummy mass. Therefore, soaps do not work in hard water.

Question 4.
Adsorption of a gas on the surface of solid is generally accompanied by a decrease in entropy still it is a spontaneous process. Why?
Answer:
According to the equation
△G = △H – T△S
For a process to be spontaneous, △G should be negative. Even though △S is negative here, △G is negative because reaction is highly exothermic, i.e., △H is negative.

PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry

Question 5.
Define the following terms:
(i) Brownian movement,
(ii) Peptization.
Answer:
(i) Brownian movement : The motion of the colloidal particles in a zig zag path due to unbalanced bombardment by the particles of dispersion medium is called Brownian movement.

(ii) Peptization : The process of converting a precipitate into colloidal sol by shaking it with dispersion medium in the presence of a small amount of suitable electrolyte is called peptization. During peptization, the precipitate absorbs one of the ions of the electrolyte on its surface. This causes development of positive or negative charge on precipitates, which ultimately break up into particles of colloidal dimension.

Question 6.
(i) Write the expression for Freundlich’s equation to describe the behaviour of adsorption from solution.
(ii) What causes charge on sol particles?
(iii) Name the promoter used in the haber’s process for the manufacture of ammonia.
Answer:
(i) \(\frac{x}{m}\) = KC\(\frac{1}{n}\)
(ii) The charge on the sol particles is due to :

  • electron capture by sol particles during electro dispersion.
  • preferential anolsorption of ions from solution.
  •  formulation of electrical double layer.

(iii) Molybdenum acts in a promoter for iron.

PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry

Long Answer Type Questions

Question 1.
Consider the adsorption isotherms given alongside and interpret the variation in the extent of adsorption (xlm) when
PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry 1
(a) (i) temperature increases at constant pressure.
(ii) pressure increases at constant temperature.
(b) Name the catalyst and the promoter used in Haber’s process for manufacture of ammonia.
Answer:
(a) (i) At constant pressure, extent of adsorption \(\left(\frac{x}{m}\right)\) decreases with increase in temperature as adsorption is an exothermic process.

(ii) At constant temperature, first adsorption \(\left(\frac{x}{m}\right)\) increases with increase in pressure up to a particular pressure and then it
At low pressure, \(\frac{x}{m}\) = kp m
At intermediate range of pressure, \(\frac{x}{m}\) = kp1/n (n > 1)
At high pressure, \(\frac{x}{m}\) = k (independent of pressure)

(b) Finely divided iron is used as a catalyst and molybdenum is used as promoter.

PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry

Question 2.
Explain the following observations:
(i) Sun looks red at the time of setting.
(ii) Cottrell’s smoke precipitator is fitted at the mouth of the chimney used in factories.
(iii) Physical adsorption is multilayered while chemical adsorption is monolayered.
Answer:
(i) At the time of setting, the sun is at horizon. The light emitted by the sun has to travel a relatively longer distance through the atmosphere. As a result, blue part of light is scattered away by the particulate in the atmosphere causing red part to be visible.

(ii) Cottrell’s smoke precipitator, neutralises the charge on unburnt carbon particles, coming out of chimney and they get precipitated and settle down at the floor of the chamber.

(iii) Physical adsorption involves van der Waals’ forces, so any number of layers may be formed one over the other on the surface of the adsorbent. Chemical adsorption takes place as a result of the reaction between adsorbent and adsorbate. When the surface of adsorbent is covered with one layer, no further reaction can take place.

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 5 Surface Chemistry Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 5 Surface Chemistry

PSEB 12th Class Chemistry Guide Surface Chemistry InText Questions and Answers

Question 1.
Distinguish between the meaning of the terms adsorption and absorption.
Give one example of each.
Answer:

Adsorption Absorption
1. It is the surface phenomenon. It is the bulk phenomenon.
2. It is the phenomenon as a result of which the species of one substance gets concentrated mainly on the surface of another substance. It is the phenomenon as a result of which one substance gets distributed uniformly throughout the total volume of another substance.
3. Adsorption is fast in the beginning then slows down due to non­availability of the surface. Absorption proceeds at uniform rate.
4. The concentration on the surface of the adsorbent is different from that in the bulk.
e.g., Water vapours on silica gel.
The concentration is same throughout the material.
e.g., Water vapours are absorbed by anhydrous CaCl2.

Question 2.
What is the difference between physisorption and chemisorption?
Answer:

Physisorption Chemisorption
1. In this type of adsorption, the adsorbate is attached to the surface of the adsorbent with weak van der Waal’s forces of attraction. In this type of adsorption, strong chemical bonds are formed between the adsorbate and the surface of the adsorbent.
2. No new compound is formed in the process. New compounds are formed at the surface of the adsorbent.
3. It is generally found to be reversible in nature. It is usually irreversible in nature.
4. Enthalpy of adsorption is low as weak van der Waal’s forces of attraction are involved. The values lie in the range of 20-40 kJ mol-1. Enthalpy of adsorption is high as chemical bonds are formed. The values lie in the range of 40-400 kJ mol-1.
5. It is favoured by low temperature conditions. It is favoured by high temperature conditions.
6. It is an example of multi-layer adsorption It is an example of mono-layer adsorption.

Question 3.
Give reason why a finely divided substance is more effective as an adsorbent.
Answer:
Adsorption is a surface phenomenon. Therefore, adsorption is directly proportional to the surface area. A finely divided substance has a large surface area. Both physisorption and chemisorption increase with an increase in the surface area. Hence, a finely divided substance behaves as a good adsorbent.

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

Question 4.
What are the factors which influence the adsorption of a gas on a solid?
Answer:
There are various factors that affect the rate of adsorption of a gas on a solid surface.

  1. Nature of the gas : Easily liquefiable gases such as NH3, HCl etc. are adsorbed to a great extent in comparison to gases such as H2, O2 etc. This is because van der Waal’s forces are stronger in easily liquefiable gases.
  2. Surface area of the solid : The greater the surface area of the adsorbent, the greater is the adsorption of a gas on the solid surface.
  3. Effect of pressure : Adsorption is a reversible process and is accompanied by a decrease in pressure. Therefore, adsorption increases with an increase in pressure.
  4. Effect of temperature : Adsorption is an exothermic process. Thus, in accordance with Le-Chatelier’s principle, the magnitude of adsorption decreases with an increase in temperature.

Question 5.
What is an adsorption isotherm? Describe Freundlich adsorption isotherm.
Answer:
PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry 1
The plot between the extent of absorption \(\left(\frac{x}{m}\right)\) against the pressure of gas (P) at constant temperature (T) is called the adsorption isotherm.

Freundlich adsorption isotherm : Freundlich adsorption isotherm gives an empirical relationship between the quantity of gas adsorbed by the unit mass of solid adsorbent and pressure at a specific temperature.
From the given plot it is clear that at pressure Ps, \(\frac{x}{m}\) reaches the maximum value. Ps is called the saturation pressure. Three cases arise from the graph now :
Case I-At low pressure
The plot is straight and sloping, indicating that the pressure is directly proportional to \(\frac{x}{m}\) i.e., \(\frac{x}{m}\) ∝ P.
\(\frac{x}{m}\) = kP (k is a constant)

Case II-At high pressure
When pressure exceeds, the saturated pressure, \(\frac{x}{m}\) becomes independent of P values.
\(\frac{x}{m}\) ∝ Po
\(\frac{x}{m}\) = kPo

Case III-At intermediate pressure
At intermediate pressure, \(\frac{x}{m}\) depends on P raised to the powers between 0 and 1. This relationship is known as the Freundlich adsorption isotherm.
\(\frac{x}{m}\) ∝ P\(\frac{1}{n}\)
\(\frac{x}{m}\) = kP1/n n > 1
Now, taking log
log\(\frac{x}{m}\) = log k + \(\frac{1}{n}\)logP
On plotting the graph between log \(\left(\frac{x}{m}\right)\) and log P, a straight line is obtained with the slope equal to \(\frac{1}{n}\) and intercept equal to log k.
PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry 2

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

Question 6.
What do you understand by activation of adsorbent? How is it achieved?
Answer:
By activating an adsorbent, we tend to increase the adsorbing power of the adsorbent. Some ways to activate an adsorbent are:

  1. By increasing the surface area of the adsorbent. This can be done by breaking it into smaller pieces or powdering it.
  2. Some specific treatments can also lead to the activation of the adsorbent. For example, wood charcoal is activated by heating it between 650 K and 1330 K in vacuum or air. It expels all the gases absorbed or adsorbed and thus, creates a space for adsorption of gases.

Question 7.
What role does adsorption play in heterogeneous catalysis?
Answer:
Heterogeneous catalysis : A catalytic process in which the catalyst and the reactants are present in different phases is known as a heterogeneous catalysis. This heterogeneous catalytic action can be explained in terms of the adsorption theory. The mechanism of catalysis involves the following steps:

  1. Adsorption of reactant molecules on the catalyst surface.
  2. Occurrence of a chemical reaction through the formation of an intermediate.
  3. Desorption of products from the catalyst surface.
  4. Diffusion of products away from the catalyst surface.

In this process, the reactants are usually present in the gaseous state and the catalyst is present in the solid state. Gaseous molecules are then adsorbed on the surface of the catalyst. As the concentration of reactants on the surface of the catalyst increases, the rate of reaction also increases. In such reactions, the products have very less affinity for the catalyst and are quickly desorbed, thereby making the surface free for other reactants.

Question 8.
Why is adsorption always exothermic?
Answer:
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) AH 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., ∆S is negative. Now for a process to be spontaneous, ∆G should be negative.
∴ ∆G – ∆H – T∆S
Since, ∆S is negative, ∆H has to be negative to make ∆G negative. Hence, adsorption is always exothermic.

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

Question 9.
How are the colloidal solutions classified on the basis of physical stjates of the dispersed phase and dispersion medium?
Answer:
One criterion for classifying colloids is the physical state of the dispersed phase and dispersion medium. Depending upon the type of the dispersed phase and dispersion medium (solid, liquid, or gas), there can be eight types of colloidal systems.

Dispersed phase Dispersion medium Type of colloid Example
Solid Solid Solid Sol Gemstones, glasses
Solid Liquid Sol Paints, cell fluids
Solid Gas Aerosol Smoke, dust
Liquid Solid Gel Cheese, butter
Liquid Liquid Emulsion Milk, hair cream
Liquid Gas Aerosol Fog, mist, cloud
Gas Solid Solid Sol Pumice stone, foam rubber
Gas Liquid Foam Froth, soap lather

Question 10.
Discuss the effect of pressure and temperature on the adsorption of gases on solids.
Answer:
Effect of pressure on adsorption : At constant temperature, the extent of adsorption of a gas (x / m) on a solid increases with pressure. A graph between x / m and the pressure p of a gas at constant temperature is called adsorption isotherm.
PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry 3
(i) At lower range of pressure, x / m is directly proportional tothe applied pressure.
\(\frac{x}{m}\) ∝ p1 or \(\frac{x}{m}\) = kp

(ii) At high pressure range, the extent of adsorption of a gas (x / m) is independent of the applied pressure, i.e.,
\(\frac{x}{m}\) ∝ po or \(\frac{x}{m}\) = k

(iii) At intermediate pressure range, the value of x / m is proportional to a fractional power of pressure, i. e.,
\(\frac{x}{m}\) ∝ p1/n or \(\frac{x}{m}\) = kp1/n
where 1 / n is a fraction. Its value may be between 0 and 1.
log\(\left(\frac{x}{m}\right)\) = log k + \(\frac{1}{n}\) log p

Effect of temperature on adsorption : Adsorption is generally temperature dependent. Mostly adsorption processes are exothermic and hence adsorption decreases with increasing temperature. However, for an endothermic adsorption process, adsorption increases with increase in temperature.

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

Question 11.
What are lyophilic and lyophobic sols? Give one example of each type. Why are hydrophobic sols easily coagulated?
Answer:
(i) Lyophilic sols : Colloidal sols directly formed by mixing substances in a suitable dispersion medium are called lyophilic sols. These sols are reversible in nature i.e., if two constituents of the sol are separated by any means (such as evaporation), then the sol can be prepared again by simply mixing the dispersion medium with the dispersion phase and shaking the mixture, e.g., gum, gelatin, starch, rubber etc.

(ii) Lyophobic sols : When substances such as metals and their sulphides etc. are mixed with the dispersion medium, they do not form colloidal sols. Their colloidal sols can be prepared only by special methods. Such sols are called lyophobic sols. These sols are irreversible in nature, e.g., gold sol, AS2O3 etc.

Now, the stability of hydrophilic sols depends on two things—the presence of a charge and the salvation of colloidal particles. On the other hand, the stability of hydrophobic sols is only because of the presence of a charge. Therefore, the latter are much less stable than the former. If the charge of hydrophobic sols is removed (by addition of electrolytes), then the particles present in them come closer and form aggregates, leading to precipitation.

Question 12.
What is the difference between multimolecular and macromolecular colloids? Give one example of each. How are associated colloids different from these two types of colloids?
Answer:
(i) In multi-molecular colloids, the colloidal particles are an aggregate of atoms or small molecules with a diameter of less than 1 nm. The molecules in the aggregate are held together by van der Waal’s forces of attraction. Examples of such colloids include gold sol and sulphur sol.

(ii) In macro-molecular colloids, the colloidal particles are large molecules . having colloidal dimensions. These particles have a high molecular mass. When these particles are dissolved in a liquid, sol is obtained. For example: starch, nylon, cellulose, etc.

(iii) Certain substances tend to behave like normal electrolytes at lower concentrations. However, at higher concentrations, these substances behave as colloidal solutions due to the formation of aggregated particles. Such colloids are called associated colloids.

Question 13.
What are enzymes? Write in brief the mechanism of enzyme catalysis.
Answer:
Enzymes are basically protein molecules of high molecular masses. These form colloidal solutions when dissolved in water. These are complex, nitrogenous organic compounds produced by living plants and animals. Enzymes are also called “biochemical catalysts’.
PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry 4
On the surface of the enzymes, various cavities are present with characteristic shapes. These cavities possess active groups such as—NH2, —COOH, etc. The reactant molecules having a complementary shape fit into the cavities just like a key fits into a lock. This leads to the formation of an activated complex. This complex then decomposes to give the product.

  1. Binding of enzyme to substrate (reactant) to form activated complex.
    E + S → ES*
  2. Decomposition of the activated complex to form product.
    ES* → E + P

Question 14.
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?
Answer:
(i) One criterion for classifying colloids is the physical state of the dispersed phase and dispersion medium. Depending upon the type of the dispersed phase and dispersion medium (solid, liquid, or gas), there can be eight types of colloidal systems.

Dispersed phase Dispersion medium Type of colloid Example
Solid Solid Solid Sol Gemstones, glasses
Solid Liquid Sol Paints, cell fluids
Solid Gas Aerosol Smoke, dust
Liquid Solid Gel Cheese, butter
Liquid Liquid Emulsion Milk, hair cream
Liquid Gas Aerosol Fog, mist, cloud
Gas Solid Solid Sol Pumice stone, foam rubber
Gas Liquid Foam Froth, soap lather

(ii) On the basis of the nature of dispersion medium, colloids 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).

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

Question 15.
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?
Answer:
(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 hydrated ferric oxide sol, it dissociates to give Na+ and Cl ions. Particles of ferric oxide sol are positively charged. Thus, they get coagulated in the presence of negatively charged Cl 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.

Question 16.
What are emulsions? What are their different types? Give example of each type.
Answer:
The colloidal solution in which both the dispersed phase and dispersion medium are liquids is called an emulsion. There are two types of emulsions:
(a) Oil in water type : Here, oil is the dispersed phase while water is the dispersion medium. For example: milk, vanishing cream, etc.

(b) Water in oil type : Here, water is the dispersed phase while oil is the dispersion medium. For example: cold cream, butter, etc.

Question 17.
What is demulsification? Name two demulsifiers.
Answer:
The process of decomposition of an emulsion into its constituent liquids is called demulsification. Examples of demulsifiers are surfactants, ethylene oxide, etc.

Question 18.
Action of soap is due to emulsification and micelle formation. Comment.
Answer:
The cleansing action of soap is due to emulsification and micelle formation. Soaps are basically sodium and potassium salts of long chain fatty acids, R-COO Na+. 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.

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

Question 19.
Give four examples of heterogeneous catalysis.
Answer:
Examples of heterogeneous catalysis
(i) Oxidation of sulphur dioxide to form sulphur trioxide. In this reaction, Pt acts as a catalyst.
PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry 5

(ii) Formation of ammonia by the combination of dinitrogen and dihydrogen in the presence of finely divided iron.
PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry 6
This process is called the Haber’s process.

(iii) Oswald’s process: Oxidation of ammonia to nitric oxide in the presence of platinum.
PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry 7

(iv) Hydrogenation of vegetable oils in the presence of Ni.
PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry 8

Question 20.
What do you mean by activity and selectivity of catalysts?
Answer:
(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. For example, by using different catalysts, we can get different products for the reaction between H2 and CO.
PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry 9

Question 21.
Describe some features of catalysis by zeolites.
Answer:
1. Zeolites are widely used as catalysts in petrochemical industries for cracking of hydrocarbons and isomerisation. They are also used for removing permanent hardness of water,
e.g., ZSM-5 is a catalyst used in petroleum industry
PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry 10
2. Zeolites are shape selective catalysts having honey comb like structure.
3. They are microporous aluminosilicates with Al—O—Si framework and general formula M x / n [(AlO2)x (SiO2)y] ∙ mH2O
4. The reactions taking place in zeolites depend upon the size and shape of the reactant and product molecules as well as upon the pores and cavities of the zeolites.

Question 22.
What is shape selective catalysis?
Answer:
A catalytic reaction which depends upon the pore structure of the catalyst and on the size of the reactant and the product molecules is called shape-selective catalysis. For example, catalysis by zeolites is a shape-selective catalysis. The pore size present in the zeolites ranges from 260-740 pm. Thus, molecules having a pore size more than this cannot enter the zeolite and undergo the reaction.

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

Question 23.
Explain the following terms:
(i) Electrophoresis
(ii) Coagulation
(iii) Dialysis
(iv) Tyndall effect.
Answer:
(i) Electrophoresis : The movement of colloidal particles under the influence of an applied electric field is known as electrophoresis. Positively charged particles move to the cathode, while negatively charged particles move towards the anode. As the particles reach oppositely charged electrodes, they become neutral and get coagulated.

(ii) Coagulation : The process of settling down of colloidal particles i.e., conversion of a colloid into a precipitate is called coagulation.

(iii) Dialysis : The process of removing a dissolved substance from a colloidal solution by the means of diffusion through a membrane is known as dialysis. This process is based on the principle that ions and small molecules can pass through animal membranes unlike colloidal particles.

(iv) Tyndall effect : When a beam of light is allowed to pass through a colloidal solution, it becomes visible like a column of light. This is known as the Tyndall effect. This phenomenon takes place as particles of colloidal dimensions scatter light in all directions.

Question 24.
Give four uses of emulsions.
Answer:
Uses of emulsions

  1. Cleansing action of soaps is based on the formation of emulsions.
  2. Digestion of fats in intestines takes place by the process of emulsification.
  3. Antiseptics and disinfectants when added to water form emulsions.
  4. The process of emulsification is used to make medicines.

Question 25.
What are micelles? Give an example of a micellers system.
Answer:
The aggregate of colloidal particles which have both hydrophobic and hydrophilic parts are called micelles. These are formed above a particular temperature called Krafts temperature (Tk)and above certain concentrations, called Critical Miceller Concentration (CMC).

These molecules are arranged radially with the hydrocarbon or non-polar part towards the centre and the polar part towards the periphery, e.g., soap solution in water is an example of micelles system.

Question 26.
Explain the terms with suitable examples:
(i) Alcosol
(ii) Aerosol
(iii) Hydrosol
Answer:
(i) Alcosol : A colloidal solution having alcohol as the dispersion medium and a solid substance as the dispersed phase is called an alcosol.
For example: colloidal sol of cellulose nitrate in ethyl alcohol is an alcosol.

(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, mist, cloud, etc.

(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 etc.

Question 27.
Comment on the statement that “colloid is not a substance but a state of substance”.
Answer:
Common salt (a typical crystalloid in an aqueous medium) behaves as a colloid in a benzene medium. Hence, we can say that a colloidal substance does not represent a separate class of substances. When the size of the solute particle lies between 1 nm and 1000 nm, it behaves as a colloid.

Hence, we can say that colloid is not a substance but a state of the substance which is dependent on the size of the particle. A colloidal state is intermediate between a true solution and a suspension.

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

Chemistry Guide for Class 12 PSEB Surface Chemistry Textbook Questions and Answers

Question 1.
Write any two characteristics of chemisorption.
Answer:

  1. Chemisorption is highly specific in nature. It occurs only if there is a possibility of chemical bonding between the adsorbent and the adsorbate.
  2. Like physisorption, chemisorption also increases with an increase in the surface area of the adsorbent.

Question 2.
Why does physisorption decrease with the increase of temperature?
Answer:
Physisorption is exothermic in nature. Therefore, in accordance with Le-Chateliere’s principle, it decreases with an increase in temperature. This means that physisorption occurs more readily at a lower temperature.

Question 3.
Why are powdered substances more effective adsorbents than their crystalline forms?
Answer:
Powdered substances are more effective adsorbents than their crystalline forms because when a substance is powdered, its surface area increases and physisorption is directly proportional to the surface area of the adsorbent.

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

Question 4.
In Haber’s process, hydrogen is obtained by reacting methane with steam in presence of NiO as catalyst. The process is known as steam reforming.
Why is it necessary to remove CO when ammonia is obtained by Haber’s process?
Answer:
Carbon monoxide acts as a poison for catalyst iron and promoter molybdenum in Haber’s process, i.e., the efficiency of catalyst and promoter is decreased. It also combines with Fe to form iron carbonyl, Fe(CO)5 which interfere in the production of ammonia. Hence, CO must be removed from the reaction mixture.

Question 5.
Why is the ester hydrolysis slow in the beginning and becomes faster after sometime?
Answer:
The chemical equation for ester hydrolysis can be represented as:
Ester + Water → Acid + Alcohol
The acid produced in the reaction acts as a catalyst and makes the reaction faster. Substances that act as catalysts in the same reaction in which they are obtained as products are known as autocatalysts.

Question 6.
What is the role of desorption in the process of catalysis?
Answer:
The role of desorption in the process of catalysis is to make the surface of the solid catalyst free for the fresh adsorption of the reactants on the surface.

PSEB 12th Class Chemistry Solutions Chapter 5 Surface Chemistry

Question 7.
What modification can you suggest in the Hardy-Schulze law?
Answer:
Hardy-Schulze law states that ‘the greater the valence of the flocculating ion added, the greater is its power to cause precipitation.’ This law takes into consideration only the charge carried by an ion, not its size. The smaller the size of an ion, the more will be its polarising power. Thus, Hardy-Schulze law can be modified in terms of the polarising power of the flocculating ion. Thus, the modified Hardy-Schulze law can be stated as ‘the greater the polarising power of the flocculating ion added, the greater is its power to cause precipitation.’

Question 8.
Why is it essential to wash the precipitate with water before estimating it quantitatively?
Answer:
When a substance gets precipitated, some ions that combine to form the precipitate get adsorbed on the surface of the precipitate. Therefore, it becomes important to wash the precipitate before estimating it quantitatively in order to remove these adsorbed ions or other such impurities.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 1 Relations and Functions Miscellaneous Exercise Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 1 Relations and Functions Miscellaneous Exercise

Question 1.
Let f: R → R be defined as f(x) = 10x + 7. Find the function g: R → R such that gof = fog = IR.
Solution.
It is given that f: R → R is defined as f(x) = 10x + 7.
One – one :
Let f(x) = f(y), where x, y ∈ R.
⇒ 10x + 7 = 10y + 7
⇒ x = y
∴ f is a one-one function.

Onto :
For y ∈ R, let y = 10x + 7.
⇒ x = \(\frac{y-7}{10}\) ∈ R
Therefore, for any y ∈ R, there exists x = \(\frac{y-7}{10}\) ∈ R such that
f(x) = f(\(\frac{y-7}{10}\))
= 10(\(\frac{y-7}{10}\)) + 7
= y – 7 + 7 = y
∴ f is onto.
Therefore f is one-one and onto.
Thus f is an invertible function.
Let us define g : R → R as g(y) = \(\frac{y-7}{10}\)
Now, we have
gof(x) = g(f(x)) = g(10x + 7)
= \(\frac{(10 x+7)-7}{10}=\frac{10 x}{10}\) = x
And, fog(y) = f(g(y))
= f(\(\frac{y-7}{10}\))
= 10(\(\frac{y-7}{10}\)) + 7
= y – 7 + 7 = y
∴ gof = IR and fog = IR
Hence, the required functiong:R → R is defined as g(y) = \(\frac{y-7}{10}\)

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise

Question 2.
Let f: W → W be defined as f(n) = n – 1, if n is odd and f(n) = n + 1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.
Solution.
It is given that
f: W → W is defined as f(n) = PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise 1

One-one :
Let f(n) = f(m)
It can be observed that if n is odd and m is even, then we will have
n – 1 = m + 1
⇒ n – m = 2
However, this is impossible.
Similarly, the possibility of n being even and m being odd can also ignored under a similar argument.
∴ Both n and m must be either odd or even.
Now, if both n and m are odd, then we have
f(n) = f(m) ⇒ n – 1 = m – 1 ⇒ n = m
Again, if both n and m are even , then we have
f(n) = f(m) ⇒ n + 1 = m+1 ⇒ n = m
∴ f is one – one.

Onto :
It is clear that any odd number 2r + 1 in co-domain W is the image of 2r in domain W and any even number 2r in co-domain W is the image of 2r + 1 in domain W.
∴ f is onto.
Hence, f is an invertible function.
Let us define g : W → W as

g(m) = PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise 2

Now, when n is odd
gof(n) = g(f(n)) = g(n – 1) = n – 1 + 1 = n
and, when n is even
gof(n) = g(f(n)) = g(n + 1) = n + 1 – 1 = n
Similarly, when m is odd
fog(m) = f(g(m)) = f(m – 1) = m – 1 + 1 = m
and when m is even
fog(m) = f(g(m)) = f(m + 1) = m + 1 – 1 = m
∴ gof = IW and fog = IW
Thus, f is invertible and the inverse of f is given by f-1 = g, which is the same as f.
Hence, the inverse of f is itself.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise

Question 3.
If f: R → R is defined by f(x) = x2 – 3x + 2, find f(f(x)).
Solution.
It is given that f: R → R is defined as f(x) = x2 – 3x + 2.
f(f(x)) = f(x2 – 3x + 2)
= (x2 – 3x + 2)2 – 3(x2 -3x + 2) + 2
= x4 + 9x2 + 4 – 6x3 – 12x + 4x2 – 3x2 + 9x – 6 + 2
= x4 – 6x3 + 10x2 – 3x.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise

Question 4.
Show that the function f: R → {x ∈ R: – 1 < x < 1} defined by f(x) = \(\frac{x}{1+|x|}\) ∈ R is one-one and onto function.
Solution.
It is given that f: R → {x ∈ R: – 1 < x < 1} is defined as f(x) = \(\frac{x}{1+|x|}\), x ∈ R.
Suppose f(x) = f(y), where x,y ∈ R ⇒ \(\frac{x}{1+|x|}=\frac{y}{1+|y|}\)
It can be observed that if x is positive and y is negative, then we have \(\frac{x}{1+x}=\frac{y}{1-y}\)
⇒ 2xy = x – y
Since x is positive and y is negative, then x > y ⇒ x – y > 0
But, 2xy is negative.
Then, 2xy ≠ x – y.
Thus, the case of x being positive and y being negative can be ruled out.
Under a similar argument, x being negative and y being positive can also be ruled out.
∴ x and y have to be either positive or negative.
When x and y are both positive, we have x y
f(x) = f(y)
⇒ \(\frac{x}{1+x}=\frac{y}{1+y}\)
⇒ x + xy = y + xy
⇒ x = y
When x and y are both negative, we have
f(x) = f(y)
⇒ \(\frac{x}{1-x}=\frac{y}{1-y}\)
⇒ x – xy = y – yx
⇒ x = y
∴ f is one-one.
Now, let y ∈ R such that – 1 < y < 1.
If y is negative, then there exists x = \(\frac{y}{1+y}\) ∈ R such that
f(x) = f(\(\frac{y}{1+y}\))
= \(\frac{\left(\frac{y}{1+y}\right)}{1+\left|\frac{y}{1+y}\right|}=\frac{\frac{y}{1+y}}{1+\left(\frac{-y}{1+y}\right)}=\frac{y}{1+y-y}\) = y
If y is positive, then there exists x = \(\frac{y}{1-y}\) ∈ R such that
f(x) = \(f\left(\frac{y}{1-y}\right)=\frac{\left(\frac{y}{1-y}\right)}{1+\left(\frac{y}{1-y}\right)}\)

= \(\frac{\frac{y}{1-y}}{1+\left(\frac{y}{1-y}\right)}=\frac{y}{1-y+y}\) = y
∴ f is onto.
Hence, f is one-one and onto.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise

Question 5.
Show that the function f: R → R given by f(x) = x3 in injective.
Solution.
f: R → R is given as f(x) = x3.
Suppose f(x) = f(y), where x, y ∈ R.
⇒ x3 = y3 …………(i)
Now, we need to show that x = y
Suppose x * y, their cubes will also not be equal.
⇒ x3 ≠ y3
However, this will be a contradiction to Eq. (i).
∴ x = y
Hence, f is injective.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercisec

Question 6.
Give examples of two functions f: N → Z and g: Z → Z such that gof is injective but g is not injective.
[Hint: consider f(x) x and g(x) = |x|].
Solution.
Define f: N → Z as f(x) – x and g: Z → Z as g(x) =|x|
We first show that g is not injective.
It can be observed that
g(- 1) = |- 1|= 1; g(1) = |1|= 1
∴ g(- 1) = g(1), but – 1 ≠ 1.
∴ g is not injective.
Now, gof: N → Z is defined as gof(x) = g(f(x)) = g(x) =|x|.
Let x, y ∈ N such that gof(x) – gof(y).
⇒ |x| = |y|
Since x and y ∈ N, both are positive.
∴ |x |= |y |=> x = y
Hence, gof is injective.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise

Question 7.
Give examples of two functions f: N → N and g: N → N such that gof is onto but f is not onto.
[Hint: consider f(x) = x + 1 and g(x) = iPSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise 3]
Solution.
Define f: N → N by f(x) = x +1
and, g: N → N by g(x) = PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise 3
We first show that f is not onto.
For this, consider element 1 in co-domain N. It is clear that this element is not an image of any of the elements in domain N.
∴ f is not onto.
Now, gof: N → N is defined by,
gof(x) = g(f(x)) = g(x + 1) = (x + 1) – 1 = x [∵ x ∈ N ⇒ (x + 1) > 1]
Then, it is clear that for y ∈ N, there exists x = y ∈ N such that gof(x) = y.
Hence, gof is onto.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise

Question 8.
Given a non-empty set X, consider P(X) which is the set of all subsets of X.
Define the relation R in P(X) as follows:
For subsets A, B in P(X), ARB if and only if A c B. Is R an equivalence relation on P(X)? Justify your answer.
Solution.
Since every set is a subset of itself, ARA for all A ∈ P(X).
∴ R is reflexive.
Let ARB ⇒ A ⊂ B.
This cannot be implied to B ⊂ A.
For instance, if A = {1, 2} and B = {1, 2,3}, then it cannot be implied that B is related to A.
∴ R is not symmetric.
Further, if ARB and BRC, then A c B and B c C.
⇒ A ⊂ C
⇒ ARC
R is transitive.
Hence, R is not an equivalence relation since it is not symmetric.

Question 9.
Given a non-empty set X, consider the binary operation *: P(X) × P(X) P(X) given by A * B = A ∩ B ∀ A, B in P(X) where P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation*.
Solution.
It is given that * : P(X) × P(X) → P(X) is defined as A * B = A ∩ B ∀ A, B ∈ P(X).
We know that A * X = A ∩ X = A = X ∩ A ∀ A ∈ P(X).
⇒ A * X = A = X * A ∀ A ∈ P (X)
Thus, X is the identity element for the given binary operation*.
Now, an element A ∈ P(X) is invertible if there exists B ∈ P(x) such that
A * B = X = B * A. (As X is the identity element)
i.e., A ∩ B = X = B ∩ A
This case in possible only when A = X = B.
Thus, X is the only invertible element in P(X) with respect to the given operation*.
Hence, the given result is proved.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise

Question 10.
Find the number of all onto functions from the set {1, 2, 3, n} to itself.
Solution.
Onto functions from the set {1, 2, 3, …, n} to itself is simply a permutation on n symbols 1, 2, …, n.
Thus, the total number of onto maps from {1, 2, 3,…, n} to itself is the same as the total number of permutations on symbols 1, 2,…, n, which is n!.

Question 11.
Let S = {a, b, c} and T = {1,2, 3}. Find F-1 of the following functions F from S to T, if it exists.
(i) F = {(o, 3), (6, 2), (c, 1)}
(ii) F = {(a, 2), (6, 1), (c, 1)}
Solution.
Given, S = {a, b, c}, and T = {1, 2, 3}
F: S → T is defined as :
F = {(a, 3), (b, 2), (c, 1)}
⇒ f(a) = 3, F(b) = 2, F(c) = 1
Therefore, F-1 : T → S is given by
F-1 = {(3, a), (2, b), (1, c)}

(ii) F: S → T is defined as
F = {(a, 2), (b, 1), (c, 1)}
Since F (b) = F (c) = 1, F is not one-one.
Hence, F is not invertible i. e., F-1 does not exist.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise

Question 12.
Consider the binary operations * : R × R → R and o: R × R → R defined as a * b = | a – b| and a o b = a, ∀ a, b ∈ R. Show that * is commutative hut not associative, o is associative but not commutative. Further, show that ∀ a, b, c ∈ R, a* (b o c) = (a * b) o (a * c). [If it is so, we say that the operation * distributes over the operation o]. Does o distribute over *? Justify your answer.
Solution.
It is given that *: R × R R and o: R × R → R is defined as a * b = |a – b| and a o b = a ∀ a, b ∈ R.
For a, b ∈ R, we have
a * b = |a – b|
b * a = |b – a| = |- (a – b)|= |a – b|
∴ a * b = b * a
Therefore, the operation * is commutative..
It can be observed that
(1 * 2) * 3 = (|1 – 2|) * 3 = 1 * 3 = |1 – 3|= 2
1 * (2 * 3) = 1 * (|2 – 3|) = 1 * 1 =|1 – 1 |= 0
∴ (1 * 2) * 3 ≠ 1 * (2 * 3) (where 1, 2, 3 ∈ R)
Therefore, the operation * is not associative.
Now, consider the operation o
It can be observed that 1 o 2 = 1 and 2 o 1 = 2.
∴ 1 o 2 ≠ 2 o 1 where 1, 2 ∈ R
Therefore, the operation o is not commutative.
Let a, b, c ∈ R. Then, we have
(a o b) o c = a o c = a
a o (b o c) = a o b = a
⇒ (a o b) o c = a o (b o c)
Therefore, the operation o is associative.
Now, let a, b, c ∈ R, then we have
a * (b o c) = a * b = |a – b|
(a * b) o (a * c) = (|a – b|) o (|a – c|) = |a – b|
Hence a * (b o c) = (a * b) o (a * c)
Now, 1 o(2 * 3) = 1 o (|2 – 3|) = 1 o 1 = 1
(1 o 2) * (1 o 3) = 1 * 1 = |1 – 1|= 0
1 o (2 * 3) ≠ (1 o 2) * (1 o 3)
where 1, 2, 3 ∈ R Therefore, the operation o does not distribute over *.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise

Question 13.
Given a non-empty set X, let *: P(X) × P(X) → P(X) be defined as A * B – (A – B) ∪ (B – A), ∀ A, B ∈ P(X). Show that the empty set Φ is the identity for the operation * and all the elements A of P(X) are invertible with A-1 = A.
[Hint: (A – Φ) ∪ (Φ – A) = A and (A – A) ∪ (A – A) = A * A = Φ]
Solution.
It is given that *: P(X) × P(X) → P(X) is defined as
A * B = (A – B) ∪ (B – A) ∀ A, B, ∈ P(X).
Let A ∈ P(X). Then, we have
A * (Φ) = (A – Φ) ∪ (Φ – A) = A ∪ Φ = A
Φ * A = (Φ – A) ∪ (A – Φ) = Φ ∪ A = A
A * Φ = A = Φ * A ∀ A ∈ P(X)
Thus, Φ is the identity element for the given operation *.
Now, an element A s P(X) will be invertible if there exists B ∈ P(X) such that
A * B = Φ = B * A. (As Φ is the identity element)
Now, we observed that
A * A = (A – A) ∪ (A – A) = Φ ∪ Φ = Φ ∀ A ∈ P(X).
Hence, all the elements A of P(X) are invertible with A-1 = A.

Question 14.
Define a binary operation * on the set {0, 1, 2, 3, 4, 5) as
a * b = PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise 4
Show that zero is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 – a being the inverse of a.
Solution.
(i) e is the identity element if a * e = e * a = a
a * 0 = a + 0, 0 * a = 0 + a = a
⇒ a * 0 = 0 * a = a
∴ 0 is the identity of the operation.

(ii) b is the inverse of a if a * b = b * a = e
Now a * (6 – a) = a + (6 – a) – 6 = 0
(6 – a) * a = (6 – a) + a – 6 = 0
Hence, each element of a of the set is invertible with inverse.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise

Question 15.
Let A = {-1, 0, 1, 2}, B = {-4,-2, 0,2} and f, g: A → B be functions defined by f(x) = x2 – x, x ∈ A and g(x) = 2|x – \(\frac{1}{2}\)| – 1, x ∈ A. Are f and g equal? Justify your answer.
[Hint: One may not be that two functions f: A → B and g: A → B
such that f(a) = g(a) ∀ a ∈ A, are called equal functions.]
Solution.
It is given that A = {- 1,0,1, 2}, B = {- 4, – 2, 0, 2).
Also, it is given that f, g: A → B are defined by f(x) = x2 – x, x ∈ A and
g(x) = 2 |x – \(\frac{1}{2}\)| – 1, x ∈ A
It is observed that
f(- 1) = (- 1)2 – (- 1) = 1 + 1 = 2
and g(- 1) = 2|(- 1) – \(\frac{1}{2}\)| – 1
= 2(\(\frac{3}{2}\)) – 1 = 3 – 1 = 2
⇒ f(- 1) = g(- 1)

⇒ f(0) = (0)2 – 0 = 0
and g(0) = 2|0 – \(\frac{1}{2}\)|
= 2(\(\frac{3}{2}\)) – 1 = 1 – 1 = 0

⇒ f(0) = g(0)
f(1) = (1)2 – 1 = 1 – 1 = 0
and g(1)= 2|1 – \(\frac{1}{2}\)|
= 2(\(\frac{1}{2}\)) – 1 = 1 – 1 = o

⇒ f(1) = g(1)
f(2) = (2)2 – 2 = 4 – 2 = 2
and g(2) = 2 |2 – \(\frac{1}{2}\)|
= 2(\(\frac{3}{2}\)) – 1 = 3 – 1 = 2
⇒ f(2) = g(2)
∴ f(a) = g(a) ∀ a ∈ A
Hence, the functions f and g are equal.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise

Question 16.
Let A = {1, 2, 3}. Then, number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive, is
(A) 1
(B) 2
(C) 3
(D) 4
Solution.
This is because relation R is reflexive as (1, 1), (2, 2), (3, 3) ∈ R.
Relation R is symmetric since (1, 2), (2 ,1) ∈ R and (1, 3), (3, 1) ∈ R.
But relationR is not transitive as (3, 1), (1, 2) ∈ R but (3, 2) ∈ R.
Now, if we add any one of the two pairs (3, 2) and (2, 3) (or both) to relation R, then relation R will become transitive.
Hence, the total number of desired relation is one.
Thus, the correct option is (A).

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise

Question 17.
Let A = {1, 2, 3}. Then, number of equivalence relations containing (1, 2) is
(A) 1
(B) 2
(C)3
(D) 4
Solution.
It is given that A = {1, 2, 3}
The smallest equivalence relation containing (1, 2) is given by,
R1 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
Now, we are left with only four pairs i. e., (2, 3), (3, 2), (1, 3), and (3, 1).
If we add any one pair [say (2, 3)] to R1 then for symmetry we must add (3, 2). Also, for transitivity, we are required to add (1, 3) and (3,1). Hence, the only equivalence relation (bigger than R1) is the universal relation.
This shows that the total number of equivalence relations containing (1, 2) is two. The correct option is (B).

Question 18.
Let f: R → R be the signum function defined as
f(x) = PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise 5
and g: R → R be the greatest integer function given by g(x) = [x], where [x] is greatest integer less than or equal to x. Then, does fog and gof coincide in (0, 1]?
Solution.
It is given that
f: R → R is defined as f(x) = PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise 5
Also, g: R → R is defined as g(x) = [x], where [x] is the greatest integer less than or equal to x .
Now, let x ∈ (0, 1]
Then, we have
[x] = 1, if x = 1 and [x] = 0 if 0 < x < 1. ∴ fog(x) = f (g(x)) = f([x]) = PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise 6 gof(x) = g(f(x))= g(1) [∵ x > 0]
= [1] = 1 .
Thus, when x ∈ (0, 1), we have fog(x) = 0 and gof(x) = 1.
But fog (1) ≠ gof (1)
Hence, fog and gof do not coincide in (0,1].

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise

Question 19.
Number of binary operations on the set {a, b} are (A) 10 (B) 16 (C) 20 (D) 8
Solution.
A binary operation * on {a, b} is a function from {a, b} × {a, b} → {a, b} i. e.,* is a function from {(a, a), (a, b), (b, a), (b, b)} → {a, b}.
Hence, the total number of binary operations on the set {a, b} is 24 i.e. 16.
Thus, the correct option is (B).

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.4

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 1 Relations and Functions Ex 1.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.4

Question 1.
Determine whether or not each of the definition of * given below gives a binary operation. In the event that * is not a binary operation, give justification for this.
(i) On Z+, define * by a * b = a – b
(ii) On Z+, define * by a * b = ab
(iii) On R, define * by a * b = ab2
(iv) On Z+, define * by a * b = |a – b|
(v) On Z+, define * by a * b = a
Sol.
(i) On Z+, * is defined by a * b = a – b.
It is not a binary operation as the image of (1, 2) under * is
1 * 2 = 1 – 2 = – 1 ∉ Z+.

(ii) On Z+, * is defined by a * b = ab.
It is seen that for each a, b ∈ Z+, there is a unique element ab in Z+.
This means that * carries each pair (a, b) to a unique element a * b = ab in Z+. Therefore, * is a binary operation.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.4

(iii) On R, * is defined by a * b = ab2.
It is seen that for each a, b ∈ R, there is a unique element ab2 in R.
This means that * carries each pair (a, b) to a unique element a * b = ab2 in R. Therefore, * is a binary operation.

(iv) On Z+, * is defined by a * b =|a – b|.
It is seen that for each a, b ∈ Z+, there is a unique element | a – b | in Z+. This means that * carries each pair (a, b) to a unique element a * b = |a – b|in Z+. Therefore, * is a binary operation.

(v) On Z+, * is defined by a * b = a.
It is seen that for each a, b ∈ Z+, there is a unique element a ∈ Z+. This means that * carries each pair (a, b) to a unique element a * b = a in Z+. Therefore, * is a binary operation.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.4

Question 2.
For each operation * defined below, determine whether * is binary commutative or associative.
(i) On Z, define a* b = a – b
(ii) On Q, define a * b = ab + 1
(iii) On Q, define a* b = \(\frac{a b}{2}\)
(iv) On Z+, define a * b = 2ab
(v) On Z+, define a * b = ab
(vi) On R – {- 1},define a * b = \(\frac{a}{b+1}\)
Solution.
(i) On Z, operation * is defined as
(a) a * b = a – b
⇒ b * a = b – a
But a – b ≠ b – a
⇒ a * b ≠ b * a
∴ Defined operation is not commutative.

(b) a – (b – c) ≠ (a – b) – c
∴ Binary operation * as defined is not associative.

(ii) On Q, operation * is defined as a * b = ab +1
(a) ab + 1 = ba + 1, a * b = b * a
∴ Defined binary operation is commutative.

(b) a * (b * c) = a * (bc + 1) = a (bc + 1) + 1 = abc + a + 1
and (a * b)* c = (ab + 1) * c = (ab + 1)c + 1
= abc + c + 1
a * (b * c) ≠ (a * b) * c
∴ Binary operation defined is not associative.

(iii) (a) On Q, operation * is defined as a * b = \(\frac{ab}{2}\)
∴ a * b = b * a
∴ Operation binary defined is commutative.

(b) a * b = a * \(\frac{b c}{2}=\frac{a b c}{4}\)
and (a * b) * c = \(\frac{b c}{2}\) * c = \(\frac{a b c}{4}\)
⇒ Defined binary operation is associative.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.4

(iv) On Z+, operation * is defined as a * b = 2ab
(a) a * b = 2ab, b * a = 2ba = 2ab
a * b = b * a
Binary operation defined is commutative.

(b) a * (b * c) = a * 2ba = 2a . bc
(a * b) * c = 2ab * c = 22ab
Thus, (a * b) * c ≠ a * (b * c)
∴ Binary operation * as defined is not associative.

(v) On Z+, a * b = ab
(a) b * a = ba
∴ ab = ba
⇒ a * b ≠ b * a
* is not commutative.

(b) (a * b) * c = ab * c
= (ab)c = abc
a * (b * c) = a * bc = abc.
This (a * b) * c ≠ (a * b * c)
∴ Operation * is not associative.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.4

(vi) On Z+ operation * is defined as
a * b = \(\frac{a}{b+1}\), b ≠ – 1
∴ b * a = \(\frac{b}{a+1}\)
(a) a * b ≠ b * a
Binary operation defined is not commutative.

(b) (a * b) * c = \(a^{*}\left(\frac{b}{c+1}\right)=\frac{a}{\frac{b}{c+1}+1}=\frac{a(c+1)}{b+c+1}\)

(a * b) * c = \(\frac{a}{b+1} * c=\frac{\frac{a}{b+1}}{c+1}=\frac{a}{(b+1)(c+1)}\)

∴ a * (b * c) ≠ (a * b) * c
⇒ Binary operation defined above is not associative.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.4

Question 3.
Consider the binary operation ^ on the set (1, 2, 3, 4, 5} defined by a ^ b = min {a, b}. Write the multiplication table of the operation ^.
Solution.
The binary operation ^ on the set {1, 2, 3, 4, 5} is defined as
a ^ b = min{a, b} for a, b ∈ {1, 2, 3, 4, 5}.
Thus, the operation table for the given operation ^ can be given as

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.4 1

Question 4.
Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table.
(i) Compute (2 * 3) * 4 and 2 * (3 * 4)
(ii) Is * commutative?
(iii) Compute (2* 3) * (4* 5).
(Hint: use the following table) (i)

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.4 2

Solution.
(i) We have (2 * 3) *4 = 1 * 4 = 1
and 2 * (3 * 4) = 2 * 1 = 1

(ii) For every a, b ∈ (1, 2, 3, 4, 5}, we have a * b = b * a. Therefore, the operation * is commutative.
(iii) We have (2 * 3) = 1 and (4 * 5) = 1 .
∴ (2 * 3) * (4 * 5) = 1 * 1 = 1.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.4

Question 5.
Let *’ be the binary operation on the set {1, 2, 3, 4, 5} is defined by a *’ b = H.C.F. of a and b. Is the operation *’ same as the operation * defined in Q. 4 above? Justify your answer.
Solution.
The binary operation *’ on the set {1, 2, 3, 4, 5} is defined as
a*’ b = HCF of a and b.
The operation table for the operation * can be given as :

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.4 3

We observe that the operation table for the operations * and *’ are the same.
Thus, the operation *’ is same as the operation *.

Question 6.
Let * be the binary operation on N given by a * b = L.C.M. of a and b.
(i) Find 5 * 7, 20 * 16
(ii) Is * commutative?
(iii) Is * associative?
(iv) Find the identity of * in N.
(v) Which elements of N are invertible for the operation *?
Solution.
The binary operation * defined as a * b = L.C.M. of a and b
(i) 5 * 7 = L.C.M. of 5 and 7 = 35
and 20 * 16 = L.C.M. of 20 and 16 = 80

(ii) a * b = L.C.M. of a and b
b * a = L.C.M. of b and a
⇒ a * b = b * a L.C.M. of a, b and b, a are equal
∴ Binary operation * is commutative.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.4

(iii) a * (b * c) = L.C.M. of a, b, c
and (a * b)* c = L.C.M. of a, b, c
⇒ a * (b * c) = (a * b) * c
⇒ Binary operation * is associative.

(iv) Identity of * in N is 1
1 * a = a * 1 = a = L.C.M. of 1 and a.

(v) Let * : N × N → N defined as a * b = L.C.M. of (a, b)
For a = 1, b = 1, a * b = 1 = b * a. Otherwise a * b ≠ 1
∴ Binary operation * is not invertible.
⇒ 1 is invertible for operaiton *.

Question 7.
Is * defined on the set {1, 2, 3, 4, 5} by a * 6 = L.C.M. of a and 6 a binary operation? Justify your answer.
Solution.
The operation * on the set A = {1, 2, 3, 4, 5} is defined as a * b = L.C.M. of a and b.
Now, 2 * 3 = L.C.M. of 2 and 3 = 6.
But 6 does not belong to the given set.
Hence, the given operation * is not a binary operation.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.4

Question 8.
Let * be the binary operation on N defined by a * 6 = H.C.F. of a and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N?
Solution.
The binary operation * on N is defined as a * b = H.C.F. of a and b It is known that
H.C.F. of a and b = H.C.F. of b and a V a, b ∈ N.
∴ a * b = b * a
Thus, the operation * is commutative.
For a, b, c ∈ N, we have
(a * b) * c = (H.C.F. of a and b) * c = H.C.F. of a, b and c
a * (b * c) = a * (H.C.F. of b and c) = H.C.F. of a, b, and c
∴ (a * b) * c = a* (b * c)
Thus, the operation * is associative.
Now, an element e ∈ N will be the identity for the operation * if a * e = a = e * a for ∀ a ∈ N.
But this relation is not true for any a ∈ N.
Thus, the operation * does not have any identity in N.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.4

Question 9.
Let * be a binary operation on the set Q of rational numbers as
(i) a * b = a – b
(ii) a * b = a2 + b2
(iii) a * b = a + ab
(iv) a * b = (a – b)2
(v) a * b = \(\frac{ab}{4}\)
(vi) a * b = ab2
Find which of the binary operations are commutative and which are associative.
Solution.
Operation is on the set Q.
(i) defined as a * b = a – b
(a) Now b * a = b – a But a – b *b – a
∴ a * b * b * a
∴ Operation * is not commutative.

(b) a* (b * c) = a * (b – c) = a – (b – c) = a – b + c
(a * b) * c = (a – b) * c = a – b – c
Thus, a * (b * c) ^ (a * b) * c
∴ The operation * as defined is not associative.

(ii) (a) a * b = a2 + b2
b * a = b2 + a2 = a2 + b2
∴ a * b = b * a
∴ This binary operation is commutative.

(b) a * (b * c) = a * (b2 + c2)
= a2 + (b2 + c2)2
(a * b) * c = (a2 + b2) * c = (a2 + b2)2 + c2
⇒ a * (b * c) * (a * b) * c
∴ The operation * given is not associative.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.4

(iii) Operation * is defined as
a * b = a + ab
(a) b* a = b + ba
∴ a * b ≠ b * a
∴ This operation is not commutative.

(b) a * (b * c) = a * (b + bc)
= a + a(b + bc)
= a + ab + abc
(a* b) * c = (a + ab) *c = a + ab + (a + ab) . c
= a + ab + ac + abc
⇒ a* (b* c)& (a* b)* c
⇒ The binary operation is not associative.

(iv) The binary operation is defined as a * b = (a – b)2
(a) b * a = (b – a)2 = (a – b)2
⇒ a * b = b * a
∴ This binary operation * is commutative.

(b) a * (b * c) = a * (b – c)2
= [a – (b – c)2]2
(a * b) * c = (a – b)2 * c = [(a – b)2 – c]2
⇒ (a * b) * c ≠ a * (b * c)
Thus, the operation given is associative.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.4

(v) Binary operation is * defined as
a * b = \(\frac{ab}{4}\)

(a) b * a = \(\frac{ba}{4}\) = \(\frac{ab}{4}\)
a* b^b* a
∴ The operation is not commutative.

(b) a * (b * c) = a * \(\frac{bc}{4}\)
= \(\frac{a}{4}\left(\frac{b c}{4}\right)=\frac{a b c}{16}\)
(a * b) * c = \(\frac{ab}{4}\) * c
= \(\frac{a b}{4} \cdot \frac{c}{4}=\frac{a b c}{16}\)
⇒ a * (b* c) = (a * b) * c
Thus, the operation given is associative.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.4

(vi) Binary operation is defined as
a * b = ab2
(a) b * a = ba2 ≠ ab2
∴ a * b ≠ b * a
∴ The operation is not commutative.

(b) a * (b * c) = a * bc2
= a(bc2)2
= ab2c4
(a * b)* c = ab2 * c
= (ab2)c2
= ab2c2
∴ a * (b * c) ≠ (a * b) * c
∴ Binary operation * given is not associative.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.4

Question 10.
Find which of the operations given above has identity.
Solution.
An element e ∈ Q will be the identity element for the operation * if
a * e = a = e * a, ∀ a ∈ Q
(i) a * b = a – b
lf a * e = a, a ≠ 0
⇒ a – e = a, a ≠ 0 ⇒ e = 0
Also, e * a = a
⇒ e – a = a ⇒ e = 2 a
e = 0 = 2a, a ≠ 0
But the identity is unique. Hence this operation has no identity.

(ii) a * b = a2 + b2
If a * e = a, then a2 + e2 = a
For a = – 2, (- 2)2 + e2 = 4 + e2 ≠ – 2
Hence, there is no identity element.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.4

(iii) a * b = a + ab
If a * e = a
⇒ a + ae a
⇒ ae = 0
⇒ e = 0, a ≠ 0
Also a * e = a
⇒ e + ae = a
⇒ e = \(\frac{a}{a+1}\), a ≠ 1
∴ e = 0 = \(\frac{a}{a+1}\), a ≠ 0
But the identity in unique. Hence this operation has no identify.

(iv) a * b = (a – b)2
If a* e = a, then (a – e)2 = a.
A square is always positive, so for a = – 2, (- 2 – e)2 ≠ – 2.
Hence, there is no identity element.

(v) a * b – ab/ 4
If a * e = a, then ae / 4 = a.
Hence, e = 4 is the identity element.
∴ a * 4 = 4 * a = 4a/4 = a.

(vi) a * b = ab2
If a * e = a
⇒ ae2 = a
⇒ e2 = 1
⇒ e = ±1
But identity is unique. Hence this operation has no identity.
Therefore only part (v) has an identity element.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.4

Question 11.
Let A = N × N and * be the binary operation on A defined by (a, b) * (c, d) = (a + c, b + d). Show that * is commutative and associative. Find the identity element for * on A, if any.
Solution.
Given that A = N × N and * is a binary operation on A and is defined by (a, b) * (c, d) = (a + c,b + d.)
Let (a, b), (c, d) ∈ A
Then, a, b, c, d ∈ N
We have (a, b) * (c, d) = (a + c, b + d)
and (c, d) * (a, b) = (c + a, d + b) = (a + c, b + d)
[Addition is commutative in the set of natural numbers]
∴ (a, b) * (c, d) = (c, d) * (a, b)
Therefore, the operation * is commutative.
Now, let (a, b), (c, d), (e, f) ∈ A
Then, a, b, c, d, e, f ∈ N
We have {(a, b) * (c, d)} * (e, f) = (a + c,b + d) * (e, f)
= (a+ c + e, b + d + f)
(a, b) * ((c, d) * (e, f)) = (a, b) * (c + e, d + f) = (a + c + e, b + d + f)
((a, b) * (c, d)) * (e, f) = (a, b) * ((c, d) * (e, f))
Therefore, the operation * is associative.
An element e = (e1, e2) ∈ A will be an identity element for the operation * if
a * e = a = e * a ∀ a = (a1, a2) ∈ A, i.e., (a1 + e1, a2 + e2)
= (a1, a2) = (e1 + a1; e2 + a2)
which is not true for any element in A.
Therefore, the operation * does not have any identity element.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.4

Question 12.
State whether the following statements are true or false. Justify.
(i) For an arbitrary binaiy operation * on a set N, a * a = a ∀ a ∈ N.
(ii) If * is a commutative binary operation on N, then a* (b* c) = (c * b) * a
Solution.
(i) Define an operation * on IV as a * b – a + b ∀ a, b ∈ N
Then, in particular, for b = a = 3, we have 3 * 3 = 3 + 3 = 6 ≠ 3
Therefore, statement (i) is false.

(ii) R.H.S. = (c * b) * a
= (b * c) * a [* is commutative]
= a * (b * c) [Again, as * is commutative]
= L.H.S.
∴ a * (b * c) = (c * b) * a
Therefore, statement (ii) is true.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.4

Question 13.
Consider a binary operation * on N defined as a * b = a3 +b3. Choose the correct answer.
(A) Is * both associative and commutative?
(B) Is * commutative but not associative?
(C) Is * associative but not commutative?
(D) Is * neither commutative nor associative?
Solution.
On N, the operation * is defined as a * b = a3 + b3.
For, a, b ∈ N, we have
a * b = a3 + b3
= b3 + a3 = b * a [Addition is commutative in N]
Therefore, the operation * is commutative.
It can be observed that
(1 * 2) * 3 = (13 + 23) * 3 = 9 * 3
= 93 + 33
= 729 + 27 = 756

1 * (2 * 3) = 1 * (23 +33)
= 1 * (8 + 27) = 1 * 35
= 13 + 353
= 1 + (35)3
= 1 + 42875 = 42876
∴ (1 * 2) * 3 ≠ 1 * (2 * 3) where 1, 2, 3 ∈ N
Therefore, the operation * is not associative.
Hence, the operation * is commutative, but not associative.
Thus, the correct answer is (B).

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.3

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 1 Relations and Functions Ex 1.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3

Question 1.
Let f:{1, 3, 4} → {1, 2, 5} and g:{ 1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5,1)}. Write down gof.
Solution.
The functions f :{1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} are defined as f = {(1, 2), (3, 5), (4,1)} and g = {(1, 3), (2, 3), (5,1)}.
gof (1) = g(f(1)) = g(2) = 3 [∵ f(1) = 2 and g(2) = 3]
gof (3) = g(f(3)) = g(5) = 1 [∵ f(3) = 5 and g(5) = 1]
gof (4) = g(f(4)) = g(1) = 3 [∵ f(4) = 1 and g(1) = 3]
∴ gof = {(1,3), (3,1), (4, 3)}.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.3

Question 2.
Let f, g and h be functions from R to R. Show that (f + g)oh = foh + goh (f . g)oh = (foh) (goh)
Solution.
To prove (f + g)oh = foh + goh Consider
((f + g)oh)(x) = (f + g)(h(x))
f(h(x)) + g(h(x)) = (foh)(x) + (goh)(x) = {{foh) + (goh)}(x)
((f + g)oh)(x) = {(foh) + (goh)} (x) ∀ x ∈ R
Hence, (f + g)oh = foh + goh.
To prove (f . g)oh = (foh) . (goh)
Consider
((f . g)oh) (x) = (f . g) (h(x)) = f(h(x)) . g(h(x))
= (foh)(x).(goh)(x)
= {(foh) . (goh)}(x)
∴ ((f . g)oh)(x) = {(foh) . (goh)}(x) ∀ x ∈ R
Hence, (f . g)oh = (/oh) . (goh)

Question 3.
Find gof and fog, if
(i) f(x) = |x| and g(x) = |5x – 2|
(ii) f(x) = 8x3 and g(x) = \(x^{\frac{1}{3}}\)
Solution.
(i) f(x) =|x| and g(x) = |5x – 2|
∴ (gof)(x) = g(f (x)) = g(| x |) =| 5| x | – 2 |
(fog(x)) = f(g(x)) = f(| 5x – 2 |) = | | 5x-2 || = |5x – 2|

(ii) f(x) = 8x3 and g(x) = \(x^{\frac{1}{3}}\)
∴ gof(x) = g(f(x))
= g(8x3)
= (8x3)\(\frac{1}{3}\)
= 8x

(fog)(x) = f(g(x))
= f(\(x^{\frac{1}{3}}\))
= 8(\(x^{\frac{1}{3}}\))3
= 8x

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.3

Question 4.
If f(x) = \(\frac{(4 x+3)}{(6 x-4)}\), x ≠ \(\frac{2}{3}\) show that fof(x) = x for all x ≠ \(\frac{2}{3}\) What is the inverse of f?
Solution.
It is given that f(x) = \(\frac{(4 x+3)}{(6 x-4)}\), x ≠ \(\frac{2}{3}\)
(fof)(x) = f(f(x)) = f(\(\frac{(4 x+3)}{(6 x-4)}\))
= \(\frac{4\left(\frac{4 x+3}{6 x-4}\right)+3}{6\left(\frac{4 x+3}{6 x-4}\right)-4}\)
= \(\frac{16 x+12+18 x-12}{24 x+18-24 x+16}=\frac{34 x}{34}\) = x
Therefore, fof(x) = x, for all x ≠ \(\frac{2}{3}\).
⇒ fof = 1.
Hence, the given function f is invertible and the inverse of f is itself.

Question 5.
State with reason whether the following functions have inverse
(i) f: {1, 2, 3, 4} → {10} with  f = {(1, 10), <2,10), (8, 10), <4, 10)}
(ii) g: {5, 6, 7,8} → {1, 2, 3, 4,} with g = {(5, 4), (6,3), (7,4), (8, 2)}
(iii) h:{2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}.
Solution.
(i) Function f:{1, 2, 3, 4} {10} defined as
f = {(1,10), (2,10), (3,10), (4,10)}
From the given definition of f, we can see that f is a many-one function as:
f(1) = f(2) = f(3) = f(4) = 10
∴ f is not one-one.
Hence, function f does not have an inverse.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.3

(ii) Function g:{5, 6, 7,8} → {1,2, 3, 4,} defined as g = {(5, 4), (6, 3), (7, 4), (8, 2)}
From the given definition of g, it is seen that g is a many-one function as : g(5) = g(7) = 4.
∴ g is not one-one,
Hence, function g does not have an inverse.

(iii) Function h:{2, 3, 4, 5,} → {7, 9,11,13} defined as h = {(2, 7), (3, 9), (4,11), (5,13)}
It is seen that all distinct elements of the set {2, 3, 4, 5} have distinct images under h.
∴ Function h is one-one.
Also, h is onto since for every element y of the set {7, 9, 11, 13}, there exists an element x in the set {2, 3, 4, 5} such that h(x) = y.
Thus, h is a one-one and onto function. Hence, h has an inverse.

Question 6.
Show that f: [- 1,1] → R, given by f(x) = \(\frac{x}{x+2}\) is one-one. Find the inverse of the function f: [- 1, 1] → Range f.
[Hint : For y ∈ R Range f, y = f(x) = \(\frac{x}{x+2}\), for some x in [- 1, 1] i.e., x = \(\frac{2 y}{1-y}\)]
Solution.
f: [- 1, 1] → R, is given as f(x) = \(\frac{x}{x+2}\)
Let f(ix) = f(y).
⇒ \(\frac{x}{x+2}=\frac{y}{y+2}\)
⇒ 2x = 2y
⇒ x = y
∴ f is one-one function.
It is clear that f: [- 1,1] Range f is onto.
∴ f: [- 1, 1] → Range f is one-one and onto and therefore, the inverse of the function :
f: [- 1, 1] → Range f exists.
Let g: Range f → [- 1, 1] be the inverse of f.
Let y be an arbitrary element of range f.
Since, f: [- 1, 1] → Range f is onto , we have
y = f(x) for some x ∈ [- 1, 1]
⇒ y = \(\frac{x}{x+2}\)
⇒ xy + 2y = x
⇒ x(1 – y) = 2y
⇒ x = \(\frac{2 y}{1-y}\), y ≠ 1
Now, let us define g: Range f → [- 1, 1] as g(y) = \(\frac{2 y}{1-y}\), y ≠ 1.
Now, (gof)(x) = g(f(x))
= g(\(\frac{x}{x+2}\)) = \(\frac{2\left(\frac{x}{x+2}\right)}{1-\frac{x}{x+2}}\)
= \(\frac{2 x}{x+2-x}=\frac{2 x}{2}\) = x

(fog)(y) = f(g(y))
= f(\(\frac{2 y}{1-y}\)) = \(\frac{\frac{2 y}{1-y}}{\frac{2 y}{1-y}+2}\)
= \(\frac{2 y}{2 y+2-2 y}=\frac{2 y}{2}\) = y
∴ gof = I[- 1, 1] and fog = IRange f
∴ f-1 = g
⇒ f-1(y) = \(\frac{2 y}{1-y}\), y ≠ 1.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.3

Question 7.
Consider f:R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
Solution.
Here, f: R → R is given by f(x) = 4x +3
Let x, y ∈ R, such that
f(x) = f(y)
⇒ 4x + 3 = 4y + 3
⇒ 4x = 4y
⇒ x = y
Therefore, f is a one-one function. .
Let y = 4x +3
⇒ There exists, x = \(\frac{y-3}{7}\) ∈ R, ∀ y ∈ R
Therefore for any y ∈ R, there exists x = \(\frac{y-3}{4}\) ∈ R such that
f(x) = f(\(\frac{y-3}{4}\)) = 4 (\(\frac{y-3}{4}\)) + 3 = y
Therefore, f is onto function.
Thus, f is one-one and onto and therefore, f-1 exists.
Let us define g: R → R by g(x) = \(\frac{x-3}{4}\)
Now, (gof)(x) = g(f(x)) = g(4x + 3)
= \(\frac{(4 x+3)-3}{4}\) = x

(fog)(y) = f(g(y))
= \(f\left(\frac{y-3}{4}\right)=4\left(\frac{y-3}{4}\right)\) + 3
= y – 3 + 3 = y
Therefore, gof = fog = IR
Hence, f is invertible and the inverse of f is given by
f-1 (y) = g(y) = \(\frac{y-3}{4}\)

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.3

Question 8.
Consider f: R → [4, ∞) given by f(x) = x2 + 4 Show that f is invertible with the inverse f-1 of f given by f-1(y) = \(\sqrt{y-4}\), where R is the set of all non-negative real numbers.
Solution.
Function f: R+ → [4, ∞) is given as f(x) = x2 + 4.

One-one :
Let f(x) = f(y).
⇒ x2 + 4 = y2 + 4
⇒ x2 = y2
⇒ x = y [as x = y ∈ R+]
∴ f is one-one function.

Onto :
For y ∈ [4, ∞), let y = x2 + 4.
⇒ x2 = y – 4 ≥ 0 [as y ≥ 4]
⇒ x = \(\sqrt{y-4}\) > 0
Therefore, for any y ∈ R, there exists x = \(\sqrt{y-4}\) ∈ R such that
f(x) = f(\(\sqrt{y-4}\))
= (\(\sqrt{y-4}\))2 + 4
= y – 4 + 4 = y
∴ f is onto.
Thus, f is one-one and onto and therefore, f-1 exists.
Let us define g:[4, ∞) → R+ by,
g(y) = \(\sqrt{y-4}\)
Now, gof (x) = g(f(x)) = g(x2 + 4)
= \(\sqrt{\left(x^{2}+4\right)-4}=\sqrt{x^{2}}\) = x

and, fog(y) = f(g(y))
= f(\(\sqrt{y-4}\))
= \((\sqrt{y-4})^{2}+4\)
= (y – 4) + 4 = y
∴ gof = fog = IR+
Hence, f is invertible and the inverse of f is given by
f-1 (y) = g(y) = \(\sqrt{y-4}\)

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.3

Question 9.
Consider f: R → [- 5, ∞) given by f(x) = 9x2 + 6x – 5. Show that f is invertible with f-1 (y) = \(\left(\frac{(\sqrt{y+6}-1}{3}\right)\)
Solution.
f: R+ → [- 5, ∞) is given as f(x) = 9x2 + 6x – 5
Let y be an arbitrary element of (- 5, ∞)
Let y = 9x2 + 6x – 5
y = (3x + 1)2 – 1 – 5
= (3x + 1)2 – 6
⇒ (3x + 1)2 = y + 6
⇒ 3x + 1 = \(\sqrt{y+6}\) [as y ≥ – 5 ⇒ y + 6 > 0]
⇒ x = \(\frac{\sqrt{y+6}-1}{3}\)
∴ f is onto, thereby range f = [- 5, ∞]
Let us define g: [- 5, ∞) → R+ as g(y) = \(\frac{\sqrt{y+6}-1}{3}\)
We now have :
(gof)(x) = g(f(x)) = g(9x2 + 6x – 5) = g((3x +1)2 – 6)
= \(\frac{\sqrt{(3 x+1)^{2}-6+6}-1}{3}=\frac{3 x+1-1}{3}\) = x
and, (fog)(y) = f(g(y))
= \(f\left(\frac{\sqrt{y+6}-1}{3}\right)=\left[3\left(\frac{(\sqrt{y+6})-1}{3}\right)+1\right]^{2}-6\)
= \((\sqrt{y+6})^{2}\) – 6 = y + 6 – 6 = y
∴ gof = IR and fog = I[ – 5, ∞]
Hence f is invertible and inverse of f is given by
f-1(y) = g(y) = \(\frac{\sqrt{y+6}-1}{3}\)

Question 10.
Let f: X → Y be an invertible function. Show that f has unique inverse.
[Hint: Suppose g1 and g2 are two inverses of f. Then for all y ∈ Y, fog1 (y) = 1, (y) = fog2 (y). Use one-one ness of f].
Solution.
Let f: X → Y be an invertible function.
Also, suppose f has two inverses (say g1 and g2).
Then, for all y ∈ Y, we have
fog1 (y) = Iy(y) = fog2(y)
⇒ f(g1(y)) = f(g2(y))
⇒ g1(y) = g2(y) [f is invertible ⇒ f is one-one, g is one-one]
⇒ g1 = g2
Hence, f has a unique inverse.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.3

Question 11.
Consider f: {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c. Find f-1 and show that (f-1)-1 = f.
Solution.
Function f: {1,2, 3} → {a, b, c} is given by f(1) = a, f(2) = b and f(3) = c
If we define g :{a, b, c} → {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3, then we
(fog)(a) = f(g(a)) = f(1) = a
(fog)(b) = f(g(b) = f(2) = b
(fog)(c) = f(g(c)) = f(3) = c
(gof)(2) = g(f(2)) = g(b) = 2
(gof)(3) = g(f(3)) = g(c) = 3
∴ gof = IX and fog = IY, where X = {1, 2, 3} and Y = {a, b, c}. Thus, the inverse of f exists and f-1 = g.
∴ f-1 : {a, b, c} → {1, 2, 3} is given by,
f-1(a) = 1, f-1(b) = 2, f-1(c) = 3
Let us now find the inverse of f-1 i.e., find the inverse of g.
If we define h:{ 1,2, 3} → {a, b, c} as h(1) = a, h(2) = b, h(3) = c, then we have
(goh)(1) = g(h(1l)) = g(a) = 1
(goh) (2) = g(h(2)) = g(b) = 2
(goh) (3) = g(h(3)) = g(c) = 3
and,(hog)(a) = h(g(a)) – h(1) = a
(hog) (b) = h(g(b)) = h(2) = b
(hog)(c) = h(g(c)) = h(3) = c
∴ goh = IX and hog = IY, where X = {1, 2, 3} and Y = {a, b, c}.
Thus, the inverse of g exists and g-1 = h
⇒ (f-1)-1 = h
It can be noted that h = f.
Hence, (f-1)-1 = f.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.3

Question 12.
Let f: X → Y be an invertible function. Show that the inverse of f1 is f, i.e., (f -1)-1 = f.
Solution.
Let f:X → Y be an invertible function.
Then, there exists a functiong:Y → X such that gof = IX and fog – IY.
Here, f-1 = g.
Now, gof = IX and fog = IY
⇒ f-1 = IX and fof-1 = IY
Hence, f-1: Y → X is invertible and f is the inverse of f-1 i-e., (f-1)-1 = f

Question 13.
If f : R → R be given by fix) = (3 – x3)\(\frac{1}{3}\), then fof(x) is
(A) x\(\frac{1}{3}\)
(B) x3
(C) x
D) (3 – x3)
Solution.
Function f: R → R is given as f(x) = {3 – x3)\(\frac{1}{3}\); f(x) = (3 – x3)\(\frac{1}{3}\)
∴ fof(x) = f(f(x)) = f(3 – x3)\(\frac{1}{3}\)
= [3 – ((3 – x3)\(\frac{1}{3}\))3]\(\frac{1}{3}\)
= [3 – (3 – x3)]\(\frac{1}{3}\)
= (x3)\(\frac{1}{3}\) = x
∴ fof(x) = x
The correct answer is (C)

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.3

Question 14.
Let f: R – {- \(\frac{4}{3}\)} R be a function defined as f(x) = \(\frac{4 x}{3 x+4}\). The inverse of f is the map g: Range f → R given by
(A) g(y) = \(\frac{3 y}{3-4 y}\)

(B) g(y) = \(\frac{4 y}{4-3 y}\)

(C) g(y) = \(\frac{4 y}{3-4 y}\)

(D) g(y) = \(\frac{3 y}{4-3 y}\)
Solution.
Given that f : R – {- \(\frac{4}{3}\)} → R is a function defined as
f(x) = \(\frac{4 x}{3 x+4}\)
i.e., y = \(\frac{4 x}{3 x+4}\)
3 xy + 4y = 4x
4y = 4x – 3xy
4 y = x(4 – 3y)
x = \(\frac{4 y}{4-3 y}\)
∴ f-1(y) = g(y) = \(\frac{4 y}{4-3 y}\)
The correct answer is (B).

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.2

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 1 Relations and Functions Ex 1.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.2

Question 1.
Show that the function F: R → R, defined by f(x) = \(\frac{1}{x}\) is one-one and onto, where R, is the set of all non-zero real numbers. Is the result true, if the domain R. is replaced by N with co-domain being same as R?
Solution.
It is given that f: R. → R. is defined by f(x) = \(\frac{1}{x}\)
One-one :
f(x) = f(y)
⇒ \(\frac{1}{x}\) = \(\frac{1}{y}\)
⇒ x = y
∴ f is one-one.

Onto :
It is clear that for y ∈ R., there exists x = \(\frac{1}{y}\) ∈ R. (Exists as y ≠ 0) such that f(x) = \(\frac{1}{\left(\frac{1}{y}\right)}\) = y.
∴ f is onto.
Thus, the given function (f) is one-one and onto.
Now, consider function g :N →R, defined by
g(x) = \(\frac{1}{x}\)
We have,
g(x1) = g(x2)
⇒ \(\frac{1}{x_{1}}=\frac{1}{x_{2}}\)
x1 = x2
∴ g is one-one.
Further, it is clear that g is^not onto as for 1.2 ∈ R, there does not exist any x in N such that g(x) = \(\frac{1}{1.2}\).
Hence, function g is one-one but not onto.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.2

Question 2.
Check the injectivity and surjectivity of the following functions
(i) f: N → N given by f(x) = x2
(ii) f: Z → Z given by f(x) = x2
(iii) f: R → R given by f(x) = x2
(vi) f: N → N given by f(x)) = x3
(v) f: Z → Z given by f(x) = x3
Solution.
(i) f: N → N is given by,
f(x) = x2
It is seen that for x, y ∈ N, f(x) = f(y)
⇒ x2 = y2
⇒ x = y
∴ f is injective.
Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x2 = 2.
∴ f is not surjective.
Hence, function f is injective but not surjective.

(ii) f: Z → Z is given by,
f(x) = x2
It is seen that f(- 1) = f(1) = 1, but = – 1 ≠ 1.
∴ f is not injective.
Now, – 2 ∈ Z. But, there does not exist any element x ∈ Z such that f(x) = x2 = – 2
∴ f is not surjective.
Hence, function f is neither injective nor surjective.

(iii) f: R → R is given by, f(x) = x2
It is seen that f(- 1) = f(1) = 1, but -1 ≠ 1.
∴ f is not injective.
Now, – 2 ∈ R. But , there does not exist any element x ∈ R such that f(x) = x2 = – 2.
∴ f is not surjective.
Hence, function f is neither injective nor surjective.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.2

(iv) f : N → N given by,
f(x) = x3
It is seen that for x, y ∈ N, f(x) = f(y)
⇒ x3 = y3
⇒ x = y
∴ f is injective.
Now, 2 ∈ N. But, there does not exist any element x in domain N such that f(x) = x3 = 2.
∴ f is not surjective
Hence, function f is injective but not surjective.

(v) f: Z → Z is given by, f(x) = x3
It is seen that for x, y ∈ Z, f(x) = f(y)
⇒ x3 = y3
⇒ x = y.
∴ f is injective.
Now, 2 ∈ Z. But, there does not exist any element x in domain Z such that f(x) = x3 = 2.
∴ f is not surjective.
Hence, function f is injective but not surjective.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.2

Question 3.
Prove that the greatest integer function f: R → R given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
Solution.
f: R → R is given by,
f(x) = [x]
It is seen that /(1.2) = [1.2] = 1,
f(1.9) = [1.9] = 1.
∴ f(1.2) = f(1.9), but 1.2 ≠ 1.9.
∴ f is not one-one.
Now, consider 0.7 ∈ R.
It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ∈ R such that f(x) = 0.7.
∴ f is not onto.
Hence, the greatest integer function is neither one-one nor onto.

Question 4.
Show that the modulus function f: R → R given by f(x) = |x|, is neither one-one nor onto, where x is x, if x is positive or 0 and |x| is – x, if x is negative.
Solution.
f: R → R is given by,
f(x) = |x| = {x, if x ≥ 0; – x if x < 0
It is seen that f(- 1) = |- 1| = 1, f(1) = |1| = 1.
∴ f(- 1) = f(1),but – 1 ≠ 1.
∴ f is not one-one.
Now, consider – 1 ∈ R.
It is known that f(x) = |x| is always non-negative,. Thus, there does not exist any element x in domain R such that f(x) = |x| = – 1.
∴ f is not onto.
Hence, the modulus function is neither one-one nor onto.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.2

Question 5.
Show that the signum function f: R → R, given by
PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.2 1
is neither one-one nor onto.
Solution.
f: R → R is given by,

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.2 1

It is seen that f(1) = f(2) = 1, but 1 ≠ 2.
∴ f is not one-one.
Now, as f(x) takes only 3 values (1, 0, or – 1) for the element – 2 in co-domain R, there does not exist any x in domain R such that f(x) = – 2.
∴ f is not onto.
Hence, the signum function is neither one-one nor onto.

Question 6.
Let A = {1, 2, 3,}, B = {4 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.
Solution.
It is given that A = {1, 2, 3}, B = {4, 5, 6, 7}.
f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}.
∴ f(1) = 4, f(2) = 5, f(3) = 6
It is seen that the images of distinct elements of A under f are distinct. Hence, function f is one-one.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.2

Question 7.
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f: R → R defined as f(x) = 3 – 4x
(ii) f: R → R defined as f(x) = 1 + x3
Solution.
(i ) f: R → R is defined as f(x) = 3- 4x.
Let x1, x2 ∈ R such that f(x1) = f(x2)
⇒ 3 – 4x1 = 3 – 4x2
⇒ – 4x1 = – 4x1
⇒ x1 = x2
∴ f is one-one.
For any real number (y) in R, there exists \(\frac{3-y}{4}\) in R such that
f(\(\frac{3-y}{4}\)) = 3 – 4(\(\frac{3-y}{4}\)) = y
∴ f is onto.
Hence, f is bijective.

(ii) f: R → R is defined as f(x) = 1 + x2
Let x1, x2 ∈ R such that f(x1) = f(x2)
⇒ 1 + x12 = 1 + x22
⇒ x12 = ± x22
⇒ x1 = x2
⇒ f(x1) = f(x2) does not imply that x1 = x2.
For instance, f(1) = f(- 1) = 2
∴ f is not one-one.
Consider an element – 2 in co-domain R.
k is seen that f(x) = 1 + x2 is positive for all x ∈ R.
Thus, there does not exist any x in domain R such that f(x) = – 2.
∴ f is not onto.
Hence, f is neither one-one nor onto.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.2

Question 8.
Let A and B be sets. Show that f: A × B – B × A such that f (a, b) (b, a) is bijective function.
Solution.
f: A × B → B × A is defined as f(a, b) = (b, a).
Let(a1, b1), (a2, b2) ∈ A × B such that f(a1, b1) = f(a2, b2)
⇒ (b1, a1) = (b2, a2)
⇒ b1 = b2 and a1 = a2
⇒ (a1, b1) = (a2, b2)
∴ f is one – one.
Now, let (b, a) ∈ B × A be any element.
Then, there exists (a, b) ∈ A × B such that f(a, b) = (b, a). [by definition of f]
∴ f is onto.
Hence, f is bijective.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.2

Question 9.
Let f: N → N be defined by

(n) = PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.2 2

State whether the function is bijective. Justify your answer.
Solution.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.2 2

It can be observed that:
f(1) = \(\frac{1+1}{2}\) = 1 amnd f(2) = \(\frac{2}{2}\) = 1 [by definition of f]
∴ f(1) = f(2), where 1 ≠ 2.
∴ f is not one-one.
Consider a natural number (n) in co-domain N.

Case I: n is odd.
∴ n = 2r + 1 for some r ∈ N. Then, there exists 4r + 1 ∈ N such that
f(4r + 1) = \(\frac{4 r+1+1}{2}\) = 2r+ 1

Case II : n is even,
∴ n – 2r for some r ∈ N. Then there exists 4r ∈ N such that 4r
f(4r) = \(\frac{4r}{2}\) = 2r.
∴ f is onto.
Hence, f is not a bijective function.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.2

Question 10.
Let A = R – {3} and B = R – {1}. Consider the function f: A → B defined by f(x) = \(\left(\frac{x-2}{x-3}\right)\). Is f one-one and onto? Justify your answer.
Solution.
Here, A = R – {3}, B = R – {1}
and f: A → B is defined as f(x) = \(\left(\frac{x-2}{x-3}\right)\)
Let x, y ∈ A such that f(x) = f(y).
⇒ \(\frac{x-2}{x-3}=\frac{y-2}{y-3}\)
⇒ (x – 2) (y – 3) = (y – 2) (x – 3)
⇒ xy – 3x – 2y + 6 = xy – 3y – 2x + 6
⇒ – 3x – 2y = – 3y – 2x
⇒ 3x – 2x = 3y – 2y
⇒ x = y
∴ f is one-one.
Let y ∈ B = R – {1}. Then, y ≠ 1.
The function f is onto if there exists x ∈ A such that f(x) = y.
Now, f(x) = y
⇒ \(\frac{x-2}{x-3}\) = y
⇒ x – 2 = xy – 3y
⇒ x(1 – y) = – 3y + 2
⇒ x = \(\frac{2-3 y}{1-y}\) ∈ A

Thus, for any y B, there exists \(\frac{2-3 y}{1-y}\) ∈ A such that
f(\(\frac{2-3 y}{1-y}\)) = \(\frac{\left(\frac{2-3 y}{1-y}\right)-2}{\left(\frac{2-3 y}{1-y}\right)-3}\)

= \(\frac{2-3 y-2+2 y}{2-3 y-3+3 y}=\frac{-y}{-1}\) = y

∴ f is onto.
Hence, function f is one-one and onto.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.2

Question 11.
Let f: R → R be defined as f(x) = x4. Choose the correct answer.
(A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto
Solution.
f : R → R is defined as f(x) = x4 Let x, yeR such that f(x) = f(y).
⇒ x4 = y4
⇒ x = ±y
∴ f(x1) = f(x2) does not imply that x1 = x2.
For instance,
f(1) = f(- 1) = 1
∴ f is not one-one.
Consider an element 2 in co-domain it. It is clear that there does not exist any x in domain R such that f(x) – 2 .
∴ f is not onto.
Hence, function f is neither one-one nor onto.
The correct answer is (D).

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.2

Question 12.
Let f: R → R be defined as f(x) = 3x. Choose the correct answer.
(A) f is one-one onto
(B) f is many-one onto
(C) f is one-one not onto
(D) f is neither one-one nor onto
Solution.
f: R → R is defined as f(x) = 3x.
Let x, y ∈ R such that f(x) = f(y).
⇒ 3x – 3y
⇒ x = y .
∴ f is one-one.
Also any real number (y) in co-domain R, there exists \(\frac{y}{3}\) in R such that
f(\(\frac{y}{3}\)) = 3(\(\frac{y}{3}\)) = y
∴ f is onto.
Hence, function f is one-one and onto.
The correct answer is (A).

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.1

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 1 Relations and Functions Ex 1.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.1

Question 1.
Determine whether each of the following relations are reflexive, symmetric and transitive:
(i) Relation R in the set A – {1, 2, 3,.. .13,14} defined as, R = {(x, y) : 3x – y = 0}
Solution:
(i) A = {1, 2, 3 ……. 13, 14};
R = {(x, y): 3x – y = 0}
∴ R = {(1, 3), (2, 6), (3, 9), (4, 12)}
R is not reflexive since (1, 1), (2, 2) … (14, 14) ∉ R.
Also, R is not symmetric as (1, 3) ∈ R, but (3, 1) ∉ R. [3(3) – 1 ≠ 0]
Also, R is not transitive as (1, 3), (3, 9) ∈ R, but (1, 9) ∉ R. [3(1) – 9 ≠ 0]
Hence, R is neither reflexive, nor symmetric, nor transitive.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.1

(ii) Relation R in the set N of natural numbers defined as, R = {(x, y): y = x + 5 and x < 4}
Solution:
R = {(x, y) : y = x + 5 and x < 4} = {(1,6), (2, 7), (3, 8)} It is seen that (1, 1) ∉ R. ∴ R is not reflexive. (1, 6) ∈ R But, (6, 1) ∉ R.
∴ R is not symmetric. Now, since there is no pair in R such that (x, y) and (y, z) ∈ R. So, we need not look for the ordered pair (x, z) in R.
R is transitive Hence, R is neither reflexive, nor symmetric but it is transitive.

(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y): y is divisible by x}
Solution:
A = {1, 2, 3, 4, 5, 6} R = {(x, y) y is divisible by x} We know that any number (x) is divisible by itself. ⇒ (x, x) ∈ R
∴ R is reflexive. Now, (2, 4) ∈ R [as 4 is divisible by 2] But, (4, 2) ∉ R. [as 2 is not divisible by 4]
∴ R is not symmetric. Let (x, y), (y, z) ∈ R. Then, y is divisible by x and z is divisible by y.
∴ z is divisible by x. ⇒ (x, z) ∈ R ∴ R is transitive.
Hence, R is reflexive and transitive but not symmetric.\

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.1

(iv) Relation R in the set Z of all integers defined as, R = {(x, y): x – y is an integer}
Solution:
R = {(x, y): x – y is an integer} Now, for every x ∈ Z, (x, x) ∈ R as x – x = 0 is an integer.
∴ R is reflexive. Now, for every x, y ∈ Z if (x, y) E R as x – y is an integer. ⇒ – (x – y) is also an integer. ⇒ (y – x) is an integer. (y, x) ∈ r ∴ R is symmetric. Now, let (x, y) and (y, z) ∈ R, where x, y, z ∈ Z. ⇒ (x – y) and (y – z) are integers. ⇒ x – z = (x – y) + (y – z) is an integer. ∴ (x, z) ∈ R ∴ R is transitive. Hence, R is reflexive, symmetric and transitive.

(v) Relation R in the set A of human beings in a town at a particular time given by, (a) R = {(x, y): x and y work at the same place} Solution: (a) R = {(x, y): x and y work at the same place} ⇒ (x, x) ∈ R R is reflexive. ⇒ y and x work at the same place. ⇒ (y, X) ∈ R. ∴ R is symmetric. Now, let (x, y), (y, z) ∈ R ⇒ x and y work at the same place and y and z work at the same place. => x and z work at the same place.
⇒ (x, z) ∈ R
∴ R is transitive.
Hence, R is reflexive, symmetric, and transitive.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.1

(b) R = {(x, y) : x and y live in the same locality}
Solution:
R = { (x, y) : x and y live in the same locality}
Clearly (x, x) ∈ R as x and x is the same human being.
∴ R is reflexive.
If (x, y) ∈ R, then x and y live in the same locality.
⇒ y and x live in the same locality.
⇒ (y, x) ∈ R
∴ R is symmetric.
Now, let (x, y) ∈ R and (y, z) ∈ R.
⇒ x and y live in the same locality and y and z live in the same locality. =} x and z live in the same locality.
⇒ (x, z) ∈ R
∴ R is transitive.
Hence, R is reflexive, symmetric, and transitive.

(c) R = {(x, y): x is exactly 7 cm taller than y}
Solution:
R = {(x, y): x is exactly 7 cm taller than y}
Now, (x, x) ∉ R
Since, human being x cannot be taller than himself.
∴ R is not reflexive.
Now, let (x, y) ∈ R.
=> x is exactly 7 cm taller than y.
Then, y is not taller than x.
∴ (y, x) ∉ R
Indeed if x is exactly 7 cm taller than y, then y is exactly 7 cm shorter than x.
∴ R is not symmetric.
Now, let (x, y), (y, z) ∈ R.
⇒ x is exactly 7 cm taller than y and y is exactly 7 cm taller than z.
⇒ x is exactly 14 cm taller than z.
∴ (x, z) ∈ R
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.1

(d) R = {(x, y) : x is wife of y}
Solution:
R = {(x, y): x is the wife of y}
Now, (x, x) ∉ R
Since, x cannot be the wife of herself.
∴ R is not reflexive.
Now, let (x, y) ∈ R ⇒ x is the wife of y.
Clearly, y is not the wife of x.
∴ (y, x) ∉ R
Indeed if x is the wife of y, then y is the husband of x.
∴ R is not symmetric.
Let (x, y),(y, z) ∈ R
⇒ x is the wife of y and y is the wife of z.
This case is not possible. Also, this does not imply that x is the wife of z.
∴ (x, z) ∉ R
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.

(e) R = {(x, y): x is father of y}
Solution:
R = {(x, y): x is the father of y}
(x, x) ∉ R
As x cannot be the father of himself.
∴ R is not reflexive.
Now, let (x, y) G R ⇒ x is the father of y.
⇒ y cannot be the father of x. Indeed y is the son or the daughter of x.
∴ (y, x) ∉ R
R is not symmetric.
Now, let (x, y) ∈ R and (y, z) ∈ R.
⇒ x is the father of y and y is the father of z.
⇒ x is not the father of z.
Indeed x is the grandfather of z.
∴ (x, z) ∉ R
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.1

Question 2.
Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b2} is neither reflexive nor symmetric nor transitive.
Solution:
R = {(a, b) : a ≤ b2}
It can be observed that (\(\frac{1}{2}\), \(\frac{1}{2}\)) ∉ R, since \(\frac{1}{2}\) > (\(\frac{1}{2}\))2 = \(\frac{1}{4}\)
∴ R is not reflexive.
Now, (1, 4) ∈ R as 1 < 42 But, 4 is not less than 12.
∴ (4, 1) ∉ a
∴ R is not symmetric.
Now, (3 2), (2, 1.5) ∈ R [as 3 < 22 = 4 and 2 < (1.5)2 = 2.25]
But, 3 > (1.5)2 = 2.25
∴ (3, 1.5) ∉ R
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.

Question 3.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b): 6 = a +1} is reflexive, symmetric or transitive.
Solution:
Let A = {1, 2, 3, 4,5,6}
A relation R is defined on set A as: R = {(a, b):b = a +1}
∴ R = {(1,2), (2, 3), (3, 4), (4, 5), (5, 6)}
∴ R is not reflexive.
It can be observed that (1, 2) ∈ R, but (2,1) £ R.
∴ R is not symmetric.
Now, (1, 2), (2, 3) ∈ R But, (1, 3) ∉ R
∴ R is not transitive
Hence, R is neither reflexive, nor symmetric, nor transitive.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.1

Question 4.
Show that the relation B in R defined as R = {(a, b):a < b}, is reflexive and transitive hut not symmetric.
Solution:
R = {(a, b) : a < b}
(i) R is reflexive : Replacing b by a, a < a ⇒ a = a is true.
(ii) R is not symmetric : a < b, and b < a which is not true. 2 < 3, but 3 is not less than 2.
(iii) R is transitive : If a < b and b < c then a < c. e.g 2 < 3, 3 < 4 ⇒ 2 < 4.

Question 5.
Check whether the relation R in R defined as R = {(a, b) : a ≤ b3} is reflexive, symmetric or transitive.
Solution:
R = {(a, b) : a ≤ b3}
Now, (\(\frac{1}{2}\), \(\frac{1}{2}\)) ∉ R as \(\frac{1}{2}\) > \(\frac{1}{2}\)3 = \(\frac{1}{8}\))
∴ R is not Reflexive.
Now, (1, 2) ∈ R (as 1 < 23 = 8)
But, (2, 1) ∉ R (as (\(\frac{4}{4}\)) > 1)
∴ R is not symmetric.
(3, \(\frac{3}{2}\)), (\(\frac{3}{2}\), \(\frac{6}{5}\)) ∈ R as 3 < (\(\frac{3}{2}\))3 and \(\frac{3}{2}\) < (\(\frac{6}{5}\))3
But (3, \(\frac{6}{5}\)) ∉ R as 3 > (\(\frac{6}{5}\))3
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.1

Question 6.
Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.
Solution.
Let A = {1, 2, 3}.
A relation R on A is defined as R = {(1, 2), (2,1)}
It is seen that (1, 1), (2, 2), (3, 3) ∉ R.
R is not reflexive.
Now, as (1, 2) ∈ R and (2, 1) ∈ R, then R is symmetric.
Now, as (1, 2) and (2,1) ∈ R – However, (1, 1) ∈ R
∴ R is not transitive.
Hence, R is symmetric but neither reflexive nor transitive.

Question 7.
Show that the relation R in the set A of all the books in a library of a college, given by
R = {(x, y) : x and y have same number of pages} is an equivalence relation.
Solution:
Set A is the set of all books in the library of a college.
R = {(x, y) : x and y have the same number of pages}
Now, R is reflexive since (x, x) ∈ R as x and x have the same number of pages.
Let (x, y) ∈ R
⇒ x and y have the same number of pages.
⇒ y and x have the same number of pages.
⇒ (y, x) ∈ R
∴ R is symmetric.
Now, let (x, y) ∈ R and (y, z) ∈ R.
⇒ x and y have the same number of pages and y and z have the same number of pages.
⇒ x and z have the same number of pages.
⇒ (x, z) ∈ R
∴ R is transitive.
Hence, R is a equivalence relation.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.1

Question 8.
Show that the relation R in the set A = {1, 2, 3, 4, ,5} given by R = {(a, b): | a – b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2,4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.
Solution.
A = {1, 2, 3, 4, 5}
R = { (a, b) : |a – b| is even}
It is clear that for any element a e∈ A, we have |a – a| = 0 (which is even).
∴ R is reflexive.
Let (a, b) ∈ R.
⇒ |a – b| is even.
⇒ |-(a – b)| = |b – a| is also even.
⇒ (b, a) ∈ R
∴ R is symmetric.
Now, let (a, b) ∈ R and (b, c) ∈ R.
⇒ | a – b | is even and | b – c | is even.
⇒ (a – b) is even and (b – c) is even.
⇒ (a – c) = (a – b) + (b – c) is even [Sum of two even integers is even] => | a – c | is even.
⇒ (a, c) ∈ R
∴ R is transitive.
Hence, R is an equivalence relation.
Now, all elements of the set {1, 2, 3} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even.

Similarly, all elements of the set {2, 4} are related to each other as all the elements of this subset are even.

Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as all elements of {1, 3, 5} are odd and elements of (2, 4} are even. Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.1

Question 9.
Show that each of the relation R in the set A = {x ∈ Z: 0 < x ≤ 12}, given by
(i) R = {(a, b) : |a – b| is a multiple of 4}
(ii) R = {(a, b) : a – b} is an equivalence relation. Find the set of all elements related to 1 in each case.
Solution.
A = {x ∈ Z : 0 < x ≤ 12} = {0,1, 2, 3, 4, 5, 6, 7, 8, 9,10,11,12} (i) R = {(a, b): |a – b | is a multiple of 4. For any element a ∈ A, we have (a, a) ∈ R as | a – a | = 0 is a multiple of 4. ∴ R is reflexive. Now, let (a, b) e R => | a – b | is a multiple of 4.
⇒ |- (a – b) | =| b – a| is multiple of 4.
⇒ (b, a) ∈ R
R is symmetric.
Now, let (a, b), (b, c) ∈ R.
⇒ | a – b | is a multiple of 4 and | b – c| is a multiple of 4.
⇒ (a – b) is a multiple of 4 and (b – c) is a multiple of 4.
⇒ (a – c) = (a – b) + (b – c) is a multiple of 4.
⇒ | a – c | is a multiple of 4.
⇒ (a, c) ∈ R
∴ R is transitive.
Hence, R is an equivalence relation.
The set of elements related to 1 is {1, 5, 9} since
|1 – 1| = 0 is a multiple of 4
| 5 – 1| = 4 is a multiple of 4, and
|9 – 1| = 8 is a multiple of 4.

(ii) R = {(a, b) : a = b}
For any element a ∈ A, we have (a, a) ∈ R, since a = a.
∴ R is reflexive.
Now, let (a, b) ∈ R.
⇒ a = b ⇒ b = a ⇒ (b, a) ∈ R
∴ R is symmetric.
Now, let (a, b) ∈ R and (b, c) ∈ R.
⇒ a = b and b = c ⇒ a = c ⇒ (a, c) ∈ R
∴ R is transitive.
Hence, R is an equivalence relation.
The elements in R that are related to 1, will be those elements form set A which are equal to 1.
Hence, the set of elements related to 1 is {1}.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.1

Question. 10.
Given an example of a relation. Which is
(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive.
Solution.
Let A – set of straight lines in a plane.
(i) R = {(a, b): a is perpendicular to b}
Let a, b be two perpendicular lines
(a) If line a is perpendicular to b then b is perpendicular to a ⇒ R is symmetric.

(b) But ‘a’ is not a perpendicular to itself.
∴ R is not reflexive.

(c) If ‘a’ is a perpendicular to to ‘b’ and ‘b’ is perpendicular to ‘o’, but ‘a’ is not perpendicular to ‘c’.
∴ R is not transitive.
Thus, R is symmetric but neither reflexive nor transitive.

(ii) Let A = set of real numbers R = {(a, b) : a>b}
(a) An element is not greater than itself .-. R is not reflexive.
(b) If a > b than b is not greater than a.
⇒ R is not symmetric.
(c) If a > b also b> c, then a > c Thus, R is transitive.
Hence, R is transitive but neither reflexive nor symmetric.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.1

(iii) The relation R in the set {1, 2, 3}, is given by
R = {(a, b) : a + b ≤ 4}
R = {(1, 1), (a, 2), (2, 1), (1, 3), (3, 1), (2, 2)}
Here (1, 1), (2, 2) ∈ R ⇒ R is reflexive.
(1, 2), (2, 1), (1, 3), (3, 1) ⇒ R is symmetric
But it is not transitive, since (2, 1) ∈ R, (1, 3) ∈ R but (2, 3) ∉ R.

(iv) The relation R in the set {1, 2, 3} given by
R = {(a, b) : a ≤ b) = {(1, 2), (2, 2), (3, 3), (2, 3), (1, 3)}
(a) (1, 1), (2, 2), (3, 3) ∈ R ⇒ R is reflexive.
(b) (1, 2) ∈ R, but (2, 1) ∈ R ⇒ R is not symmetric.
(c) (1, 2) ∈ R, (2, 3) ∈ R, Also (1, 3) e R ⇒ R is transitive.

(v) The relation R in the set {1, 2, 3} given by
R = {(a, b): 0 < |a – b | ≤ 2} = {(1, 2), (2,1), (1, 3), (3,1), (2, 3), (3, 2)}
(a) R is not reflexive, (1, 1), (2, 2), (3, 3) do not belong to R.
(b) R is symmetric. (1, 2), (2,1), a, 3), (3,1), (2, 3), (3, 2) ∈ R
(c) R is transitive (1, 2) ∈ R, (2, 3) ∈ R, Also (1, 3) ∈ R.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.1

Question. 11.
Show that the relation R in the set A of points in a plane given by R = {(P, Q): distance of the point P form the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.
Solution.
R – {(P, Q) : distance of point P from the origin is the same as the distance of point Q from the origin}
Clearly, (P, P) ∈ R, since the distance of point P from the origin is always the same as the distance of the same point P from the origin.
∴ R is reflexive.
Now, let (P, Q) ∈ R.
⇒ The distance of point P from the origin is the same as the distance of point Q from the origin. y’
⇒ The distance of point Q from the origin! is the same as the distance of point P from the origin.
⇒ (Q, P) ∈ P
⇒ R is symmetric.
Now, let (P, Q), (Q, S) ∈ R.
⇒ The distance of points P and Q from the origin is the same and also, the distance of points Q and S from the origin is the same.
⇒ The distance of points P and S from the origin is the same.
⇒ (P, S) ∈ P
∴ R is transitive.
Therefore, R is an equivalence relation.
The set of all points related to P ≠ (0 ,0) will be those points whose distances from the origin are the same as the distance of point P from the origin.
In other words, if O (0, 0) is the origin and OP = k, then the set of all points related to P is at a distance of k from the origin.
Hence, this set of points forms a circle with the centre as the origin and this circle passes through point P.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.1

Question. 12.
Show that the relation R defined in the set A of all triangles as R = {(T1, T2) : T1 is similar to T2}, is equivalence relation. Consider three right angled triangles T1 with sides (3, 4, 5), T2 with sides (5, 12, 13) and T3 with sides (6, 8, 10). Which triangles among T1, T2 and T3 are related?
Solution.
R = { (T1, T2): T1 is similar to T2}
R is reflexive since every triangle is similar to itself.
Further, if (T1, T2) ∈ R, then T1 is similar to T2.
⇒ T2 is similar to T1.
⇒ (T2, T1)E R R is symmetric.
Now, let (T1, T2), (T2, T3) ∈ R.
⇒ T1 is similar to T2 and T2 is similar to T3.
⇒ T1 is similar to T3.
⇒ (T1, T3) ∈ R
∴ R is transitive.
Thus, R is an equivalence relation.
Now, we can observe that : \(\frac{3}{6}=\frac{4}{8}\) = \(\frac{5}{10}=\frac{1}{2}\)
∴ The corresponding sides of triangles T1 and T3 are in the same ratio.
Then, triangle T1 is similar to triangle T3.
Hence, T1 is related to T3.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.1

Question 13.
Show that the relation R defined in the set A of all polygons as R = {(P1, P2) : P1 and P2 have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angled triangle T with sides 3, 4 and 5?
Solution.
R = {(P1, P2) : Px and P2 have same number of sides.}
R is reflexive since (P1, P1) e R as the same polygon has the same number of sides with itself.
Let (P1, P2) ∈ R.
⇒ P1 and P2 have the same number of sides.
⇒ P2 and P1 have the same number of sides.
⇒ (P2, P1) ∈ R
∴ R is symmetric.
Now, let (P1, P2), (P2, P3) e R.
⇒ P1 and P2 have the same number of sides.
Also, P2 and P3 have the same number of sides.
⇒ P1 and P3 have the same number of sides.
⇒ (P1, P3) ∈ R
∴ R is transitive. Hence, R is an equivalence relation.
The elements in A related to the right-angled triangle (T) with sides 3, 4 and 5 are those polygons which have 3 sides (since T is a polygon with 3 sides).
Hence, the set of all elements in A related to triangle T is the set of all triangles.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.1

Question. 14.
Let L be the set of all lines in XY-plane and R be the relation in L defined as R = {(L1, L2): L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.
Solution.
R = {(L1, L2) : L1 is parallel to L2}
R is reflexive as any line L1 is parallel to itself i.e., (L1, L1) ∈ R.
Now, let (L1, L2) ∈ R
⇒ L1 is parallel to L2.
⇒ L2 is parallel to L1.
⇒ (L2, L1) ∈ R R is symmetric.
Now, let (L1, L2), (L2, L3) ∈ R.
⇒ L1 is parallel to L2. Also, L2 parallel to L3.
⇒ L1 is parallel to L3.
∴ R is transitive.
Hence, R is an equivalence relation.
The set of all lines related to the line y = 2x + 4 is the set of all lines that are parallel to the line y – 2x + 4.
Slope of line y = 2x + 4 is m = 2. It is known that parallel lines have the same slopes.
The line parallel to the given line is of the form y = 2x + c, where c ∈ R. Hence, the set of all lines related to the given line is given by y = 2x + c, where c ∈ R.

PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Ex 1.1

Question 15.
Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer.
(A) R is reflexive and symmetric but not transitive.
(B) R is reflexive and transitive but not symmetric.
(C) R is symmetric and transitive but not reflexive.
(D) R is an equivalence relation.
Solution.
R = {(1, 2), (2, 2), (1,1), (4, 4), (1, 3), (3, 3), (3, 2)}
It is seen that (a, a) ∈ R, for every a ∴ {1, 2, 3, 4}
∴ R is reflexive.
It is seen that (1, 2) ∈ R, but (2, 1) $ R.
∴. R is not symmetric.
Also, it is observed that (a, b), (b, c) ∈ R
⇒ (a, c) ∈ R for all a, b, c ∈ {1, 2, 3, 4}
∴ R is transitive.
Hence, R is reflexive and transitive but not symmetric.
Thus, the correct answer is (B).

Question 16.
Let R be the relation in the set N given by
R = {a, b): a = b – 2, b > 6}. Choose the correct answer.
(A) (2, 4) ∈ R
(B) (3, 8) ∈ R
(C) (6, 8) ∈ R
(D) (8, 7) ∈ R
Solution.
R = {(a, b): a = b – 2, b > 6}
Now, since b > 6, (2, 4) ∉ R
Also, as 3 ≠ 8 – 2, (3, 8) ∉ R
And, as 8 ≠ 7 – 2
∴ (8, 7) ∉ R
Now, consider (6, 8). We have 8 > 6 and also, 6 = 8 – 2.
∴ (6, 8) ∈ R
Thus, the correct answer is (C).

PSEB 12th Class Sociology Solutions स्रोत आधारित प्रश्न

Punjab State Board PSEB 12th Class Sociology Book Solutions स्रोत आधारित प्रश्न.

PSEB Solutions for Class 12 Sociology स्रोत आधारित प्रश्न

स्त्रोत आधारित प्रश्न (Source Based Questions) :

प्रश्न 1.
निम्न दिए स्रोत को पढ़ें व साथ दिए प्रश्नों के उत्तर दें
जनजातियाँ भारत के विभिन्न हिस्सों में पूरे देश में विविध समानुपातों में निवास करती हैं। जनजातीय जनसंख्या का उच्चतम अनुपात केन्द्रीय भारत में है। भारत के उत्तरी-पूर्वी हिस्से में भी इनकी संख्या अधिक है। उनमें से कुछ जनजातियां यथा-गोंड, भील, संथाल, ओरायोन्स आदि हैं जो केन्द्रीय भारत में निवास करती हैं। भारत के जनजातीय समुदाय जो जंगलों, पहाड़ियों तथा प्राकृतिक रूप से पृथक क्षेत्रों में रहते हैं उन्हें विभिन्न नामों से जाना जाता है यथा वन्य जाति, वनवासी, पहाड़ी, आदिमजाति, आदिवासी, जनजाति, अनुसूचित जनजाति इत्यादि। इन सब में आदिवासी ज्यादा प्रमुख हैं तथा अनुसूचित जनजाति इन सबका संवैधानिक नाम है।

(i) भारत के किन क्षेत्रों में जनजातियां सबसे अधिक हैं ?
(ii) जनजातियों को किन नामों से पुकारा जाता है ?
(iii) जनजाति किसे कहते हैं ?
उत्तर-
(i) वैसे तो मध्य भारत में जनजातीय जनसंख्या काफी अधिक है परन्तु अगर हम जनसंख्या में उनके प्रतिशत की बात करें तो उत्तर-पूर्वी भारत में इनकी जनसंख्या सबसे अधिक है।
(ii) जनजातियों को अलग-अलग क्षेत्रों में अलग-अलग नामों से पुकारा जाता है जैसे कि आदिवासी, वनजाति, आदिमजाति, पहाड़ी, वनवासी, जनजाति, अनुसूचित जनजाति इत्यादि।
(iii) एक जनजाति ऐसे लोगों का समूह होता है जो हमारी सभ्यता से दूर किसी जंगल, पहाड़ या घाटी में रहता है, जिसके सदस्य आपस में रक्त संबंधी होते हैं, जो अन्तर्वैवाहिक होता है तथा जिसकी भाषा, धर्म तथा अन्य विशेषताएं अन्य जनजातीय समूहों से अलग होते हैं।

PSEB 12th Class Sociology Solutions स्रोत आधारित प्रश्न

प्रश्न 2.
निम्न दिए स्त्रोत को पढ़ें व साथ में दिए प्रश्नों के उत्तर दें
पर्यावरण के असंतुलन के प्रमुख कारणों में से एक जंगलों का काटा जाना है। वृक्षों का काटा जाना इसमें शामिल है। इसके अतिरिक्त कृषि क्षेत्रों का विस्तार व चरागाह भी वन कटाव के प्रमुख कारण हैं। प्रारंभिक दौर में जंगलों तथा प्राकृतिक संसाधनों की सुविधा के कारण जनजातियां अपना जीवन निर्वाह कर रही थीं। अपनी आज आजीविका हेतु वे पूर्णरूपेण जंगलों पर निर्भर थीं। परंतु औद्योगीकरण, नगरीकरण, कृषि, जनसंख्या वृद्धि, व्यापारिक लाभ, ईंधन लकड़ी के संग्रह के कारण जंगलों का कटाव भारी मात्रा में हुआ जिसने प्रत्यक्ष व परोक्ष रूप से जनजातीय आजीविका को प्रभावित किया है। वनों के काटे जाने से मौसम पर भी प्रभाव पड़ा है जो जैव विभिन्नता के क्षरण के रूप में सामने आया है।
(i) वन कटाव का क्या अर्थ है ?
(ii) वन कटाव के क्या कारण हैं ?
(iii) वन कटाव का जनजातियों के जीवन पर क्या प्रभाव पड़ता है ?
उत्तर-
(i) जब अलग-अलग कारणों के कारण वनों में प्राकृतिक रूप से उत्पन्न हुए पेड़ों को काटा जाता है तो इसे वन कटाव कहा जाता है।
(ii) (a) कृषि क्षेत्र तथा चारागाह का क्षेत्र बढ़ाने के लिए वन काटे जाते हैं।
(b) बढ़ती जनसंख्या के लिए घर बनाने तथा बाँध बनाने के लिए जंगलों का सफाया कर दिया जाता है।
(c) ईंधन की लकड़ी तथा फर्नीचर बनाने के लिए लकड़ी की आवश्यकता होती है जिस कारण वन काट दिए जाते हैं।

(iii) (a) इससे जनजातियों के लिए रहने के स्थान की कमी हो जाती है।
(b) वनों से आदिवासी काफी कुछ प्राप्त करते थे जो अब वह नहीं कर सकते हैं।
(c) वनों पर जनजातीय अर्थव्यवस्था निर्भर होती थी परन्तु इनके कटने से वह अर्थव्यवस्था खत्म हो गई।

प्रश्न 3.
निम्न दिए स्त्रोत को पढ़ें व साथ में दिए प्रश्नों के उत्तर दें’ग्राम’ शब्द ‘नगर’ का विपरीत शब्द है। ‘ग्रामीण समाज’ पद का अंतर परिवर्तनीय रूप से ‘ग्राम’ शब्द रूप में ही प्रयोग किया जाता है। 2011 की मतगणना के अनुसार 121 करोड़ भारतीयों में 68 प्रतिशत जनसंख्या ग्रामीण क्षेत्रों में रहती है। ग्रामीण समुदाय का अपना लंबा इतिहास है। यह कृषि तथा सहायक व्यवसायों पर निर्भर करने वाला लगभग 5000 लोगों का समूह है जो स्थायी रूप से विशिष्ट भौगोलिक क्षेत्र में रहता है तथा सांझे सामाजिक, आर्थिक तथा सांस्कृतिक कार्यों में भाग लेते हैं।
(i) ग्राम शब्द का अर्थ बताएं।
(ii) ग्राम की तीन विशेषताएं बताएं।
(iii) ग्राम तथा नगर में तीन अंतर बताएं।
उत्तर-
(i) ग्राम एक ऐसे क्षेत्र को कहते हैं जो प्राकृतिक वातावरण के नज़दीक होता है, जिसकी अधिकतर जनसंख्या कृषि आधारित कार्यों में लिप्त होती है तथा जो अपनी कुछ विशेषताओं के कारण नगरीय क्षेत्र से अलग होता है।

(ii) (a) ग्राम के लोगों के बीच प्रत्यक्ष व प्राथमिक संबंध होते हैं।
(b) ग्राम की अधिकतर जनसंख्या कृषि अथवा संबंधित कार्यों में लगी होती है।
(c) ग्राम का आकार छोटा होता है तथा यहां सामाजिक एकरूपता होती है।

(iii) (a) ग्राम का आकार छोटा जबकि नगर का आकार काफी बड़ा होता है।
(b) ग्राम के लोगों के बीच प्रत्यक्ष व प्राथमिक संबंध होते हैं जबकि नगर के लोगों के बीच अप्रत्यक्ष व द्वितीय संबंध होते हैं।
(c) ग्राम की अधिकतर जनसंख्या कृषि अथवा आधारित कार्यों में लगी होती है जबकि नगरों में 75% से अधिक जनसंख्या उद्योगों अथवा गैर-कृषि कार्यों में लगी होती है।

प्रश्न 4.
निम्न दिए स्त्रोत को पढ़ें व साथ दिए प्रश्नों के उत्तर दें
ग्रामीण समाज की प्रमुख समस्याओं में से एक ऋणग्रस्तता है। ऐसी चिरकालीन ऋणग्रस्तता का कारण निर्धनता एवं घाटे की अर्थव्यवस्था है। यह समस्या केवल एक व्यक्ति से सम्बद्ध नहीं है बल्कि एक पीढ़ी से दूसरी पीढ़ी तक स्थानांतरित होती है। कृषि उत्पादन के लिए ऋण लेना वास्तव में आवश्यक है क्योंकि कृषि का उत्पादन देश के उत्पादनों में से महत्त्वपूर्ण होता है। फिर भी ग्रामीण लोग गैर उत्पादन उद्देश्यों यथापरिवार की ज़रूरतों को पूरा करने हेतु, सामाजिक समारोहों (विवाह, जन्म तथा मृत्यु से संबंधित) को पूरा करने, मुकद्दमेबाज़ी इत्यादि के लिए भी ऋण (कर्ज) ले लेते हैं इस प्रकार, ऋण पर ली गई धनराशि उत्पादन के बजाय उपभोक्ता पर व्यय हो जाती है। यह स्थिति ग्रामीण लोगों को ऋणग्रस्तता की ओर धकेल देती है। इस प्रकार, इन ऋणों की अदायगी असंभव बन जाती है। वे लोभी साहूकारों तथा दलालों के शोषण के आसान शिकार बन जाते हैं जो स्थिति का लाभ उठाकर बहुत उच्चतर दर से ब्याज वसूलते हैं। परिणामः साहूकार उनकी जो भी पूँजी यथा-घर अथवा भूमि इत्यादि छीन लेते हैं। यह व्यवस्था देश के अधिकांश भागों में प्रचलित है।
(i) ऋणग्रस्तता का क्या अर्थ है ?
(ii) ऋणग्रस्तता के क्या कारण हैं ?
(iii) ऋणग्रस्तता के तीन प्रभाव बताएं।
उत्तर-
(i) जब कोई व्यक्ति किसी दूसरे व्यक्ति, साहूकार अथवा बैंक से ऋण ले तथा उसे समय पर वापिस न कर पाए तो इसे ऋणग्रस्तता कहते हैं।

(ii) लोग कई कारणों की वजह से ऋण लेते हैं जैसे कि परिवार की आवश्यकताएं पूर्ण करने के लिए, कानूनी झगड़ों का निपटारा करने के लिए, कृषि करने के लिए, विवाह तथा मृत्यु का खर्चा करने के लिए इत्यादि।

(iii) (a) ऋण के कारण व्यक्ति साहूकारों के शोषण का शिकार बन जाता है। .. (b) व्यक्ति की सम्पूर्ण भूमि पर साहूकार कब्जा कर लेता है तथा वह बेघर हो जाता है।
(c) उसके रहने व जीवन जीने के साधन उससे छीन लिए जाते हैं तथा कई बार ग्रामीण लोग आत्महत्या भी कर लेते हैं।

प्रश्न 5.
निम्न दिए स्रोत को पढ़ें व साथ दिए प्रश्नों के उत्तर दें
नगरवाद नगरीय समाज का वह महत्त्वपूर्ण तत्व है जो उनकी पहचान अथवा व्यक्तित्व को ग्रामीण एवम् जनजाति समाज से अलग करती है। यह एक जीवन शैली का प्रतिनिधित्व करती है। यह नगरीय संस्कृति के प्रसार तथा नगरीय समाज के विकास को प्रकट करती है। यह जटिल श्रम विभाजन, उच्च तकनीकी स्तर, तीव्र गतिशीलता तथा आर्थिक कार्यों की पूर्ति के लिए सदस्यों की अन्तर्निर्भरता तथा सामाजिक सम्बन्धों में बेरूखी के रूप में समाज के संगठन को दर्शाता है। लूइस वर्थ ने नगरवाद की चार विशेषताओं का उल्लेख किया है। अस्थायीपन (अल्पकालता), प्रदर्शन (दिखावापन), गुमनामी, वैयक्तिकता।
(i) नगरीकरण का क्या अर्थ है ?
(ii) नगरीकरण के तीन मापदण्ड बताएं।
(iii) लुइस वर्थ ने नगरवाद की कौन-सी चार विशेषताओं का उल्लेख किया है ?
उत्तर-
(i) जब लोग गाँव को छोड़ कर नगरों की तरफ रहने के लिए अथवा कार्य की तलाश में चले जाएं तो इसे नगरीकरण का नाम दिया जाता है।

(ii) नगरीकरण के निम्नलिखित मापदण्ड हैं-
(a) जनसंख्या का 5000 से अधिक होना।
(b) प्रति वर्ग किलोमीटर 400 तक का जनसंख्या घनत्व।
(c) 75% जनसंख्या का गैर कृषि कार्यों में लगे होना।

(iii) (a) अस्थायीपन (अल्पकालता)
(b) प्रदर्शन (दिखावापन)
(c) गुमनामी
(d) वैयक्तिकता।

PSEB 12th Class Sociology Solutions स्रोत आधारित प्रश्न

प्रश्न 6.
निम्न दिए स्त्रोत को पढ़ें व साथ दिए प्रश्नों के उत्तर दें-
नगरों में जनसंख्या की वृद्धि इतनी तीव्र है कि सभी को आवास सुविधा प्रदान करना असम्भव सा हो गया है। अतः नगर में स्थापित होने के लिए आवास समस्या अथवा आवासहीनता नगरीय समाज की एक अत्यन्त गंभीर समस्या बन चुकी है। नगरों में स्थान की समस्या इतनी अधिक है कि नगरों में अधिकाँश लोग सड़कों, बस स्टैण्ड, रेलवे स्टेशन व टूटे-फूटे सुविधाविहीन घरों में रहने के लिए विवश हैं। यह कहा जा सकता है कि भारत की आधी नगरीय जनसंख्या या तो अस्वच्छ घरों में रहती है अथवा अपनी आय का 20% से अधिक घर के किराए के रूप में देती है। मुम्बई, कलकत्ता, दिल्ली व चेन्नई जैसे महानगरों में आवास की समस्या और भी विकट है।
(i) नगरों की जनसंख्या क्यों बढ़ रही है ?
(ii) नगरों की बढ़ती जनसंख्या के क्या नुकसान हैं ?
(iii) क्या नगरों की जनसंख्या का बढ़ना एक गंभीर समस्या है ?
उत्तर-
(i) ग्रामीण लोगों में यह धारणा होती है कि नगरों में अधिक सुविधाएं होती हैं तथा वहां पर पेशों की भरमार होती है व इस कारण लोग नगरों की तरफ भाग रहे हैं। इस कारण नगरों की जनसंख्या में लगातार बढ़ौत्तरी हो रही है।

(ii) (a) नगरों में रहने के स्थान की काफी कमी हो रही है। (b) बहुत से लोग खुले आकाश के नीचे या झुग्गियों में रहने को बाध्य होते हैं। (c) लोगों की आय का 20% से अधिक भाग घर के किराए के रूप में निकल जाता है।

(iii) यह सत्य है कि नगरों की जनसंख्या का बढ़ना एक बहुत ही गंभीर समस्या बन रही है। लोग अधिक सुविधाओं तथा पेशों की तलाश में नगरों में आते हैं परन्तु जब उन्हें यह सब नहीं मिल पाता तो वह हताश हो जाते हैं तथा मानसिक तौर पर परेशान हो जाते हैं। इस कारण नगरों में अपराध भी बढ़ रहे हैं जो स्वयं में एक बड़ी समस्या बन रही है।

प्रश्न 7.
निम्न दिए स्त्रोत को पढ़ें व साथ दिए प्रश्नों के उत्तर दें
कार्ल मार्क्स जिन्होंने मजदूरों के हित में समर्थन दिया है। उन्होंने श्रमिकों की वर्ग सम्बन्धी चेतना को महत्त्व दिया है। मार्क्स के अनुसार, वर्ग चेतना का उदय मजदूरों में उनके वर्ग पहचान, वर्ग एकता तथा वर्ग संघर्ष को प्रस्तुत करता है। अतः उन्होंने श्रमिकों को अन्तर्राष्ट्रीय तौर पर यह कह कर इकट्ठे होने के लिए कहा कि “विश्व के मज़दूरो एकत्रित हो जाओ तुम्हारे पास गुलामी की जंजीरों के अलावा खोने के लिए कुछ नहीं है परन्तु एकता द्वारा तुम विश्व को जीत सकते हो।” वर्ग जागरूकता को किसी माध्यम के द्वारा समूह गतिविधि में तबदील किया जा सकता है और राजनैतिक दल इसी प्रकार का एक अंग है। अतः लेनिन ने इसमें यह विस्तार जोड़ा है कि एक दल का विचार मार्क्सवाद में मजदूरों को वर्ग संघर्ष के लिए तैयार करना है। ‘वर्ग’ के संबंध में विभिन्न समाजशास्त्रियों की विभिन्न विचारधाराएं हैं।
(i) कार्ल मार्क्स कौन थे ?
(ii) वर्ग चेतना का क्या अर्थ है ?
(iii) कार्ल मार्क्स के वर्ग संघर्ष के सिद्धांत को संक्षेप में बताएं।
उत्तर-
(i) कार्ल मार्क्स एक जर्मन दार्शनिक थे जिन्होंने समाजशास्त्र की प्रगति में काफी योगदान दिया। उनके दिए संकल्पों के कारण ही 1917 में रूसी क्रान्ति हुई तथा मजदूर वर्ग की सरकार स्थापित हुई।

(ii) जब एक वर्ग अपने अस्तित्व, विशेषताओं के प्रति चेतन हो जाए तथा स्वयं को अन्य वर्गों से अलग समझने लग जाए तो इसे वर्ग चेतना कहा जाता है। मार्क्स के अनुसार, वर्ग चेतना वर्ग एकता व वर्ग की पहचान करवाती है।

(iii) मार्क्स के अनुसार, समाज में दो प्रकार के वर्ग होते हैं-पूँजीपति व मज़दूर। इन दोनों के बीच संघर्ष चलता रहता है। पूँजीपति अपने पैसे के कारण मजदूरों का शोषण करता रहता है। वह कम पैसे देकर मजदूरों से अधिक कार्य करवाना चाहता है तथा मज़दूर कम कार्य करके अधिक पैसा लेना चाहता है। इस कारण दोनों वर्गों में संघर्ष चलता रहता है जिसे वर्ग संघर्ष कहा जाता है।

प्रश्न 8.
निम्न दिए स्त्रोत को पढ़ें व साथ दिए प्रश्नों के उत्तर दें
ग्रामीण भारत में बड़े ज़मींदार, भूमिहीन ज़मींदार, ऊँचे व मध्यम स्तर के किसान व पूंजीपति किसान मूलतः उच्च व मध्यम वर्ग से सम्बंधित होते हैं जबकि नीचे स्तर के किसान मध्यम किसान व भूमि विहीन किसान निम्न जाति से सम्बन्धित होते हैं। ग्रामीण भारत में धन उधार देने वाले वर्ग, विशेषतः वे जातियाँ हैं जो वैश्य वर्ग से सम्बंधित होती हैं। इस प्रकार, सामान्य रूप से उच्च, मध्यम व निम्न जाति भारत में उच्च, मध्यम व निम्न वर्ग हैं। इस तरह, यह भी सत्य है कि संरक्षात्मक भेदभाव (आरक्षण) के कारण, नए अवसर मिलने के कारण शिक्षाएं व उद्योग के क्षेत्र में गतिशीलता आई है। निम्न जाति के कुछ खण्डों ने मध्यम तथा उच्च जातियों के क्षेत्रों में आना शरू कर दिया है। यद्यपि वर्ग स्थिति को अर्जित होने के कारण बदला जा सकता है लेकिन जाति की स्थिति को प्रदत्त प्रकृति के कारण बदला नहीं जा सकता।
(i) ग्रामीण भारत में कौन से वर्ग होते हैं ?
(ii) किन कारणों ने अलग-अलग समूहों को आगे बढ़ने के मौके दिए ?
(iii) क्या वर्ग में स्थिति को परिवर्तित किया जा सकता है ?
उत्तर-
(i) ग्रामीण भारत में बड़े ज़मींदार, उच्च तथा मध्यम वर्ग के किसान, बड़े पूँजीपति किसान, भूमिहीन मज़दूर रहते हैं तथा उन्हें भूमि के अनुसार उच्च श्रेणी, मध्यम श्रेणी तथा निम्न श्रेणी में रखा जा सकता है।
(ii) वैसे तो आजकल समाज में. बहुत सी सुविधाएं मौजूद हैं तथा व्यक्ति स्वयं परिश्रम करके आगे बढ़ सकता है परन्तु कई समूहों को आरक्षण तथा सुरक्षा के नए मौकें प्रदान किए हैं जिसके कारण वह काफी तेजी से आगे बढ़ रहे हैं।
(iii) जी हाँ, वर्ग में स्थिति को परिवर्तित किया जा सकता है। अगर व्यक्ति में योग्यता हो तो वह परिश्रम करके बहुत सा पैसा कमा सकता है अथवा कोई उच्च पद प्राप्त कर सकता है। इससे उसका वर्ग व स्थिति दोनों ही बढ़ जाते हैं।

प्रश्न 9.
निम्न दिए स्रोत को पढ़ें व साथ दिए प्रश्नों के उत्तर देंलिंग वर्ग से संबंधित सम्बन्ध पुरुष व स्त्री के उन सम्बन्धों की व्याख्या करता है जिनका आधार वैज्ञानिक, सांस्कृतिक, राजनैतिक व आर्थिक होता है। लिंग वर्ग में सम्बन्धों में हम लिंग वर्ग अधीनता का परीक्षण करते हैं। स्त्री सशक्तिकरण व स्त्रियों के शोषण की प्रकृति के मुद्दे से संबंधित विभिन्न समाजों में भिन्न-भिन्न रूप पाए जाते हैं। लिंग वर्ग सम्बन्धों में, यह महत्त्वपूर्ण है कि विवाह की संस्था परिवार, शादी से पूर्व, शादी एवम् शादी के बाद के सम्बन्धों, समलैंगिकता का मुद्दे, तीसरे लिंग के मुद्दों व मानवीय सम्बन्धों की प्रकृति आदि की बात करना अति महत्त्वपूर्ण है। प्रायः यह भी स्वीकार किया जाता है कि पुरुष व स्त्री, प्राकृतिक तौर पर शारीरिक भिन्नताओं के कारण भिन्न स्वभाव रखते हैं। परन्तु ये जैविक अथवा शारीरिक भिन्नताएं समाज तथा संस्कृति की संरचना के द्वारा सामाजिक भिन्नताओं में बदल जाती हैं। मानवशास्त्रीय तथा ऐतिहासिक प्रमाणों ने यह सिद्ध कर दिया है कि सांस्कृतिक पुनर्स्थापना ने इन भिन्नताओं को सामाजिक प्रतिक्रिया की महत्त्वपूर्ण भूमिका के परिप्रेक्ष्य में स्थापित तथा पुनर्स्थापित किया है।
(i) लिंग वर्ग संबंध का क्या अर्थ है ?
(ii) लिंग स्थिति तथा लिंग वर्ग में अंतर बताएं।
(iii) लिंग अंतर कैसे सामाजिक अंतरों में बदल जाते हैं ?
उत्तर-
(i) लिंग वर्ग संबंध पुरुष व स्त्री के उन संबंधों के बारे में बताता है जिनका आधार सांस्कृतिक, वैज्ञानिक, आर्थिक व राजनीतिक होता है।
(ii) लिंग स्थिति को जैविक अर्थों में समझा जाता है कि कौन पुरुष है तथा कौन स्त्री है जबकि लिंग वर्ग की भिन्नता का अर्थ उनके व्यवहार से है जो सामाजिक क्रियाओं से बनती है तथा जिसके अनुसार पुरुष व स्त्री अपनी योग्यता के अनुसार अपनी सामाजिक भूमिका को निभाते हैं।
(iii) वैसे तो यह माना जाता है कि पुरुष तथा स्त्री प्राकृतिक तथा शारीरिक अंतरों के कारण अपना स्वभाव अलग रखते हैं परन्तु जैविक व शारीरिक अंतर समाज व संस्कृति की सहायता से सामाजिक अंतरों में बदल जाते हैं। शारीरिक अंतर सांस्कृतिक अंतरों को परस्पर सामाजिक क्रियाओं के रूप में स्थापित करते हैं।

PSEB 12th Class Sociology Solutions स्रोत आधारित प्रश्न

प्रश्न 10.
निम्न दिए स्रोत को पढ़ें व साथ दिए प्रश्नों के उत्तर दें___ 19वीं शती के दरम्यान, अंग्रेजों ने धीरे-धीरे आधुनिक राज्य की नींव रखी। उस समय भूमि का सर्वेक्षण किया गया व लगान निश्चित किया गया। यह युग नई नौकरशाही के उदय का सूचक कहा जा सकता है। उस समय सेना, पुलिस एवम् कानून न्यायालय स्थापित किए गए, जिसने सभी जातियों के लिए नौकरियों के नए राह खोल दिए जहाँ योग्यता के आधार पर भर्ती को आधार बनाया गया। तत्पश्चात्, स्कूल व कालेजों की स्थापना हुई जिसने सभी जातियों के लिए शिक्षा के रास्ते खोल दिए। रेलवे, डाक तथा तार सेवा की, सड़कें व नहरें भी स्थापित की। प्रिंटिग प्रैस, जिसने भारतीय समाज पर गहन प्रभाव डाला यह भी ब्रिटिश साम्राज्य द्वारा विकसित किए गए। इससे स्पष्ट होता है कि इन परिवर्तनों ने भारत के आधुनिक तथा परम्परागत ज्ञान में परिवर्तनशीलता की नीवं रखी है। ज्ञान अब कुछ विशेष अधिकार सम्पन्न लोगों तक सीमित नहीं था। संचार के सर्वोत्तम साधन, समाचार-पत्रों ने लोगों को अहसास करवाया कि देश के विशाल भू-भाग से उनका दृढ़ सम्बन्ध है इस प्रकार, विश्व के किसी भी भाग में होने वाली घटनाओं ने लोगों पर अच्छा या बुरा प्रभाव डालना आरम्भ कर दिया।
(i) पश्चिमीकरण का संकल्प किसने दिया था ?
(ii) पश्चिमीकरण का क्या अर्थ है ?
(iii) पश्चिमीकरण के क्या कारण थे ?
उत्तर-
(i) पश्चिमीकरण का संकल्प प्रसिद्ध भारतीय समाजशास्त्री एम० एन० श्रीनिवास ने दिया था।
(ii) श्रीनिवास के अनुसार, “पश्चिमीकरण उस परिवर्तन का नाम है जिसने अंग्रेजों के भारत पर किये 150 वर्ष के राज्य के समय भारतीयों के सामाजिक जीवन के अलग-अलग स्तरों पर धीरे-धीरे प्रभाव डाला गया जैसे कि विचारधारा, करें कीमतें इत्यादि।”
(iii) अंग्रेजों ने भारतीय समाज में काफ़ी परिवर्तन किए। उन्होंने सेना, पुलिस तथा न्यायालय स्थापित किए जिससे सभी जातियों के लोग वहां पर कार्य करने लग गए। उन्होंने रेल सेवा, डाक तथा तार सेवा, सड़कों, नदियों का निर्माण किया। प्रिंटिंग प्रैस भारत में आई, कारखाने स्थापित किए। लोगों ने अंग्रेजों के जीवन स्तर की नकल करनी शुरू की जिससे पश्चिमीकरण की प्रक्रिया तेज़ी से बढ़ी।

प्रश्न 11.
निम्न दिए स्रोत को पढ़ें व साथ दिए प्रश्नों के उत्तर दें
संदर्भ समूह का अभिप्राय है एक ऐसा समूह जिससे हम अपने समूह की तुलना करते हैं। एक अनुकरणीय समूह होता है, जिसके अनुसार, कोई व्यक्ति अथवा समूह अपनी विचारधारा, व्यवहार, दृष्टिकोण व विश्वास बदलता है। उदाहरणतः राम अपनी कक्षा में निम्न औसत छात्र है। वह अपनी कक्षा के होशियार विद्यार्थियों से प्रभावित हो कर अपने आप में सुधार करना चाहता है। वह उनके व्यवहारों व लक्षणों का अवलोकन करता है तथा उन्हें अपना संदर्भ समूह समझता है। वह समय का पाबंद, अनुशासन को अपनाकर शिक्षा में बेहतर प्रदर्शन करता है। अपने दैनिक जीवन में हम कितने ही संदर्भ समूह पर विश्वास करते हैं हमारे परिवार के सदस्य, मित्र समूह व अभिनेता भी हो सकते हैं।
(i) संदर्भ समूह का संकल्प किसने दिया था ? (ii) संदर्भ समूह का क्या अर्थ है ? (iii) क्या प्रत्येक व्यक्ति का कोई संदर्भ समूह होता है ? यदि हां तो क्यों ? उत्तर-(i) संदर्भ समूह का संकल्प हरबर्ट हाईमैन (Herbert Hymen) ने 1942 में अपनी पुस्तक में दिया था।
(ii) संदर्भ समूह ऐसा समूह होता है जिसमें हम स्वयं की तुलना उससे करते हैं। यह हमारे लिए एक आदर्श समूह होता है जिसमें हम अपनी विचारधारा, व्यवहार तथा विश्वास को उस आदर्श समूह के अनुसार बदलने का प्रयास करते हैं।
(iii) जी हाँ, प्रत्येक व्यक्ति का कोई-न-कोई आदर्श समूह अथवा संदर्भ समूह अवश्य होता है। वास्तव में यह मानवीय प्रकृति है कि हम जीवन में प्रगति करना चाहते हैं तथा हम अपने सामने किसी समूह को रख लेते हैं। वह समूह ही हमारे लिए संदर्भ समूह बन जाता है।

प्रश्न 12.
निम्न दिए स्रोत को पढ़ें व साथ दिए प्रश्नों के उत्तर दें
आधुनिकीकरण का अर्थ आधुनिक जीवन के तरीकों तथा मूल्यों को अपनाना होता है। प्राचीन तौर पर, इस अवधारणा का प्रयोग मुख्यतः अर्थव्यवस्था में हो रहे परिवर्तनों तथा इसके सामाजिक मूल्यों पर पड़ रहे प्रभावों को मापना होता था। परन्तु आज आधुनिकीकरण का क्षेत्र व्यापक हो गया है यह पूरी तरह से कृषि से औद्योगिक अर्थव्यवस्था तक परिवर्तन ले आया है। इसका उन लोगों पर भी प्रभाव पड़ा है जो कि किसी प्रथा में बँधे हैं। इसने आधुनिकीकरण में लोगों को वर्तमान समय व स्थिति के अनुसार बदलने के लिए विवश किया है। परिणामतः आधुनिकीकरण द्वारा लोगों के विचारों, प्राथमिकताओं, मनोरंजनात्मक सुविधाओं में धीरे-धीरे परिवर्तन हुआ है। दूसरे शब्दों, में, वैज्ञानिक और तकनीकी आविष्कारों ने सामाजिक सम्बन्धों में अभूतपूर्व परिवर्तन किए हैं और परम्परागत रूप को नई विचारधारा में आत्मसात् कर दिया है।
(i) आधुनिकीकरण का संकल्प किसने दिया था ? (ii) आधुनिकीकरण का क्या अर्थ है ? (iii) आधुनिकीकरण से क्या परिवर्तन आते हैं ?
उत्तर-
(i) आधुनिकीकरण शब्द का प्रयोग पहली बार डेनियल लर्नर ने दिया था परन्तु इसका व्यापक प्रयोग योगेन्द्र सिंह ने किया था।
(ii) डेनियल लर्नर के अनुसार, यह परिवर्तन की ऐसी प्रक्रिया है जो गैर पश्चिमी देशों में पश्चिमी देशों से प्रत्यक्ष अथवा अप्रत्यक्ष सम्पर्क के द्वारा आई है तथा इससे लोग प्राचीन प्रथाओं को छोड़ कर नई प्रथाओं व विचारों को अपना लेते हैं।
(iii) (a) आधुनिकीकरण से लोगों के विचारों व व्यवहार के तरीकों में परिवर्तन आए हैं। – (b) लोगों ने नई तकनीक को अपनाना शुरू कर दिया तथा उनके जीवन में काफ़ी तेज़ी आ गई। (c) देश में इससे औद्योगिकरण व नगरीकरण में बढ़ौतरी हुई तथा आधुनिक सुविधाएं आनी शुरू हो गई।

प्रश्न 13.
निम्न दिए स्रोत को पढ़ें व साथ दिए प्रश्नों के उत्तर दें वर्ग आधारित आन्दोलनों में श्रमिक व किसान आन्दोलनों को शामिल किया जाता है, जिनकी मुख्य माँग आर्थिक शोषण से मुक्ति थी। भारत में श्रमिक संघ आन्दोलन, श्रमिक व किसान वर्ग की स्थिति, उनकी मांगों, उनके मालिकों के व्यवहार एवम् सरकार द्वारा इस दिशा में उठाए गए प्रयासों को दर्शाते हैं। सूती मिलों, पटसन मिलों व चाय उद्योग की वृद्धि के साथ भारत में ग़रीब लोगों को इन कारखानों में श्रमिक के तौर पर रोज़गार मिला। कम मजदूरी, लम्बा कार्यकाल, अस्वस्यकारी हालातों एवम् स्वदेशी व विदेशी पूंजीपतियों द्वारा शोषण ने उनकी स्थिति को दयनीय बना दिया। भिन्न-भिन्न समय पर कई ‘फैक्ट्री अधिनियम’ आए। परन्तु इससे कार्यरत श्रमिकों की दशा में कोई सुधार नहीं हो सका। इसके पश्चात, किसानों का भी आर्थिक शोषण हुआ। पंजाब में, बंगाल के किसानों का नील उत्पादन के विरुद्ध आन्दोलन तथा पंजाब का किसान आन्दोलन, इनमें विशेष रूप से उल्लेखनीय हैं।
(i) वर्ग आधारित आन्दोलन का क्या अर्थ है ?
(ii) भारतीय उद्योगों में वर्ग आधारित आन्दोलन क्यों शुरू हुए थे ?
(iii) वर्ग आधारित आन्दोलन की उदाहरण दें।
उत्तर-
(i) जब कोई आन्दोलन किसी विशेष वर्ग की मांगों को सामने रख कर शुरू किया जाए तो उसे वर्ग आधारित आन्दोलन कहा जाता है।
(ii) भारतीय उद्योगों में मजदूरों की स्थिति काफ़ी खराब थी। उन्हें कम पैसा दिया जाता था, कार्य का समय अधिक था, गंदगी भरे हालात थे, देशी तथा विदेशी पूँजीपति उनका शोषण करते थे जिस कारण मजदूरों की स्थिति काफ़ी दयनीय थी। इसलिए भारतीय उद्योगों में मज़दूर आन्दोलन शुरू किए गए थे।
(iii) वैसे तो वर्ग आन्दोलनों में ट्रेड यूनियन संगठनों का आन्दोलन, कृषकों के आन्दोलन शामिल हैं। परन्तु हम पंजाब के कृषक आन्दोलन, मुम्बई की मिलों के मज़दूर आन्दोलन, बंगाल के नील आन्दोलन इत्यादि को मुख्य वर्ग आन्दोलन कह सकते हैं।

प्रश्न 14.
निम्न दिए स्रोत को पढ़ें व साथ दिए प्रश्नों के उत्तर दें
नशीली दवाओं के व्यसन की समस्या आज हमारे समाज में बहुत तीव्रता से बढ़ रही है। युवा पीढ़ी अपनी कमज़ोर विचार शक्ति, कम अकादमिक उपलब्धियों, पारिवारिक पृष्ठभूमि एवम् अपने मित्रों के दबाव के कारण इसका अधिक शिकार है। कई बार वे महसूस करते हैं कि वे इतने बुद्धिमान, शक्तिशाली नियंत्रित हैं कि वे नशे के आदी नहीं हो सकते। परन्तु वे फिर भी इस व्यसन का शिकार हो जाते हैं। अतः नशीली दवाओं के व्यसन की यह आदत किसी को भी अपने जाल में ग्रसित कर सकती है। यह व्यक्ति के स्वास्थ्य को हानि पहँचाने, पारिवारिक संरचना में समस्या पैदा करने व समाज में अपराधों की वृद्धि का कारण बनती है। व्यसनी लोग प्रायः जीवन की अन्य गतिविधियों में अपनी रुचि खो देते हैं। वे अपना दायित्व निभाने के योग्य नहीं रहते व अपने परिवार तथा समाज पर बोझ बन जाते हैं।
(i) नशीली दवाओं के व्यसन का क्या अर्थ है ?
(ii) लोग नशा क्यों करते हैं ?
(iii) नशीली दवाओं के व्यसन का परिणाम क्या होता है ? .
उत्तर-
(i) जब कोई व्यक्ति मदिरा, अफीम अथवा किसी अन्य प्रकार के नशे का प्रयोग करने का इतना आदि हो जाए कि उसके बिना रह न सके तो इसे नशीली दवाओं का व्यसन कहा जाता है।
(ii) कई लोग शौक के कारण नशा करते हैं, कई लोग तनाव को दूर करने के लिए नशा करते हैं, कई लोग दुःखों को भुलाने के लिए नशा करते हैं। धीरे-धीरे वह नशे के इतने आदी हो जाते हैं कि इसके बिना नहीं रह सकते तथा नशीली दवाओं के चक्र में फँस जाते हैं।
(iii) (a) नशे के कारण स्वास्थ्य ख़राब हो जाता है तथा व्यक्ति कुछ करने लायक नहीं रहता। (b) नशा करने से व्यक्ति का सम्पूर्ण पैसा खत्म हो जाता है तथा उसकी आर्थिक व्यवस्था भी बुरी हो जाती है। (c) इससे समाज की प्रगति पर भी सीधा प्रभाव पड़ता है तथा प्रगति कम होना शुरू हो जाती है।

PSEB 12th Class Sociology Solutions स्रोत आधारित प्रश्न

प्रश्न 15.
निम्न दिए स्त्रोत को पढ़ें व साथ दिए प्रश्नों के उत्तर दें वृद्धावस्था में पाचक प्रक्रियाएँ धीमी पड़ जाती हैं। लोग शारीरिक व मानसिक रूप से कमजोर हो जाते हैं। उनमें विभिन्न प्रकार के रोगों के शिकार होने का भय बढ़ जाता है। व्यक्ति की रोगक्षमता कम हो जाने के कारण वृद्ध लोग अधिकतर असंक्रामक रोगों के शिकार हो जाते हैं। ज्यादातर वृद्ध व्यक्तियों को अच्छी गुणवत्ता वाली तथा संवेदनशील स्वास्थ्य देख रेख न मिलने के कारण उनकी बढ़ रही आयु के कारण उनके स्वास्थ्य स्तर में गिरावट और भी जटिलता से आती है। इसके साथ-साथ स्वास्थ्य सम्बन्धी अच्छी जानकारी व ज्ञान की कमी के साथ-साथ इलाज का अत्यन्त महंगा होने के कारण वृद्धावस्था की संभाल करना विशेषकर उनके लिए जो निर्धन व साधनविहीन वर्ग से सम्बन्धित हैं, की पहुँच से बाहर हो गया है।
(i) वृद्ध व्यक्ति कौन होता है ?
(ii) वृद्ध व्यक्ति कौन-सी तीन समस्याओं का सामना करते हैं ?
(iii) वृद्ध व्यक्तियों की समस्याओं को कैसे दूर किया जा सकता है ?
उत्तर-
(i) जो व्यक्ति रिटायर हो चुका हो अथवा 60 वर्ष की आयु से अधिक का हो चुका हो, उसे वृद्ध व्यक्ति कहा जाता है।
(ii) (a) वृद्ध व्यक्ति को कई बीमारियां लग जाती हैं। (b) उसके रिटायर होने के पश्चात् उसकी आय खत्म हो जाती है तथा वह आर्थिक रूप से अपने बच्चों पर निर्भर हो जाता है।
(c) उसके शरीर में बीमारियों से लड़ने का सामर्थ्य भी कम हो जाता है। उसे कम दिखना व सुनना आम हो जाता है।
(iii) (a) सरकारी कानूनों को कठोरता से लागू किया जाना चाहिए ताकि कोई भी वृद्ध व्यक्तियों को तंग न करे।
(b) सरकार को वृद्ध व्यक्तियों को बढ़िया तथा निःशुल्क स्वास्थ्य सुविधाएं देनी चाहिए।
(c) सरकार को वृद्ध व्यक्तियों को अच्छी बुढ़ापा पैंशन देनी चाहिए ताकि वह अपने जीवन के अन्तिम पड़ाव पर बच्चों के ऊपर आर्थिक रूप से निर्भर न रहें।

प्रश्न 16.
निम्न दिए स्रोत को पढ़ें व साथ दिए प्रश्नों के उत्तर देंविश्व में एक बिलियन से भी अधिक लोग हैं जो किसी-न-किसी प्रकार की असर्मथता के साथ जीवन व्यतीत कर रहें हैं। हममें से बहुत से लोग ऐसे हैं जिनके मित्र मण्डली या परिवार में ऐसे व्यक्ति होंगे जिन्हें दैनिक जीवन में इस समस्या के कारण बहुत सारी कठिनाइयों का सामना करना पड़ता है। असमर्थ व्यक्तियों की मूलभूत सेवाओं तक सीमित पहुँच होने के कारण, कई सुविधाओं जैसे शिक्षा, रोजगार, पुनर्वास सुविधाओं आदि से वंचित रखा जाता है। इसके अतिरिक्त असमर्थता, सामाजिक कलंक के रूप में उनकी सामान्य सामाजिक व आर्थिक ज़िन्दगी में रुकावटें पैदा करती है।

असमर्थता शब्द का अभिप्राय, एक प्रकार या बहु प्रकार की कुशलता का अभाव है जो मानसिक, शारीरिक तथा संवेदना से संबंधित हो सकती है। इसे प्राथमिक रूप से एक चिकित्सा सम्बन्धी कमी के रूप में भी परिभाषित किया जा सकता है। अतः असमर्थता शब्द, स्वयं में एक प्रकार की नहीं बल्कि अपने आप में बहु प्रकार की कमियों को समेटे हए है। हालांकि, असमर्थता शब्द, अपने आप में सजातीय श्रेणी नहीं है क्योंकि इसमें विभिन्न प्रकार की शारीरिक विभिन्नताएं, शारीरिक अवरोध (कमियाँ), संवेदनशीलता में त्रुटि तथा मानसिक अथवा शिक्षण सम्बन्धी असमर्थताएँ आती हैं जो कि जन्मजात या फिर जन्म के बाद भी हुई हो सकती हैं।
(i) असमर्थता का क्या अर्थ है ?
(ii) असमर्थ व्यक्तियों को किन समस्याओं का सामना करना पड़ता है ?
(iii) असमर्थता के प्रकार बताएं।
उत्तर-
(i) जो व्यक्ति किसी शारीरिक अकुशलता के कारण रोज़ाना जीवन जीने के लिए संघर्ष करते हैं उसे असमर्थता कहा जाता है।
(ii) (a) असमर्थता के कारण व्यक्ति ठीक ढंग से शिक्षा नहीं ले पाता।
(b) उसके नौकरी करने के मौके सीमित हो जाते हैं।
(c) वह किसी भी कार्य को उतनी तेजी से नहीं कर सकते जिस तेज़ी से एक समर्थ व्यक्ति कर सकता है।
(iii) असमर्थता कई प्रकार की होती है जैसे कि(a) संचालन की असमर्थता (Locomotor Disability) (b) देखने की असमर्थता (Visual Disability)
(c) सुनने की असमर्थता (Hearing Disability) (d) मानसिक असमर्थता (Mental Disability) (e) बोलने की असमर्थता (Speech Disability)।

PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 11 Alcohols, Phenols, and Ethers Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols, and Ethers

PSEB 12th Class Chemistry Guide Alcohols, Phenols and Ethers Intext Questions and Answers

Question 1.
Write IUPAC names of the following compounds:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 1
Answer:
(i) 2, 2, 4-Trimethylpentan-3-ol
(ii) 5-Ethylheptane-2,4-diol
(iii) Butane-2, 3-diol
(iv) Propane-1, 2, 3-triol
(v) 2-Methylphenol
(vi) 4-Methylphenol
(vii) 2, 5-Dimethylphenol
(viii) 2, 6-Dimethylphenol
(ix) l-Methoxy-2-methylpropane
(x) Ethoxybenzene
(xi) 1-Phenoxyheptane
(xii) 2-Ethoxybutane

Question 2.
Write structures of the compounds whose IUPAC names are as follows:
(i) 2-Methylbutan-2-ol
(ii) l-Phenylpropan-2-ol
(iii) 3,5-Dimethylhexane -1, 3, 5-triol
(iv) 2,3 – Diethylphenol
(v) 1 -Ethoxypropane
(vi) 2-Ethoxy-3-methylpentane
(vii) Cyclohexylmethanol (viii) 3-Cyclohexylpentan-3-ol
(ix) Cyclopent-3-en-l-ol
(x) 3-Chloromethylpentan-l-ol.
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 3

PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 4

Question 3.
(i) Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names.
(ii) Classify the isomers of alcohols in question 3(i) as primary, secondary, and tertiary alcohols.
Answer:
(i) Eight isomers are possible. These are :
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 5
(ii) Primary : (a), (d), (e) arid (g); Secondary : (b), (c), (h); Tertiary : (f).

Question 4.
Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
Answer:
Propanol undergoes intermolecular H-bonding because of the presence of -OH group. On the other hand, butane does not.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 6
Therefore, extra energy is required to break hydrogen bonds. For this reason, propanol has a higher boiling point than hydrocarbon butane.

Question 5.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer:
Alcohols form H-bonds with water due to the presence of -OH group. However, hydrocarbons cannot form H-bonds with water.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 7
As a result, alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses.

Question 6.
What is meant by the hydroboration-oxidation reaction? Illustrate it with an example.
Answer:
The addition of diborane to alkenes to form trialkyl boranes followed by their oxidation with alkaline hydrogen peroxide to form alcohols is called hydroboration-oxidation reaction.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 8
The alcohol obtained by this process is contrary to the Markovnikov’s rule.

Question 7.
Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O.
Answer:
The three isomers are given as follows :
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 9

Question 8.
While separating a mixture of ortho and para-nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
Answer:
o-Nitrophenol is steam volatile as it exists as discrete molecules due to intramolecular H-bonding and hence can be separated by steam distillation from p-nitrophenol which is less volatile as it exists as associated molecules because of intermolecular H-bonding.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 10

Question 9.
Give the equations of reactions for the preparation of phenol from cumene.
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 11

Question 10.
Write chemical reaction for the preparation of phenol from chlorobenzene.
Answer:
Chlorobenzene is fused with NaOH (at 623 K and 300 atm pressure) to produce sodium phenoxide, which gives phenol on acidification.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 12

Question 11.
Write the mechanism of hydration of ethene to yield ethanol.
Answer:
The mechanism of hydration of ethene to yield ethanol involves three steps.
Step 1:
Protonation of ethene to form carbocation by electrophilic attack of H3O+:
H2O+H+ → H3O+
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 13

Step 2 :
Nucleophilic attack of water on carbocation:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 14

Step 3:
Deprotonation to form ethanol:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 15

Question 12.
You are given benzene, cone. H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents.
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 16

Question 13.
Show how will you synthesize:
(i) 1-phenyl ethanol from a suitable alkene.
(ii) cyclohexyl methanol using an alkyl halide by an SN2 reaction.
(iii) pentane-l-ol using a suitable alkyl halide?
Answer:
(i) By acid-catalyzed hydration of ethylbenzene (styrene), 1-phenyl ethanol can be synthesized.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 17
(ii) When chloro methylcyclohexane is treated with sodium hydroxide, cyclohexyl methanol is obtained.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 18
(iii) When 1-chloromethane is treated with NaOH, pentane-l-ol is produced.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 19

Question 14.
Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
Answer:
The reactions showing acidic nature of phenol are :
(i) Reaction with sodium: Phenol reacts with active metals like sodium to liberate H2 gas.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 20
(ii) Reaction with NaOH: Phenol dissolves in NaOH to form sodium phenoxide and water.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 21
Comparison of acidic character of phenol and ethanol:
Phenol is more acidic than ethanol. This is due to the reason that phenoxide ion left after the loss of a proton from phenol is stabilized by resonance while ethoxide ion left after loss of a proton from ethanol is not stabilized by resonance.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 22

Question 15.
Explain why is ortho nitrophenol more acidic than ortho methoxy phenol?
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 23
The nitro-group is an electron-withdrawing group. The presence of this group in the ortho position decreases the electron density in the O-H bond. As a result, it is easier to lose a proton. Also, the o-nitro phenoxide ion formed after the loss of protons is stabilized by resonance. Hence, ortho nitrophenol is a stronger acid. On the other hand, methoxy group is an electron-releasing group. Thus, it increases the electron density in the O-H bond and hence, the proton cannot be given out easily. For this reason,ortho-nitrophenol is more acidic than ortho- methoxy phenol.

Question 16.
Explain how does the -OH group attached to a carbon of benzene ring activates it towards electrophilic substitution?
Answer:
Phenol may be regarded as a resonance hybrid of following structures:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 24
Thus, due to +R effect of the -OH group, the electron density in the benzene ring increases thereby facilitating the attack of an electrophile. In other words, presence of -OH group, activates the benzene ring towards electrophilic substitution reactions. Now, since the electron density is relatively higher at the two o-and one p-position, electrophilic substitution occurs mainly at o-and p-positions.

Question 17.
Give equations of the following reactions:
(i) Oxidation of propane-1-ol with alkaline KMnO4 solution.
(ii) Bromine in CS2 with phenol.
(iii) Dilute HNO3 with phenol.
(iv) Treating phenol with chloroform in presence of aqueous NaOH.
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 25

Question 18.
Explain the following with an example.
(i) Kolbe’s reaction.
(ii) Reimer-Tlemimnn reaction.
(iii) WiIHamon ether synthesis.
(iv) Unsymmetrical ether.
Answer:
(i) Kolbe’s reaction: When phenol is treated with sodium hydroxide, sodium phenoxide is produced. This sodium phenoxide when treated with carbon dioxide, followed by acidification, undergoes electrophilic substitution to give ortho-hydroxybenzoic acid as the main product. This reaction is known as Kolbe’s reaction.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 26

(ii) Reimer-Telemann reaction: When phenol is treated with chloroform (CHCl3 ) in the presence of sodium hydroxide, a -CHO group is introduced at the ortho position of the benzene ring.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 27
This reaction is known as the Reimer-Tiemann reaction. The intermediate is hydrolyzed in the presence of alkalis to produce salicylaldehyde.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 28

(iii) Williamson ether synthesis: Williamson ether synthesis is a laboratory method to prepare symmetrical and unsymmetrical ethers by allowing alkyl halides to react with sodium alkoxides.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 29
This reaction involves SN2 attack of the alkoxide ion on the alkyl halide. Better results are obtained in case of primary alkyl halides.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 30
If the alkyl halide is secondary or tertiary, then elimination competes over substitution.

(iv) Unsymmetrical ether: An unsymmetrical ether is an ether where two groups on the two sides of an oxygen atom differ (i.e., have an unequal number of carbon atoms). For example: ethyl methyl ether (CH3-O-CH2CH3), methyl phenyl ether (CH3 -0 – C6H5), etc.

Question 19.
Write the mechanism of acid-catalyzed dehydration of ethanol to yield ethene.
Answer:
The mechanism of acid-catalyzed dehydration of ethanol to yield ethene involves the following three steps:
Step 1:
Protonaüon of ethanol to form ethyl oxonium ion:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 31

Step 2 :
Formation of carbocation (rate-determining step):
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 32

Step 3 :
Elimination of a proton to form ethene:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 33
The acid consumed in Step 1 is released in Step 3. After the formation of ethene, it is removed to shift the equilibrium in a forward direction.

Question 20.
How are the following conversions carried out?
(i) Propene → Propan-2-ol
(ii) Benzyl chloride → Benzyl alcohol
(iii) Ethyl magnesium chloride → Propan-1-ol.
(iv) Methyl magnesium bromide → 2-Methylpropan-2-ol.
Answer:
(i) If propene is allowed to react with water in the presence of an acid as a catalyst, then propane-2-ol is obtained.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 34

(ii) If benzyl chloride is treated with NaOH (followed by acidification) then benzyl alcohol is produced.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 35

(iii) When ethyl magnesium chloride is treated with methanal, an adduct is produced which gives propane-1-ol on hydrolysis.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 36

(iv) When methyl magnesium bromide is treated with propanone, an adduct is produced which gives 2- methylpropane-2-ol on hydrolysis.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 37

Question 21.
Name the reagents used in the following reactions:
(i) Oxidation of a primary alcohol to carboxylic acid.
(ii) Oxidation of a primary alcohol to aldehyde.
(iii) Bromination of phenol to 2,4,6-tribromophenol.
(iv) Benzyl alcohol to benzoic acid.
(v) Dehydration of propane-2-ol to propene.
(vi) Butan-2-one to butan-2ol.
Answer:
(i) Acidified potassium dichromate (K2Cr2O7/H2SO4)
(ii) Pyridinium chlorochromate (PCC)
(iii) Bromine water (Br2/H2O)
(iv) Acidified potassium permanganate (KMnO4/H2SO4)
(v) 85% phosphoric acid (H3PO4)
(vi) NaBH4 or LiAlH4.

Question 22.
Give reason for the higher boiling point of ethanol in comparison to methoxymethane.
Answer:
Ethanol undergoes intermolecular H-bonding due to the presence of -OH group, resulting in the association of molecules. Extra energy is required to break these hydrogen bonds. On the other hand, methoxymethane does not undergo H-bonding. Hence, the boiling point of ethanol is higher than that of methoxymethane.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 38

Question 23.
Give IUPAC mimes of the following ethers:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 39
(ii) CH3OCH2CH2Cl
(iii) O2N- C6H4 – OCH3(p)
(iv) CH3CH2CH2OCH3
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 40
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 41
Answer:
(i) 1-Ethoxy-2-methylpropane
(ii) 2-Chioro- 1 -methoxyethane
(iii) 4-Nitroanisole
(iv) 1 -Methoxypropane
(v) 4-Ethoxy- 1, 1 -dime thylcyclohexane
(vi) Ethoxybenzene.

Question 24.
Write the names of reagents and equations for the preparation of the following ethers by WilÌianlÑon’s synthesis:
(i) 1-Propoxypropane
(ii) Ethoxybenzene
(iii) 2.Methoxy-2-methylpropane
(iv) 1 -Methoxyethane
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 42

Question 25.
Illustrate with examples the limitations of Willi9mon synthesis for the preparation of certain types of ethers.
Answer:
The reaction of Williamson synthesis involves SN2 attack of an alkoxide ion on a primary alkyl halide.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 43
But if secondary or tertiary alkyl halides are taken in place of primary alkyl halides, then elimination would compete over substitution. As a result, alkenes would be produced. This is because alkoxides are nucleophiles as well as strong bases. Hence, they react with alkyl halides, which results in an elimination reaction.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 44

Question 26.
How is 1-propoxy propane synthesized from propane-l-ol? Write mechanism of this reaction.
Answer:
1-propoxy propane can be synthesized from propane-1-ol by dehydration Propan- 1 -ol undergoes dehydration in the presence of protic acids (such as H2SO4, H3PO4) to give 1-propoxy propane.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 45
The mechanism of this reaction involves the following three steps:
Step 1: Protonation
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 46
Step 2: Nucleophilic attack
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 47
Step 3: Deprotonanon
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 48

Question 27.
Preparation of ethers by acid-catalyzed dehydration of secondary or tertiary alcohols is not a suitable method. Give
reason.
Answer:
Acid-catalyzed dehydration of primary alcohols to ethers occurs by SN2 reaction involving nucleophilic attack of the alcohol molecule on the protonated alcohol molecule.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 49
Under these conditions, secondary and tertiary alcohols, however, give alkenes rather than ethers. This is because due to steric hindrance, nucleophilic attack of the alcohol molecule on the protonated alcohol molecule does not occur. Instead, protonated secondary and tertiary alcohols lose a molecule of water to form stable secondary and tertiary carbocations. These carbocations prefer to lose a proton to form alkenes rather than undergoing nucleophilic attack by alcohol molecules to form ethers.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 50
Similarly, tertiary alcohols give alkenes rather than ethers.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 51

Question 28.
Write the equation of the reaction of hydrogen iodide with:
(i) 1-propoxy propane
(ii) Methoxybenzene and
(iii) Benzyl ethyl ether
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 52
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 53

Question 29.
ExplaIn the fact that in aryl alkyl ethers
(i) the alkoxy group activates the benzene ring towards electrophilic substitution and
(ii) it directs the incoming substituents to ortho and para positions in benzene ring.
Answer:
(i) In aryl alkyl ethers, the +Reffect of the alkoxy group (OR) increases the electron density in the benzene ring thereby activating the benzene ring towards electrophilic substitution reactions.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 54

(ii) Since the electron density increases more at the two orthos and one para position as compared to meta-positions, electrophilic substitution reactions mainly occur at ortho-and para-positions. For example,
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 55

Question 30.
Write the mechanism of the reaction of III with methoxymethane.
Answer:
When equimolar amounts of HI and methoxymethane are taken; a mixture of methyl alcohol and iodomethane are formed.
Mechanism
Step I:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 56

Step II :
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 57
If HI is present in excess, CH3OH formed in step II is further converted into CH3I.

Step III:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 58
Step IV:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 59

Question 31.
Write equations of the following reactions:
(i) Friedel-Crafts reaction-alkylation of anisole.
(ii) Nitration of anisole.
(iii) Bromlnation of anisole in ethanoic acid medium.
(iv) Friedel-Craft’s acetylation of anisole.
Answer:
(i) Friedel-Crafts reaction (Alkylation of anisole)
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 60

(ii) Nitration of anisole
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 61

(iii) Bromination of anisole in ethanoic acid medium
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 62

(iv) Friedel-Crafts acetylation of anisole
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 63

Question 32.
Show ho would you synthesize the following alcohols from appropriate alkenes?
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 64
Answer:
The given alcohols can be synthesized by applying Markovnikov’s rule of acid-catalyzed hydration of appropriate alkenes. –
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 65

Question 33.
When 3-methyl butan-2-ol is treated with HBr, the following reaction takes place:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 66
Give a mechanism for this reaction.
(Hint: The secondary carbocation formed in step It rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)
Answer:
Mechanism: The reaction takes place through the following mechanism :
Step I: Formation of protonated alcohol.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 67
Step II: Formation of carbocatlon.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 68
2° carbocation being less stable undergoes hydride shift to form more stable 3° carbocation.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 69
Step III: Attack of nucleophile
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 70

Chemistry Guide for Class 12 PSEB Alcohols, Phenols, and Ethers Textbook Questions and Answers

Question 1.
Classify the following as primary, secondary and tertiary alcohols:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 71
Answer:
Primary alcohols (i), (ii), (iii)
Secondary alcohols (iv) and (v)
Tertiary alcohols (vi)

Question 2.
Identify allylic alcohols in the above examples.
Answer:
Allylic alcohols (ii) and (vi).

Question 3.
Name the following compounds according to IUPAC system :
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 72
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 73
Answer:
(i) 3-Chloromethyl-2-isopropylpentan-l-ol _
(ii) 2, 5-Dimethylhexane-l,3-diol .
(iii) 3-Bromocyclohexanol
(iv) Hex-l-en-3-ol
(v) 2-Bromo-3-methylbut-2-en-l-ol.

Question 4.
Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 74
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 75
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 76

Question 5.
Write structures of the products of the following reactions:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 77
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 78

Question 6.
Give structures of the products you would expect when each of the following alcohol reacts with
(a) HCl-ZnCl2
(b) HBr and
(c) SOCl2.
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 79
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 80
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 81
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 82

Question 7.
Predict the major product of acid-catalyzed dehydration of
(i) 1-methyl cyclohexanol and
(ii) butane-l-ol.
Answer:
(i) Acid-catalyzed dehydration of 1-methyl cyclohexanol can give two products, I and II. Since product (I) is more substituted, according to Saytzeff rule, it is the major product.
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 83

Question 8.
Ortho and para-nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.
Answer:
The resonance structures of o – and p -nitrophenoxide ions and phenoxide ion are given as follows :
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 85

PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 86
It is clear from the above structures that due to -R effect of NO2 group, o-and p-nitro phenoxide ions are more stable than phenoxide ions. Consequently, o-and p-nitrophenols are more acidic than phenol.

Question 9.
Write the equations involved in the following reactions :
(i) Reimer-Tiemann reaction
(ii) Kolbe’s reaction –
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 87

Question 10.
Write the reactions of Williamson synthesis of 2- ethoxy-3-methyl pentane starting from ethanol and 3- methyl pentane-2-ol.
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 88

Question 11.
Which of the following is an appropriate set of reactants for the preparation of l-methoxy-4- nitrobenzene and why ?
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 89
Answer:
(ii) because inset (i), sodium methoxide (CH3ONa) is a strong nucleophile as well as a strong base. So, elimination reaction predominates over substitution reaction.

Question 12.
Predict the products of the following reactions :
(i) CH3 – CH2 – CH2 – O – CH3 + HBr →
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 90
Answer:
PSEB 12th Class Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 91

PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 12 Aldehydes, Ketones, and Carboxylic Acids Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones, and Carboxylic Acids

PSEB 12th Class Chemistry Guide Aldehydes, Ketones, and Carboxylic Acids InText Questions and Answers

Question 1.
What is meant by the following terms? Give an example of the reaction in each case.
(i) Cyanohydrin
(ii) Acetal
(iii) Semicarbazone
(iv) Aldol
(v) Hemiaeetal
(vi) Oxime
(vii) Ketal
(viii) Imine
(ix) 2, 4-DNP derivative
(x) Schiffs base.
Answer:
(i) Aldehydes and ketones react with hydrogen cyanide (HCN) to yield cyanohydrins.
It is catalyzed by a base and the generated cyanide ion (CN) being a stronger nucleophile readily adds to carbonyl compounds to yield corresponding cyanohydrin.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 1
(ii) gem-Dialkoxy compounds in which the two alkoxy groups are present on the terminal carbon atom are called acetals. These are produced by the action of an aldehyde with two equivalents of a monohydric alcohol in the presence of dry HCl gas.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 2
These are easily hydrolyzed by dilute mineral acids to regenerate the original aldehydes. Therefore, these are used for the protection of aldehydic group in organic synthesis.

(iii) Semicarbazones are derivatives of aldehydes and ketones and are produced by the action of semicarbazide on them in weak acidic medium.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 3
These are used for identification and characterization of aldehydes and ketones.
(iv) Aldehydes and ketones having at least one α -hydrogen undergo a reaction in the presence of dilute alkali as catalyst to form f3-hydroxy aldehydes (aldol) or β-hydroxy ketones (ketol), respectively. This is known as aldol condensation.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 4
(v) gem-Alkoxyalcohols are called hemiacetals. These are produced by addition of one molecule of a monohydric alcohol to an aldehyde in presence of dry HCl gas.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 5
(vi) Oximes are produced when aldehydes or ketones react with hydroxylamine in weak acidic medium.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 6
(vii) gem-Dialkoxyalkanes are called ketals. In ketals, the two alkoxy groups are present on the same carbon within the chain. These are produced when a ketone is heated with ethylene glycol in presence of diy HCl gas or p-toluene sulphonic acid (PTS).
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 7
These are easily hydrolyzed by dilute mineral acids to regenerate the original ketones. Therefore, ketals are used for protection of keto groups in organic synthesis.

(viii) Compounds containing PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 8 group are called imines. These are produced when aldehydes and ketones react with ammonia derivatives.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 9
Z = Alkyl, aryl – NH2,-OH,-C6H5NH,-NHCONH2 etc.

(ix) 2, 4-Dinitrophenythydrazones (i.e., 2, 4-DNP derivatives) are produced when aldehydes or ketones react with 2, 4-dinitrophenyl hydrazine in weakly acidic medium.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 10
2, 4-DNP derivatives are used for identification and characterization of aldehydes and ketones.
(x) Aldehydes and ketones react with primary aliphatic or aromatic amines to form azomethines or Schiffs bases.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 11

Question 2.
Name the following compounds according to IUPAC system of nomenclature:
(i) CH3CH(CH3)CH2CH2CHO
(ii) CH3CH2COCH(C2H5)CH2CH2Cl
(iii) CH3CH=CHCHO
(iv) CH3COCH2COCH3
(v) CH3CH(CH3)CH2C(CH3)2COCH3
(vi) (CH3)3CCH2COOH
(vii) OHCC6H4CHO-p
Answer:
(i) 4-Methylpentanal
(ii) 6-Chloro-4-ethylhexan-3-one
(iii) But-2-en-l-al
(iv) Pentane-2,4-dione
(v) 3,3,5-Trimethylhexan-2-one
(vi) 3,3-Dimethylbutanoic acid
(vii) Benzene-1,4-dicarbaldehyde

PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question 3.
Draw the structures of the following compounds.
(i) 3-Methylbutanal
(ii) p-Nitropropiophenone
(iii) p-Methylbenz aldehyde
(iv) 4-Methylpent-3-en-2-one
(v) 4-Chloropentan-2-one
(vi) 3-Bromo-4-phenylpentanoic acid
(vii) p,p’-Dihydroxybenzophenone
(viii) Hex-2-en-4-ynoic acid
Answer:
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 12

Question 4.
Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.
(i) CH3CO(CH2)4CH3
(ii) CH3CH2CH Br CH2CH(CH3) CHO
(iii) CH3(CH2)5CHO
(vi) PhCOPh
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 13
Answer:

IUPAC Name Common Name
(i)Heptane-2-one Methyl n-propyl ketone
(ii) 4-Bromo-2-methyl hexanal γ -Bromo- α –methyl caproaldehyde
(iii) Heptanal n-heptyl aldehyde
(iv) 3-Phenyprop-2-enal Β- Phenylacrole in
(v) Cyclopentane-carbaldehyde Cyclopentane carbaldehyde
(vi) Diphenyl methanone Benzophenone

Question 5.
Draw structures of the following derivatives.
(i) The 2,4-dinitrophenylhydrazone of benzaldehyde
(ii) Cyclopropanone oxime
(iii) Acetaldehydedimethylacetal
(iv) The semicarbazone of cyclobutanone
(v) The ethylene ketal of hexan-3-one
(vi) The methyl hemiacetal of formaldehyde
Answer:
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 14

Question 6.
Predict the products formed when cyclohexane carbaldehyde reacts with following reagents.
(i) PhMgBr and then H3O+
(ii) Tollens’ reagent
(iii) Semicarbazide and weak acid
(iv) Excess ethanol and acid
(v) Zinc amalgam and dilute hydrochloric acid
Answer:
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 15

PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 15

Question 7.
Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.
(i) Methanal
(ii) 2-Methylpentanal
(iii) Benzaldehyde
(iv) Benzophenone
(v) Cyclohexanone
(vi) 1-Phenyipropanone
(vii) Phenylacetaldehyde
(viii) Butan.1-oI
(ix) 2, 2-Dimethylbutanal
Answer:
Aldol condensation
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 17
Cannizzaro reaction
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 19
(iv) Benzophenone: It is a ketone, so it does not undergo Cannizzaro’s reaction. Without a-hydrogen, it cannot participate in aldol condensation.
(viii) Butan-1-ol: It is an alcohol. So, it cannot participate in any of the above 2 reactions.

PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question 8.
How will you convert ethanal into the following compounds?
(i) Butane-1, 3-diol
(ii) But-2-enal
(iii) But-2-enoic acid
Answer:
(i) On treatment with dilute alkali, ethanal produces 3-hydroxybutanal gives butane-1, 3-diol on reduction.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 21
(ii) On treatment with dilute alkali, ethanal gives 3-hydroxybutanal which on heating produces but-2-enal.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 22
(iii) When treated with Tollen’s reagent, but-2-ena produced in the above reaction produces but-2-enoic acid.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 23

Question 9.
Write structural formulas and names of four possible aldol condensation products from propanal and butanal.
In each case, indicate which aldehyde acts as nucleophile and which as electrophile.
(i) Taking two molecules of propanal, one which acts as a nucleophile and the other as an electrophile.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 24
(ii) Taking two molecules of butanal, one which acts as a nucleophile and the other as an electrophile.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 25
(iii) Taking one molecule each of propanal and butanal in which propanal acts as a nucleophile and butanal acts as an electrophile.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 26
(iv) Taking one molecule each of propanal and butanal in which propanal acts as an electrophile and butanal acts as a nucleophile.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 27

Question 10.
An organic compound with the molecular formula C9H10O forms 2, 4-DNP derivative, reduces Tollen’s reagent, and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzene dicarboxylic acid. Identify the compound.
Answer:
It is given that the compound (with molecular formula C9H10O) forms 2,4-DNP derivative and reduces Tollen’s reagent. Therefore, the given compound must be an aldehyde. Again, the compound undergoes Cannizzaro reaction and on oxidation gives 1, 2-benzene dicarboxylic acid. Therefore, the -CHO group is directly attached to a benzene ring and this benzaldehyde is ortho-substituted. Hence, the compound is 2-ethyl benzaldehyde.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 28
The given reactions can be explained by the following equations:
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 29

Question 11.
An organic compound (A) (molecular formula C8H16O2) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-l-ene. Write equations for the reactions involved.
Answer:
As the ester (A) with molecular formula C8H16O2 upon hydrolysis gives carboxylic acid (B) and the alcohol (C) and oxidation of (C) with chromic acid produces the acid (B), therefore, both the carboxylic acid (B) and alcohol (C) must contain the same number of carbon atoms. Now ester (A) contains eight carbon atoms, therefore, both the carboxylic acid (B) and the alcohol (C) must contain four carbon atoms each. As the alcohol (C) on dehydration gives but-l-ene, therefore, (C) must be a straight chain alcohol, i.e., butan-l-ol.

If C is butane-lol, then the acid (B) which it gives on oxidation must be butanoic acid and the ester (A) must be butyl butanoate. The relevant equations for all the reactions involved may be explained as follow :
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 30

Question 12.
Arrange the following compounds in increasing order of their property as indicated:
(i) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN)
(ii) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH,
(CH3)2CHCOOH, CH3CH2CH2COOH (acid strength)
(iii) Benzoic acid, 4-Nitrobenzoic acid, 3, 4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)
Answer:
(i) When HCN reacts with a compound, the attacking species is a nucleophile, CN. Therefore, as the negative charge on the compound increases, its reactivity with HCN decreases. In the given compounds, the +I effect increases as shown below. It can be observed that steric hindrance also increases in the same.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 32
Hence, the given compounds can be arranged according to their increasing reactivities toward HCN as:
Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde

(ii) After losing a proton, carboxylic acids gain a negative charge as shown: R – COOH →R – COO + H+ Now, any group that will help stabilise the negative charge will increase the stability of the carboxyl ion and as a result, will increase the strength of the acid. Thus, groups having +I effect will decrease the strength of the acids, and groups having – I effect will increase the strength of the acids.

In the given compounds, -CH3 group has +I effect and Br group has – I effect. Thus, acids containing Br are stronger. Now, the +I effect of isopropyl group is more than that of n-propyl group. Hence, (CH3)2CHCOOH is a weaker acid than CH3CH2CH2COOH.

Also, the -I .effect grows weaker as distance increases. Hence, CH3CH(Br)CH2COOH is a weaker acid than CH3CH2CH(Br)COOH.
Hence, the strengths of the given acids increase as:
(CH3)2CHCOOH < CH3CH2CH2COOH < CH3CH(Br) CH2COOH < CH3CH2CH(Br) COOH

(iii) As we have seen in the previous case, electron-donating groups decrease the strengths of acids, while electron-withdrawing groups increase the strengths of acids. As methoxy group is an electron-donating group, 4- methoxy benzoic acid is a weaker acid than benzoic acid.

Nitro group is an electron-withdrawing group and will increase the strengths of acids. As 3,4-di nitrobenzoic acid i.ontains two nitro groups, it is a slightly stronger acid than 4-nitrobenzoic acid. Hence, the strengths of the given acids increase as:
4-Methoxybenzoic acid <Benzoic acid <4-Nitrobenzoic acid <3,4-Dinitrobenzoic acid.

PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question 13.
Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Propanal and Propanone
(ii) Acetophenone and Benzophenone
(iii) Phenol and Benzoic acid
(iv) Benzoic acid and Ethyl benzoate
(v) Pentan-2-one and Pentan-3-one
(vi) Benzaldehyde and Acetophenone
(vil) Ethanal and Propanal
Answer:
(i) Propanal and propanone:
lodoform test: This test is given by propanone and not by propanal.
Propanone on reacting with hot NaOH/I2 gives a yellow precipitate of CHI3 while propanal does not.
2NaOH + I2 → NaI + NaOI + H2O
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 33

(ii) Acetophenone and benzophenone: Acetophenone responds to iodoform test, but benzophenone does not.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 34

(iii) Phenol and benzoic acid: Benzoic acid reacts with NaHCO3 giving CO2 gas with effervescence, whereas phenol does not.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 35

(iv) Benzoic acid and ethyl benzoate: Benzoic acid on reaction with sodium hydrogen carbonate give out CO2 gas with effervescence, while ethyl benzoate does not.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 36

(v) Pentan-2-one and pentane-3-one: Pentan-2-one, when treated with NaOI(I2/NaOH), gives yellow precipitate of iodoform but pentane-3-one does not give this test.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 37

(vi) Benzaldehyde and acetophenone: Benzaldehyde “being an aldehyde gives silver mirror with Tollen’s reagent but acetophenone being a ketone does not give this test.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 38

(vii) Ethanal and propanal: Ethanal responds to iodoform test, while propanal does not.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 39

Question 14.
How will you prepare the following compounds from, benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom.
(i) Methyl benzoate
(ii) m -Nitrobenzoic acid
(iii) p-Nitrobenzoic acid
(iv) Phenylacetic acid
(v) p-Nitrobenz aldehyde.
Answer:
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 40
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 41
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 42

PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 43

Question 15.
How will you bring about the following conversions in not more than two steps?
(i) Propanone to Propene
(ii) Benzoic acid to Benzaldehyde
(iii) Ethanol to 3-Hydroxybutanal
(iv) Benzene to m-Nitroacetophenone
(y) Benzaldehyde to Benzophenone
(vi) Bromobenzene to 1-Phenylethanol
(vii) Benzaldehyde to 3-Phenylpropan-1-oI
(viii) Benzaldebyde to a -Hydroxy phenylacetic acid
(ix) Benzoic acid to m-Nitro benzyl alcohol
Answer:
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 44
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 47

PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question 16.
Describe the following:
(i) Acetylation
(ii) Cannizzaro reaction
(iii) Cross aldol condensation
(iv) Decarboxylation
Answer:
(i) Acetylation: The introduction of an acetyl functional group into an organic compound is known as acetylation. It is usually carried out in the presence of a base such as pyridine, dimethylaniline, etc. This process involves the substitution of an acetyl group for an active hydrogen atom. Acetyl chloride and acetic anhydride are commonly used as acetylating agents.
For example, acetylation of ethanol produces ethyl acetate.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 49

(ii) Cannizzaro reaction: The self oxidation-reduction (disproportionation) reaction of aldehydes having no a-hydrogens on treatment with concentrated alkalis is known as the Cannizzaro reaction. In this reaction, two molecules of aldehydes participate where one is reduced to alcohol and the other is oxidized to carboxylic acid.
For example, when ethanol is treated with concentrated potassium hydroxide, ethanol and potassium ethanoate are produced.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 50
(iii) Cross-aldol condensation: When aldol condensation is carried out between two different aldehydes, or two different ketones, or an aldehyde and a ketone, then the reaction is called a cross-aldol condensation. If both the reactants contain a-hydrogens, four compounds are obtained as products. For example, ethanol and propanal react to give four products. CH3CHO + CH3CH2CHO
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 51
(iv) Decarboxylation: Decarboxylation refers to the reaction in which carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with soda-lime (a mixture of NaOH and CaO in ratio 3:1).
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 52
Decarboxylation also takes place when aqueous solutions of alkali metal salts of carboxylic acids are electrolyzed. This electrolytic process is known as Kolbe’s electrolysis.

Question 17.
Complete each synthesis by giving missing starting material, reagent or products
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 53
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 54
Answer:
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 55

PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 56

Question 18.
Give plausible explanation for each of the following:
(i) Cyclohexanone forms cyanohydrin in good yield but 2, 2, 6 trimethyl cyclohexanone does not.
(ii) There are two -NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
(iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.
Answer:
(i) Cyclohexanones form cyanohydrins according to the following equation :
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 57
In this case, the nucleophile CN can easily attack without any steric hindrance. However, in the case of 2, 2, 6- trimethyl cyclohexanone, methyl groups at α-positions offer steric hindrances and as a result, CN cannot attack effectively.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 58
For this reason, it does not form a cyanohydrin.
(ii) Semicarbazide undergoes resonance involving only one of the two -NH2 groups, which is attached directly to the carbonyl carbon atom.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 59
Therefore, the electron density on -NH2 group involved in the resonance also decreases. As a result, it cannot act as a nucleophile.
Since the other -NH2 group is not involved in resonance; it can act as a nucleophile and can attack carbonyl-carbon atoms of aldehydes and ketones to produce semicarbazones.
(iii) Ester along with water is formed reversibly from a carboxylic acid and an alcohol in presence of an acid.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 60
If either water or ester is not removed as soon as it is formed, then it reacts to give back the reactants as the reaction is reversible. Therefore, to shift the equilibrium in the forward direction i.e., to produce more ester, either of the two should be removed.

PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question 19.
An organic compound contain 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound Is 86. It does not reduce Tollens’ reagent but forms an additional compound with sodium hydrogen sulphite and give positive iodoform test. On vigorous oxidation, it gives ethanoic and propanoic acid. Write the possible structure of the compound.
Solution:
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 61
Empirical formula of the compound A = C5H10O
Molecular formula of the compound A = n (Empirical formula)
n = \(\frac{\text { Molecular mass of compound } A}{\text { Empirical formula mass of compound } A} \)
Molecular mass of compound A =86

Empirical formula mass of compound
A = 5 x 12 + 1 x 10+1 x 16
= 60+10+16 = 86
n = \(\frac{86}{86}\) = 1

Molecular formula of the compound
A = 1 (C5H10O) = C5H10O
As the compound A forms addition compound with NaHSO3 therefore it must be either an aldehyde or ketone. As it does not reduce Tollen’s reagent and give positive iodoform test therefore it must be a methyl ketone. As on oxidation, the compound A gives a mixture of ethanoic acid and propane acid, therefore compound A is
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 62
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 63

Question 20.
Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger
acid than phenol. Why?
Answer:
(i) Resonance structures of phenoxide ion are:
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 64
It can be observed from the resonance structures of phenoxide ion that in II, III and IV, less electronegative carbon atoms carry a negative charge. Therefore, these three structures contribute negligibly towards the resonance stability of the phenoxide ion. Hence, these structures can be eliminated. Only structures I and V carry a negative charge on the more electronegative oxygen atom.

(ii) Resonance structures of carboxylate ion are :
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 65
In the case of carboxylate ions, resonating structures I and II contain a charge carried by a more electronegative oxygen atom. Further, in resonating structures I and II, the negative charge is delocalized over two oxygen atoms. But in resonating structures I and V of the phenoxide ion, the negative charge is localized on the same oxygen atom. Therefore, the resonating structures of carboxylate ion contribute more towards its stability than those of phenoxide ion. As a result, carboxylate ion is more resonance-stabilized than phenoxide ion. Hence, carboxylic acid is a stronger acid than phenol.

Chemistry Guide for Class 12 PSEB Aldehydes, Ketones, and Carboxylic Acids Textbook Questions and Answers

Question 1.
Write the structures of the following compounds :
(i) α-Methoxypropionaldehyde
(ii) 3-Hydroxybutanal
(iii) 2-Hydroxycyclopentane carbaldehyde
(iv) 4-Oxopentanal
(v) Di-sec. butyl ketone
(vi) 4-Fluoroacetophenone
Answer:
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 66

Question 2.
Write the structures of products of the following reactions;
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 67
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 68
Answer:
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 69

PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question 3.
Arrange the following compounds in increasing order of their boiling points.
CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3
Answer:
Hydrocarbons are non-polar having weakest attractive forces; Ethers are polar (dipolar forces); Aldehydes have strong dipolar interactions;
Alcohols have maximum intermolecular forces due to hydrogen bonding.
CH3CH2CH3 < CH3OCH3 < CH3CHO< CH3CH2OH.

Question 4.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions.
(i) Ethanal, Propanal, Propanone, Butanone.
(ii) Benzaldehyde, p-Tolualdehyde, p -Nitrobenz aldehyde, Acetophenone.
(Hint: Consider steric effect and electronic effect)
Answer:
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 70
The +I effect of the alkyl group increases in the order:
Ethanal < Propanal < Propanone < Butanone
The electron density at the carbonyl carbon increases with the increase in the +I effect. As a result, the chances of attack by a nucleophile decrease. Hence, the increasing order of the reactivities of the given carbonyl compounds in nucleophilic addition reactions is:
Butanone < Propanone < Propanal < Ethanal

PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 71
The + I effect is more in ketone than in aldehyde. Hence, acetophenone is the least reactive in nucleophilic addition reactions. Among aldehydes, the +1 effect is the highest in p-total dehyde because of the presence of the electron-donating -CH3 group and the lowest in p-nitrobenzaldehyde because of the presence of the electron-withdrawing 2 groups. Hence, the increasing order of the reactivities of the given compounds is :
Acetophenone < p – tolualdehyde <Benzaldehyde <p – Nitrobenzaldehyde

Question 5.
Predict the products of the following reactions:
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 72
Answer:
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 73

Question 6.
Give the IUPAC names of the following compounds :
(i) PhCH2CH2COOH
(ii) (CH3)2C = CHCOOH
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 74
Answer:
(i) 3-Phenylpropanoic acid
(ii) 3-Methylbut-2-enoic acid
(iii) 2-Methylcyclopentanecarboxylic acid
(iv) 2,4,6-Trinitrobenzoic acid

PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question 7.
Show how each of the following compounds can be converted to benzoic acid.
(i) Ethylbenzene
(ii) Acetophenone
(iii) Bromobenzene
(iv) Phenylethene (Styrene)
Answer:
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 75

Question 8.
Which acid of each pair shown here would you expect to be stronger?
(i) CH3CO2H or CH2FCO2H
(ii) CH2FCO2H or CH2CICO2H
(iii) CH2FCH2CH2CO2H or CH3CHFCH2CO2H
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 76
Answer:
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 77
The +I effect of the -CH3 group increases the electron density on the O-H bond. Therefore, the release of protons becomes difficult. On the other hand, the – I effect of F decreases the electron density on the O-H bond. Therefore, protons can be released easily. Hence, CH2FCO2H is a stronger acid than CH3CO2H.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 78
F has a stronger – I effect than Cl. Therefore, CH2FCO2H can release protons more easily than CH2ClCO2H.
Hence, CH2FCO2H is a stronger acid than CH2ClCO2H.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 79
The inductive effect decreases with an increase in distance. Hence, the +I effect of F in CH3CHFCH2CO2H is more than it is in CH2FCH2CH2CO2H.
Hence, CH3CHFCH2CO2H is a stronger acid than CH2FCH2CH2CO2H.
PSEB 12th Class Chemistry Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 80
Due to the – I effect of F, it is easier to release proton in the case of compound (A). However, in the case of compound (B), release of proton is difficult due to the +1 effect of -CH3 group. Hence, (A) is a stronger acid than (B).