Inter 2nd Year

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material 4th Lesson Surface Chemistry Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material 4th Lesson Surface Chemistry

Very Short Answer Questions

Question 1.
What is an interface? Give one example.
Answer:
The interface is normally a few molecules thick but its area depends on the size of the particles of bulk.
Interface is represented by separating the bulk phases by hyphen (-) (or) a slash (/).
Eg : Interface between a solid and a gas may be represented by solid-gas (or) solid/gas.

Question 2.
What is adsorption ? Give one example.
Answer:
Adsorption : The accumulation (or) concentration of a substance on the surface rather than in the bulk of solid (or) liquid is known as adsorption.
Eg : Adsorption of gases like O₂, H₂, Cl₂ etc., on charcoal.

Question 3.
What is absorption ? Give one example.
Answer:
Absorption : The uniform distribution of a substance through out the bulk of the solid substance is known as absorption.
Eg : Chalk stick dipped in ink.

Question 4.
Distinguish between adsorption and absorption. Give one example of each.
Answer:
Adsorption
1. The accumulation (or) concentration of a substance on the surface rather than in the bulk of solid (or) liquid is known as adsorption.
Eg :Adsorption of gases like O₂, H₂, Cl₂ etc., on charcoal.

Absorption
1) The uniform distribution of a susbstance through out the bulk of the solid substance is known as absorption.
Eg : Chalk stick dipped in ink.

Question 5.
The moist air becomes dry in the presence of silica gel. Give reason for this.
Answer: The moist air becomes dry in presence of silica gel because the water molecules present in air get adsorbed on the surface of the gel.

Question 6.
Methylene blue solution when shakes with animal charcoal gives a colourless filtrate on filteration. Give the reason.
Answer:
Methylene blue solution (organic dye) when shaken with animal charcoal gives a colourless filtrate on filtration. The molecules of the dye from the solution are adsorbed on the charcoal.

Question 7.
A small amount of silica gel and a small amount of anhydrous calcium chloride are placed separately in two corners of a vessel containing water vapour. What phenomena will occur ?
Answer:
A small amount of silica gel and a small amount of anhydrous CaCl₂ are placed separately in two comers of a vessel of water vapour. Water vapours are is absorbed by anhydrous CaCl₂ but adsorbed by silical gel.

Question 8.
What is desorption ?
Answer:
Desorption: The process of removing aa adsorbed substance from a surface on which it is adsorbed is called desorption.

Question 9.
What is sorption ?
Answer:
In case of some substances both adsorption and absorption takes place. This phenomenon is called sorption.

Question 10.
Amongst adsorption, absorption which is a surface phenomena and why ?
Answer:

1. Adsorption is a surface phenomenon.
2. In adsorption the concentration of the adsorbate increases only at the surface of the adsorbent. While in absorption the concentration is uniformly distributed throughout the bulk of the compound.

Question 11.
What is the name given to the phenomenon when both absorption and adsorption take place together ?
Answer:
The name of the phenomenon when both adsorption and absorption take place together is sorption.

Question 12.
Chalk stick dipped in an ink solution exhibits the following :
a) The surface of the stick retains the colour of the ink.
b) Breaking the chalk stick,-it is found still white from inside.Explain the above observations.
Answer:
a) Chalk stick dipped in an ink solution, the surface of the stick retains the colour of the ink due to adsorption of coloured molecules in the ink.
b) Chalk stick dipped in an ink solution, the surface of the stick retains the colour of the ink due to adsorption of coloured molecules in the ink while the solvent of the in k goes deep into the stick due to absorption so on breaking the chalk stick, it is found to be white inside.

Question 13.
What are the factors which influence the adsorption of a gas on a solid ?
Answer:
Adsorption of a gas on solids is influenced by the following factors.
a) Surface area of the adsorbent, b) Nature of the gas, c) Pressure of the adsorbate, d) Temperature.

Question 14.
Why is adsorption always exothermic ?
Answer:
During the adsorption, there is always a decrease in residual forces of the surface. There is decrease in surface energy which appears as heat.
∴ Adsorption is always Exothermic.

Question 15.
Give the signs of ∆H & ∆S, when ammonia gas gets adsorbed on charcoal.
Answer:
When NH₃ gas gets adsorbed on charcoal the sign of ∆H is negative and the sign of ∆S is also negative.

Question 16.
How many types of adsorption are known ? What are they ?
Answer:
Adsorption process is divided into two types.
1) Physisorption 2) Chemisorption

Question 17.
What types of forces are involved in physisorption of a gas on solid ?
Answer:
Weak Vander waals forces are involved in physisorption of a gas on solid.

Question 18.
What type of interaction occurring between gas molecules and a solid surface is responsible for chemisorption of the gas on solid.
Answer:
Chemical bonds occur between gas molecules and a solid surface, is responsible for chemisorption of the gas on solid.

Question 19.
Why chemisorption is called activated adsorption ?
Answer:
Chemisorption involves a high energy of activation. Hence it is referred as activated adsorption.

Question 20.
What is the difference between physisorption and chemisorption ?
Answer:
1) In physisorption the forces present between adsorbate and adsorbent are weak vander . waal’s forces.
2) In chemisorption the forces present between adsorbate and adsobent are chemical bonds (valency forces).

Question 21.
Out of physisorption and chemisorption, which can be reversed ?
Answer:
Out of physisorption and chemisorption, phyisorption is reversed (reversible).

Question 22.
How is adsorption of a gas related to its critical temperature ?
Answer:
Higher the critical temperature, greater is the case of liquefaction of a gas. Then the extent of adsorption will be high.

Question 23.
The critical temperature of SO₂ is 630 K and that of CH₄ is 190 K. Which is adsorbed easily on activated charcoal ? Why ?
Answer:
Given the critical temperature of SO₂ is 630 K and that of CH₄ is 190 K.
SO₂ gas is adsorbed easily on activated charcoal because of higher critical temperature. Higher the critical temperature of gas, extent of adsorption is high.

Question 24.
Easily liquefiable gases aye readily adsorbed on solids. Why ?
Answer:
Easily liquefiable gases are readily adsorbed on solids because vander waal’s forces are stronger near the critical temperatures.

Question 25.
Amongst SO₂, H₂ which will be adsorbed more readily on the surface of charcoal and why ?
Answer:
Among SO₂, H₂ gases, SO₂ gas adsorb more readily on the surface of charcoal because SO₂ has high critical temperature (630 K) than dihydrogen (33K).

Question 26.
Compare the enthalpy of adsorption for physisorption and chemisorption.
Answer:

1. In physical adsorption enthalpy of adsorption is low (20 – 40 KJ/mole)
2. In case of chemical adsorption enthalpy of adsorption is high (80 – 240 KJ/mole)

Question 27.
What is the magnitude of enthalpy of physical adsorption ? Give reason for this magnitude.
Answer:
The magnitude of enthalpy of physical adsorption is low 20 – 40 KJ ,/mole. This is due to the presence of weak vander waal’s forces between gas molecules and solid surface.

Question 28.
What is the magnitude of enthalpy of chemisorption ? Give reason for this magnitude.
Answer:
The magnitude of enthalpy of chemical adsorption is high 80 – 240 KJ /mole. This is due to the presence of chemical bonds between gas molecules and solid surface.

Question 29.
Give any two applications of adsorption.
Answer:
Applications of adsorption :
a) Separation of inert gases : Different noble gases adsorb at different temperatures on coconut charcoal. By this principle (adsorption) mixture of noble gas is separated by adsorption on coconut charcoal.
b) Gas masks : Gas mask is a device which consists of activated charcoal (or) mixture of adsorbents is used by coal miners to adsorb poisonous gases during breathing.

Question 30.
Why physisorption suffers from lack of specificity ?
Answer:
Physisorption suffers from lack of specificity.
Explanation: A given surface of an adsorbent doesnot show any preference of a particular gas as the vander waal’s forces are universal.

Question 31.
What is an adsorption isotherm ? Write the equation of Freundlich adsorption isotherm.
Answer:
Adsorption Isotherm : The variation in the amount of gas adsorbed by the adsorbent with pressure at constant temperature can be expressed by means of a curve known as adsorption isotherm.
Freundlich adsorption isotherm equation is \(\frac{x}{m}\) = k. P^1/n
x = mass of the gas adsorbed
m = mass of the adsorbent
P, k and n are constants.

Question 32.
In the Freundlich adsorption isotherm, mention the conditions under which, following graph will be true ?

Answer:
In the above graph extent of adsorption does not depends on the pressure.
When \(\frac{1}{n}\) = 0, \(\frac{x}{m}\) = constant, the adsorption is independent of pressure.

Question 33.
What role does adsorption play in heterogenous catalysis ?
Answer:
In heterogeneous catalysis, adsorption of reactants on the solid surface of the catalysts increases the rate of reaction.
Eg : Manufacturing of NH₃ using Fe-as catalyst [Haber’s process].

Question 34.
What is the role of MnO₂ in the preparation of O₂ from KClO₃ ?
Ans. The chemical equation for the preparation of O₂ from KClO₃ is

MnO₂ increases the rate of reaction. Hence it is a catalyst for the above reaction.

Question 35.
Define “promoters” and “poisons” in the phenamenon of catalysis ?
Answer:
Promoters : The substances with enhance the activity of catalyst are known as promoters. Poisons : The substances which decrease the activity of a catalyst are known as poisons.

Question 36.
What is homogeneous catalysis ? How is it different from heterogeneous catalysis ?
Answer:
Homogeneous Catalysis : The catalysis in which reactants and catalyst are in same phase is called Homogeneous catalysis.
In case of heterogeneous catalysis, catalyst and reactants are present in different phases where as in case of homogeneous catalysis catalyst and reactants are present in same phase.

Question 37.
Give two examples for homogeneous catalytic reactions.
Answer:
The following are the examples for homogeneous catalytic reactions.

Question 38.
Give two examples for heterogeneous catalysis.
Answer:
The following are the examples of heterogeneous catalytic reactions.

Question 39.
Give two examples which indicate the selectivity of heterogeneous catalysis.
Answer:
The following reactions indicate the selectivity of heterogeneous catalysis.
Starting with H₂ and CO, and using different catalysts, we get different products.

The action of a catalyst is highly selective in nature. A substance which acts as a catalyst in one reaction may fail to catalyse another reaction.

Question 40.
Why zeolites are treated as shape selective catalysts ?
Answer:

• The catalytic reaction that depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape selective catalysts.
• Zeolites are shope selective catalysts because of their honey comb-like structure. They are microporous alumino silicates with 3-dimensional net work of silicates.

Question 41.
Which zeolite catalyst is used to convert alcohols directly into gasoline ?
Answer:
Zeolite ZSM – 5 is used to convert alcohols directly into gasoline (petrol) by dehydrating them to give a mixture of hydrocarbons.

Question 42.
What are enzymes ? What is their role in human body ?
Answer:
Enzymes are complex nitrogenous organic compounds which are produced by living plants and animals.
These act as specific catalysts in biological reactions.
These catalyse the numerous reactions that occur in the bodies of animals and plants to maintain the life process.

Question 43.
Can catalyst increase the yield of reaction ?
Answer:
Catalyst does not increase the yield of reaction. Catalyst just speed up the product formation.

Question 44.
Name any two enzyme catalyzed reactions. Give the reactions.
Answer:
1) Inversion of Cane Sugar :

2) Decomposition of urea into ammonia and CO₂:

Question 45.
Name the enzymes obtained from soyabean source.
Answer:
The enzyme obtained from soyabean source is urease.

Question 46.
Name the enzymes used in
a) Decomposition of urea into ammonia.
b) Conversion of proteins into peptides in stomach.
Answer:
a) The enzyme used in the decomposition of urea into ammonia is urease.
b) The enzyme used in conversion of proteins into peptides in stomach is pepsin.

Question 47.
What enzymes are obtained from yeast ?
Answer:
The enzymes obtained from yeast are invertase, zymase and maltase.

Question 48.
At what ranges of temperature and pH, enzymes are active ?
Answer:
The optimum temperature range for enzymatic activity is 298 – 310 K. Human body temperature being 310 K is suited for enzyme catalysed reactions. The rate of an enzyme – catalysed reaction is maximum at a particular pH called optimum pH, which lies between 5-7.

Question 49.
Represent diagrammatically the mechanism of the enzyme catalyis.
Answer:

Question 50.
Name any two industrially important heterogeneous catalytic reactions mentioning the catalysts used.
Answer:
i) Manufacturing of NH₃ by Haber’s process

ii) Hydrogenation of vegetable oils in presence of finely divided nickel as catalyst.

The above mentioned are industrially important heterogeneous catalytic reactions.

Question 51.
What is a colloidal solution ? How is it different from a true solution with respect to dispersed particle size and homogeneity ?
Answer:
A heterogeneous system in which one substance is dispersed as large particles in another substance is called colloidal solution.

• In colloidal solution the particle size of the dispersed phase is of the order of 1 mμ – 1 μ where as in true solution the particle size of the solute are the order of mp or less.
• Colloidal solution is a heterogeneous binary system where as true solution is a homogeneous binary system.

Question 52.
Name the dispersed phase and a dispersion medium in the following colloidal systems

1. fog
2. smoke
3. milk.

Answer:

1. Fog : Dispersed phase → liquid
   Dispersion medium → gas
2. Smoke : Dispersed phase → carbon particles (solid)
   Dispersion medium → Air
3. Milk : Dispersed phase → liquid fat
   Dispersion medium → water

Question 53.
What are lyophilic and lyophobic sols ? Give one example for each type.
Answer:
Lyophilic colloid : The colloidal solution in which the dispersed phase has great affinity to the dispersion medium is called a Lyophilic colloid or Lyophilic solution.
Ex : Starch solution.

The starch paste when dissolved in hot water, with stirring, the starch solution is formed. The starch particles (dispersed phase) has great affinity to water molecules (dispersion medium). So starch solution is a lyophilic solution or lyophilic colloid.

Lyophobic colloid : The colloidal solution in which there exists not much affinity between the dispersed phase and dispersion medium, it is called a Lyophobic colloid or Lyophobic solution.
Ex: Gold solution.

Gold rods are placed in water containing alkali. Electric arc is applied between gold rods. The gold particles dissolves in water, to give gold solution.

Gold particles (dispersed phase) have not much affinity towards water (dispersion medium). So this is a Lyophobic solution or Lyophobic colloid.

Question 54.
Explain the terms with suitable examples .

1. aerosol
2. hydrosol.

Answer:

1. Aerosol: The colloidal solution in which dispersed phase is solid and dispersion medium is gas is called an aerosol.
   Eg : Smoke, dust.
2. Hydrosol: In a colloidal solution dispersion medium is water them it is called aquasol (or) hydrosol.
   Eg : Milk, Gold sol.

Question 55.
Explain why lyophilic colloids are relatively more stable than lyophobic colloids ?
Answer:

• Lyophilic colloids are reversible and are quite stable. These are not coagulated.
• Lyophobic sols are irreversible and need stabilizing agents for their stabilization. These sols are readily precipitated on the addition of small amounts of electrolytes.

Question 56.
Give two examples of colloidal solutions of liquids dispersed in solid. What is the name given to the colloidal solution ?
Answer:
Cheese, butter, jellies are examples of colloidal solutions of liquids dispersed in solid. The name of the colloidal solution given to this type is Gels.

Question 57.
What is the difference between multimolecular and macromolecular colloids ? Give one example for each.
Answer:

• In multimolecular colloids, a large number of atoms (or) small molecules of the dispersed phase aggregate together to form species in the colloidal range.
• Macro molecules in suitable solvents form solutions in which the sizes of the macromolecules are in the colloidal range.

Question 58.
What are micelles ? Give one example.
Answer:
Micelles : Some substances which at low concentrations behave as normal strong electrolytes, but at high concentrations exhibit colloidal behaviour due to formation of aggregates. The aggregated particles thus formed are called micelles.
Eg : Stearate ions (C₁₇H₃₅COO^–) associate together in high concentration, in a solution of soap in water and forms a micelle.

Question 59.
How do micelles differ from a normal colloidal solutions ?
Answer:
Some substances which at low concentrations behave as normal strong electrolytes, but at high concentrations exhibit colloidal behaviour due to formation of aggregates. The aggregated particles thus formed are called micelles.

• hese are also known as associated colloids.
• These colloids have both lyophilic and lyophobic parts.
• Micelles may contain as many as 100 or more of normal molecules.
• On dilution these colloids revert back to individual electrolytes.

Question 60.
Give two examples of associated colloids.
Answer:
Surface active agents such as soaps and synthetic detergents are examples of associated colloids.

Question 61.
Can the same substance act both as colloid and crystalloid ?
Answer:
Yes, Micelles, (associated colloids) act as normal strong electrolytes at low concentrations and exhibit colloid behavior at high concentrations.

Question 62.
Give two examples of lyophobic sols.
Answer:
Metal sols and metal suiphide sols are examples of lyophobic colloids

• Gold sol is lyophobic colloid

Question 63.
Give examples of’colloidal system of

1. Liquid In solid
2. Gas in solid.

Answer:

1. Cheese, butter, jellies are examples of liquid in solid type colloidal system.
2. Foam rubber, pumice ‘stone are examples of gas in solid type of colloidal system.

Question 64.
What type of substances form lyophobic sols?
Answer:
Substances like metals, their sulphides.donot form colloidal sol simply by mixing with the dispersion medium. These form colloids by special methods, These sols are lyophobic colloids.

Question 65.
What is critical micelle concentration (CMC) and kraft temperature (T_k) ?
Answer:
The formation of micelles takes place only above a particular temperature called Kraft temperature (T_k) and above a particular concentration called Critical micelle concentration (CMC).

Question 66.
Why lyophobic colloids are called irreversible colloids ?
Answer:

• Lyophobic colloids are coagulated by the addition of small amounts of electrolytes.
• The precipitates does not give back the colloidal sol by same addition of the dispersion medium.
• Hence lyophobic sols are called irreversible sols.

Question 67.
How a colloidal sol of arsenous sulphide is prepared ?
Answer:
Colloidal sol of arsenous sulphide is prepared by the double decompostion of As₂O₃ and H₂S as follows.

Question 68.
What is Peptization ?
Answer:
Peptization : The process of converting a precipitate into colloidal sol by shaking it with the dispersion medium in the presence of a small amount of electrolyte is called Peptization.

Question 69.
What is dialysis ? How is dialysis can be made fast ?
Answer:
Dialysis : The process of removing a dissolved substance from a colloidal solution using a suitable membrane is called dialysis.
Dialysis is made faster by applying an Emf. This is known as Electrodialysis.

Question 70.
What is collodion solution ?
Answer:
Collodion solution : Collodion is a 4% solution of nitro cellulose in a mixture of alcohol and ether.

Question 71.
How an ultrafilter paper is prepared from ordinary filter paper ?
Answer:
Ultra filter paper is prepared by soaking the filter paper in collodion solution, hardening by formaldehyde and then finally dried.

Question 72.
What is Tyndall effect ?
Answer:
Tyndall effect : When light enters a colloidal solution, it is scattered by the large sized colloidal dispersed phase particles. Therefore when light passes through a solution we will be able to see the path of the light as a luminous beam. This is called Tyndall effect.

This is an optical property exhibited by colloidal solution. This phenomenon is clearly seen with a microscope placed at right angles to the path of light. Colloidal particles become self luminous due to absorption of light. A true solution does not show Tyndall effect.

Question 73.
What conditions is tyndall effect observed ?
Answer:
Tyndall effect is observed only when

1. The diameter of the dispersed particles is not much smaller than the wave length of the light used.
2. The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.

Question 74.
Can Tyndall effect be used to distinguish between a colloidal solution and a true solution ? Explain.
Answer:
Tyndall effect is used to distinguish between a colloidal solution and a true solution.

• True solution placed in dark and is observed in the direction of the light passed beam passing through it. It appears clear and it is observed from a direction at right angles to the direction of the light beam, it appears perfectly dark.
• Colloidal solution viewed in the same way may appear reasonably clear in the direction of light but they show a mild to strong opalescence when viewed at right angles to the direction of light.

Question 75.
Sky appears blue in colour. Explain.
Answer:
Dust particles along with water vapour suspended in air scatter blue light which reaches our eyes and hence the sky looks blue to us.

Question 76.
What is Brownian movement.
Answer:
Brownian movement: This is a kinetic property of colloidal solution.

When a colloidal solution is examined by ultramicroscope, the colloidal particles are seem to be moving in a rapid zig-zag motion.
This rapid motion of colloidal particles is called Brownian movement.

This motion is due to unequal bombardment of colloidal particles by molecules of dispersion medium. Smaller the colloidal particles the more rapid is the Brownian movement.

Question 77.
What is the main cause for charge on a colloidal solution ?
Answer:
The charge on the colloidal particles is due to electron capture by sol particles during electro dispersion of metals , and due to preferential adsorption ions from solution (or) due to formulation of electrical double layer.

Question 78.
What is electrokinetic potential or zeta potential ?
Answer:
In a colloidal sol the charges of opposite signs on the fixed and diffused parts of the double layer results in a difference in potential between these layers. The potential difference between the fixed layer and the diffused layer of opposite charge is called electro kinetic potential (or) zeta potential.

Question 79.
Write the chemical formula of positively charged and negatively charged hydrated ferric oxide colloidal solutions.
Answer:

• Formula of positively charged hydrated ferric oxide is Fe₂O₃.xH₂O/Fe⁺³
• Formula of negatively charged hydrated ferric oxide is Fe₂O₃.xH₂0/OH^–

Question 80.
Give the order of coagulating power of Cl, SO₄^2-, PO₄^3- in the coagulation of positive sols.
Answer:
The order of coagulating ability’of given ions with positive sols is PO₄^3- > SO₄^2- > Cl^–

Question 81.
Amongst Na⁺, Ba²⁺, Al³⁺, which coagulates negative sol readily and why ?
Answer:
The order of coagulating ability of given ions with negative sols is Al⁺³ > Ba⁺² > Na⁺

Question 82.
A colloidal solution of AgI is positively charged when prepared from a solution containing excess of Ag⁺ ions and negatively charged when prepared from a solution containing excess of I^– ions Explain.
Answer:
When a dilute AgNO₃ solution is added to a dilute KI solution taken in excess, the precipitated Agl adsorbs I^– ions present in excess and a negatively charged AgI colloidal solution is formed. However when dilute KI solution is added to dilute AgNO₃ solution taken in excess, AgI precipitate adsorbs Ag⁺ ions present in excess and a positively charged AgI colloidal solution is formed. Generally the ion common to dispersed particle is adsorbed. .
AgI/I^–
Negatively charged
AgI/Ag⁺
Positively charged.

Question 83.
What is electrophoresis ?
Answer:
When electric potential is applied across two platinum electrodes dipping in a colloidal solution,-the colloidal particles move towards one or the other electrode. The movement of colloidal particles under an applied emf is called “electrophoresis”.

Question 84.
What is electro osmosis ?
Answer:
If the movement of colloidal particles is arrested by some suitable means, the dispersion medium moves in opposite direction. This phenomenon is termed “electro osmosis”.

Question 85.
What is coagulation ?
Answer:
The stability of the lyophilic sols is due to the presence of charge on the colloidal particles. If this charge is neutralised the particles will come nearer to each other to form aggregates (or coagulate) and settle down under the force of gravity. This process of settling down of colloidal particles is called coagulation (or) flocculation (or) precipitation.

Question 86.
Define flocculation value.
Answer:
The minimum concentration of an electrolyte in millimoles per litre required to cause coagulation of a sol in two hours is called “coagulating value” (or) flocculation value.

Question 87.
State Hardy – Schulze rule.
Answer:
Greater the valence of the coagulating ion added, the greater is its power to cause coagulation. This is known as Hardy-Schulze rule.

Questions 88.
Coagulation takes place when sodium chloride solution is added to a colloidal solution of hydrated ferric oxide. Explain.
Answer:
When NaCl solution is added to the colloidal solution of hydrated ferric oxide coagulation takes place (precipitation formed). .
The reason is that colloidal particles interact with ions (Na⁺, Cl^–) carrying charge opposite to that present on themselves. This causes neutralisation of charges leading to their coagulation.

Question 89.
How are lyophobic solutions protected from phenomenon of coagulation.
Answer:
Lyophobic sols are protected from the phenomenon of coagulation by adding suitable lyophilic sol.

Question 90.
What is protective colloid ?
Answer:
Lyophilic colloids used for the prevention of coagulation of lyophobic colloids are called protective colloids.

• Lyophilic colloids protect the lyophobic colloids.

Question 91.
What is an emulsion ? Give two examples. [A.P. Mar. 17]
Answer:
Emulsion: The colloidal system in which a dispersion of finely divided droplets of a liquid in another liquid medium is called emulsion.
Eg : Milk, Vanishing cream, Cold cream.

Question 92.
How emulsions are classified ? Give one example for each type of emulsion. [A.P. Mar. 17]
Answer:
Emulsion : A dispersion of finely divided droplets of a liquid in another liquid medium is called’emulsion’.
Ex: Milk.
In Milk, the droplets of liquid fat are dispersed in water. This is an example for oil in water type emulsion.
Classification of emulsions : Emulsions are “classified into two classes. These are
a) oil in water (O/W) and
b) water in oil (W/O), (O = Oil; W = Water).

These emulsions are classified as such depending on which is dispersed phase and which is dispersion medium.
a) Oil in Water (O/W) type emulsions : In this type of emulsions, the dispersed phase is oil and the dispersion medium is water.
Ex : Milk, liquid, fat (oil) in water.
Vanishing cream; fat in water.

b) Water in Oil (W/O) type emulsions : In this type of emulsions, the dispersed phase is water and the dispersion medium is oil.
Ex : Stiff greases : water in lubrication oils
Cod liver oil : water in cod liver oil
Cold cream : water in fat.

Question 93.
What is an emulsifying agent ?
Answer:
Emulsifying agent: The third substance which is added in small amounts to an emulsion to stabilize the emulsion is called emulsifying agent.

Question 94.
What is demulsification ? Name two demulsifiers.
Answer:
Demulsification : The separation of an emulsion into constituent liquids is known as demulsification.

Question 95.
How is artificial rain produced ?
Answer:
Artificial rain is produced by throwing electrified sand (or) spraying a sol carrying charge opposite to the one on clouds from an aeroplane.

Question 96.
Bleeding from fresh cut can be Stopped by applying alum. Give reasons.
Answer:
Bleeding from fresh cut can be stopped by applying alum. This is due to styptic action of alum which coagulates the blood and stops further bleeding.

Question 97.
Deltas are formed at the points where river enters the sea. Why ?
Answer:
Deltas are formed at the points where river enters the sea.
Explantation : River water is a colloidal solution of clay. Sea water contains a number of electrolytes. When river water meets the sea water, the electrolytes present in sea water coagulate the colloidal solution of clay resulting in its deposition of clay with the formation of delta.

Question 98.
Name any two applications of colloidal solutions.
Answer:
Applications of colloidal solutions :
Rubber: Plant latex is a colloidal solution of rubber particles which are negatively charged. Rubber is obtained from latex by coagulation.
Industrial Products : Paints, inks, synthetic plastics, rubber, graphite, lubricants, cement, etc., are all colloidal in nature.

Question 99.
How can aerial pollution by colloidal particles of smoke be prevented ? Explain.
Answer:
Aerial pollution by colloidal particles of smoke be prevented by electrical precipitation of smoke. Smoke is a colloidal solution of solid particles such as carbon, arsenic compounds, dust, etc., in air. The smoke, before it comes out from the chimney, is led through a precipitator containing plates having a charge opposite to that carried by smoke particles. The particles on coming in contact with these plates lose their charge and get precipitated. The particles thus settle down on the floor of the chamber. The precipitator is called cottrell precipitator.

Question 100.
Alum is used to purify water obtained from natural sources. Explain.
Answer:
The water obtained from natural sources often contains suspended impurities. Alum is added to such water to coagulate the suspended impurities and make water fit for drinking purposes.

Question 101.
Why medicines are more effective in colloidal state ?
Answer:
Colloidal medicines are more effective because they have large surface area and are therefore easily assimilated.

Question 102.
How rubber is obtained from latex ?
Answer:
Plant latex is a colloidal solution of rubber particles which are negatively charged. Rubber is obtained from latex by coagulation.

Question 103.
Name the type of emulsion to which milk belongs.
Answer:
Milk is a oil in water type of emulsion.
Dispersed phase : liquid fat.
Dispersion medium : water.

Short Answer Questions

Question 1.
What is adsorption ? Discuss the mechanism of adsorption of gases on solids.
Answer:
Adsorption : The accumulation (or) concentration of a substance on the surface rather than in the bulk of solid (or) liquid is known as adsorption.
Eg : Adsorption of gases like O₂, H₂, Cl₂ etc., on charcoal.
Mechanism of Adsorption :

• Adsorption arises due to the fact that the surface particles of the adsorbent are not in the same environment as the particles inside the bulk.
• Adsorbent possess unbalanced or residual attractive forces. These forces are responsible for attracting the adsorbate molecules on adsorbent surface.
• The extent of adsorption increases with increase in surface area of adsorbent at a given temperature and pressure.
• During the adsorption, there is always a decrease in residual forces of the surface. There is decrease in surface energy which appears as heat. So, Adsorption is always Exothermic.
• Adsorption is accompanied by decrease in enthalpy as well as decrease in entropy of system. .
• As the adsorption proceeds AH becomes less and less negative, ultimately, AH becomes equal to TAS and AG becomes zero. At this state equilibrium is attained.

Question 2.
What are different types of adsorption ? Give any four differences between characteristics of these different types. [T.S. Mar. 19, 15; A.P. Mar. 15]
Answer:
Adsorption process is divided into two types.
1) Physisorption
2) Chemisorption.
Distinguishing characteristics of Physisorption and Chemisorption are given in the following table:
Physisorption

1. This process is weak, due to vander Waal’s forces.
2. The process is reversible.
3. This is a quick process, i.e., takes place quickly.
4. The process decreases with increase of temperature.
5. This is a multilayered process.
6. The process depends mainly on the nature of the adsorbent.

Chemisorption

1. This process is strong, due to chemical forces.
2. The process is irreversible.
3. This is a slow process.
4. The process increases with increase of temperature.
5. This is a unilayered process.
6. The process depends both on the nature of adsorbent and adsorbate.

Question 3.
What do you understand by the terms given below (a) absorption (b) Adsorption (c) Adsorbent and Adsorbate
Answer:
Absorption : The uniform distribution of a susbstance through out the bulk of the solid .substance is known as absorption. Eg: Chalk stick dipped in ink.
Adsorption : The accumulation (or) concentration of a substance on the surface rather than in the bulk of solid (or) liquid is known as adsorption.
Eg: Adsorption of gases like O₂, H₂, Cl₂ etc., on charcoal.

Adsorbent: The substance on whose surface the adsorption occurs is known as adsorbent.

Adsorbate : The substance whose molecules get adsorbed on the surface of the adsorbent is known as adsorbate.
Eg: In the adsorption of acetic acid by charcoal, acetic acid is adsorbate and the charcoal is adsorbent.

Question 4.
Adsorption of a gas on the surface of solid is generally accompanied by decrease in entropy. Still it is a spontaneous process. Explain.
Answer:
Adsorption is accompanied by decrease in enthalpy as well as decrease in entropy of the system.

• For a process to be spontaneus there is a decrease in Gibbs Energy. .
• On the basis of equation ∆G = ∆H – T∆S; ∆G = -Ve
•  If AH has sufficiently high negative value and T∆S is positive.
• In an adsorption process which ¡s spontaneous ∆G becomes negative by combining above.
• As the adsorption proceeds ∆H becomes less and less negative ultimately, ∆H becomes equal to T∆S and ∆G becomes zero. At this state equilibrium is attained.

Question 5.
How can the constants k and n of the Freundlich adsorption equation be calculated?
Answer:
Freundlich adsorption isotherm equation is
\(\frac{x}{m}\) = k. P =, x/m = Extent of adsorption ⇒ P = Pressure
k and n are constants which depend on the nature of the adsorbent and the gas at a particular temperature.

Applying logarithm to the above equation
log \(\frac{x}{m}\) = log k + \(\frac{1}{n}\) log P.

• A graph is plotted taking log \(\frac{x}{m}\) on y – axis and log P on x – axis. 1f the graph is a straight line then Freundlich isotherm is valid.
• The slope of the straight line gives \(\frac{1}{n}\) value.
• The intercept on the y-axis gives value of log k.
• \(\frac{1}{n}\) has values between 0 and 1
• When \(\frac{1}{n}\), = 0, \(\frac{x}{m}\) = constant, the adsorption is independent of pressure
  \(\frac{1}{n}\) = 1, \(\frac{x}{m}\) = kP i.e., \(\frac{x}{m}\) ∝ p.

Question 6.
How does the extent of adsorption depend upon
a) Increasing the surface area per unit mass of adsorbent.
b) Increasing temperature of the system.
c) Increasing pressure of the gas.
Answer:
a) The extent of adsorption increases by increasing the surface area per unit mass of adsorbent.
b) The extent of adsorption decrease with an increase in temperature.
c) The extent of adsorption increase with increasing pressure of the gas.

Question 7.
What is catalysis ? How is catalysis classified ? Give two examples for each type of catalysis. [A.P. Mar. 16] [Mar. 14]
Answer:
Catalysis : A substance which alters the rate of a chemical reaction without itself being consumed in the process, is called a catalyst.
The action of catalyst in altering the rate of a chemical reaction is called catalysis.

Types of catalysis: Catalysis is classified into two types as
a) Homogeneous catalysis and b) Heterogeneous catalysis.

a) Homogeneous catalysis : The catalytic process in which the catalyst is present in the same phase as that of reactants, is known as homogeneous catalysis.

b) Heterogeneous catalysis : The catalytic process in which the catalyst is present in a phase different from that of the reactants is known as heterogeneous catalysis.

Question 8.
Discuss the mechanism involved in adsorption of heterogeneous catalysis.
Answer:
Adsorption theory of heterogeneous catalysis :
Adsorption theory explains the mechanism of heterogeneous catalysis.

• The modern adsorption theory is the combination of the intermediate compound formation theory and the old adsorption theory.
•  The mechanism of catalysis involves five steps.
• Diffusion of the reactants to the surface of the catalyst.
• Adsorption of the reactant molecules on the surface of the catalyst.
• Occurrence of chemical reaction on the catalyst’s surface through formation of an . intermediate.
• Desorption of reaction products from the catalyst surface, and thereby making the surface available again for more reaction to occur.
• Diffusion of reaction products away from the catalyst’s surface.

Question 9.
Discuss some features of catalysis by zeolites.
Answer:
The catalytic reaction that depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape selective catalysts.

• Zeolites are shope selective catalysts because of their honey comb-like structure. They are microporous alumino silicates with 3-dimensional net work of silicates.
• Zeolites are microporous alumino silicates with dimensional network of silicates in which some Si-atoms are replaced by aluminium atoms giving Al-o-Si frame work.
• The reactions taking place in zeolites depend upon the size and shape of reactant and product molecules as well as on the pores and cavities of the zeolites.
• Zeolites are found in nature and also synthesized for catalytic selectivity.

Uses:

• Zeolites are used as catalysts in petrochemical industries for cracking.
• Zeolite ZSM – 5 is used to convert alcohol directly into gasoline (petrol) by dehydrating them to give a mixture of hydrocarbons.

Question 10.
Give brief account of mechanism of enzyme catalysis with suitable diagrams.
Answer:
Mechanism of enzyme catalysis :

The enzyme – catalysed reactions may be considered to proceed in two steps.
Step 1 : Binding of substrate to enzyme to form an activated complex (ES^≠).
E + S → E S^≠
Step 2 : Decomposition of the activated complex to form product.
E S^≠ → E + P

Question 11.
Discuss the factors that influence the catalytic activity of enzymes.
Answer:
Factors influencing the catalytic activity of enzymes :

• Highly specific nature: One catalyst catalyses only one reaction. Each enzyme is specific ‘ for a given reaction.
  Eg : The enzyme urease catalyses the hydrolysis of urea only and not any other amide.
• Highly active under optimum temperature: The rate of an enzyme reaction is maximum at a temperature called optimum temperature.
• The optimum temperature range for enzymatic activity is 298 – 310 K.
• Highly active under optimum pH : The rate of an enzyme – catalysed reaction is maximum at a particular pH called optimum pH which lies between 5 – 7.
• Increasing activity in presence of activators and co-enzymes : The enzymatic activity is increased in presence of co-enzymes (or) in presence of activators such as Na⁺, Mn⁺² etc.,
•  Influence of inhibitors and poisons: The inhibitors (or) poisons interact with the active functional groups on the enzyme surface and reduce or completely destroy the catalytic activity.

Question 12.
Name any six enzyme catalysed reactions. [A.P. Mar. 19]
Answer:
i) Inversion of Cane sugar : Enzyme : Invertase

Question 13.
What do you mean by activity and selectivity of catalysts ?
Answer:
Activity :
The ability of a catalyst in increasing the rate of reaction is defined as activity of catalyst.

• The activity of a catalyst depends upon the strength of chemisorption to a large extent.
• The reactants must get adsorbed reasonably strongly on to the catalyst to become reactive.
  Eg : The catalystic activity increases from Group – 5 to Group – 11 for hydrogenation reactions.
  The maximum activity being shown by 7 – 9 group metals.
  

Selectivity:
The selectivity of a catalyst is its-ability to direct a reaction to form specific products. The following reactions indicate the selectivity of heterogeneous catalysis.
Starting with H₂ and CO, and using different catalysts, we get different products.

The action of a catalyst is highly selective in nature. A substance which acts as a catalyst in one reaction may fail to catalyse another reaction.

Question 14.
How are colloids classified on the basis of physical states of components ?
Answer:
On the basis of physical states of dispersed phase and dispersion medium colloidal solutions are classified into eight types.

Question 15.
How are colloids classified on the basis of nature of the dispersion medium ?
Answer:
On the basis of nature of dispersion medium colloids are classified as follows.

•  If the dispersion medium is ‘air’, the sols are called aerosols.
  Eg : Smoke.
• If the dispersion medium is ‘water’ the sols are called Hydrosol (or) aquasol.
  Eg : starch solution.
• If the dispersion medium is ‘alcohol’ then the sols are called alcosols.

Question 16.
How are colloids classified on the basis of interaction between dispersed phase and dispersion medium ?
Answer:
Lyophilic colloid : The colloidal solution in which the dispersed phase has great affinity to the dispersion medium is called a Lyophilic colloid or Lyophilic solution.
Ex: Starch solution.
The starch paste when dissolved in hot water, with stirring, the starch solution is formed. The starch particles (dispersed phase) has great affinity to water molecules (dispersion medium). So starch solution is a lyophilic solution or lyophilic colloid.

Lyophobic colloid : The colloidal solution in which there exists not much affinity between the dispersed phase and dispersion medium, it is called a Lyophobic colloid or Lyophobic solution.
Ex: Gold solution.

Gold rods are placed in water containing alkali. Electric arc is applied between gold rods. The gold particles dissolves in water, to give gold solution.

Gold particles (dispersed phase) have not much affinity towards water (dispersion medium). So this is a Lyophobic solution or Lyophobic colloid.

Question 17.
What is the difference between a colloidal sol, gel, emulsion and a foam?
Answer:

Question 18.
What are lyophilic and lyophobic sols? Compare the two terms in terms of stability and reversibility.
Answer:
Lyophilic colloid : The colloidal solution in which the dispersed phase has great affinity to the dispersion medium is called a Lyophilic colloid or Lyophilic solution.
Ex : Starch solution.
The starch paste when dissolved in hot water, with stirring, the starch solution is formed. The starch particles (dispersed phase) has great affinity to water molecules (dispersion medium). So starch solution is a lyophilic solution or lyophilic colloid.

Lyophobic colloid : The colloidal solution in which there exists not much affinity between the dispersed phase and dispersion medium, it is called a Lyophobic colloid or Lyophobic solution.
Ex: Gold solution.

Gold rods are placed in water containing alkali. Electric arc is applied between gold rods. The gold particles dissolves in water, to give gold solution.

Gold particles (dispersed phase) have not much affinity towards water (dispersion medium). So this is a Lyophobic solution or Lyophobic colloid.

Question 19.
Name a substance whose molecules consist of lyophilic as well as lyophobic parts. Give its use in our daily life.
Answer:

• Associated colloids (or) micelles contains both lyophobic and lyophillic parts.
• Examples are soaps and synthetic detergents.
• The cleaning action of soap is due to the fact that soap molecule form a micelle around the oil droplet.

Question 20.
Describe Bredig’s arc method of preparation of colloids with a neat diagram.
Answer:
Electrical disintegration or Bredig’s Arc method : This process involves dispersion as well as condensation. Colloidal sols of metals such as gold, silver, platinum, etc., are prepared by this method. In this method, electric arc is struck between electrodes of the metal immersed in the dispersion medium. The intense heat produced vapourises the metal, which then condenses to form particles of colloidal size.

Question 21.
Name any four examples of preparation of colloids by chemical methods with necessary chemical equations.
Answer:
Chemical methods : Colloidal solutions are prepared by chemical reactions leading to the formation of species by double decompostion, oxidation, reduction or hydrolysis. These species then aggregate leading to the formation of sols.

Question 22.
Describe the purification of colloidal solutions by the phenomenon of dialysis with a neat diagram. [A.P. Mar. 18]
Answer:
Colloidal solutions generally contain excess amount of electrolytes and some other soluble impurities. It is necessary to reduce the concentration of soluble impurities to a requisite minimum. The impurities presence required in traces.
“The process used for reducing the amount of impurities to a requisite minimum is known as purification of colloidal solution”.

Purification of colloidal solution by Dialysis :
Dialysis : The process of removing a dissolved substance from a colloidal solution using a suitable membrane is called dialysis.

• In a true solution particles can pass through animal membrane (or) cellophane sheet (or) parchment paper but not colloidal particles.
• The apparatus used for the dialysis is called dialyser.
• A bag of suitable membrane containing the colloidal solution is suspended in a vessel containing a continuously flowing water.
• The molecules and ions diffuse through the membrane into the water and pure colloidal solution is left behind in the bag.

Question 23.
Explain the formation of micelles with a neat sketch.
Answer:
Mechanism of micelle formation: Let us take the example of soap solution. Soap is sodium or potassium salt of a higher fatty acid and may be represented as RCOC^– Na⁺ (e.g., sodium strearate CH₃ (CH₂)₁₆ COO^– Na⁺, which is a major component of many bar soaps). When dissolved in water, it dissociates into RCOO^– and Na⁺ ions. The RCOO^– ions, however, consist of two parts – a long hydrocarbon chain R (also called non-polar ‘tail’) which is hydrophobic (water repelling), and a polar group COO^–(also called polar-ionic ‘head’), which is hydrophilic (water loving).

The RCOO^– ions are, therefore, present on the surface with their COO^– groups in water and the hydrocarbon chains (R) staying away from it and remain at the surface. But at critical micelle concentration, the COO^– irons are pulled into the bulk of the solution and are aggregated to form a spherical shape with their hydrocarbon chains pointing towards the centre of the . sphere with COO^– part remaining outward on the surface of the sphere. The aggregate thus formed is known as ‘ionic micelle’. These micelles may contain as many as 100 simple molecules.

Similarly, in case of detergents, e.g., sodium laurylsulphate. CH₃ (CH₂)₁₁ SO₃^–Na⁺, the polar group is -SO₃^– along with the long hydrocarbon chain. Hence, the mechanism of micelle formation here also is same as that of soaps.

Question 24.
Action of soap is due to emulsification and micelle formation. Comment.
Answer:
Soap is sodium stearate, C₁₇H₃₅COONa, in water soap gives the ions stearate anion and sodium ion.

Cleaning action of soap : Soap anions form a micelle. The grease or dirt of the cloth are absorbed into the interior of the micelle. The tails of the anion are pegged into micelle and these micelle are washed away with the soap solution.

Soap thus, functions as an emulsifying agent for the water dirt emulsion. The emulsified grease or dirt is then washed away with soap solution.

Question 25.
Explain the phenomenon of Brownian movement giving reasons for the occurrence of this phenomena.
Answer:
Brownian movement: This is a kinetic property of colloidal solution.

When a colloidal solution is examined by ultramicroscope, the colloidal particles are seem to be moving in a rapid zig – zag motion.
This rapid motion of colloidal particles is called Brownian movement. This motion is due to unequal bombardment of colloidal particles by molecules of dispersion medium. Smaller the colloidal particles the more rapid is the Brownian movement.

Question 26.
Name the four positively charged sols. .
Answer:
The following are the positively charged sols.

• Hydroated metallic oxide sols. Eg : Al₂O₃.xH₂O, CrO₃.xH₂O etc.
• Basic dye stuffs Eg : Methylene blue sol.
• Hemoglobin (blood).
• Oxides Eg : TiO₂ sol.

Question 27.
Name the four negatively charged sols.
Answer:
The following are the negatively charged sols.

• Metal sols Eg : Ag, Au-sols.
• Metallic suphides sols Eg : ArS₃, CdS sols.
• Acid dye stuffs Eg: eosinol.
• Sols of starch, gum, clay etc.

Question 28.
Explain the terms helmholtz electrical double layer and zeta potential. What are their significancies in the colloidal solutions ?
Answer:
Helmholtz electrical double layer :
The combination of the two layers of opposite charges around the colloidal particle is called Helmholtz electrical double layer.

• In a colloidal sol the charges of opposite signs on the fixed and diffused parts of the double layer results in a difference in potential between these layers. The potential difference between the fixed layer and the diffused layer of opposite charge is called electro kinetic potential (or) zeta potential.
• The above concepts applicable in colloidal solution in which the solid particles carry one kind of charge while liquid medium carries opposite charge to that of solids.

Question 29.
Explain with a neat sketch the phenomenon of electrophoresis.
Answer:
Electrophoresis : The existence of charge on colloidal particles is confirmed by electro-phoresis experiment.
When electric potential is applied across two platinum electrodes dipping in a colloidal solution, the colloidal articles move towards one or the other electrode. The movement of colloidal particles under an applied emf is called “electrophoresis”.

Positive charged particles move towards the cathode while negative charged particles towards anode.

Question 30.
Explain the following terms, i) Electrophoresis ii) Coagulation iii) Tyndall effect.
Answer:
Electrophoresis: When electric potential is applied across two platinum electrodes dipping in a colloidal solution, the colloidal particles move towards one or the other electrode. The movement of colloidal particles under an applied emf is called “electrophoresis”.

Coagulation : The stability of the lyophilic sols is due to the presence of charge on the colloidal particles. If this charge is neutralised the particles will come nearer to each other to form aggregates (or coagulate) and settle down under the force of gravity. This process of settling down of colloidal particles is called coagulation (or) flocculation (or) precipitation.
Coagulation of lyophobic sols can be carried out by
i) Electrophoresis, ii) Boiling, iii) Adding Electrolytes, iv) Prolonged dialysis etc.

Tyndall effect : When light enters a colloidal solution, it is scattered by the large sized colloidal dispersed phase particles. Therefore when light passes through a solution we will be able to see the path of the light as a luminous beam. This is called Tyndall effect.

Question 31.
Explain the phenomenon observed
i) When a beam of light is passed through a colloidal sol.
Answer:
Tyndall effect : When light enters a colloidal solution, it is scattered by the large sized colloidal dispersed phase particles. Therefore when light passes through a solution we will be able to see the path of the light as a luminous beam. This is called Tyndall effect.

This is an optical property exhibited by colloidal solution. This phenomenon is clearly seen with a microscope placed at right angles to the path of light. Colloidal particles become self luminous due to absorption of light. A true solution does not show Tyndall effect.

ii) An electrolyte, NaCl is added to hydrated ferric oxide.
Answer:
When NaCl solution is added to the colloidal solution of hydrated ferric oxide coagulation takes place (precipitation formed).

The reason is that colloidal particles interact with ions (Na⁺, Cl^–) carrying charge opposite to that present on themselves. This causes neutralisation of charges leading to their coagulation.

iii) An electric current is passed through a colloidal solution.
Answer:
Electrophoresis : the existence of charge on colloidal particles is confirmed by electrophoresis experiment.

When electric potential is applied across two platinum electrodes dipping in a colloidal solution, the colloidal particles move towards one or the other electrode. The movement of colloidal particles under an applied emf is called “electrophoresis”.

Positive charged particles move towards the cathode while negative charged particles towards anode.

Question 32.
Describe cottrell smoke precipitator with a neat diagram.
Answer:
Aerial pollution by colloidal particles of smoke be prevented by electrical precipitation of smoke. Smoke is a colloidal solution of solid particles such as carbon, arsenic compounds, dust, etc., in air. The smoke, before it comes out from the chimney, is led through a precipitator containing plates having a charge opposite to that carried by smoke particles. The particles on coming in contact with these plates lose their charge and get precipitated. The particles thus settle down on the. floor of the chamber. The precipitator is called conttrels precipitator.

Question 33.
Among NaCl, Na₂SO₄, Na₃PO₄ electrolytes, which is more effective for coagulation of hydrated ferric oxide sol and why ?
Answer:
Among NaCl, Na₂SO₄, Na₃PO₄ electrolytes, Na₃PO₄ is more effective for coagulation of hydrated ferric oxide sol. This is explained by Hardy-Schulze rule.
Greater the valence of the coagulating ion added, the greater is its power to cause coagulation. This is known as Hardy-Schulze rule.

Coagulating ability order of the anions in the above salts is
PO₄^-3 > SO₄^-2 > Cl^–.

Question 34.
Discuss how a lyophilic colloids protect a lyophobic colloids.
Answer:
Lyophilic colloids are much stable than lyophobic colloids. A very small amount of electrolyte can precipitate the lyophobic colloids. But their precipitation can be prevented by the addition of small amounts of lyophilic colloids. This lyophilic colloids form a protecting layer around the lyophobic colloids and protect them from precipitation (coagulation). Lyophilic colloids used for. this purpose are called protective colloids.
Ex : Gelatin, gum, starch sols are generally used as protective colloids.

Question 35.
Discuss the use of colloids in
i) Purification of drinking water ii) Tanning iii) Medicines.
Answer:
i) Purification of drinking water: The water obtained from natural sources often contains suspended impurities. Alum is added to such water to coagulate the suspended impurities and make water fit for drinking purposes.

ii) Tanning: Animal skins are colloidal in nature. When a skin, which has positively charged particles, is soaked in tannin, which contains negatively charged colloidal particles, mutual coagulation takes place. This results in the hardening of skin (leather). This process is termed as tanning. Chromium salts are also used in place of tannin.

iii) Medicines : Most of the medicines are colloidal in nature. For example argyrol is a silver sol used as an eye lotion. Colloidal antimony is used in curing kalaazar. Colloidal gold is used as intramuscular injection. Milk of magnesia, an emulsion, is used for stomach disorders. Colloidal medicines are more effective because they have large surface area and are therefore easily assimilated.

Question 36.
Define Gold Number.
Answer:
The capacity of the lyophilic colloid in protecting a lyophobic colloid is measured in terms of Gold Number, introduced by “Zigmondy”.

Gold Number:” It is the number of milligrams of lyophilic colloid that is to be added to 10 ml of standard gold sol to prevent its precipitation by the addition of 1ml of a 10% sodium chloride solution”.
Lesser in the gold number the greater is the protecting capacity. Gold number of a few colloids are given below.
Starch = 25 .
Gelatin = 0.005 – 0.01
Albumin =0.1 – 0.2

Question 37.
How do emulsifiers stabilize emulsion ? Name two emulsifiers.
Answer:
Emulsifying agent: The third substance which is added in small amounts to an emulsion to stabilize the emulsion is called emulsifying agent.
—> The emulsifying agent forms an interfacial film between suspended particles and the medium.
Eg : 1) Carbon black stabilizes water in oil type emulsions.
2) Casein and silica stabilize oil in water type emulsions.

Long Answer Questions

Question 1.
Explain the terms absorption, adsorption and sorption. Describe the different types of adsorption.
Answer:
Absorption : The uniform distribution of a substance through out the bulk of the solid substance is known as absorption.
Eg : Chalk stick dipped in ink.

Adsorption : The accumulation (or) concentration of a substance on the surface, rather than in the bulk of solid (or) liquid is known as adsorption.
Eg : Adsorption of gases like O₂, H₂, Cl₂, etc., on charcoal

Sorption : In case of some substances both adsorption and absorption takes place. This phenomenon is called sorption.

Types of adsorption : On the basis of forces present between adsorbate and adsorbent molecules, adsorption classified into two types :
1) Physisorption, 2) Chemisorption.

1) Physisorption: If the forces that are responsible for the adsorption of adsorbate molecules on the surface of the adsorbent are weak vander waal’s for the adsorption is known as physisorption.
Eg : Adsorption of H₂, on O₂, on charcoal.
Characteristics :

• It is a weak adsorption.
• Enthalpy of adsorption is low (20 – 40 KJ / mole).
• It is reversible and occurs rapidly.
• It decrease with increase in temperature.
• It increase with increase in pressure.
• It is multilayered and not specific.

2) Chemisorption : If the forces that are responsible for the formation of adsorbate molecules on the source of adsorbent are chemical forces, the adsorption is called chemisorption.
Eg : Adsorption of H₂ on Ni – metal surface.
Characteristics :

• It is a strong adsorption. ,
• Enthalpy of adsorption of high (80 – 240 KJ / mole).
• It is irreversible and occurs slowly.
• It increases with increase of temperature and finally decrease.
• Pressure has no effect on chemisorption.
• It is unilayered and specific.

Question 2.
Discuss the characteristics of physical adsorption.
Answer:
Characteristics of physical adsorption :

• Lack of specificity : Physisorption is not specific. In this adsorption the given surface of adsorbent doesnot show any preference for a particular gas as the forces present are variderwaal’s forces (Universal).
• Nature of adsorbate : The amount of gas adsorbed by a solid depends on the nature of gas. Easily liquefiable gases are readily adsorbed.
• The gas with high critical temperature value is easily liquefied and gets readily adsorbed.
• Reversible Nature : Physical adsorption is reversible and occurs rapidly at low temperatures.
• Surface area of adsorbent: The extent of adsorption increase with increase in surface area of adsorbent. .
• Finely divided metals and porous substances having large surface area are good adsorbents.
• Enthalpy of adsorption : Physisorption is an exothermic process but the enthalpy of adsorption is low (20 – 40 KJ/mole). This is due to the presence of Vander waal’s forces.

Question 3.
Discuss the characteristics of chemisorption. .
Answer:

1. High specificity: Chemisorption is highly specific and it will only occur if there is some possibility of chemical bonding between adsorbate and adsorbent.
2. Irreversibility : Chemisorption involves compound formation and is irreversible in nature.
   • Chemisorption is exothermic and increases with increase of temperature.
3. Surface area : Chemisorption increase with increase of surface area of the adsorbent.
4. Enthalpy of adsorption : Enthalpy of chemisorption is high (80 – 240 KJ/mole).

Question 4.
Compare and contrast the phenomenon of physisorption and chemisorption.
Answer:
Characterstics of Physical adsorption :
Physisorption: If the forces that are responsible for the adsorption of adsorbate molecules on the surface of the adsorbent are weak vander waal’s for the adsorption is known as physisorption.
Eg : Adsorption of H₂ on O₂ on charcoal.

• Lack of specificity : Physisorption is not specific. In this adsorption the given surface of adsorbent doesnot show any preference for a particular gas as the forces present are vander waal’s forces (Universal). ‘
• Nature of adsorbate : The amount of gas adsorbed by a solid depends on the nature of gas.
• Easily liquefiable gases are readily adsorbed:
  The gas with high critical temperature value is easily liquefied apd gets readily adsorbed.
• Reversible Nature : Physical adsorption is reversible and occurs rapidly at low temperatures.
• Surface area of adsorbent: The extent of adsorption increase with increase in surface area of adsorbent.
• Finely divided metals and porous substances having large surface area are good adsorbents. .
• Enthalpy of adsorption : Physisorption is an exothermic process but the enthalpy of adsorption is low (20 – 40 KJ/mole). This is due to the presence of Vander waal’s forces.

2) Chemisorption : If the forces that are responsible for the formation of adsorbate molecules on the source of adsorbent are chemical forces, the adsorption is called chemisorption.
Eg : Adsorption of H₂ on Ni – metal surface.

1. High specificity: Chemisorption is highly specific and it will only occur if there is some possibility of chemical bonding between adsorbate and adsorbent.
2.  Irreversibility : Chemisorption involves compound formation and is irreversible in nature.
   • Chemisorption is exothermic and increases with increase of temperature.. ,
3. Surface area : Chemisorption increase with increase of surface area of the adsorbent.
4. Enthalpy of adsorption : Enthalpy of chemisorption is high (80 – 240 KJ/mole).

Question 5.
What is an adsorption isotherm ? Discuss the phenomenon of adsorption of gases on solids with the help of Freundlich adsorption isotherm.
Answer:
Adsorption Isotherm : The variation in the amount of gas adsorbed by the adsorbent with pressure at constant temperature can be expressed by means of a curve known as adsorption Isotherm.

• Freundlich adsorption isotherm equation is
  \(\frac{\mathrm{x}}{\mathrm{m}}\) = k, P^1/n
  x = mass of the gas adsorbed
  m = mass of the adsorbent
  P, k and n are constants.
• k and n depend on the nature of the adsorbent and the gas at a particular temperature.
• The relation ship is generally represented in the form of a curve where x/m is plotted against pressure.
• These curves indicate that at a fixed pressure there is a decrease in physical adsorption with rise of temperature.
  
  \(\frac{\mathrm{x}}{\mathrm{m}}\) = k.P^1/4
  Applying logarithm
  Log \(\frac{\mathrm{x}}{\mathrm{m}}\) = log k + \(\frac{\mathrm{1}}{\mathrm{n}}\) log p.
• The validity of Freundlich isotherm can be verified by plotting log \(\frac{\mathrm{x}}{\mathrm{m}}\) on y-axis and log P on x-axis. If it is a straight line then the isotherm is valid.
• The slope of the straight line give the value of \(\frac{\mathrm{1}}{\mathrm{n}}\)
• The intercept on the y-axis gives the value of log k.

Question 6.
Give a detailed account of applications of adsorption.
Answer:
Applications of adsorption : The phenomenon of adsorption funds a number of applications. Important ones are listed here.

1. Production of high vacuum: The traces of air still remaining in the vessel evacuated by a vacuum pump to give high vacuum can be adsorbed by charcoal.
2. Gas masks : Gas mask, a device which consists of activated charcoal or mixture of adsorbents is usually used by coal miners to adsorb poisonous gases during breathing.
3. Control of humidity : Silica gel and alumina gel are used as adsorbents for removing mixture and controlling humidity of air in rooms.
4. Removal of colouring matter from solutions : Animal charcoal removes colours of impure coloured solutions by adsorbing impurities responsible for the colour.
5. Separation of inert gases : Due to the difference in degree of adsorption of gases by charcoal, a mixture of noble gases can be separated by adsorption on coconut charcoal at different temperatures.
6. In curing diseases : A number of drugs kill germs by getting themselves adsorbed on germs.
7. Froth floatation process : A low grade sulphide ore is concentrated by separating silica and other earthy matter by this method using pine oil and frothing agent.
8. Adsorption indicators : Surfaces of certain precipitates such as silver halides have the property of adsorbing some dyes like eosin, fluorescein, etc. and thereby producing a characteristic colour change at the end point in argentometric titrations.
9. Chromatographic analysis : Chromatographic analysis based on the phenomenon of adsorption finds a number of applications in analytical and industrial methods.

Question 7.
What is catalysis ? How is catalysis classified ?* Give four examples for each type of catalysis.
Answer:
Catalysis: A substance which alters the rate of a chemical reaction without itself being consumed in the process, is called a catalyst.
The action of catalyst in altering the rate of a chemical reaction is called catalysis.
Types of catalysis : Catalysis is classified into two types as .
a) Homogeneous catalysis and
b) Heterogeneous catalysis.
Homogeneous catalysis : The catalytic process in which the catalyst is present in the same phase as that of reactants, is known as homogeneous catalysis.

Heterogeneous catalysis: The catalytic process in which the catalyst is present in a phase different from that of reactarts is known as heterogeneous catalysis.

Question 8.
Discuss the mechanism of heterogeneous catalysis.
Answer:
Adsorption theory of heterogeneous catalysis :
Adsorption theory explains the mechanism of heterogeneous catalysis.

• The modern adsorption theory is the combination of the intermediate compound formation theory and the old adsorption theory.
• The mechanism of catalysis involves five steps.
• Diffusion of the reactants to the surface of the catalyst.
• Adsorption of the reactant molecules on the surface of the catalyst.
• Occurrence of chemical reaction on the catalyst’s surface through formation of an intermediate.
• Desorption of reaction products from the catalyst surface, and thereby making the surface available again for more reaction to occur.
• Diffusion of reaction products away from the catalyst’s surface.

Question 9.
What are enzymes ? Explain in detail the enzyme catalysis, with necessary examples.
Answer:
Enzymes are complex nitrogenous organic compounds which are produced by living plants and animals.

• These act as specific catalysts in biological reactions.
• These catalyse the numerous reactions that occur in the bodies of animals and plants to maintain the life process.
  Mechanism of enzyme catalysis :
  
  The enzyme – catalysed reactions may be considered to proceed in two steps.
  Step 1 : Binding of substrate to enzyme to form an activated complex (ES^≠)
  E + S → ES^≠
  Step 2 : Decomposition of the activated complex to form product.
  ES^≠ → E + P

Factors influencing the catalytic activity of enzymes :

• Highly specific nature: One catalyst catalyses only one reaction. Each enzyme is specific for a given reaction.
  Eg : The enzyme urease catalyses the hydrolysis of urea only and not any other amide
• Highly active under optinum temperature: The rate of an enzyme reaction is maximum at a temperature called optimum temperature.
• The optimum temperature range for enzymatic activity is 298 – 310 K.
•  Highly active under optimum pH : The rate of an enzyme – catalysed reaction is maximum at a particular pH called optimum pH which lies between 5-7
•  Increasing activity in presence of activators and co-enzymes : The enzymatic activity is increased in presence of co-enzymes (or) in presence of activators such as Na⁺, Mn⁺² etc.,
• Influence of inhibitors and poisons: The inhibitors (or) poisons interact with the active functional groups on the enzyme surface and reduce or completely destroy the catalytic activity.
  i) Inversion of Cane sugar: Enzyme : Invertase
  

Question 10 & 11.
What are colloidal solutions ? How are they classified ? Give examples.
Answer:
A heterogeneous system in which one substance is dispered as large particles in another substance is called colloidal solution.

• In colloidal solution the particle size of the dispersed phase is of the order of 1 mμ – 1 μ where as in true solution the particle size of the solute are the order of mp or less.
• Colloidal solution is a heterogeneous binary system where as true solution is a homogeneous binary system.
  On the basis of physical states of dispersed phase and dispersion medium colloidal solutions are classified into eight types.
  

On the basis of nature of dispersion medium colloids are classified as follows.
If the dispersion medium is ‘air’, the sols are called aerosols.
Eg : Smoke.

If the dispersion medium is ‘water’ the sols are called Hydrosol (or) aquasol.
Eg: starch solution.
If the dispersion medium is ‘alcohol’ then the sols are called alcosols.

The colloidal solution in which the dispersed phase has great affinity to the dispersion medium is called a Lyophilic colloid or Lyophilic solution.
Ex : Starch solution.
The starch paste when dissolved in hot water, with stirring, the starch solution is formed. The starch particles (dispersed phase) has great affinity to water molecules (dispersion medium). So starch solution is a lyophilic solution or lyophilic colloid.

Lyophobic colloid : The colloidal solution in which there exists not much affinity between the dispersed phase and dispersion medium, it is called a Lyophobic colloid or Lyophobic solution. Ex : Gold solution.

Gold rods are placed in water containing alkali. Electric arc is applied between gold rods. The gold particles dissolves in water, to give gold solution.

Gold particles (dispersed phase) have not much affinity towards water (dispersion medium). So this is a Lyophobic solution or Lyophobic colloid.

Question 12.
How are colloids classified on the basis of the nature of interaction between a dispersed phase and a dispersion medium ? Describe an important characteristic of each class. Which of the sols need stabilising agents for preservation ?
Answer:
The colloidal solution in which the dispersed phase has great affinity to the dispersion medium is called a Lyophilic colloid or Lyophilic solution.
Ex : Starch solution.

The starch paste when dissolved in hot water, with stirring, the starch solution is formed. The starch particles (dispersed phase) has great affinity to water molecules (dispersion medium). So starch solution is a lyophilic solution or lyophilic colloid.

Lyophobic colloid : The colloidal solution in which there exists not much affinity between the dispersed phase and dispersion medium, it is called a Lyophobic colloid or Lyophobic solution.
Ex: Gold solution.

Gold rods are placed in water containing alkali. Electric arc is applied between gold rods. The gold particles dissolves in water, to give gold solution.

Gold particles (dispersed phase) have not much affinity towards water (dispersion medium). So this is a Lyophobic solution or Lyophobic colloid.

Lyophilic colloids are much stable than lyophobic colloids. Avery small amount of electrolyte can precipitate the lyophobic colloids. But their precipitation can be prevented by the addition of small amounts of lyophilic colloids. This lyophilic colloids form a protecting layer around the lyophobic colloids and protect them from precipitation (coagulation). Lyophilic colloids used for this purpose are called protective colloids.
Ex : Gelatin, gum, starch sols are generally used as protective colloids.

Question 13.
What are micelles ? Discuss the mechanism of micelle formation and cleaning action of soap.
Answer:
Micelles : Some substances which at low concentrations behave as normal strong electrolytes, but at high concentrations exhibit colloidal behaviour due to formation of aggregates. The aggregated particles thus formed are called micelles.
Eg : Stearate ions (C₁₇H₃₅COO^–) associate together in high concentration, in a solution of soap in water and forms a micelle.

Mechanism of micelle formation : Let us take the example of soap solution. Soap is sodium or potassium salt of a higher fatty acid and may be represented as RCOC^– Na⁺ (E.g., sodium strearate CH₃ (CH₂)₁₆ COO^– Na⁺, which is a major component of many bar soaps). When dissolved in water, it dissociates into RCOO^– and Na⁺ ions. The RCOO^– ions, however, consist of two parts – a long hydrocarbon chain R (also called non-polar ’tail) which is hydrophobic (water repelling), and a polar group COO^–(also called polar-ionic ‘head), which is hydrophilic (water loving).

The RCOO^– ions are, therefore, present on the surface with their COO^– groups in water and the hydrocarbon chains (R) staying away from it and remain at the surface. But at critical micelle concentration, the COO^– irons are pulled into the bulk of the solution and are aggregated to form a spherical shape with their hydro¬carbon chains pointing towards the centre of the sphere with COO- part remaining outward on the surface of the sphere. The aggregate thus formed is known as ‘ionic micelle’. These micelles may contain as many as 100 simple molecules.

Similarly, in case of detergents, E.g., sodium laurylsulphate. CH₃ (CH₃)₁₁ SO₃^–Na⁺, the polar group is -SO₃^–along with the long hydrocarbon chain. Hence, the mechanism of micelle formation here also is same as that of soaps.
Soap is sodium stearate, C₁₇H₃₅COONa, in water soap gives the ions stearate anion and sodium ion.
C₁₇H₃₅COONa → C₁₇H₃₅COO^– + Na⁺ (Stearate ion)
The anion of soap = C₁₇H₃₅COO^–

Cleaning action of soap : Soap anions form a micelle. The grease or dirt of the cloth are absorbed into the interior of the micelle. The tails of the anion are pegged into micelle and these micelle are washed away with the soap solution.

Soap thus, functions as an emulsifying agent for the water dirt emulsion. The emulsified grease or dirt is then washed away with soap solution.

Question 14.
Describe the properties of colloids with necessary diagrams wherever necessary.
Answ:
i) Tyndall effect : When light enters a colloidal solution, it is scattered by the large sized colloidal dispersed phase particles.. Therefore when light passes through a solution we will be able to see the path of the light as a luminous beam. This is called Tyndall effect.

This is an optical property exhibited by colloidal solution. This phenomenon is clearly seen with a microscope placed at right angles to the path of light. Colloidal particles become self luminous due to absorption of light.
A true solution does not show Tyndall effect.

ii) Brownion movement: This is a kinetic property of colloidal solution.

When a colloidal solution is examined by ultramicroscope, the colloidal particles are seem to be moving in a rapid zig-zag motion.
This rapid motion of colloidal particles is called Brownian movement.

This motion is due to unequal bombardment of colloidal particles by molecules of dispersion medium. Smaller the colloidal particles the more rapid is the Brownian movement.

iii) Charge on colloidal particles : The charge on the colloidal particles is due to electron capture by sol particles during electro dispersion of metals and due to preferrential adsorption ions from solution (of) due to formulation of electrical double layer.

When a dilute AgNO₃ solution is added to a dilute KI solution taken in excess the precipitated AgI adsorbs I^– ions present in excess and a negatively charged AgI colloidal solution is formed. However when dilute KI solution is added to dilute AgNO₃ solution taken in excess, AgI precipitate adsorbs Ag⁺ ions present in excess and a positively charged Agl colloidal solution is formed. Generally the ion common to dispersed particle is adsorbed.
AgI/I^–
Negatively charged
AgI/Ag⁺
Positively charged.

iv) Zeta Potential: In a colloidal sol the charges of opposite signs on the fixed’and diffused parts of the double layer results in a difference in potential between these layers. The potential difference between the fixed layer and the diffused layer of opposite charge is called electro kinetic potential (or) zeta potential.

v) electrophoresis : When electric potential is applied across two platinum electrodes dipping in a colloidal solution, the colloidal particles move towards one or the other electrode. The movement of colloidal particles under an applied emf is called “electrophoresis”.
If the movement of colloidal particles is arrested by some suitable means, the dispersion medium moves in opposite direction. This phenomenon is termed “electro osmosis”.

vi) Coagulation : The stability of the lyophilic sols is due to the presence of charge on the colloidal particles. If tins charge is neutralised the particles will come nearer to each other to form aggregates (or coagulate) and settle down under the force of gravity. This process of settling down of colloidal particles is called coagulation (or) flocculation (or) precipitation.

Question 15.
What are emulsions ? How are they classified ? Describe the applications of emulsion. [T.S. Mar. 18, 16]
Answer:
Emulsion: The colloidal system in which a dispersion of finely divided droplets of a liquid in another liquid medium is called Emulsion.
Ex: Milk. .
In Milk, the droplets of liquid fat are dispersed in water. This is an example for oil in water type emulsion.
Classification of emulsions : Emulsions are classified, into two classes. These are
a) oil in water (O/W) and ‘
b) water in oil (W/O), (O = Oil; W = Water).
These emulsions are classified as such depending on which is dispersed phase and which is dispersion medium.

a) Oil in Water (O/W) type emulsions : In this type of emulsions, the dispersed phase is oil and the dispersion medium is water.
Ex : Milk, liquid, fat (oil) in water.
Vanishing cream; fat in water.

b) Water in Oil (W/O) type emulsions : In this type of emulsions, the dispersed phase is water and the dispersion medium is oil.
Ex : Stiff greases : water in lubrication oils
Cod liver oil : water in cod liver oil
Cold cream : water in fat.

Applications of Emulsions : Emulsions are useful

• In the digestion of fats in intestines.
•  In washing processes of clothes and crockery.
•  In the preparation of lotions, creams, ointments in pharmaceutical and cosmetics.
• In the extraction of metals (froth flotation).
• In the conversion of cream into butter by churning.
• To break oil and water emulsions in oil wells.
• In the preparation of oily type of drugs for easy adsorption to the body.

Intext Questions

Question 1.
Write any two characteristics of Chemisorption. .
Solution:

1. High specificity: Chemisorption; is highly specific and it will only occur when adsorbent and adsorbate molecules can chemically react with each other.
   Eg.: oxygen is adsorbed on metals by oxide formation.
2. Surface area : Chemisorption increases with increase in surface area of the adsorbent.

Question 2.
Why does physisorption decrease with the increase of temperature ?
Solution:
Physisorption is an exothermic process :
Solid (Adsorbent) + Gas (Adsorbate) ⇌ Gas adsorbed on solid + Heat
When temperature is increased, the equilibrium shifts towards the backward direction to neutralise the effect of the change (Le-Chatelier’s principle). So, physisorption decreases with increase in temperature.

Question 3.
Why are finely powdered substances more effective adsorbents than their non powdered crystal forms ? .
Solution:
Finally divided substances provide increased surface area for adsorption which is not available to such extent in their crystalline forms. That’s why the powdered forms are more effective than crystalline forms for the purpose of adsorbents.

Question 4.
Hydrogen used in Haber’s process 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 formed in steam reforming when ammonia is obtained by Haber’s process ?
Solution:
Carbon monoxide (CO) 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 ester hydrolysis slow in the beginning but is fast after sometime ?
Solution:
The chemical equation for ester hydrolysis is as follows :

Carboxylic acid produced during hydrolysis releases H+ ions in the solution which act as catalyst (auto-catalysis) for the reaction. Therefore, ester hydrolysis is slow in the beginning but becomes faster after sometime.

Question 6.
What is role of desorption in the process of adsorption catalysis.
Solution:
The reaction products formed on the catalyst surface get detached from the surface as a result of desorption, thereby making the surface available again for more reactant particles.

Question 7.
What modification can you suggest in the Hardy-Schulze law ?
Solution:
According to Hardy Schulze law, the ions carrying charge opposite to the charge on sol particles neutralise their charge and thus cause their coagulation or precipitation. But actually, the sol carrying these ions also get coagulated since the ions neutralize their charge. So, Hardy Schulze law can be modified as :

“When equimolar proportions of two oppositely charged sols are mixed, they mutually neutralize their charge and both get coagulated”.

Question 8.
Why is it essential to wash the precipitate in gravimetric chemical analysis with wash liquid before drying and weighing it quantitatively ?
Solution:
Some of the reactant ions my be adsorbed or may adhere to the surface of the particles of the precipitate formed during an ionic reaction. In order to remove these reactant ions, the precipitate should.be washed with water. If this is not done, an error may be produced during quantitative analysis.

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 1(a) Digestion and Absorption Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 1(a) Digestion and Absorption

Very Short Answer Questions

Question 1.
Give the dental formula of an adult human being.
Answer:
The arrangement of different types of teeth in each half of both the jaws in the order I, C, PM, M is represented by the dental formula.
In adult it is = \(\frac{2123}{2123}\) = 32.

Question 2.
Bile juice contains no digestive enzymes, yet it is important for digestion. How?
Answer:
Bile juice doesn’t contain enzymes, but it contains bile salts such as sodium/potassium glycocholates and taurocholates, which help in the digestion and absorption of lipids. Bile salts emulsify fats and also render them water-soluble. Bile salts activate the lipases. These lipases act on emulsified fats and convert them into fatty acids and glycerols.

Question 3.
Describe the role of chymotrypsin. Name two other digestive enzymes of the same category secreted by the same gland.
Answer:
Chymotrypsin plays an important role in digestion of proteins, proteoses and peptones and convert them into tripeptides and dipeptides. Chymotrypsin, trypsin and carboxy peptidase are ehdopeptidases, produced by the pancreas arid involved in the digestion of proteins.

Question 4.
What would happen if, HCl were not secreted in the stomach?
Answer:
HCl is secreted by the glands present on the stomach walls. It provides acidic pH which is optimal for the action of pepsin. HCl activates the pepsinogen into pepsin. Pepsin plays an important role in digestion of proteins. Therefore, if HCl were not secreted in the stomach, then pepsin would not be activated. This would affect on protein digestion.

Question 5.
Explain the terms thecodont and diphyodont dentitions.
Answer:
Thecodont:
Teeth of human beings are embedded in the sockets of the jaw bones is called thecodont.

Diphyodont:
Majority of mammals including human beings form two sets of teeth during their life time, a set of temporary / milk teeth replaced by a set of permanent teeth. This type of dentition is called diphyodont dentition.

Question 6.
What is Autocatalysis? Give two examples.
Answer:
Autocatalysis is the catalysis of a reaction in which the catalyst is one of the product of the reaction (or) catalysis caused by a catalytic agent formed during a chemical reaction.

Question 7.
What is chyme?
Answer:
Semi fluid mass of partly digested acidic food formed in the stomach is called chyme.

Question 8.
Name the different types of Salivary glands of man, and their locations in the human body?
Answer:
There are three pairs of Salivary glands in man.
1. Parotid glands — present below the pinna / inner surface of the cheeks.
2. Sub maxillary (or) Sub mandibular glands —.located at the angles of lower jaw.
3. Sublingual glands — present below the tongue.

Question 9.
Name different types of papillae present on the tongue of man.
Answer:
The upper surface of the tongue has small projections called papillae. In humans the tongue bears 3 (three) types of papillae namely 1) fungi form 2) filiform 3) Circumvallate papillae.

Question 10.
What is the hardest substance in the human body? What is its origin?
Answer:
Enamel of tooth is the hardest substance in the human body, which is secreted by ameloblasts of ectodermal origin.

Question 11.
Name the structure of gut which is vestigial in human beings, but well developed in herbivores. And mention the type of tissue with which it is mostly formed.
Answer:
Appendix is vestigial part in human beings. It is a narrow finger like tubular projection, arises from the caecum. In herbivores it is a functional part and useful in the digestion of cellulose materials. The appendix contain a high concentration of lymphoid tissues. These are highly specialized structures which are a part of the immune system.

Question 12.
Distinguish between deglutition and mastication.
Answer:
Deglutition :
Deglutition is the swallowing of food and involves a complex and coordinated process. It is divided into three phases.

Phase one :
The collection and swallowing of masticated food.

Phase two :
Passage of food through the pharynx into the beginning of the esophagus.

Phase three :
The passage of food into the stomach.

Mastication :
The mastication process includes the biting and tearing of food into manageable pieces. This usually involves using the incisors and canines teeth. The grinding of food is usually performed by the molars and premolars. During the mastication process, food is moistened and mixed with saliva.

Question 13.
Distinguish between diarrhoea and constipation.
Answer:
Diarrhoea :
The abnormal frequency of bowel movement and increased liquidity of the faecal discharge is known as diarrhoea. It reduces the absorption of fqod arid results in loss of water.

Constipation :
A condition ip which the faeces are retained within the rectum as it is hard due to low content of water and the movement of bowel occurs irregularly.

Question 14.
Name two hormones secreted by the duodenal mucosa.
Answer:
The epithelium of duodenum secretes the hormones namely secretin and cholecystokinin (cck).

Question 15.
Distinguish between absorption and assimilation.
Answer:
Absorption :
Absorption is the process by which the end products of digestion pass through the intestinal mucosa into blood (or) lymph. It is carried out by passive, active (or) facilitated transport mechanisms.

Assimilation:
The absorbed substances finally reach the tissues, where food materials become integral components of the living protoplasm and used for the production of energy, growth and repair. This process is called assimilation.

Short Answer Questions

Question 1.
Draw a neat labelled diagram of L.S of a tooth. Ans.
Answer:

Question 2.
Describe the process of digestion of proteins in the stomach.
Answer:
Protein digestion begins in the stomach. The food entered into the stomach is mixed thoroughly with the gastric juice of the stomach by the churning movements of its muscular wall and the product is called chyme. The main components of gastric juice are protein digestive enzymes, hydrochloric acid and mucus.

HCl provides the acidic pH (1.8) which is optimal for the action of pepsin. The proenzymes of gastric juice, the pepsinogen and prorennin, on exposure to hydrochloric acid are convened into the active enzymes, pepsin and rennin respectively. Pepsin converts proteins into proteoses and peptones. Rennin is a proteolytic enzyme found in the gastric juice of infants. It acts on the milk protein, the casein in the presence of calcium ions and converts it into calcium paracaseinate and proteoses. Pepsin acts on calcium paracaseinate and converts it into peptones. The entire process of protein digestion in the stomach takes about 4 hours.

Question 3.
Explain the role ofpancreatic juice in the digestion of proteins.
Answer:
Pancreatic juice is secreted by the pancreas and it plays an important role in protein ‘digestion. Pancreatic juice contains protein hydrolysing enzymes like trypsinogen, chymotrypsinogen and pro carboxy peptidases, but they are inactive enzymes.

Trypsinogen is activated by the enzyme enterokinase, secreted by the intestinal mucosa into active trypsin, which in turn activates the other enzymes in the pancreatic juice. Trypsin itself can similarly activate trypsinogen into trypsin.

Chymotrypsin, Trypsin and Carboxy Peptidase of pancreatic juice act upon proteins, proteoses and peptones in the chyme, result in the formation of tri and dipeptides. Which in turn hydrolysed into aminoacids by the action Of tri aruj di peptidases.

Question 4.
How are polysaccharides and disaccharides digested?
Answer:
Dietary carbohydrates principally consist of polysaccharides :
Starch and glycogen. It contains disaccharides and small amounts of monosaccharides.

Digestion in mouth:
Digestion of carbohydrates starts at the mouth, where they come in contact with saliva during mastication. Saliva contains carbohydrate-splitting enzyme called Salivay amylase (ptyalin). This enzyme hydrolyses the starch into disaccharides (maltose).

Digestion in stomach:
Ptyalin action stops in stomach when pH falls to 3.0. No carbohydrate splitting enzymes are available in gastric juice. Some dietary sucrose may be hydrolysed by HCl.

Digestion in small intestine :
Chyme reaches the duodenum from stomach where it meets pancreatic juice. Carbohydrates in the chyme are hydrolysed by the pancreatic amylase into disaccharides. Intestinal disaccharidases act on the disaccharides and convert them into monosaccharides.

Question 5.
If you take butter in your food how does it get digested and absorbed in the body? Explain.
Answer:
Butter contains fat. Fats remain mostly undigested in stomach.

Digestion of fat in the small intestine :
The major site of fat digestion is the small intestine. This is due to the presence of a powerful lipase/(steapsin) in the pancreatic juice and bile juice. Bile juice contains bile salts such as Sodium/Potassium glycocholates and taurocholates, which helps in the emulsification of fat i.e., break down of fats into very small micelles. Bile also activates lipases of pancreatic juice (steapsin) and intestinal lipases. These lipases act on emulsified fats and convert them into fatty acids and glycerols.

Absorption :
Fatty acids and glycerol being insoluble in water cannot be absorbed into the blood directly. They are first modified into small droplets called micelles, which move into intestinal mucosal cells. They are reformed into very small protein coated fat globules called chylomicrons, which are transported into the lymph capillaries in the villi by exocytosis. Then they are ultimately released into blood stream through left subclavian vein via the thoracic duct. These chylomicrons are broken down to fatty acids and glycerol by the action of an enzyme lipoprotein lipase and they diffuse into the adipocytes of the adipose tissue and liver for storage.

Question 6.
What are the functions of liver?
Answer:
Liver performs a variety of functions such as synthesis, storage and secretion of various substances.

1. Liver secretes bile juice, it contains bile salts such as sodium / potassium glycocholates and taurocholates, which helps in digestion and absorption of lipids.
2. Liver plays the key role in carbohydrate metabolism.
   a) Glycogenesis : formation of glycogen from glucose.
   b) Glycogenolysis : breakdown of glycogen into glucose.
   c) Gluconeogenesis : Synthesis of glucose from certain amino acids, lactate (or) glycerol.
3. Liver also plays an important role in synthesis of cholesterol and production of triglycerides.
4. Deamination of proteins occurs in the liver.
5. Liver is the chief organ of detoxification of toxic substances that enter the gut along with food.
6. Liver acts as thermoregulatory organ.
7. Liver acts as a haemopoietic organ in the foetus and erythroclastic organ in the adult.
8. The liver synthesizes the plasma proteins such as albumin, globulins, blood clotting factors such as fibrinogen / prothrombin, etc., and the anticoagulant called heparin.
9. The lactic acid formed during anaerobic muscle contraction is converted into glycogen (gluconeogenesis) in the liver by Cori cycle.
10. Kupffer cells are the largest phagocytic cells which remove unwanted substances and microbes that attack the liver by phagocytosis.

Long Answer Questions

Question 1.
Describe the physiology of digestion of various types of food in the human digestive system
Answer:
Digestion is the process of convertion of complex non-diffusible food substances into simple diffusible forms. The process of digestion is accomplished by mechanical and biochemical process.

I. Digestion in the buccal cavity :
Buccal cavity performs two major functions, mastication of food and facilitation of swallowing. Teeth and tongue with the help of saliva masticate and mix up the food thoroughly. Mucus in saliva helps in lubricating and adhering the masticated food particles into bolus. The saliva secreted into oral cavity contains electrolytes such as Na⁺, K⁺, Cl^–, HCO₃^– arid enzymes like salivary amylase (ptyalin) and lysozyme. Carbohydrates digestion starts in the buccal cavity, about 30% of starch is hydrolysed here into a disaccharide called maltose by the enzyme amylase (ptyaline). Lysozyme acts as antibacterial agent that prevents infections.

II. Digestion in the stomach :
As the bolus enters into stomach starch digestion stops and protein digestion begins. The food entered into stomach is mixed thoroughly with gastric juice of the stomach by the churning movement of its muscular wall and the product is called chyme. The mucus and bicarbonates present in the gastric juice act as lubricant and protect the mucosal epithelium from HCl. HCl in the stomach provides the acidic pH (1.8) which is optimal for the action of pepsin.

The proenzymes of gastric juice, the pepsinogen and prorennin on exposure to HCl are converted into the active enzymes, pepsin and rennin respectively. Pepsin converts proteins into proteoses and peptones. Rennin found in gastric juice of infants. It acts on the milk protein, the casein in the presence of calcium ions convert into calcium paracaseinate and proteoses. Pepsin acts on paracaseinate and convert it into peptones. The entire process of protein digestion in stomach takes about 4 hours.

III. Digestion in the small intestine :
Various types of movements are generated by the muscular external layer of small intestine. These movements help in thorough mixing of the food with bile, pancreatic juice and intestinal juice in the intestine and thereby facilitate digestion. The duodenal cells of the proximal part produces large amount of bicarbonates to completely neutralize any gastric acid that passes further down into the digestive tract.

i) Digestion of proteins :
Pancreatic juice contains protein hydrolysing enzymes like trypsinogen, chymotrypsin and procarboxy peptidases, but they are inactive enzymes.

Trypsinogen is activated by the enzyme enterokinase secreted by the intestinal mucosa into active trypsin which intum activate the other enzymes in the pancreatic juice. Trypsin itself can similarly activate trypsinogen into trypsin.

Chymotrypsin, trypsin and carboxy peptidase of pancreatic juice act upon proteins, proteoses and peptones in the chyme, result in the formation of tri and dipeptides which inturn hydrolysed into amino acids by the action of tri and dipeptidases.

ii) Digestion of fats:
Bile salts of bile help in the emulsification of fat i.e., breakdown of fats into very small micelles. Bile also activates lipases of pancreatic juice (steapsin) and intestinal lipases. These lipases act on emulsified fats and convert them into fatty acids and glycerols.

iii) Digestion of Carbohydrates :
Carbohydrates in the chyme are hydrolysed by the pancreatic amylase into disaccharides. Intestinal disaccharidases act on the disaccharides and convert them into monosacharides.

iv) Digestion of nucleic acids :
Nucleases of the pancreatic juice act on the nucleic acids to form nucleotides and nucleosides. Nucleotidases and nucleosidases of the intestinal juice convert the nucleotides and nucleosides into pentose sugar and nitrogen bases.

The end products of digestion pass through the intestinal mucosa into blood (or) lymph is carriedout by passive, active (or) facilitated transport mechanisms.

Question 2.
Explain the digestive system of man with neat labelled diagram.
Answer:

The digestive system is a group of organs and tissues involve in the breaking down of ingested food in the alimentary canal into a form that can be absorbed ai a assimilated by the tissues of the body.

Human digestive system consists of the alimentary canal and the associated glands. Alimentary canal / Digestive tract:

The alimentary canal of man begins with the anterior opening, the mouth and ends with the posterior opening, the anus.

Parts of the alimentary canal / digestive tract:

1. Mouth and Buccal (oral) cavity
2. Pharynx
3. Oesophagus
4. Stomach
5. Small intestine
6. Large intestine

1. Mouth and Buccal (oral) cavity:
Mouth is the first part of the alimentary canal. It is formed by the cheek on either side and boardered by the movable upper and lower lips, leads into the buccal (or) oral cavity. The palate separate the ventral buccal cavity from the dorsal nasal chamber and facilitates chewing and breathing simultaneously. The jaw bones bear teeth and tongue occurs at the base of the buccal cavity.

i) Teeth :
These are ecto-mesodermal in origin. An adult human has 32 permanent teeth, which are of four different types namely, incisors (I), canines (C), premolars (PM), and molars (M). These are useful in cutting, tearing and grinding of food. The arrangement’ of teeth is represented by dental formula. In adult human
it is = \(\frac{2123}{2123}\) = 32

ii) Tongue :
It is a freeely movable muscular sense organ, attached to the floor of the oral cavity by a fold of tissue called frenulum. The upper surface of the tongue has small projections called papillae, some of which bear taste buds. The tongue acts as universal toothbrush and helps in mixing saliva with food, taste detection, deglutition and speaking.

2. Pharynx:
It is a muscular tube connecting the oral cavity and oesophagus and trachea. It is a common passage for food and air. It is divided into nasopharynx, oropharynx and laryngo pharynx. Oesophagus and trachea open into the laryngopharynx. The trachea open into the laryngopharynx through the glottis. A cartilaginous flap called epiglottis prevents the entry of food into glottis during swallowing.

3. Oesophagus:
It is a thin long muscular tube (9 to 12 inches). The semisolid digested food from pharynx enters the oesophagus. Oesophagus is separated by the Cardiac sphincter from stomach. When the food reaches lower end of Oesophagus the cardiac sphincter opens allowing the food to enter the stomach.

4. Stomach :
It is a wide ‘J’ shaped muscular sac, located iii the upper left portion of the abdominal cavity just below the diaphragm. It has three major parts, an anterior cardiac portion into which the oesophagus opens, a middle large fundic region and a posterior pyloric portion which opens into the first part of the small intestine through the pyloric aperture which is guarded by the pyloric sphincter.

5. Small intestine :
The small intestine is the longest part of alimentary canal. It has three regions namely proximal duodenum middle long coiled jejunum and distal highly coiled ileum. Duodenum receives the hepato-pancreatic duct. Ileum opens into the large intestine.

6. Large intestine :
It consist of caecum, colon and rectum. Caecum is a small blind sac. A narrow finger like vestigial tubular organ arises from caecum called appendix. The caecum opens into colon which is‘divided into an ascending, a transverse, a descending parts and a sigmoid colon that continues behind into rectum. Rectum is a small dilated sac which leads into anal canal that opens out through the anus.

Digestive glands :
1. Salivary glands : There are three pairs of glands in man.
i) Parotid glands
ii) Sub-maxillary glands
iii) Sub-lingual glands
They secrete saliva, which mainly contains salivary amylase and lysozyme.

2. Gastric glands :
These are located in the wall of the stomach beneath the surface
epithelium, Gastric glands are of three types namely
i) Cardiac glands – secrete mucus
ii) Pyloric glands – secrete mucus and hormone gastrin
iii) Fundic / Oxyntic glands – secrete mucus, proenzymes like pepsinogen and prorennin, HCl, intrensic factor and some amount of gastric lipase.

3. Intestinal glands :
They are of two types
i) Brunner’s glands
ii) Crypts of lieberkuhn
which secrete intestinal juice contains peptidases, disaccharidases, enterokinase and lysozyme.

4. Liver :
Liver is the largest gland in human. Liver secretes bile juice, contains bile salts, which play a very important role in lipid digestion.

5. Pancreas :
The pancreas is the second largest gland in human. Exocrine part of pancreas secretes pancreas juice contains sodium bicarbonates, trypsinogen, chymotrypsinogen, carboxy peptidase, steapsin, -pancreatic amylase and nucleases such as DNAase and RNAase.

Students get through AP Inter 2nd Year Chemistry Important Questions Lesson 6(c) Group-17 Elements which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions Lesson 6(c) Group-17 Elements

Very Short Answer Questions

Question 1.
Interhalogen compounds are more reactive than the constituent halogens except fluorine – Explain.
Answer:
Inter halogen compounds are more reactive than halogens. This is because X – X’ bond in interhalogens is weaker than X — X bond in halogens except F – F bond.

Question 2.
What is the use of ClF₃?
Answer:
ClF₃ is very useful fluorinating agent and it is used for the production of VF₆ in the enrichment of U²³⁵.
U_(s) + 3ClF_3(l) → UF_6(g) + 3ClF_(g)

Question 3.
Why are halogens coloured?
Anšwer:
Halogens are coloured due to the absorption of radiations in visible region which results in the excitation of outer electrons to higher energy level. Halogens absorb different quanta of radiation and display different colours.

Question 4.
Write the reactions of F₂ and Cl₂ with water. (TS Mar. 17) (IPE Mar. ‘14)
Answer:
Fluorine produces O₂ and O₃ on passing through water.
3F₂ + 3H₂O → 6HF + O₃
2F₂ + 2H₂O → 4HF + O₂

Question 5.
Electron gain enthalpy of fluorine is less than that of chlorine – explain.
Answer:
The negative electron gain enthalpy of the fluorine is less than that of chlorine. This is due to small size of fluorine atom which results in strong inter electronic repulsions in the relatively small 2p orbitals of fluorine and thus the incoming electron does not experience much attraction.

Question 6.
Bond dissociation enthalpy of F₂ is less than that of Cl₂ – Explain.
Answer:
Bond dissociation enthalpy of F₂ is less than that of Cl₂
Explanation:
In F₂ molecule electron repulsions are greater among lone pairs because these lone pairs are much closer to each other than in case of Cl₂.

Question 7.
Write the formulae of the compounds, in which oxygen has positive oxidation states and mention the oxidation states of oxygen in them.
Answer:
In OF₂ and O₂F₂ oxygen has positive oxidation states.

• In OF₂, oxygen oxidation state is + 2.
• In O₂F₂ oxygen oxidation state is + 1.

Question 8.
What is the use of O₂F₂ and I₂O₅?
Answer:
Uses of O₂F₂

• O₂F₂ is a fluorinating agent. O₂F₂ oxidises plutonium to PUF₆ and the reaction is used in removing plutonium as PUF₆ from spent nuclear fuel.

Use of I₂O₅:

• I₂O₅ is a good oxidising agent and is used in the estimation of carbon monoxide (CO).

Question 9.
Explain the reactions of Cl₂ with NaOH. (IPE – 2015 (AP), 2016 (TS), (AP))
Answer:
i) Reaction with cold dilute NaOH: Chlorine reacts with cold dilute NaOH to give sodium hypochlorite and sodium chloride.

ii) Reaction with hot concentrated NaOH : Chlorine reacts with hot concentrated NaOH to give sodium chlorate and sodium chloride.

Question 10.
What happens when Cl₂ reacts with dry slaked lime ? (AP Mar. 17: IPE 16, 15 (AP))
Answer:
Chlorine reacts with dry slaked lime and forms bleaching powder.
Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O.

Question 11.
What is aqua regia ? Write its reaction with gold and platinum.
Answer:
A mixture of 3 parts of conc. HCl and one part of Conc. HNO₃ constitutes aqùa regia. It is used for dissolving noble metals.
It’s reaction with gold:
Au + 4H⁺ + \(\mathrm{NO}_3^{-}\) + 4Cl^– → \(\mathrm{AuCl}_4^{-}\) + NO + 2 H₂O
It’s reaction with Platinum:
3Pt + 16H⁺ + \(4 \mathrm{NO}_3^{-}\) + 18Cl^– → 3PtC\(l_6^{-2}\) + 4NO + 8 H₂O

Question 12.
How is chlorine manufactured by Deacon’s method? (AP Mar.’ 17: IPE ’16 (TS))
Answer:
Deacon’s process: [n Deacon’s process. Chlorine is obtained by the oxidation of hydrogen chloride gas by atmospheric oxygen in the presence of CuCl₂ (catalyst) at 723 K.

Question 13.
Why is dry chlorine cannot act as a bleaching agent.
Answer:
Dry chlorine cannot produce nascent oxygen. Hence it cannot act as a bleaching agent.

Question 14.
HF is a liquid while HCl is a gas — explain.
Answer:
HF is a liquid due to the presence of inter molecular hydrogen bonding where as HCl is a gas as there is no such type of bonding in it.

Question 15.
Write two uses of hydrogen chloride.
Answer:
Uses of hydrogen chloride:

• It is used in medicines and as a laboratory reagent.
• It is used in the manufacturing of Cl₂, NH₄Cl and glucose (from corn starch).
• It is used in extracting glue from bones and purifying bone black.

Question 16.
Chlorine acts as an oxidizing agent – explain with two examples.
Answer:
Chlorine acts as oxidising agent.
Example – 1: Cl₂ oxidises Iodine to Iodate.
I₂ + 6 H₂O + 5 Cl₂ → 2HIO₃ + 10 HCl

Example – 2: Cl₂ oxidises Sodium Sulphite to Sodium Sulphate.
Cl₂ + Na₂SO₃ + H₂O → Na₂SO₄ + 2 HCl

Question 17.
Write the reaction of chlorine with hypo (Na₂S₂O₃). (IPE Mar.2015 (AP, TS), 2016 (TS), (AP))
Answer:
Reaction of chlorine with hypo (Na₂S₂O₃). Na₂S₂O₃ + Cl₂ + H₂O → Na₂SO₄ + 2HCl + S. In this reaction hypo acts an “antichlor”.

Question 18.
Give the bond dissociation order of halogens.
Answer:
Bond dissociation order of halogens is, Cl₂ > Br₂ > F₂ > I₂.

Question 19.
I₂ is more soluble in KI give reason.
Answer:
I₂ forms soluble KI₃ with KI solution. I₂ + KI → KI₃.

Question 20.
Chlorine acts as a bleaching agent only in the presence of moisture — explain.
Answer:
Moist chlorine is a powerful bleaching agent. This bleaching property is due to oxidation.
Cl₂ + H₂O → 2HCl + (O)
Ex: Coloured substance + (0) → colourless substance.

Question 21.
The decreasing order of acidic character among hypohalogen acids is HClO > HBrO > HIO. Give reason.
Answer:
Given the decreasing order of acidic character among hypohalogen acids is
HClO > HBrO > HIO

From the above mentioned Ka values the given order of hypohalogen acids – acid strength order is
HClO > HBrO > HIO

Question 22.
fluorine exhibits only-1 oxidation state whereas other halogens exhibit +1, +3, +5 and +7 oxidation states also. Explain.
Solution:
Fluorine is the most electronegative element and cannot exhibit any positive oxidation state. Other halogens have d- orbitais and therefore, can expand their octets and show +1, +3, +5 and +7 oxidation states also.

Question 23.
What are the inter halogen compounds ? Give two examples.
Answer:
The compounds formed between different halogens are called inter halogen compounds.
Ex: ClF₃, BrF₃, IF₇, ICl₃.

Short Answer Questions

Question 1.
Explain the structures of
a) BrF₅ and
b) IF₇.
Answer:
a) Structure of BrF₅.

• Central atom in BrF₅ is ‘Br’
• ‘Br’ undergoes sp³d² hybridisation in 2^nd excited state.
  
• Shape of the molecule is octahedral with one position occupied by a lone pair (or) square pyramidal.
  

b) Structure of IF₇:

• Central atom in IF₇ is 1’.
• ‘I’ undergoes sp³d³ hybridisation iñ 3^rd excited state.
  
• Shape of the molecule is Pentagonal bipyramid structure.
  

Question 2.
How is chlorine obtained in the laboratory? How does it react with the following?
a) cold dil. NaOH
b) excess NH₃
c) KI (IPE Mar & May – 2015 AP TS), BMP)
Answer:
In the laboratory chlorine is prepared by the oxidation of HCl with MnO₂.
4 HCl + MnO₂ → MnCl + 2 H₂O + Cl₂↑
a) Chlorine reacts with cold dil. NaOH to form sodium hypochlorite.
Cl₂ + 2 NaOH → NaCl + NaOCl + H₂O
b) Cl₂ reacts with excess of NH₃, Nitrogen and ammonium chloride are formed.
8 NH₃ + 3 Cl₂ → 6 NH₄Cl + N₂↑
c) Cl₂ reacts with KI to liberate iodine.
Cl₂ + 2KI → 2KCl + I₂↑

Question 3.
How is ClF₃ prepared? How does it react with water? Explain its structure.
Answer:
Preparation of ClF₃: Chlorine reacts with excess of fluorine to form ClF₃.

Reaction with H₂O : ClF₃ reacts with water explosively and oxidises water to give oxygen or in controlled quantitieš oxygen diflouride (OF₂) as well as HF and HCl.
ClF₃ + 2 H₂O → 3 HF + HCl + O₂
ClF₃ + H₂O → HF + HCl + OF₂

Structure of ClF₃:

• Central atom in ClF₃ is ‘Cl’
• Excited state electronic configuration of ‘Cl’ is
  
• Cl’ atom undergoes sp³d hybridisation.
• It is a bent T-shaped molecule (or) trigonal bipyramidal with 2 – positions occupied by lone pairs.
  

Question 4.
How can you prepare Cl₂ from HCl and HCl from Cl₂ ? Write the reactions.
Answer:
Preparation of Cl₂ from HCl:

• On heating MnO₂ with conc. HCl, Cl₂ gas is liberated.
  MnO₂ + 4 HCl → MnCl₂ + Cl₂ + 2 H₂O
• By the oxidation of HCl gas by atmospheric oxygen in presence of CuCl₂ catalyst at 723K.
  
  Preparation of HCl from Cl₂:
• Cl₂ when reacted with H₂ to form HCl,
  H_2(g) + Cl_2(g) → 2HCl_(g)

Question 5.
How is chlorine prepared by electrolytic method? Explain Its reaction with
a) NaOH and
b) NH₃ under different conditions. (IPE Mar & May – 2015 (AP, TS))
Answer:
Preparation of chlorine by electrolytic method: Chlorine is obtained by the electrolysis of brine solutions (Cone. NaCl). Cl₂ gas is liberated at anode.
2 NaCl → 2 Na⁺ + 2Cl^–
2 H₂O + 2e^– → 2 OH^– + H₂ (cathode)
2 Cl^– → Cl₂ + 2e^– (anode) .

a)

i)NaOH:
Chlorine reacts with cold dil. NaOH to form sodium hypochlorite.
Cl₂ + 2 NaOH → NaCl + NaOCl + H₂O
ii) Cl₂ reacts with hot conc. NaOH to form sodium chloride and sodium chlorate.
3 Cl₂ + 6 NaOH → 5 NaCl + NaClO₃ + 3 H₂O

b)

i) Cl₂ reacts with excess of NH₃, Nitrogen and ammonium chlorate are formed.
8 NH₃ + 3 Cl₂ → 6 NH₄Cl + N₂↑
ii) NH₃ reacts with excess of Cl₂ to form NCl₃ and HCl.

Question 6.
What are interhalogen compounds? Give some examples to illustrate the definition. How are they classified?
Answer:
The binary diamagnetic compounds of halogens which are formed by the reaction of halogens among themselves are called interhalogen compounds.
E.g.: IF₇, ClF₃, BrF₃, ClF, IF₃ etc.
The above examples are binary diamagnetic compounds and formed by combination of halogens only.
Inter halogen compounds are classified into four types.

1. AX – Týpe :Eg: ClF, BrF
2. AX₃ – Tÿpe : Eg : ClF₃, IF₃
3. AX₅ – Type : Eg: ClF₅, BrF₅
4. AX₇ – Type : Eg : IF₇
   • ‘A’ is less electronegative halogen.
   • X is more electronegative halogens.

Question 7.
Write the names and formulae of the oxoacids of chlorine. Explain their structures and relative acidic nature.
Answer:
Four oxyacids of chlorine are known. They are
Hypochlorous acid – HOCl
Chlorous acid — HClO₂
Chloric acid – HClO₃
Perchloric acid – HClO₄
Structure of HClO: In this chlorine atom is sp³ hybridised. Outer electronic configuration of Cl in ClO^– after sp³ hybridisation.

Shape is tetrahedral with 3 lone pairs (or) linear
No π — bonds.
Chlorous acid: (HClO₂) : Chlorine is in sp³ hybrid state, in first excited state. Shape is tetrahedral with 2 lone pairs (or) angular one π_d-p bond is present.
First excited state

Chloric acid (HClO₃) : The central chlorine atom undergoes sp³ hybridisation in second excited state.
Second excited state

Shape is tetrahedral with one lone pair (or) pyramidal.
Two π_d-p bonds present.

Perchloric acid (HClO₄) : The central chlorine atom undergoes sp³ hybridisation in third excited state.

Shape is perfect tetrahedral. No lone pairs.
Three π_d-p bonds present.

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 12(a) Alcohols, Phenols, and Ethers Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material Lesson 12(a) Alcohols, Phenols, and Ethers

Very Short Answer Questions

Question 1.
Explain why propanol has a higher boiling point than that hydrocarbon-butane.
Answer:
Propanol has a higher boiling point (391 K) than that hydrocarbon butane (309K).

Reason: In propanol, strong intermolecular hydrogen bonding is present between the molecules. But in the case of butane weak van der Waals force of attractions are present.

Question 2.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer:
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses.
Explanation: .

• Alcohols and water are both polar solvents. Alcohol is dissolves in water, due to formation of hydrogen bonding with water molecules.
• Hydro carbons are non polar and these dorv.t form hydrogen bonds with water molecules. So alcohols are soluble in water where as hydrocarbons are not soluble in water.

Question 3.
Give the structures and IUPAC names of monohydric phenols of molecular formula, C₇H₈O.
Answer:
Given molecular formula of monnhydric phenols is C₇H₈O. The no. of possible isomers with molecular formula C₇H₈O are three.

Question 4.
Give the reagents used for the preparation of phenol from chiorobenzene.
Answer:
Phenol is prepared from chiorobenzene as follows. Reagents required are

1. NaOH, 623K, 300 atm
2. HCl.
   Chemical reaction :
   

Question 5.
Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Answer:
Only 1° – alcohols form Ethers on acid dehydration. But not 2° or 3°-alcohols.

Reason : In case of 2° or 3° alcohols steric hindrance arises. Due to this steric hindrance alkenes are formed but not Ethers.

Question 6.
Write the mechanism of the reaction of HI with methoxymethane.
Answer:
Case – I: When methoxy methane reacts with cold.dil. HI then methyl alcohol and methyl iodide are formed.
Mechanisms:

Case – II : When methoxy methane reacts with hot.conc.HI then only methyl iodide is formed.
Mechanisms:

Question 7.
Name the reagents used in the following reactions.

1. Oxidation of primary alcohol to carboxylic acid
2. Oxidation of primary alcohol to aldehyde.

Answer:

1. The reagents used for the oxidation of 1° – alcohols to carboxylicacid are acidified K₂Cr₂O₇7 (or) Acidic/alkaline KMnO₄ (or) Neutral KMnO₄
2. The reagents used for the oxidation of 1°- alcohols to aldehyde are pyridine chloro chromate (PCC) in CH₂Cl₂.

Question 8.
Write the equations for the following reactions.
i) Bromination of phenol to 2,4, 6-tribromophenol
ii) Benzyl alcohol to benzoic acid.
Answer:
i) Bromination of phenól to 2, 4, 6 tribromophenol.

Question 9.
IdentIfy the reactant needed to form t-.butylalcohol from acetone.
Answer:
When acetone reacts with methyl magnesium bromide followed by the hydrolysis forms t-butyl alcohol.

Question 10.
Write the structures for the following compounds

1. Ethoxyethane
2. Ethoxybutane
3. Phenoxyethane

Answer:

1. Ethoxyethane → CH₃ – CH₂ – O – CH₂ – CH₃
2. Ethoxybutane → CH₃ – CH₂ – O – CH₂ – CH₂ – CH₂ – CH₃
3. Phenoxyethane → C₆H₅ – O – CH₂ – CH₃

Short Answer Questions

Question 1.
Draw the structures of all isomeric alcohols of molecular formula C₅H₁₂O₂ and give their IUFAC names and classify them as primary, secondary and tertiary alcohols.
Answer:

• Given molecular formula of compound is C₅H₁₂O.
• It has eight isomeric alcohols.
  
  

In the above isomeric alcohols (i), (ii), (iii), (iv) and 1°-alcohols; (v), (vi) and (viii) are 2°- alcohols, (vii) is 3°-alcohol.

Question 2.
While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give rèason.
Answer:
While separating a mixture of ortho and para nitrophenols by steam distillation, ortho nitrophenol is steam volatile.

Reason: In ortho nitrophenol intra molecular hydrogen bonding is present and in case of para nitrophenol inter molecular hydrogen bonding is present. So O-nitrophenol is steam volatile.

Question 3.
Give the equations for the preparation of phenol from Cumene. [Mar. 14]
Answer:
Phenol.is prepared from Cumene as follows.

1. Oxidation of Cumene to Cumene hydroperoxide.
2. Cumene hydroperoxide on acidic hydrolysis to form phenol.
   

Question 4.
Write the mechanism of hydration of ethene to yield ethanol.
Answer:
Hydration of Ethene to yield ethanol involves 3—step mechanism.
Step – 1: In step – 1 formation of carbocation takes place by the protonation of ethene.

Step – 2 : In step – 2 carbocation formed in the above step attacked by water.

Step – 3 : Ethyl alcohol is formed by he deprotonation in step -3

Question 5.
Explain the acidic nature of phenols and compare with that of alcohols.
Answer:
The reaction of phenol with sodium metal and with aq.NaOH indicates the acidic nature of phenol.
i) Phenol reacts with sodium metal to form sodium phenoxide.

•  In phenol hydroxyl group is attached to the Sp² hydridised carbon of benzene ring which acts as electron with drawing group. The formed phenoxide ion from phenol is more stabilised due to delocalisation of negative charge.

Comparison of acidic character of Phenol and Ethanol:

• The reaction of phenol with aq. NaOH indicates that phenols are stronger acids than alcohols.
• The hydroxyl group attached to an aromatic ring is more acidic than in hydroxyl group is attached to an alkyl group.
• Phenol forms stable phenoxide ion stabilised by resonance but ethoxide ion is not.
  

Question 6.
Write the products formed by the reduction and oxidation of phenol.
Answer:
a) Reduction of phenol: Phenol undergo reduction in presence of zinc dust to form benzene.

b) Oxidation of phenol : Phenol undergo oxidation with chromicacid and forms a conjugated diketone known as benzoquinone.

Question 7.
Ethanol with H₂SO₄, at 443K forms ethene while at 413 K it forms ethoxy ethane. Explain the mechanism.
Answer:
Case – 1: Ethanol reacts with Cone. H₂SO₄ at 443K forms ethene

Mechanism:
Step – 1: Formation of protonated alcohol

Step 2 : Formation of carbo cation

Step 3 : Formation of ethene by elimination of a proton

Case – II : Ethanol reacts with Cone. H₂SO₄ at 413 K to form ethoxy ethane.

Mechanism:
In the above reaction ether formation is a SN reaction. This involve attack of alcohol molecule on a protonated alcohol.

Question 8.
Account for the statement: Alcohols boil at higher temperature than hydrocarbons and ethers of comparable molecular masses.
Answer:
Alcohols boil at higher temperature than hydrocarbons and ethers of comparable molecular masses.

Explanation : Consider ethanol, propane and methoxy methane which are having comparable molecular masses. The boiling points, molecular masses and structures of the above compounds mentioned below.

The higher boiling points of alcohols are due to the presence of intermolecular hydrogen bonding in them which is lacking in ethers and hydrocarbons.

Question 9.
Explain why in anisole electrophilic substitution takes place at ortho and para positions and not at meta position.
Answer:
Anisole is an aryl alkyl ether. In anisole the group -OCH₃ influences + R effect. This increases the electron density in the benzene ring and it leads to the activation of benzene ring towards electrophihic substitution reactions.

In Anisole eletron density is more at O-and P-Positions but not at m—position. So 0-and P-products are mainly formed during electrophillic substitution reactions.

Question 10.
Write the products of the following reactions :

Answer:

Long Answer Questions

Question 1.
Write the IUPAC name of the following compounds :

Answer:

Question 2.
Write structures of the compounds whose IUPAC names are as follows:
i) 2, Methyl butan—ol
ii) 1-Phenylprpan-2-ol
iii) 3, 5-Dhuethylhexane-1, 3, 5-triol
iv) 2, 3-Diethylphenol
v) 1-Ethoxypropane
vi) 2-Ethoxy-3-methylpentane
vii) Cyclohexylmethanol
viii) 3-Chloromethylpentan-1-ol
Answer:

Question 3.
Write the equations for the preparation of phenol using benzene, conc. H₂SO₄ and NaOH. [Mar. 14]
Answer:
The equations for the preparation of phenol using conc.H₂SO₄ and NaoH as follows

Question 4.
Illustrate hydroboration-oxidation reaction with a suitable example.
Answer:
When alkenes undergo addition reaction with diborane to form tri alkyl boranes. These followed by the oxidation by alkaline H₂O₂ to form alcohols. This reaction is called as hydroboration-oxidation reaction.

Question 5.
Write the IUPAC name of the following compounds:

Answer:

Question 6.
How will you synthesise:
i) 1 – Phenylethanol from a suitable alkene
ii) Cyclohexylmethanol using an alkyl halide by an SN² reaction.
iii) Pentan-1-ol using a suitable alkyl halide.
Answer:
i) Synthesis of 1-phenylethanol from a suitable alkene : When styrene undergo hydrolysis in presence of dil.H₂SO₄ to form 1-phenyl ethanol. It is an example of Marknowni koff’s rule.

ii) Synthesis of cyclohexyl methanol using an alkyl halide by SN² reaction : When cyclohexyl methyl bromide reacts with aq. NaOH to form cyclohexyl methanol.

iii) Synthesis of 1-pentanol using a suitable alkyl halide: When 1-Bromo pentane reacts with aq.KOH to form 1-pentanol.

Question 7.
Explain why-
i) Ortho nitrophenol is more acidic than Ortho methoxyphenol.
ii) OH group attached to benzene ring activates it towards electrophilic substitution.
Answer:
i) Ortho nitrophenol is more acidic than Ortho methoxyphenol.
Explanation: .

• -NO₂ is an electron withdrawing group and -OCH₃ is electron releasing group.
• By the presence of electron withdrawing group the phenoxide ion is more stabilized. By the presence of electron releasing group the phenoxide ion is less stabilized.
• Due to high stability of phenoxide ion, acidic nature increases.
  

ii) The -OH group attached to benzene ring activates it towards electrophilic substitution.

Explanation : When an electrophile is attacked, the – OH group exerts +R effect on the benzene ring. So electrodensity in the ring increases at ortho and para positions. When an electrophile attacks, substitution takes place at O and p-positions. So benzene ring activates towards electrophilic substitution.

Question 8.
Wth a suitable example write equations for the following:. [T.S. MAr. 19, 18; A.P. Mar. 18, 16] [A.P. Mar. 16]
i) Kolbe’s reaction
ii) Reimer-Tiemann reaction
iii) Williamsons ether synthesis
Answer:
i) Kolbes reaction: Phenol reacts with sodium hydroxide to form sodium phenoxide. This undergoes electrophilhic substitution with CO₂to form salicylic acid.

ii) Relmer-Tlemann reactIon : Phenol reacts with chloroform in presence of NaOH to form salicylaldehyde (O-Hydroxy benzaldehyde). This reaction is known as Reimer-Tiemann reaction. .

iii) Wilhiamsons ether synthesis:

• This method is used for the preparation of symmetrical and unsymmetrical ethers.
• The reaction of an alkyl halide with sodium alkoxide to form ethers is known as Williamsons Synthesis.
  

Question 9.
How are the following conversions carried out?
i) Benzyl chloride to Benzyl alcohol
ii) Ethyl magnesium bromide to Propan-1-ol
iii) 2-Butanone to 2-Butanol
Answer:
i) Conversion of Benzyl chloride to Benzyl alcohol : Ben.zyl chloride reacts with aq. NaOH to form benzyl alcohol.

ii) Conversion of Ethyl magnesium bromide to Propan-1-ol : Ethyl magnesium bromide reacts with form aldehyde followed by hydrolysis to form 1-propanol.

iii) Conversion of 2-Butanone to 2-Butanol: 2-Butanone undergo reduction in presence of LiA/H₄ to form 2-Butanol.

Question 10.
Write the names of the reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
i) 1-Propoxypropane
ii) Ethoxybenzene
iii) 2-Methoxy-2-methylpropane
iv) 1 -Methoxyethane
Answer:
i) Preparation of 1-propoxy propane :

Question 11.
How is 1-propoxypropane synthesized from propan-1-ol ? Write mechanism of this reaction.
Answer:
1 – Proponal reacts with Conc. H₂SO₄ at 413 K to form 1-propoxy propane.

Question 12.
Explain the fact that in aryl alkyl ethers the alkoxy group activates the benzene ring towards electrophilic substitution.
Answer:
Anisole is an aryl alkyl ether. In anisole the group -OCH₃ influences +R effect. This increases the electron density in the benzene ring and it leads to the activation of benzene ring towards electrophillic substitution reactions.

In Anisole eletron density is more at O-and P-Positions but not at m-position. So O-and P-products are mainly formed during electrophillic substitution reactions.

Question 13.
Write equations of the below given reactions:
i) Alkylation of anisole
ii) Nitration of anisole
iii) Friedel-Crafts acetylation of anisole
Answer:
i) Friedel crafts Alkylation of anisole :

ii) Nitration of anisole

Question 14.
Show how you would synthesize the following alcohols from appropriate alkenes?

Answer:

Question 15.
Explain why phenol with bromine water forms 2,4,6-tribromophenol while on reaction with bromine in CS₂ at low temperatures forms para-bromophenol as the major product.
Answer:
a) Phenol under goes Bromination in presence of CS₂ to form p-bromophenol as major product.

b) Phenol undergoes bromination in aqueous medium form 2,4,6 -tribromo phenol (white ppt).

Explanation: In bromination of phenol, the polarisation of Br₂ molecule takes place even in the absence of Lewis acid. This is due to the highly activating effect of -OH group attached to the benzene ring.

Textual Examples

Question 1.
Give IUPAC names of the following compounds:

Solution:
i) 4-Chloro-2, 3-dimethylpentan-1-ol
ii) 2-Ethoxypropane
iii) 2, 6-Dimethyiphenol
iv) 1-Ethoxy-2-nitrocyclohexane

Question 2.
Give the structures and IUPAC names of the products expected from the following reactions :
a) Catalytic reduction of butanal.
b) Hydration of propene in the presence of dilute sulphuric acid.
c) Reaction of propanone with methylmagnesium bromide followed by hydrolysis.
Solution:

Question 3.
Arrange the following sets of compounds in order of their increasing boiling points :
a) Pentan-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol.
b) Pentan-1-ol, n-butane, pentanal, ethoxyethane.
Solution:
a) Methanol, ethanol, propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol.
b) n-Butane, ethoxyethane, pentanal and pentan-1-ol.

Question 4.
Arrange the following compounds in increasing order of their acid strength:
Propan-1-ol, 2, 4, 6-trinitrophenol, 3-nitrophenol, 3, 5-dinitrophenol, phenol, 4-methylphenol.
Solution:
Propan-1-ol, 4-methylphenol, phenol, 3-nitrophenol, 3, 5-dinitrophenol, 2, 4, 6-trinitrophenol.

Question 5.
Write the structures of the major products expected from the following reactions:
a) Mononitratlon of 3-methylphenol .
b) Dinitratlon of 3methylphenol
c) Mononitration of phenyl methanoate.
Solution:
The combined influence of -OH and -CH₃ groups determine the position of the incoming group.

Question 6.
The following is not an appropriate reaction for the preparation of t-butyl ethyl ether.

i) What would be the major product of this reaction?
ii) Write a suitable reaction for the preparation of t-butylethyl ether.
Solution:
i) The major product of the given reaction is 2-methylprop-1-ene.
It is because sodium ethoxide is a strong nucleophile as well as a strong base. Thus elimination reaction predominates over substitution.

Question 7.
Give the major products that are formed by heating each of the following ethers with HI.

Solution:

Intext Questions

Question 1.
Classify the following as primary, secondary and tertiary alcohols:

Answer:
Primary alcohols (i), (ii), (iii)
Secondary alcohols (iv) and (y)
Tertiary alcohols (vi)

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

Question 3.
Name the following compounds according to IUPAC system.


Answer:
i) 3-Chioremethyl 2-isopropylpentan-1-ol
ii) 2, 5-Dimethylhexane-1, 3-dial
iii) 3-Bromocyclohexanol
iv) Hex-1-en-3-ol
v) 2-Bromo-3-methylbut-2-en-1-ol.

Question 4.
Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent of methanol?

Answer:

Question 5.
Write structures of the products of the following reactions :


Answer:

Question 6.
Predict the major product of acid catalysed dehydration of

1. 1-methylcyclohexanol and
2. butan-1-ol.

Answer:

1. 1-Methylcyclohexene
2. A mixture of but-1-ene and but-2-ene. But-1-ene is the major product formed due to rearrangement to give,secondary carbocation.

Question 7.
Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.
Answer:

Question 8.
Predict the products of the following reactions:

Answer:

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 2 System of Circles to solve questions creatively.

Intermediate 2nd Year Maths 2B System of Circles Formulas

Definition:
→ The angle between two intersecting circles is defined as the angle between the tangents at the point of intersection of the two circle’s. If θ is the angle between the circles, then
cos θ = \(\frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2 r_{1} r_{2}}\)
Here d = distance between the centres, r₁, r₂ be their radii.

→ If θ is angle between the two circles.
x² + y² + 2gx + 2fy + c = 0 and x² + y² + 2g’x + 2f’y + c’ = 0, then
cos θ = \(\frac{-2 g g^{\prime}-2 f f^{\prime}+c+c^{\prime}}{2 \sqrt{g^{2}+f^{2}-c} \sqrt{g^{\prime 2}+f^{\prime 2}-c^{\prime}}}\)

→ Circles cut orthogonally if
2g’g + 2f’f = c + c’ [∵ cos 90° = 0]
(or) d² = r²₁ + r²₂ then also two circles cut orthogonally.

Theorem:
If d is the distance between the centers of two intersecting circles with radii r₁, r₂ and θ is the angle between the circles then cos θ = \(\frac{\mathrm{d}^{2}-\mathrm{r}_{1}^{2}-\mathrm{r}_{2}^{2}}{2 \mathrm{r}_{1} \mathrm{r}_{2}}\).

Proof:
Let C₁, C₂ be the centre s of the two circles S = 0, S’ = 0 with radii r₁, r₂ respectively. Thus C₁C₂ = d. Let P be a point of intersection of the two circles. Let PB, PA be the tangents of the circles S = 0, S’ = 0 respectively at P.

Now PC₁ = r₁, PC₂ = r₂, ∠APB = θ
Since PB is a tangent to the circle S = 0, ∠C₁PB = π/2
Since PA is a tangent to the circle S’ = 0, ∠C₂PA = π/2
Now ∠C₁PC₂ = ∠C₁PB + ∠C₂PA – ∠APB = π/2 + π/2 – θ = π – θ
From ∆C₁PC₂, by cosine rule,
C₁²C₂² = PC₁² + PC₂² – 2PC₁ . PC₂ cos ∠C₁PC₂ ⇒ d² = r₁² + r₂² – 2r₁r₂ cos(π – θ) ⇒ d² = r₁² + r₂² + 2r₁r₂ cos θ
⇒ 2r₁r₂ cos θ = d² – r₁² – r₂² ⇒ cos θ = \(\frac{\mathrm{d}^{2}-\mathrm{r}_{1}^{2}-\mathrm{r}_{2}^{2}}{2 \mathrm{r}_{1} \mathrm{r}_{2}}\)

Corollary:
If θ is the angle between the circles x² + y² + 2gx + 2fy + c = 0, x² + y² + 2g’x + 2f’y + c’= 0 then cos θ = \(\frac{c+c^{\prime}-2\left(g g^{\prime}+f f^{\prime}\right)}{2 \sqrt{g^{2}+f^{2}-c} \sqrt{g^{2}+f^{\prime 2}-c^{\prime}}}\)
proof:
Let C₁, C₂ be the centre s and r₁, r₂ be the radii of the circles S = 0, S’ = 0 respectively and C₁C₂ = d.
∴ C₁ = (- g, – f), C₂ = (- g’, – f’),
r₁ = \(\sqrt{g^{2}+f^{2}-c}\), r₂ = \(\sqrt{g^{2}+f^{2}-c}\)
Now cos θ = \(\frac{\mathrm{d}^{2}-\mathrm{r}_{1}^{2}-\mathrm{r}_{2}^{2}}{2 \mathrm{r}_{1} \mathrm{r}_{2}}\) = \(\frac{\left(g-g^{\prime}\right)^{2}+\left(f-f^{\prime}\right)^{2}-\left(g^{2}+f^{2}-c\right)-\left(g^{\prime 2}+f^{\prime 2}-c^{\prime}\right)}{2 \sqrt{g^{2}+f^{2}-c} \sqrt{g^{2}+f^{\prime 2}-c^{\prime}}}\)
= \(\frac{\mathrm{g}^{2}+\mathrm{g}^{\prime 2}-2 \mathrm{gg}^{\prime}+\mathrm{f}^{2}+\mathrm{f}^{\prime 2}-2 \mathrm{ff} \mathrm{f}^{\prime}-\mathrm{g}^{2}-\mathrm{f}^{2}+\mathrm{c}-\mathrm{g}^{2}-\mathrm{f}^{\prime 2}+\mathrm{c}^{\prime}}{2 \sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}} \sqrt{\mathrm{g}^{2}+\mathrm{f}^{\prime 2}-\mathrm{c}^{\prime}}}\)
= \(\frac{c+c^{\prime}-2\left(g g^{\prime}+f f^{\prime}\right)}{2 \sqrt{g^{2}+f^{2}-c} \sqrt{g^{\prime 2}+f^{\prime 2}-c^{\prime}}}\)

Note: Let d be the distance between the centers of two intersecting circles with radii r₁, r₂. The two circles cut orthogonally if d² = r₁² + r₂².
Note: The condition that the two circles

S = x² + y² + 2gx + 2fy + c = 0, S’ = x² + y² + 2g’x + 2f’y + c’ = 0 may cut each other orthogonally is 2gg’ + 2ff’ = c + c’.
Proof: Let C₁, C₂ be the centers and r₁, r₂ be the radii of the circles S = 0, S’ = 0 respectively.
∴ C₁ = (- g, – f), C₂ = (- g’, – f’)
r₁ = \(\sqrt{g^{2}+f^{2}-c}\), r₂ = \(\sqrt{g^{2}+f^{2}-c^{\prime}}\)
Let P be point of intersection of the circles.
The two circles cut orthogonally at P
⇔ ∠C₁PC₂ = 90°.
⇒ C₁C₂² = C₁P² + C₂P² ⇔ (g – g’)² + (f – f’)² = r₁² + r₂²;
⇔ g² + g’² – 2gg’ + f² + f’² – 2ff’ = g² + f² – c + g’² + f’² + c’
⇔ – (2gg’ + 2ff’) = – (c + c’) ⇒ 2gg’+ 2ff’ = c + c’

Note:

• The equation of the common chord of the intersecting circles s = 0 and s¹ = 0 is s – s¹ = 0.
• The equation of the common tangent of the touching circles s = 0 and s¹ = 0 is s – s¹ = 0
• If the circle s = 0 and the line L = 0 are intersecting then the equation of the circle passing through the points of intersection of s = 0 and L = 0 is S + λL = 0.
• The equation of the circle passing through the point of intersection of S = 0 and S’ = 0 is s + λS’ = 0.

Theorem: The equation of the radical axis of the circles S = 0, S’ = 0 is S – S’ = 0.

Theorem: The radical axis of two circles is perpendicular to their line of centers.
Proof:
Let S = x² + y² + 2gx + 2fy + c = 0, S’ = x² + y² + 2g’x + 2f’y + c’= 0 be the given circles.

The equation of the radical axis is S – S’ = 0
⇒ 2(g – g’)x + 2(f – f’)y + (c – c’) = 0
⇒ a₁x + b₁y + c₁ = 0 where
a₁ = 2(g – g’), b₁ = 2(f – f’), c₁ = e – e’
The centers of the circles are (- g, – f), (- g’, – f’)
The equation to the line of centers is:
(x + g) (f – f’) = (y + f) (g – g’)
⇒ (f – f’)x – (g – g’)y – gf’ + fg’= 0
⇒ a₂x + b₂y + c₂ = 0 where
a₂ = f – f’, b₂ = – (g – g’), c₂ = fg’ – gf’
Now a₁a₂ + b₁b₂ = 2(g – g’) (f – f’) – 2(f – f’) (g – g’) = 0.

Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B System of Circles Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2B System of Circles Important Questions

Question 1.
x² + y² + 4x + 8 = 0, x² + y² – 16y + k = 0 [A.P. & T.S. Mar. 16]
Solution:
g₁ = 2; f₁ = 0; c₁ = 8
g₂ = 0; f₂ = – 4; c₂ = k
Two circles are said to be orthogonal
2g₁g₂ + 2f₁f₂ = c₁ + c₂
2(2) (0) + 2(0) (-8) = 8 + k
0 + 0 = 8 + k
⇒ k = -8

Question 2.
(x – a)² + (y – b)² = c², (x – b)² + (y – a)² = c² (a ≠ b) [A.P. Mar. 15]
Solution:
(x² + y² – 2xa – 2yb – c²)
– (x² + y² – 2xb – 2ya – c²) = 0
– 2x (a – b) – 2y(b – a) = 0
(or) x – y = 0

Question 3.
Find the angle between the circles given by the equations. [T.S. Mar.17]
i) x² + y² – 12x – 6y + 41 = 0, x² + y² + 4x + 6y – 59 = 0.
Solution:
C₁ = (6, 3)
r₁ =(36 + 9 – 41)^1/2
r₁ = 2

C₂ = (-2, -3)
r₂ =(4 + 9 + 59)^1/2
r₂ = (72)^1/2 = 6\(\sqrt{2}\)

C₁C₂ = d = \(\sqrt{(6+2)^{2}+(3+3)^{2}}\)
= \(\sqrt{64+36}\) = 10
cos θ = \(\frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2 r_{1} r_{2}}\)
= \(\frac{100-4-72}{2 \times 2 \cdot \sqrt{72}}=\frac{24}{4 \times 6 \sqrt{2}}=\frac{1}{\sqrt{2}}\)
θ = 45°

Question 4.
Find the equation of the circle which cuts the circles x² + y² – 4x – 6y + 11 = 0 and x² + y² – 10x – 4y + 21 = 0 orthogonally and has the diameter along the straight line 2x + 3y = 7. [A.P. Mar. 16; May 07]
Solution:
Let circle be x² + y² + 2gx + 2fy + c = 0 ……………….. (i)
Orthogonal to circle
2g (-2) +2f(-3) = 11 + c ……………………. (ii)
2g (-5) + 2f(-2) = 21 + c ……………………. (iii)
Subtracting it we get
-6g + 2f = 10 ……………………… (iv)
Circles centre is on 2x + 3y = 7
∴ -2g – 3f = 7 ……………………. (v)
Solving (iv) and (y)
f = -1, g = -2, c = 3
Equation of circle be x² + y² – 4x – 2y + 3 = 0

Question 5.
Show that the angle between the circles x² + y² = a², x² + y² = ax + ay is \(\frac{3\pi}{4}\) [Mar. 14]
Solution:
Equations of the circles are
S ≡ x² + y² – a² = 0
S’ ≡ x² + y² – ax – ay = 0
C₁ (0, 0), C₂ (\(\frac{a}{2}\), \(\frac{a}{2}\))
∴ C₁C₂² = (0 – \(\frac{a}{2}\))² + (0 – \(\frac{a}{2}\))²

Question 6.
If x + y = 3 ¡s the equation of the chord AB of the circle x² + y² – 2x + 4y – 8 = 0, find the equation of the circle having \(\overline{\mathrm{AB}}\) as diameter. [A.P. Mar. 15]
Solution:
Required equation of circle passing through
intersection S = 0 and L = 0 is S + λL = 0
(x² + y² – 2x + 4y – 8) + λ(x + y – 3) = 0
(x² + y² + x(-2 + λ)+ y(4 + λ) – 8 – 3λ = 0 ………………. (i)
x² + y² + 2gx + 2fy + c = 0 ……………………. (ii)
Comparing (i) and (ii) we get
g = \(\frac{(-2+\lambda)}{2}\), f = \(\frac{(4+\lambda)}{2}\)
Centre lies on x + y = 3
∴ \(-\left(\frac{-2+\lambda}{2}\right)-\left(\frac{4+\lambda}{2}\right)\) = 3
2 – λ – 4 – λ = 6
-2λ = 8 ⇒ λ = – 4
Required equation of circle be
(x² + y² – 2x + 4y – 8) – 4(x + y – 3) = 0
x² + y² – 6x + 4 = 0

Question 7.
If two circles x² + y² + 2gx + 2fy = 0 and x² + y² + 2g’x + 2f’y = 0 touch each other then show that f’g = fg’. [T.S. Mar. 16]
Solution:
C₁ = (-g, -f)
r₁ = \(\sqrt{g^{2}+f^{2}}\)
C₂ = (-g^-1, -f^-1)
r₂ = \(\sqrt{g^{\prime 2}+r^{\prime 2}}\)
C₁C₂ = r₁ + r₂
(C₁C₂)² = (r₁r₂)²
(g’ – g)² + (f’ – f)² = g² + f² + g’² + f’² + \(2 \sqrt{g^{2}+f^{2}} \sqrt{g^{2}+f^{1^{2}}}\)
-2(gg’ + ff’) = 2{g²g’² + g²f’² + f²g’²}^1/2
Squaring again
(gg’ + ff’)² = g²g’² + f²f’² + g²f’² + g’²f²
g²g’² + f²f’² + 2gg’ff’ = g²g’² + f²f’² + g²f’² + g’²f’²
2gg’ff’ = g²f’² + f²g’²
⇒ g²f’² + g’²f’² – 2gg’ff’ = 0
(or) (gf’ – fg’)² = 0 (or) gf’ = fg’

Question 8.
Show that the circles
S ≡ x² + y² – 2x – 4y – 20 = 0 …………………… (1)
and S ≡ x² + y² + 6x + 2y – 90 = 0 …………………(2)
touch each other internally. Find their point of contact and the equation of common tangent. [T.S. Mar. 15]
Solution:
Let C₁, C₂ be the centres and r₁, r₂ be the radii of the given circles (1) and (2). Then
C₁ = (1, 2); C₂ = (-3, -1); r₁ = 5; r₂ = 10.
C₁C₂ = distance between the centres = 5
|r₁ – r₂| = |5 – 10| = 5 = C₁C₂
∴ The given two circles touch internally. In
this case, the common tangent is nothing but the radical axis. Therefore its equation is S – S’ = 0.
i.e., 4x + 3y – 35 = 0
Now we find the point of contact. The point of contact divides \(\overline{\mathrm{C}_{1} \mathrm{C}_{2}}\) in the ratio 5 : 10
i.e., 1: 2 (externally)
∴ Point of contact
= \(\left(\frac{(1)(-3)-2(1)}{1-2}, \frac{(1)(-1)-2(2)}{1-2}\right)\)
= (5, 5)

Question 9.
Find the angle between the circles x² + y² + 4x – 14y + 28 = 0 and x² + y² + 4x – 5 = 0
Solution:
Equations of the given circles are .
x² + y² + 4x – 14y + 28 = 0
x² + y² + 4x – 5 = 0
Centres are C₁ (-2, 7), C₂ (-2, 0)
C₁C₂ = \(\sqrt{(-2+2)^{2}+(7-0)^{2}}\)
= \(\sqrt{0+49}\) = 7
r₁ = \(\sqrt{4+49-28}\) = \(\sqrt{25}\) = 5
r₂ = \(\sqrt{4+5}\) = \(\sqrt{9}\) = 3
If θ is the angle between the given circles,
then cos θ = \(\frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2 r_{1} r_{2}}\)
cos θ = \(\frac{49-25-9}{2(5)(3)}\) = \(\frac{15}{2.5 .3}\) = \(\frac{1}{2}\) = cos 60°
Angle between the circles = θ = \(\frac{\pi}{3}\)

Question 10.
If the angle between the circles x² + y² – 12x – 6y + 41 = 0 and x² + y² + kx + 6y – 59 = 0 is 45° find k.
Solution:
Suppose θ is the angle between the circles
x² + y² – 12x – 6y + 41 = 0
and x² + y² + kx + 6y – 59 = 0
g₁ = -6, f₁ = -3, c₁ = 41,
g₂ = \(\frac{k}{2}\), f₂ = 3, c₂ = -59

\(\frac{1}{\sqrt{2}}=\frac{6 k}{4 \cdot \sqrt{\frac{k^{2}}{4}+68}}\)
Squaring and cross – multiplying
4(\(\frac{k^{2}}{4}\) + 68) = 18k²
\(\frac{2\left[k^{2}+272\right]}{4}\) = 9k²
k² + 272 = 18 k²
17k² = 272
k² = \(\frac{272}{17}\)
k² = 16
k = ±4.

Question 11.
Find the equation of the circle which passes through (1, 1) and cuts orthogonally each of the circles.
x² + y² – 8x – 2y + 16 = 0 and …………….. (1)
x² + y² – 4x – 4y – 1 = 0. ………………. (2)
Solution:
Let the equation of the required circle be
x² + y² + 2gx + 2fy + c = 0 ……………. (3)
Then the circle (3) is orthogonal to (1) and (2).
∴ By applying the condition of orthogonality give in x² + y² + 2gx + 2fy + c = 0 we get
2g(-4) + 2f(-1) = c + 16 and …………………. (4)
2g (-2) + 2f(-2) = c – 1 ……………….. (5).
Given that the circle (3) is passing through (1, 1)
∴ 1² + 1² + 2g(1) + 2f(1) + c = 0
2g + 2f + c + 2 = 0 ………………….. (6)
Solving (4), (5) and (6) for g, f and c, we get
g = –\(\frac{7}{3}\), f = \(\frac{23}{6}\), c = -5
Thus the equation of the required circle is
3(x² + y²) – 14x + 23y – 15 = 0.

Question 12.
Find the equation of the circle which is orthogonal to each of the following three circles.
x² + y² + 2x + 17y + 4 = 0 ………………… (1)
x² + y² + 7x + 6y + 11 = 0 ………………. (2)
and x² + y² – x + 22y + 3 = 0 ………………… (3)
Solution:
Let the equation of the required circle be
x² + y² + 2gx + 2fy + c = 0 ……………… (4)
Since this circle is orthogonal to (1), (2) and (3). by applying the condition of orthogonality given in x² + y² + 2gx + 2fy + c = 0
we have
2(g)(1) + 2(f) (\(\frac{17}{2}\)) = c + 4 ……………….. (5)
2(g) (\(\frac{7}{2}\)) + 2(f)(3) = c + 11 ………………. (6)
and 2(g) (-\(\frac{1}{2}\)) + 2(f)(11) = c + 3 ………………….. (7)
Solving (5), (6) and (7) for g, f, c we get
g = -3, f = -2 and c = -44
Thus the equation of the required circle is
x² + y² – 6x – 4y – 44 = 0.

Question 13.
If the straight line is represented by
x cos α + y sin α = p ………………….. (1)
intersects the circle
x² + y² = a² ……………… (2)
at the points A and B, then show that the equation of the circle with \(\overline{\mathrm{AB}}\) as diameter is(x² + y² – a²) – 2p(x cos α + y sin α – p) = 0.
Solution:
The equation of the circle passing through the points A and B is S ≡ x² + y² + 2gx + 2fy + c = 0
(x² + y² – a²) + λ(x cos α + y sin α – p) = 0 ……………… (3)
The centre of this circle is
\(\left(-\frac{\lambda \cos \alpha}{2},-\frac{\lambda \sin \alpha}{2}\right)\)
If the circle given by (3) has \(\overline{\mathrm{AB}}\) as diameter then the centre of it must lie on (1)
∴ \(-\frac{\lambda \cos \alpha}{2}\) (cos α) – \(\frac{\lambda \sin \alpha}{2}\) (sin α) = p
i.e., –\(\frac{\lambda}{2}\) (cos² α + sin² α) p
i.e., λ = -2p
Hence the equation of the required circle is
(x² + y² – a²) – 2p(x cos α + y sin α – p) = 0.

Question 14.
Find the equation of the circle passing through the points of intersection of the circles.
x² + y² – 8x – 6y + 21 = 0 ……………….. (1)
x² + y² – 2x – 15 = 0 ……………….. (2)
and (1, 2).
Solution:
The equation of circle passing through the points of intersection of (1) arid (2) is
(x² + y² – 8x – 6y + 21) + λ (x² + y² – 2x – 15) = 0 ……………….. (3)
If it passes through (1, 2). we obtain
(1 + 4 – 8 – 12 + 21) + λ(1 + 4 – 2 – 15) = 0
i.e., 6 + λ(-12) = 0
i.e., λ = \(\frac{1}{2}\)
Hence the equation of the required circle is
(x² + y² – 8x – 6y + 21) + \(\frac{1}{2}\) (x² + y² – 2x – 15) = 0
i.e., 3(x² + y²) – 18x – 12y + 27 = 0.

Question 15.
Let us find the equation the radical axis of the circles S ≡ x² + y² – 5x + 6y + 12 = 0 and S’ ≡ x² + y² + 6x – 4y – 14 = 0
Solution:
The given equations of circles are in general form. Therefore their radical axis is (S – S’ = 0)
i.e., 11x – 10y – 26 = 0

Question 16.
Let us find the equation of the radical axis of the circles
2x² + 2y² + 3x + 6y – 5 = 0 ………………….. (1)
and 3x² + 3y² – 7x + 8y – 11 = 0 ……………….. (2)
Solution:
Hence the given equations are not in general form, we get :
x² + y² + \(\frac{3}{2}\)x + 3y – \(\frac{5}{2}\) = 0 and
x² + y² – \(\frac{7}{3}\)x + \(\frac{8}{3}\) y – \(\frac{11}{3}\) = 0
Now the radical axis equation of given circles is
(\(\frac{3}{2}\) + \(\frac{7}{3}\))x + (3 – \(\frac{8}{3}\))y + (-\(\frac{5}{2}\) + \(\frac{11}{3}\)) = 0
i.e., 23x + 2y + 7 = 0

Question 17.
Let us find the radical centre of the circles
x² + y² – 2x + 6y = 0 ………………… (1)
x² + y² – 4x – 2y + 6 = 0 …………………. (2)
and x² + y² – 12x + 2y + 3 = 0 ………………. (3)
Solution:
The radical axis of (1) and (2) and (3)
x + 4y – 3 = 0 ………………… (4)
8x – 4y + 3 = 0 …………………. (5)
10x + 4y – 3 = 0 ……………… (6)
Solving (4) and (5) for the point of intersection we get (0, \(\frac{3}{4}\)) which is the required radical centre. Observe that the co-ordinates of this point satisfies (6) also.

Question 18.
Find the equation and length of the common chord of the two circles
S ≡ x² + y² + 3x + 5y + 4 = 0
and S’ ≡ x² + y² + 5x + 3y + 4 = 0
Solution:
Equations of the given circles are
S ≡ x² + y² + 3x + 5y + 4 = 0 ………………… (1)
S’ ≡ x² + y² + 5x + 3y + 4 ………………. (2)
Equations of the common chord is S – S’ = 0
-2x + 2y = 0
L ≡ x – y = 0 …………….. (3)

Question 19.
Show that the circles
S ≡ x² + y² – 2x – 4y – 20 = 0 ……………….. (1)
and S’ ≡ x² + y² + 6x + 2y – 90 = 0 ……………. (2)
touch each other internally. Find their point of contact and the equation of common tangent. [T.S. Mar. 15]
Solution:
Let C₁, C₂ be the centres and r₁, r₂ be the radii of the given circles (1) and (2). Then
C₁ = (1, 2); C₂ = (-3, -1); r₁ = 5; r₂ = 10
C₁C₂ = distance between the centres = 5
|r₁ – r₂| = |5 – 10| = 5 = C₁C₂
∴ The given two circles touch internally. In this case, the.common tangent is nothing but the radical axis. Therefore its equation is
S – S’ = 0
i.e., 4x + 3y – 35 = 0
Now we find the point of contact. The point of contact divides in the ratio 5 : 10 i.e., 1 : 2 (externally)
∴ Point, of contact
= \(\left(\frac{(1)(-3)-2(1)}{1-2}, \frac{(1)(-1)-2(2)}{1-2}\right)\)
= (5, 5).

Question 20.
Find the equation of the circle whose diameter is the common chord of the circles
S ≡ x² + y² + 2x + 3y + 1 = 0 ……………….. (1)
and S’ ≡ x² + y² + 4x + 3y + 2 = 0 ……………….. (2)
Solution:
Here the common chord is the radical axis of (1) and (2). The equation of the radical axis is S – S’ = 0.
i.e., 2x + 1 = 0 …………………… (3)
The equation of any circle passing through
the points of intersection of (1) and (3) is (S + λL = 0)
(x² + y² + 2x + 3y + 1) + λ(2x + 1) = 0
x² + y² + 2(λ + 1)x + 3y + (1 + λ) = 0 ……………………. (4)
The centre of this circle is (-(λ + 1), \(\frac{3}{2}\)).
For the circle (4), 2x + 1 = 0 is one chord. This chord will be a diameter of the circle (4) if the centre of (4) lies on (3).
∴ 2{-(λ + 1)} + 1 = 0
⇒ λ = –\(\frac{1}{2}\)
Thus equation of the circle whose diameter is the common chord (1) and (2)
(put λ = \(\frac{1}{2}\) in equation (4))
2(x² + y²) + 2x + 6y + 1 = 0

Question 21.
Let us find the equation of a circle which cuts each of the following circles orthogonally
S’ ≡ x² + y² + 3x + 2y + 1 = 0 ………………… (1)
S” ≡ x² + y² – x + 6y + 5 = 0 …………….. (2)
and S” ≡ x² + y² + 5x – 8y + 15 = 0 ……………………. (3)
Solution:
The centre of the required circle is radical centre of (1), (2) and (3) and the radius is the length of the tangent from this point to any one of the given three circles. First we shall find the radical centre. For, the radical axis of (1) and (2) is
x – y = 1 ………………… (4)
and the radical axis of (2) and (3) is
3x – 7y = -5 …………………… (5)
The point of intersection (3, 2) of (4) and (5) is the radical centre of the circles (1), (2) and (3). The length of tangent from (3. 2) to the circle (1)
= \(\sqrt{3^{2}+2^{2}+3(3)+2(2)+1}=3 \sqrt{3}\)
Thus the required circle is
(x – 3)² + (y – 2)² = (3\(\sqrt{3}\))²
x² + y² – 6x – 4y – 14 = 0.

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 10 యాదృచ్ఛిక చలరాశలు, సంభావ్యత విభాజనాలు to solve questions creatively.

AP Intermediate 2nd Year Maths 2A యాదృచ్ఛిక చలరాశలు, సంభావ్యత విభాజనాలు Formulas

→ యాదృచ్ఛిక చలరాశి : ఒక యాదృచ్ఛిక ప్రయోగం శాంపిల్ ఆవరణం S అనుకుందాం. ఏదైనా ప్రమేయం X : S → R ను యాదృచ్ఛిక చలరాశి అంటాం.

→ సంభావ్యతా విభాజన ప్రమేయం : X ఒక యాదృచ్ఛిక చలరాశి అప్పుడు F : R → R ప్రతి X ∈ Rకు F(x) = P(X ≤ x) తో నిర్వచితమైన ప్రమేయాన్ని X కు సంభావ్యతా విభాజన ప్రమేయం అంటాం.

→ X : S → R ఒక యాదృచ్ఛిక చలరాశి. X వ్యాప్తి పరిమితం లేదా అపరిమిత గణ్యసమితి అయితే X ను విచ్ఛిన్న చలరాశి అని, అట్లా కాకపోతే అవిచ్ఛిన్న యాదృచ్ఛిక చలరాశి అని అంటాం.

→ X : S → R ఒక విచ్ఛిన్న యాదృచ్ఛిక చలరాశి. దాని వ్యాప్తి = {X₁, X₂, ….} అయిన \(\sum_{r=1}^n P\left(X_r\right)\) = 1, P(X_r) ≥ 0.

→ X ఒక విచ్ఛిన్న యాదృచ్ఛిక చలరాశి. దాని వ్యాప్తి {X₁, X₂, ……} అనుకుందాం. ప్రతి n కు P(X = x_n) తెలిసి, Σx_n P(X = x_n) అనే మొత్తం పరిమితమైతే, ఆ మొత్తాన్ని X కు మధ్యమం (లేదా సగటు) అంటాం. దీన్ని µ తో సూచిస్తాం. (i.e.,) µ = Σx_n P(X = x_n)
Σ(x_n – µ)² P(X = x_n) అనేది ఒక పరిమిత సంఖ్య అయితే ఆ మొత్తాన్ని X కు విస్తృతి అంటాం.

→ X విస్తృతిని σ² తో సూచిస్తే, σ ను X కు క్రమ విచలనం అంటారు.
∴ μ = Σx_n P(X = x_n); σ² = Σ(x_n – μ)² P(X = x_n) = \(\Sigma x_n^2 P\left(X=x_n\right)\) – μ²

→ ద్విపద విభాజనం: n ఒక ధన పూర్ణాంకం. p వాస్తవ సంఖ్య మరియు 0 ≤ p ≤ 1. యాదృచ్ఛిక చలరాశి వ్యాప్తి {0, 1, 2, 3, …. n}. X ద్విపద చలరాశి లేదా ద్విపద విభాజనాన్ని పాటిస్తూ, n, p లు పరామితులుగా గల్గి వుంటే, P(X = r) = ⁿC_r . p^r q^n-r; r = 0, 1, 2, ….. n మరియు q = 1 – p అవుతుంది.

→ ద్విపద విభాజనాన్ని X ~ B(n, p) గా లేదా P(X = r) = ⁿC_r p^r q^n-r . p^r, r = 0, 1, 2, 3, …. n లేదా (q + p)ⁿతో సూచిస్తాం.

→ ద్విపద విభాజనం యొక్క మధ్యమం = np. ద్విపద విభాజనం యొక్క విస్తృతి = npq, క్రమవిచలనం = \(\sqrt{n p q}\).

→ పాయిజాన్ విభాజనం : λ > 0 ఒక స్థిరరాశి. యాదృచ్ఛిక చలరాశి X యొక్క వ్యాప్తి {0, 1, 2, ….}
P(X = k) = \(\frac{\lambda^k}{k !} e^{-\lambda}\), (k = 0, 1, 2, …….) అనుకుంటే, λ పరామితిగా, X పాయిజాన్ విభాజనాన్ని అనుసరిస్తుందని అంటాం. X ను పాయిజాన్ యాదృచ్ఛిక చలరాశి అంటాం. పాయిజాన్ విభాజనానికి మధ్యమము = λ, విస్తృతి = λ, క్రమ విచలనం = √λ

→ ఈ క్రింది షరతులకు లోబడి పాయిజాన్ విభాజనాన్ని, ద్విపద విభాజనపు సమతాస్థితి (limiting case) గా ఉజ్జాయింపు చేయవచ్చు.

• యత్నాల సంఖ్య n అనిశ్చితమైనంత పెద్దది, అంటే n → ∞
• ప్రతి యత్నంలో గెలుపు సంభావ్యత (స్థిరం) అతిస్వల్పం, అంటే p → 0

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 9 సంభావ్యత to solve questions creatively.

AP Intermediate 2nd Year Maths 2A సంభావ్యత Formulas

→ యాదృచ్ఛిక ప్రయోగం : ఒక ప్రయోగంలో ఏ ఫలితం వస్తుందో ముందే చెప్పలేనిదై, ఆ ప్రయోగ ఫలితాల జాబితా ముందే తెలిసి ఉండి, ఒకే విధమైన పరిస్థితుల్లో ఆ ప్రయోగాన్ని ఎన్నిసార్లైనా చేయడానికి వీలుంటే ఆ ప్రయోగాన్ని యాదృచ్ఛిక ప్రయోగం అంటాం.

→ లఘుఘటన, ఘటన ఒక యాదృచ్ఛిక ప్రయోగంలోని ఫలితాన్ని లఘుఘటన అంటాం. కొన్ని లఘు ఘటనల సమూహాన్ని ఘటన అంటాం.

→ పరస్పర వివర్జిత ఘటనలు: రెండు లేదా అంతకంటే ఎక్కువ ఘటనల్లో ఏదైనా ఒక ఘటన సంభవించడం, మిగతా ఘటనల సంభవాన్ని నిరోధించేటట్టుంటే, అటువంటి ఘటనలను పరస్పర వివర్జిత ఘటనలు అంటారు.

→ సమ సంభవ ఘటనలు: రెండు లేదా అంతకంటే ఎక్కువ ఘటనల్లో ఏ ఘటన అయినా మిగతా ఘటనల కంటే ఎక్కువగా సంభవిస్తుందనడానికి కారణమేమి లేకపోతే, అటువంటి ఘటనలను సమసంభవ ఘటనలు అంటారు.

→ పూర్ణ ఘటనలు : ఒక యాదృచ్ఛిక ప్రయోగంలోని రెండు లేదా అంతకంటే ఎక్కువ ఘటనల్లో ఆ ప్రయోగం ఫలితం వాటిలో ఒక్కదానికైనా చెందేటట్లుంటే, అటువంటి ఘటనలను పూర్ణ ఘటనలు అంటాం.

→ సంభావ్యత సాంప్రదాయిక (లేదా గణితాత్మక) నిర్వచనం : ఒక యాదృచ్ఛిక ప్రయోగంలో n పూర్ణ, పరస్పర వివర్జిత, సమసంభవ ఘటనలుండి వాటిలో ఏదైనా ఘటన E జరగడానికి m అనుకూల ఫలితాలుంటే, ఆ ఘటన సంభావ్యతను P(E) తో సూచిస్తూ, P(E) = \(\frac{m}{n}\) గా నిర్వచిస్తాం. 0 ≤ P(E) ≤ 1

→ శాంపిల్ ఆవరణ : ఒక యాదృచ్ఛిక ప్రయోగపు ఫలితాన్ని లఘుఘటన అంటాం. ఒక యాదృచ్ఛిక ప్రయోగంలోని ఫలితాలన్నింటితో కూడిన సమితిని ఆ యాదృచ్ఛిక ప్రయోగానికి సంబంధించి శాంపిల్ ఆవరణ అంటాం. దీనిని S తో సూచిస్తాం. S లోని మూలకాలను శాంపిల్ బిందువులు అంటాం. S లో ప్రతిమూలకం, యాదృచ్ఛిక ప్రయోగం యొక్క ఒక ఫలితం S. ఒక ఉపసమితిని ఘటన అంటాం. అంటే, ఒక లఘుఘటనల సమితినే ఘటన అంటాం.

→ S శాంపిల్ ఆవరణ E ⊂ S.E లో ఒకే ఒక మూలకం ఉంటే E ని లఘుఘటన అంటాం. ఒక ప్రయోగ ఫలితం ఘటన E సంభవించింది లేదా జరిగింది అంటాం. అలాకాకపోతే ఘటన E సంభవించలేదు అంటాం.

→ Φ, S లు S కి ఉపసమితులు. వాటిని వరసగా అసంభవ ఘటన, నిశ్చిత ఘటన అంటాం.

→ ఘటన E కి పూరక ఘటనను E^C తో సూచిస్తాం. E^C = S – E శాంపిల్ ఆవరణ S కు E₁, E₂ లు రెండు ఘటనలు. అంటే E₁ ⊆ S, E₂ ⊆ S మరియు E₁ ∩ E₂ = Φ అయితే E₁, E₂ లను పరస్పర వివర్జిత ఘటనలు అంటాం.

→ ఘటనలు E₁, E₂, ……., E_n లు i ≠ j లకు 1 ≤ i, j ≤ n లకు E_i ∩ E_j = Φ అయ్యేటట్లుంటే, వాటిని పరస్పర వివర్జిత ఘటనలు అంటాం.

→ ఘటనలు E₁, E₂, …… E_k లు E₁ ∪ E₂ ∪ E₃ ∪ E_k = S అయితే వాటిని పూర్ణ ఘటనలు అంటాం. శాంపిల్ ఆవరణ S లో E₁, E₂లు రెండు ఘటనలు పూరక ఘటనలైన E₁ ∪ E₂ = S, E₁ ∩ E₂ = Φ.

→ ఒక యాదృచ్ఛిక ప్రయోగంకి శాంపిల్ ఆవరణ S. ఈ ప్రయోగపు అన్ని ఘటనల సమితి P(S) తో సూచిస్తాం. ఇచ్చట P(S), Sకు ఘాత సమితి.

→ సంభావ్యతా స్వీకృత నిర్వచనం: ఒక యాదృచ్ఛిక ప్రయోగపు శాంపిల్ ఆవరణ S. ప్రమేయం P : P(S) → R

→ క్రింది స్వీకృతాలను ధ్రువపరిస్తే Pని సంభావ్యతా ప్రమేయం అంటాం.

• P(E) ≥ 0 ∀ E ∈ P(S) (ధన స్వీకృతం)
• P(S) = 1 (పూరణ స్వీకృతం)
• E₁, E₂ ∈ P(S), E₁ ∩ E₂ = Φ అయితే P(E₁ ∪ E₂) = P(E₁) + P(E₂) (సమ్మేళును స్వీకృతం)
• ప్రతి E ∈ P(S) కు P(E) ను ఘటన E సంభావ్యత అంటాం.

→ శాంపిల్ ఆవరణ Sలో E ఒక ఘటన అయితే 0 ≤ P(E) ≤ 1. శాంపిల్ ఆవరణ Sలో E ఒక ఘటన అయితే

• P(E) : P(\(\bar{S}\)) ను E కు అనుకూలత అని
• P(E) : P(E) ను E కు ప్రతికూలత అంటాం.

→ సంభావ్యతపై సంకలన సిద్ధాంతం: ఒక యాదృచ్ఛిక ప్రయోగంలో E₁, E₂ లు రెండు ఘటనలైతే P(E₁ ∪ E₂) = P(E₁) + P(E₂) – P(E₁ ∩ E₂)

→ శాంపిల్ ఆవరణ Sకు E₁, E₂ లు రెండు ఘటనలు.
P(E₂ – E₁) = P(E₂) – P(E₁ ∩ E₂) మరియు P(E₁ – E₂) = P(E₁) – P(E₁ ∩ E₂)

→ E₁, E₂, E₃ లు శాంపిల్ ఆవరణ Sలో మూడు ఘటనలు అయిన P(E₁ ∪ E₂ ∪ E₃) = P(E₁) + P(E₂) + P(E₃) – P(E₁ ∩ E₂) – P(E₂ ∩ E₃) – P(E₃ ∩ E₁) + P(E₁ ∩ E₂ ∩ E₃)

→ ఒక యాదృచ్ఛిక ప్రయోగపు ఘటనలు A, Bలు అనుకుందాం. అప్పుడు “A జరిగిన తరువాత B జరగడం” అనే ఘటనను నియత ఘటన అంటాం. దీనిని \(\frac{B}{A}\) తో సూచిస్తాం. ఇట్లే \(\frac{A}{B}\) అనే ఘటన “B జరిగిన తరువాత A జరగడం” అనే ఘటనను సూచిస్తుంది.

→ నియత సంభావ్యత : ఘటన A జరిగిందని ఇస్తే, ఘటన B జరిగే సంభావ్యతను P(B/A) తో సూచిస్తాం. P(B/A) ని నియత సంఖ్య అంటాం.
దీనిని P(B/A) = \(\frac{P(B \cap A)}{P(A)}\), P(A) > 0 గా నిర్వచిస్తాం. ఇట్లే P(A/B) = \(\frac{P(B \cap A)}{P(B)}\), P(B) > 0.
సూచన : P(A/B) = \(\frac{n(A \cap B)}{n(B)}\); P(B/A) = \(\frac{n(B \cap A)}{n(A)}\)

→ నియత సంభావ్యతకు గణన సిద్ధాంతం ఒక శాంపిల్ ఆవరణం S లోని ఘటనలు A, Bలు ; P(A) > 0, P(B) > 0, అయితే P(A ∩ B) = P(A) . P(B/A) = P(B) . P(A/B).

→ స్వతంత్ర ఘటనలు : రెండు ఘటనలు A, Bలు P(A ∩ B) = P(A) . P(B) అయితే వాటిని స్వతంత్ర ఘటనలు అంటాం. అలాకాకపోతే A, B లను అస్వతంత్ర ఘటనలు అంటాం.

→ బేయీ సిద్ధాంతం: ఒక ప్రయోగంలో E₁, E₂, ……., E_n లు పరస్పర వివర్జిత, పూర్ణ ఘటనలవుతూ, P(E_i) > 0, i = 1, 2, …. n అయినప్పుడు k = 1, 2, 3, ….. n లకు
\(P\left(\frac{E_k}{A}\right)=\frac{P\left(E_k\right) \cdot P\left(\frac{A}{E_k}\right)}{\sum_{i=1}^n P\left(E_i\right) \cdot P\left(\frac{A}{E_i}\right)}\)

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 8 విస్తరణ కొలతలు to solve questions creatively.

AP Intermediate 2nd Year Maths 2A విస్తరణ కొలతలు Formulas

→ అవర్గీకృత దత్తాంశానికి మధ్యమం = \(\frac{విచలనాల మొత్తం}{పరిశీలనల సంఖ్య}\) = \(\frac{\Sigma x_i}{n}\)

→ అవర్గీకృత దత్తాంశానికి మధ్యగతం: ముందుగా దత్త n పరిశీలనలను పరిమాణపరంగా అవరోహణ లేదా ఆరోహణ క్రమంలో వ్రాయవలెను.

→ n బేసి సంఖ్య అయితే \(\frac{n+1}{2}\) వ పరిశీలనల అవర్గీకృత దత్తాంశం యొక్క మధ్యగతం అగును.

→ n సరి సంఖ్య అయితే \(\frac{n}{2}\) మరియు \(\frac{n+2}{2}\) వ పరిశీలనల సరాసరిను అవర్గీకృత దత్తాంశం యొక్క మధ్యగతం అగును.

→ వర్గీకృత మరియు అవర్గీకృత దత్తాంశానికి వ్యాప్తి, మధ్యమ విచలనం, విస్తృతి మరియు ప్రామాణిక విచలనం కొన్ని విస్తరణ కొలతలు

→ వ్యాప్తిని దత్తాంశ గరిష్ట విలువకు, కనిష్ఠ విలువకు మధ్యగల భేదంగా నిర్వచిస్తారు.

→ అవర్గీకృత విభాజనానికి మధ్యమ విచలనం

• మధ్యమం నుంచి మధ్యమ విచలనం = \(\frac{\sum\left|x_i-\bar{x}\right|}{n}, \bar{x}\) మధ్యమం
• మధ్యగతం నుంచి మధ్యమ విచలనం = \(\frac{\sum \mid x_i-\text { మధ్యగతం } \mid}{n}\)

→ వర్గీకృత దత్తాంశానికి మధ్యమ విచలనం

• మధ్యమం నుంచి మధ్యమ విచలనం = \(\frac{1}{N} \sum f_i\left|x-\bar{x}_i\right|\); N = Σf_i, మరియు \(\bar{x}\) మధ్యమం
• మధ్యగతం నుంచి మధ్యమ విచలనం = \(\frac{1}{N} \sum f_i \mid x_i\) – మధ్యగతం|, N = Σf_i

→ అవర్గీకృత దత్తాంశానికి విస్తృతి, σ² = \(\frac{1}{n}\) = \(\sum\left(x_i-\bar{x}\right)^2\), ప్రామాణిక విచలనం σ = \(\sqrt{\frac{1}{n} \sum\left(x_1-\bar{x}\right)^2}\)

→ విచ్ఛిన్న పౌనఃపున్య విభాజనానికి విస్తృతి, σ² = \(\frac{1}{N}\) = \(\sum f_i\left(x_i-\bar{x}\right)^2\), \(\bar{x}\) మధ్యమం

→ ప్రామాణిక విచలనం σ = \(\sqrt{\frac{1}{N} \sum f_i\left(x_i-\bar{x}\right)^2}\)

→ అవిచ్ఛిన్న పౌనఃపున్య విభాజనానికి ప్రామాణిక విచలనం σ = \(\frac{1}{N} \sqrt{N \sum f_i x_i^2-\left(\sum f_i x_i\right)^2}\) (లేదా) σ = \(\frac{h}{N} \sqrt{N \Sigma f_i y_i^2-\left(\Sigma f_i y_i\right)^2}, y_i=\frac{x_i-A}{h}\)

→ విచలనాంకం = \(\frac{\sigma}{x} \times 100(\bar{x} \neq 0)\)

→ దత్తాంశంలోని ప్రతి పరిశీలనను ఒక స్థిరరాశి K చే గుణించినపుడు ఫలితంగా వచ్చే పరిశీలనల విస్తృతి, తొలిపరిశీలనల విస్తృతికి K² రెట్లు ఉంటుంది.

→ పరిశీలనలు x₁, x₂, …….., x_n లలో ప్రతిదానిని K కి పెంచితే లేదా కలిపితే (K ఒక ధనాత్మక లేదా ఋణాత్మక సంఖ్య), వచ్చే పరిశీలనల విస్తృతి మారదు.

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 7 పాక్షిక భిన్నాలు to solve questions creatively.

AP Intermediate 2nd Year Maths 2A పాక్షిక భిన్నాలు Formulas

→ f(x), Φ(x) లు రెండు బహుపదులు. Φ(x) ఒక శూన్యేతర బహుపది (i.e.,) Φ(x) ≠ 0 అయితే \(\frac{f(x)}{\phi(x)}\) ను అకరణీయ భిన్నం అంటాం.

→ f(x) తరగతి Φ(x) తరగతి కంటే తక్కువ అయితే \(\frac{f(x)}{\phi(x)}\) ను క్రమభిన్నమని అంటారు.

→ f(x) తరగతి ≥ Φ(x) తరగతి అయిన \(\frac{f(x)}{\phi(x)}\) ను అపక్రమ భిన్నమని అంటారు.

→ \(\frac{f(x)}{\phi(x)}\) క్రమభిన్నం

→ Φ(x) కు పునరావృతం కాని ఏకఘాత కారణాంకాలున్నప్పుడు Φ(x) లో (ax + b) రూపంలో ఉండే ప్రతి కారణాంకానికి సంబంధించి \(\frac{A}{a x+b}\) అనే పాక్షిక భిన్నం ఉంటుంది A వాస్తవ సంఖ్య.

→ Φ(x) కు పునరావృతం అయ్యేవి, కానివి ఏకఘాత కారణాంకాలున్నప్పుడు, Φ(x) కు (ax + b)ⁿ, n ∈ N రూపంలో వుండే ప్రతి పునరావృత కారణాంకానికి సంబంధించి \(\frac{A_1}{a x+b}+\frac{A_2}{(a x+b)^2}\) + …… + \(\frac{A_n}{(a x+b)^n}\) అనే n పాక్షిక భిన్నాలుంటాయి. ఇక్కడ A₁, A₂, ……., A_n లు వాస్తవ స్థిరాంకాలు.

→ Φ(x) కు ax² + bx + c రూపంలో పునరావృతం కాని అవిభాజ్య కారణాంకం ఉన్నప్పుడు Φ(x) కు ax² + bx + c రూపంలో ఉన్న ప్రతి కారణాంకానికి సంబంధించి \(\frac{A x+B}{a x^2+b x+c}\); A, B లు వాస్తవ స్థిరరాసులు, రూపంలో ఒక పాక్షిక భిన్నం ఉంటుంది.

→ Φ(x) కు ax² + bx + c రూపంలో పునరావృతం కాని అవిభాజ్య కారణాంకం ఉన్నప్పుడు Φ(x) కు (ax² + bx + c)ⁿ రూపంలో ఉన్న ప్రతి కారణాంకానికి సంబంధించి n పాక్షిక భిన్నాలు \(\frac{A_1 x+B_1}{a x^2+b x+c}+\frac{A_2 x+B_2}{\left(a x^2+b x+c\right)^2}+\ldots . .+\frac{A_n x+B_n}{\left(a x^2+b x+c\right)^n}\), n ధన పూర్ణాంకం, A₁, A₂, ……, A_n, B₁, B₂, ………, B_n లు స్థిరరాసుల రూపంలో వుంటాయి.

→ Φ(x) కు ax² + bx + c రూపంలో పునరావృతం అయ్యేవి. కానివి అవిభాజ్య కారణాంకాలున్నప్పుడు, థ(x) కు (ax² + bx + c)ⁿ రూపంలో ఉన్న ప్రతి కారణాంకానికి సంబంధించి n పాక్షిక భిన్నాలు.

→ \(\frac{f(x)}{\phi(x)}\) అపక్రమ భిన్నమైతే \(\frac{f(x)}{\phi(x)}=q(x)+\frac{R(x)}{\phi(x)}\) అని వ్రాయవచ్చు. ఇచ్చట q(x) అనేది f(x) ను Φ(x) చే భాగించగా వచ్చే భాగఫలం, R(x) శేషం, R(x) తరగతి Φ(x), తరగతి కన్నా తక్కువ \(\frac{R(x)}{\phi(x)}\) ను పాక్షిక భిన్నాల మొత్తంగా వ్రాయటానికి పై పద్ధతులను ఉపయోగిస్తారు.

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 6 ద్విపద సిద్ధాంతం to solve questions creatively.

AP Intermediate 2nd Year Maths 2A ద్విపద సిద్ధాంతం Formulas

→ n ఒక ధన పూర్ణాంకం, x, a లు వాస్తవ సంఖ్యలు అయితే
(x + a)ⁿ = ⁿC₀ . xⁿ . a⁰ + ⁿC₁ . x^n-1 . a¹ + ⁿC₂ . x^n-2 . aⁿ + …… + ⁿC_r . x^n-r a^r +……. + ⁿC_n . x⁰ . aⁿ = \(\sum_{r=0}^n{ }^n C_r \cdot x^{n-r} \cdot a^r\)

→ (x + a)ⁿ విస్తరణలో (n + 1) పదాలున్నాయి.

→ (x + a)ⁿ విస్తరణలోని rవ పదాన్ని T_r తో సూచిస్తే T_r = ⁿC_(r-1) x^n-r+1 a^r-1, 1 ≤ r ≤ n+1

→ (x + a)ⁿ విస్తరణలో (r + 1)వ పదాన్ని ‘సాధారణ పదం’ (General Term) అంటాం.
i.e., T_r+1 = ⁿC_r . x^n-r. a^r ; r = 0, 1, 2, ….., n.

→ (a + b + c)ⁿ విస్తరణలో పదాల సంఖ్య = \(\frac{(n+1)(n+2)}{2}\)

→ n సరిసంఖ్య (ధన పూర్ణాంకం) అయితే (x + a)ⁿ విస్తరణలో మధ్య పదం = \(T_{\left(\frac{n}{2}+1\right)}\)

→ n బేసిసంఖ్య (ధన పూర్ణాంకం) అయితే (x + a)ⁿ విస్తరణలో రెండు మధ్య పదాలుంటాయి. అవి \(T_{\left(\frac{n+1}{2}\right)}\), \(T_{\left(\frac{n+3}{2}\right)}\)

→ \(\frac{(n+1)|x|}{|x|+1}\) = p, ధన పూర్ణాంకమైన, (1 + x)ⁿ విస్తరణలో pవ, (p + 1)వ పదాలు సంఖ్యాత్మకంగా గరిష్ఠ పదాలు అవుతాయి. మరియు |T_p| = |T_p+1|

→ \(\frac{(n+1)|x|}{|x|+1}\) = p + F; p ధన పూర్ణాంకం, 0 < F < 1 అయిన (1 + x)ⁿ విస్తరణలో (p + 1)వ పదం సంఖ్యాత్మకంగా గరిష్ఠ పదం అవుతుంది.

→ C₀, C₁, C₂, …….., C_n లు ద్విపద గుణకాలు అంటాం. ఇచ్చట C_r = ⁿC_r, r = 0, 1, 2, ….. n

• C₀ + C₁ + C₂ + …… + C_n = \(\sum_{r=0}^n c_r\) = 2ⁿ
• C₀ – C₁ + C₂ – C₃ + ……. (-1)ⁿ C_n = 0
• C₀ + C₂ + C₄ + …… = C₁ + C₃ + C₅ + …….. = 2^n-1
• \(\sum_{r=0}^n r \cdot{ }^n C_r\) = n . 2^n-1
• \(\sum_{r=2}^n(r)(r-1) \cdot{ }^n C_r\) = (n) (n – 1) . 2^n-2
• \(\sum_{r=1}^n r^2 \cdot{ }^n C_r\) = (n) (n + 1) . 2^n-2
• a . C₀ + (a + d) . C₁ + (a + 2d) . C₂ + ……. + (a + nd) . C_n = (2a + nd) 2^n-1
• C₀C_r + C₁C_r+1 + C₂C_r+2 + C_n-rC_n = ⁽²ⁿ⁾C_(n-r) = ⁽²ⁿ⁾C_(n+r) = \(\frac{(2 n) !}{(n+r) !(n-r) !}\)

→ f(x) = a₀ + a₁x + a₂x² + ….. + a_nxⁿ అయిన

• గుణకాల మొత్తం = f(1)
• x యొక్క సరి ఘాతాల గుణకాల మొత్తం = \(\frac{f(1)+f(-1)}{2}\)
• x యొక్క బేసి ఘాతాల గుణకాల మొత్తం = \(\frac{f(1)-f(-1)}{2}\)

→ m అకరణీయ సంఖ్య, x ఒక వాస్తవ సంఖ్య, |x| < 1 అయితే
(1 + x)^m = \(1+\frac{m}{1} x+\frac{(m)(m-1)}{1.2} x^2+\frac{(m)(m-1) \ldots .(m-r+1)}{1.2 .3 \ldots . .(r)} \cdot x^r+\ldots \ldots\)
= \(1+\sum_{r=1}^{\infty} \frac{(m n)(m-1) \ldots .(m-r+1)}{1.2 \cdot 3 \ldots . . r^{\prime}} x^r\)

→ (1 – x)^-n = \(1+n x+\frac{(n)(n+1)}{1.2} x^2+\ldots+\frac{(n)(n+1) \ldots(n+r-1)}{1: 2.3 \ldots \ldots(r)} x^r+\ldots\)

→ |x| < 1, p, q ∈ N అయిన
\((1-x)^{-p / q}=1+\frac{p}{1}\left(\frac{x}{q}\right)+\frac{(p)(p+q)}{1.2}\left(\frac{x}{q}\right)^2+\ldots .+\frac{(p)(p+q) \ldots .[p+(r-1) q]}{1.2 .3 \ldots . . r}\left(\frac{x}{q}\right)^r+\ldots .\)

→ \((1+x)^{-p / q}=1-p\left(\frac{x}{q}\right)+\frac{(p)(p+q)}{1.2}\left(\frac{x}{q}\right)^2+\ldots \ldots\) \(+(-1)^r \frac{(p)(p+q) \ldots(p+(\overline{r-1}) q)}{1.2 \cdot 3.4 \ldots .(r)} \cdot\left(\frac{x}{q}\right)^r+\ldots .\)

→ \((1+x)^{p / q}=1+\frac{p}{1 !}\left(\frac{x}{q}\right)+\frac{(p)(p-q)}{1.2}\left(\frac{x}{q}\right)^2+\ldots\) \(+\frac{(p)(p-q)(p-2 q) \ldots \ldots(p-(\overline{r-1}) q)}{r !}\left(\frac{x}{q}\right)^r+\ldots .\)

→ \((1-x)^{p / q}\) లో Tr+1 = \((-1)^r \frac{(p)(p-q)(p-2 q) \ldots \ldots(p-(\overline{r-p}) q)}{r !}\left(\frac{x}{q}\right)^r\)

→ n ధన పూర్ణాంకం, x ఒక వాస్తవ సంఖ్య ; |x| < 1 అయితే

• (1 + x)^-n = \(1-\frac{n}{1 !} x+\frac{(n)(n+1)}{2 !} x^2+\ldots . .+\ldots .+(-1)^r \frac{(n)(n+1) \ldots .(n+r-1)}{r !} x^r\) \(+\ldots \infty=\sum_{r=0}^{\infty}(-1)^{r(n+r-1)} C_r \cdot x^r\)
• (1 – x)^-n = \(1+\frac{n}{1 !} x+\frac{(n)(n+1)}{2 !} x^2+\ldots .+\ldots .+\frac{(n)(n+1)(n+2) \ldots \ldots(n+r-1)}{r !} x^r\) \(+\ldots \infty=\sum_{r=0}^{\infty}{ }^{(n+r-1)} C_r \cdot x^r\)

→ x², ఆపై x ఘాతాలు ఉపేక్షించేంతగా, |x| చిన్నదయితే (1 + x)ⁿ = 1 + nx

→ x³, ఆపై x ఘాతాలు ఉపేక్షించేంతగా, |x| చిన్నదయితే, (1 + x)ⁿ = 1 + nx + \(\frac{(n)(n-1)}{1.2} x^2\)

→ x⁴, ఆపై x ఘాతాలు ఉపేక్షించేంతగా, |x| చిన్నదయితే, (1 + x)ⁿ = 1 + nx + \(\frac{(n)(n-1)}{2 !} x^2+\frac{(n)(n-1)(n-2)}{3 !} x^3\)

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 5 ప్రస్తారాలు-సంయోగాలు to solve questions creatively.

AP Intermediate 2nd Year Maths 2A ప్రస్తారాలు-సంయోగాలు Formulas

→ ప్రాథమిక సూత్రం: ఒక పని W₁ ను ‘m’ విభిన్న విధాలుగానూ, మరొక పని W₂ ను ‘n’ విభిన్న విధాలుగానూ చేయగలిగితే, ఈ రెండు పనులునూ ఒకేసారి ‘mn’ విభిన్న విధాలుగా చేయవచ్చు.

→ n అనేది ఋణేతర పూర్ణాంకం అయిన (i) 0! = 1 (ii) n! = n(n – 1)!

→ ఇచ్చిన వస్తువుల నుంచి (ఒకే విధంగా లేక విభిన్లు) కొన్ని లేదా అన్నీ ఎంచుకొని ఒక వరసలో (సరళరేఖలో) అమర్చడాన్ని ఒక ‘రేఖీయ ప్రస్తారం’ లేదా ‘ప్రస్తారం’ అంటాం.

→ n, r లు ధన పూర్ణాంకాలు, r ≤ n అయినప్పుడు ‘n’ విభిన్న వస్తువుల నుంచి ఒక్కొక్కసారి ‘r’ వస్తువుల చొప్పున తీసుకొంటే వచ్చే ప్రస్తారాల సంఖ్యను ⁿP_r తో సూచిస్తాం.
ⁿP_r = \(\frac{n !}{(n-r) !}\), 0 ≤ r ≤ n

→ n, r లు ధన పూర్ణాంకాలు, r ≤ n అయిన

• ⁿP_r = n . ^(n-1)P_(r-1), r ≥ 1
• ⁿP_r = (n) . (n – 1) ^(n-2)P_(r-1), r ≥ 2
• ⁿP_r = ^(n-1)P_r + r . ^(n-1)P_(r-1)

→ n అసరూప వస్తువుల నుంచి ‘r’ వస్తువుల చొప్పున తీసుకొంటే వచ్చే ప్రసారాలలో

• నిర్దేశించిన ఒక వస్తువు ఉండే ప్రస్తారాల సంఖ్య (r) ^(n-1)P_(r-1)
• నిర్దేశించిన ఒక వస్తువు లేని ప్రస్తారాల సంఖ్య ^(n-1)P_r
• నిర్దేశించిన ఒక వస్తువు నిర్ధేశించిన స్థానంలో ఉండే ప్రస్తారాల సంఖ్య = ^(n-1)P_(r-1)

→ A లో వున్న విభిన్న మూలకాల సంఖ్య n(A); B లో వున్న విభిన్న మూలకాల సంఖ్య n(B) అయితే n(A) ≤ n(B) అయినపుడు A నుండి B కు గల విభిన్న అన్వేక ప్రమేయాల సంఖ్య ^n(B)P_n(A)

→ n(A) = n(B) అయినపుడు A నుండి B కు గల ద్విగుణప్రమేయాల సంఖ్య n(A)!

→ A నుండి B కు గల ప్రమేయాల సంఖ్య [n(B)]^n(A)

→ పునరావృతాన్ని అనుమతించినపుడు n విభిన్న వస్తువులతో, r వస్తువులుండేటట్లు ఏర్పరచగల ప్రస్తారాల సంఖ్య n^r.

→ n విభిన్న శూన్యేతర 1, 2, 3, ……, 9 అంకెలను ఉపయోగించి పునరావృతం లేకుండా ఏర్పరచగల r స్థానాలు గల సంఖ్యల మొత్తం ^(n-1)P_(r-1) × (దత్త అంకెల మొత్తం) × (111…. r సార్లు).

→ పైన చెప్పిన అంశంలోని n విభిన్న పూర్ణాంకాలలో ‘0’ కూడా ఉన్నప్పుడు ఏర్పరచగల 7 స్థానాలున్న సంఖ్యల మొత్తం = ^(n-1)P_(r-1) × దత్త అంకెల మొత్తం × (111……. r సార్లు) ^(n-2)P_(r-2) × (దత్త అంకెలమొత్తం × (111 – (r – 1) సార్లు)

→ n విభిన్న వస్తువుల నుంచి వచ్చే వృత్తాకార ప్రస్తారాల సంఖ్య = (n – 1)!

→ n విభిన్న వస్తువులతో ఏర్పడే పువ్వుల దండలు, పూసల గొలుసులు వంటి వేలాడే రకం వృత్తాకార ప్రస్తారాల సంఖ్య \(\frac{1}{2}\) (n – 1)!

→ ఇచ్చిన n వస్తువులలో p వస్తువులు ఒక రకంగానూ, q వస్తువులు మరొక రకంగానూ, r వస్తువులు వేరొక రకంగానూ ఉంటూ మిగిలిన వస్తువులు విభిన్నంగా ఉంటే, ఈ n వస్తువులను అమర్చడం ద్వారా వచ్చే ప్రస్తారాల సంఖ్య \(\frac{n !}{p ! q ! r !}\)

→ n విభిన్న వస్తువుల నుండి ‘r’ వస్తువుల వంతున తీసుకొంటే వచ్చే సంయోగాల సంఖ్యను ⁿC_r తో సూచిస్తాం. మరియు ⁿC_r = \(\frac{n !}{(n-r) ! r !}\), 0 ≤ r ≤ n

→ n, r లు ధన పూర్ణాంకాలు, 0 ≤ r ≤ n అయితే ⁿC_r = ⁿC_(n-r)

→ n, r, s ధన పూర్ణాంకాలు 0 ≤ r ≤ n, 0 ≤ s ≤ n మరియు ⁿC_r = ⁿC_s అయితే r = s లేదా r + s = n అవుతుంది.

→ (m ≠ n) అయినపుడు (m + n) విభిన్న వస్తువుల నుండి m, n వస్తువులు ఉన్న రెండు భాగాలుగా విభజించే విధానాల సంఖ్య ^{(m + n)}C_m = ^{(m + n)}C_n = \(\frac{(m+n) !}{m ! n !}\)

→ ఇట్లే m, n, p లు విభిన్న ధన పూర్ణాంకాలయినప్పుడు (m + n + p) వస్తువులను m, n, p వస్తువులున్న 3 భాగాలుగా విభజించే విధానాల సంఖ్య \(\frac{(m+n+p) !}{m ! n ! p !}\)

→ ‘mn’ విభిన్న వస్తువులను ‘m’ సమభాగాలుగా (ఒక్కొక్క భాగంలో n వస్తువులుండే విధంగా) విభజించే విధాల సంఖ్య \(\frac{(m n) !}{(n !)^m m !}\)

→ (mn) విభిన్న వస్తువులను m వ్యక్తులకు సమానంగా పంచే విధాల సంఖ్య \(\frac{(m n) !}{(n !)^n}\)

→ ఒక రకం సరూప వస్తువులు p, మరొక రకం సరూప వస్తువులు q, వేరొక రకం సరూప వస్తువులు r ఇచ్చినపుడు వాటి నుంచి ఒకటి లేదా అంత కంటే ఎక్కువ వస్తువులను ఎంచుకొనే విధానాల సంఖ్య (p + 1) (q + 1) (r + 1) – 1.

→ m ధన పూర్ణాంకాం మరియు \(p_1^{\alpha_i} \cdot p_2^{\alpha_2} \ldots \ldots p_k^{\alpha_k}\) అవుతూ, p₁, p₂, ……., p_k లు విభిన్న ప్రధాన సంఖ్యలు α₁, α₂, ….., α_k ధన పూర్ణాంకాలు అయినపుడు m కు గల ధన భాజకాల సంఖ్య (α₁ + 1) (α₂ + 1) …… (α_k + 1), (1, m లతో కలిపి)

→ n విభిన్న వస్తువుల నుండి 0, 1, 2,…… లేదా n వస్తువులతో ఏర్పడే సంయోగాల సంఖ్య ⁿC₀ + ⁿC₁ + ⁿC₂ + …… + ⁿC_r = 2ⁿ

→ n విభిన్న వస్తువుల నుండి ఒకటి లేదా అంతకంటే ఎక్కువ వస్తువులతో ఏర్పడే సంయోగాల సంఖ్య = 2ⁿ – 1

→ n భుజాలున్న బహుభుజిలో కర్ణాల సంఖ్య = \(\frac{n(n-3)}{2}\)