Mahesh

Students can go through TS Inter 2nd Year Chemistry Notes 8th Lesson Polymers will help students in revising the entire concepts quickly.

TS Inter 2nd Year Chemistry Notes 8th Lesson Polymers

→ Polymers are the backbone of four major industries namely, plastics, elastomers, fibres, paints & varnishes.

→ Polymer is defined as very large molecule having high molecular mass.

→ Polymers are also called macromolecules.

→ The repeating structural unit of a macro-molecule is called a monomer.

→ The process of formation of polymers from respective monomers is called polymerisation.

→ Based on source polymers are classified into natural polymers, semisynthetic polymers and synthetic polymers.

→ Based on structure, polymers are classified into linear polymers, branched chain polymers and cross linked or network polymers.

→ Cellulose, starch, rubber are examples for natural polymers.

→ Cellulose acetate (rayon) and cellulose nitrate are examples for semisynthetic polymers.

→ Buna – S, polythene, nylon 6,6 are examples for synthetic polymers.

→ HDP, PVC etc., are examples for linear polymers.

→ LDP is an example for branched chain polymers.

→ Bakelite, melamine etc., are examples for cross linked polymers.

→ On the basis of mode of polymerisation polymers are classified into addition polymers and condensation polymers.

→ An addition polymer is formed by the repeated addition of monomer molecules containing double bonds.

→ Polymers formed by the polymerisation of a single monomeric species are called homo-polymers. Ex.: Polythene.

→ Polymers formed by the polymerisation of two different monomeric species are called copolymers. Ex.: Buna – S, Buna – N.

→ Polymers formed by the repeated condensation reaction between two different monomeric species having two or three functional groups are called condensation polymers. Ex.: Nylon 6,6. ’

→ On the basis of inter molecular forces, polymers are classified into

• Elastomers,
• Fibres,
• Thermoplastic polymers and
• Thermosetting polymers.

→ Elastomers are rubber-like solids with elastic properties. Ex.: Buna – S, Buna – N etc.

→ Fibres are thread forming solids with high tensile strength and high modulus.
Ex.: Terylene, Nylon 6,6.

→ Thermoplastic polymers are those which soften on heating and become rigid again on cooling. Ex.: Polythene, polystyrene.

→ Thermosetting polymers are those which become hard on heating. They cannot be softened on heating. Ex. : Bakelite, urea- formaldehyde resins.

→ Addition polymerisation takes place through the formation of either free radicals or ionic species.

→ Alkyl aluminium and titanium chloride is called Ziegler-Natta catalyst.

→ The earliest Ziegler – Natta catalysts were combinations of titanium chloride, TiCl₄ and diethyl aluminium chloride [(C₂H₅)₂ Al Cl].

→ There are two types of polythene.

• Low density polythene (LDP) and
• High density polythene (HDP).

→ Polytetra fluoroethene (Teflon) is chemically inert and resists attack by corrosive reagents. It is used in making oil seals, gaskets etc.

→ Polyacrylonitrile is a substitute for wool in making commercial fibres, such as orlon.

→ Condensation polymerisation is also called step growth polymerisation.

→ Terylene or dacron is formed by the interac-tion of ethylene glycol and terephthalic acid.

→ Nylon 6,6 is obtained by the condensation polymerisation of hexamethylene diamine with adipic acid under high pressure and at high temperature. It is used in making sheets, bristles for brushes, textiles etc.

→ Nylon 6 or perlan – L is obtained by heating caprolactam with water at high temperature. It is used in the manufacture of tyre cords, fabrics and ropes.

→ Phenol – formaldehyde polymers are obtained by the condensation reaction of phenol with formaldehyde in the presence of either an acid or a base catalyst. The initial product is a linear product – Novolac used in paints.

→ Novolac on heating with formaldehyde undergoes cross linking to form an infusible solid mass called bakelite.

→ Bakelite is used in making combs, electrical switches etc.

→ Melamine formaldehyde polymer is formed by the condensation polymerisation of melamine and formaldehyde. It is used in making unbreakable crockery.

→ Buna – S is a copolymer formed by the polymerisation reaction of 1, 3 – butadiene and styrene.

→ Buna – S is used in the manufacture of autotyres, footwear components, cable insulation etc.

→ Natural rubber is a linear polymer of isoprene.It is also called Cis-1,4-polyisoprene.

→ The process of heating a mixture of raw rubber with sulphur and an appropriate additive at a temperature range between 373 K to 415 K is called vulcanisation. On vulcanisation sulphur forms cross links at the reactive sites of double bonds and thus rubber becomes hard.

→ Neoprene is a synthetic rubber obtained by the free radical polymerisation of chloroprene.

→ Buna – N is obtained by the co-polymerisation of 1, 3 – butadiene and acrylonitrile in the presence of a peroxide catalyst. It is used in making oil seals, tank lining etc.

→ The average molecular masses of polymers are expressed as

• Number average molecular mass (M̅_n) and
• Weight average molecular mass (M̅_w)

→ Both M̅n and M̅w have no units.

→ The ratio between weight average molecular mass (M̅_w) and number average molecular mass (M̅_n) of a polymer is called Poly Dispersity Index (PDI).

→ Poly β – hydroxybutyrate – co-β-hydroxy valerate (PHBV) and Nylon 2 – nylon 6 are examples for biodegradable polymers.

→ PHBV is obtained by the copolymerisation of 3 – hydroxybutanoic acid and 3 – hydroxy- pentanoic acid. PHBV is used in speciality packaging, orthopaedic devices and in con¬trolled release of drugs.

→ Nylon 2 – nylon 6 is an alternating polyamide copolymer of glycine and amino ceproic acid.

→ Polypropene is used in the manufacture of ropes, toys, pipes etc.

→ Polyvinyl chloride (PVQ is used in the manufacture of rain coats, hand bags, water pipes etc.

→ Urea-formaldehyde resin is used in making unbreakable cups and laminated sheets.

→ Glyptal obtained by the polymerisation of ethylene glycol and phthalic acid is used in the manufacture of paints and lacquers.

→ Bakelite obtained by the polymerisation of phenol and formaldehyde is used in making combs, electrical switches, computer discs etc.

Students can go through TS Inter 2nd Year Chemistry Notes 7th Lesson d and f Block Elements & Coordination Compounds will help students in revising the entire concepts quickly.

TS Inter 2nd Year Chemistry Notes 7th Lesson d and f Block Elements & Coordination Compounds

→ Elements with partially filled d-orbitals in the penultimate shell in their atoms or in the ions in principal oxidation state are called transition elements.

• 3d series contain Sc (21) to Zn (30).
• 4d series contain Y(39) to Cd (48).
• 5d series contain La (57) to Hg (80).
• The fourth row of 6d is still incomplete.

→ General electronic configuration of d-block elements is (n – 1) d^1-10 ns^1-2.

→ Transition elements exhibit different oxidation states, formation of aqueous coloured ions, and ability to form complex compounds with a variety of ligands.

→ Transition metals have high melting points. The high melting points are attributed to the involvementtsf greater number of electrons in the interatomic metallic bonding.

→ As shielding effect of d electrons is poor, net electrostatic attraction between the nuclear charge and the outermost electron increases and the ionic radius decreases.

→ There is a gradual size decrease from La (57) to Lu (71) which is known as lanthanoid contraction.

→ Transition elements show variable oxidation states due to involvement of ns² electrons and some or all electrons in (n – 1) d ele¬ctrons as the energy difference between ns and (n – 1) d sublevels is less.

→ Paramagnetism arises from the presence of unpaired electrons. The magnetic moment is determined by spin only formula.
μ = \(\sqrt{n(n+2)}\)
where n is the no. of unpaired electrons and μ is the magnetic moment in Bohr magneton.

→ The colour exhibited by transition metal ions is due to d-d transitions.

→ Transition metal ions form complex compounds. This is due to the comparatively smaller sizes of the metal ions, their high ionic charges and the availability of d-orbi- tals for bond formation.

→ Zn++ is colourless due to non-availability of unpaired d-electrons.

→ Tetrahedral complexes do not show geometrical isomerism because the relative positions of unidentate ligands attached to central metal atom are the same with respect to each other.

→ When (n – 1) d orbitals participate in hybridisation, inner orbital complex form. They are called low spin complexes. When d- orbitals of outer shell participate, outer orbital complexes form. They are high spin complexes.

→ Valence bond explains the formation of complexes. Central metal ion makes available vacant d-orbitals. Ligands donate electrons to form coordinate covalent bonds.
(n – 1) d, ns, np or ns, np, nd orbitals hybridise to form of set of equivalent orbitals of definite geometry such as octahedral, square planar and so on. These hybridised orbitals overlap with ligand orbitals that can donate electron pairs for bonding.

→ In the presence of ligands, d-orbitals are split into two groups.

→ Ligands which cause large splitting are called strong field ligands.

→ Ligands which cause low value for splitting are called weak field ligands.

→ The metal-carbon bond in metal carbonyls possess both σ and π character. The M – C σ bond is formed by the donation of the pair of electrons on the carbonyl carbon into a vacant orbital of the metal. The M – C π bond is formed by the donation of a pair of electrons from a filled d-orbital of metal into the vacant antibonding π* orbital of carbon monoxide.

→ The instability constant is the reciprocal of formation constant

→ Hardness of water is estimated using simple titration with Na₂EDTA. The Ca²⁺ and Mg²⁺ ions form stable complexes with EDTA.

→ Coordination compounds are used as catalysts for many industrial processes.
Ex : Rhodium complex [(Ph₃P)₃ RhCl], is used for hydrogenation of alkenes:

Students can go through TS Inter 2nd Year Chemistry Notes 6th Lesson P-Block Elements will help students in revising the entire concepts quickly.

TS Inter 2nd Year Chemistry Notes 6th Lesson P-Block Elements

→ Group 15 elements include Nitrogen, Phos-phorus, Arsenic, Antimony and Bismuth.

→ Nitrogen and Phosphorous are non-metals, arsenic and antimony are metalloids. Bismuth is a typical metal.

→ Nitrogen is inert due to high bond dissociation enthalpy of N = N.

→ Stability and basicity of hydrides decreases from NH₃ to BiH₃.
NH₃ > PH₃ > ASH₃ > SbH₃ > BiH₃

→ Pentoxides are more acidic than trioxides N₂O₅ > N₂O₃.

→ Pentahalides are more covalent than trihalides because the element in the higher oxidation state exerts more polarising power.

→ Ammonia is prepared by Haber s process. N₂, and H₂, combine to give ammonia. Low tem¬perature and high pressure are required for high yield.

→ White phosphorus consists of tetrahedral P₄ molecules held by van der Waal’s forces.

→ Holme’s signal uses calcium carbide and calcium phosphide.

→ Oxoacids of phosphorus are

• • Orthophosphoric acid H₃PO₄
  • Pyrophosphoric acid H₄P₂O₇
  • Orthophosphorus acid H₃PO₃
  • Hypophosphorus acid H₃PO₂

→ Oxygen, Sulphur, Selenium, Tellurium, and Polonium constitute Group 16 of the periodic table. They are known as chalcogens as they are ore forming.

→ Oxygen cannot show higher oxidation state due to non-availability of d-orbitals in outermost shell.

→ Stability of hydrides decreases H₂O > H₂S > H₂Se > H₂Te > H₂Po.

→ The acidic nature of hydrides of Group 16 increases. H₂O < H₂S < H₂Se < H₂Te < H₂Po.

→ Molecular oxygen is paramagnetic. It contains two unpaired electrons in antibonding π* orbitals.

→ Ozone is prepared by passing silent electric discharge through pure and dry oxygen.

→ Stable form of sulphur at room temperature is rhombic sulphur.

→ Monoclinic sulphur is stable above 369K. At 369K both Rhombic and Monoclinic sulphur are stable. This temperature is called transition temperature.

→ H₂SO₄ is prepared by contact process.

→ H₂SO₄ is needed for manufacture of hundreds of compounds and also in manufacture of fertilisers.

→ Fluorine, Chlorine, Bromine, Iodine and Astatine are members of Group 17. They are known as halogens.

→ All these elements have ns2np5 electronic configuration. .

→ All the halogens are highly reactive. They behave as oxidising agents.

→ Stability of Hydrogen halides decreases down the group due to decrease in bond dissociation enthalpy.
HF > H – Cl > H-Br > H – I

→ The acidic strength of hydrogen halides varies in the order HF < HCl < HBr < HI. → HF has highest boiling point and it is the only liquid. Others are gases. Order of B.Ps. HF > HI > HBr > HCl
The high boiling point of HF is due to hydrogen bonds among HF molecules.

→ In OF₂ and O₂F₂ oxygen has + 2 and +1 oxida¬tion states. They are termed oxygen fluorides. O₂F₂ oxidises plutonium to P₄F₆. It is used in removing plutonium as P4F6 from spent nuclear fuel.

→ Chlorine forms Cl₂O, ClO₂, Cl₂O₆ and Cl₂O₇.
ClO₂ is used as a bleaching agent for paper Pulp.

→ I₂O₅ is a very good oxidising agent and is used In the estimation of carbon monoxide.
I₂O₅ + 2CO → Ir₅CO₂

→ Halogens combine amongst themselves to form a number of compounds known as Interhalogens. They are classified as :
AX, AX₃, AX₅, AX₇ where A is bigger halogen.

→ Interhalogens are more reactive than halogens. A – X bond is weaker than X – X bond.

→ ClF₃ is T-shaped. BrF₅ is square pyramidal and IF. has pentagonal bipyramidal.

→ Among the different oxoacids of chlorine, the acidic character follows the order HOCl < HClO₂, < HClO₃, < HClO₄.

→ The acidic character of hypohalous acids varies as HClO > HBrO > HIO.

• ICl + H₂O → HCl + HOI
• ICl₃ + 2H₂O → 3HCl + HIO₂
• BrF₅ + 3H₂O → 5HF + HBrO₃

→ Group 18 consists of six elements. They are Helium, Neon, Argon, Krypton, Xenon and Radon.

→ All noble gases occur in atmosphere except radon.

→ Their atmospheric abundance in dry air is – 1 % by volume of which argon is the major constituent.

→ The commercial source of Helium is natural gas.

→ Xenon and Radon are the rarest elements of the group.

→ Radon is obtained as a decay product of
²²⁶₈₈Ra → ²²⁶₈₆Rn+ ⁴₂He

→ All noble gases have general electronic con-figuration ns np6 except Helium which has 1s².

→ They have large positive values of electron gain enthalpy.

→ Helium has the lowest boiling point (4.2K).

→ Among a few compounds of Krypton (KrF₂) has been studied in detail.

→ Xenon forms three binary fluorides XeF₂, XeF₄ and XeF₆ by the direct reaction under appropriate experimental conditions.

→ Helium is non-inflammable and light gas. It is used in filling balloons for meteorological observations. It is also used in gas cooled nuclear reactors.

→ Neon is used in discharge tubes and fluoroscent bulbs.

→ Argon is used mainly t0 provide an inert atmosphere in high temperature metallurgical processes and for filling electric bulbs. It is also used in labs for handling air sensitive substances.

→ There are no significant uses of Xenon and Krypton. They are used in light bulbs designed for special purposes.

Students can go through TS Inter 2nd Year Chemistry Notes 5th Lesson General Principles of Metallurgy will help students in revising the entire concepts quickly.

TS Inter 2nd Year Chemistry Notes 5th Lesson General Principles of Metallurgy

→ The naturally occurring chemical substances in the earth’s crust which are obtained by mining are known as minerals.

→ Very few minerals are chemically and com-mercially viable to be used as sources of extraction are known as ores.

→ Ores are usually associated with earthy materials called gangue.

→ The process of extraction and isolation of the metal from its naturally occurring com-pounds is called metallurgy.

→ Extraction involves the following major steps.

• Concentration of the ore.
• Isolation of the metal from the concentrated ore by chemical or electrochemical methods.
• Purification of the metal.

→ Removal of the gangue from the ore is known as concentration, dressing or benefaction.

→ Sulphide ores are concentrated by froth floatation process.

→ If either the ore or the gangue is a magnetic substance, magnetic separation method is used.

→ Isolation of metals from concentrated ore involves calcination or roasting and reduction.

→ Reduction of the metal oxide involves heating it with reducing agents.
C or CO or A/ or any other active metal.

→ Ellingham diagram consists of plots of Δ_fG° Vs T for formation of oxides of elements.

→ Such diagrams help us in predicting the feasibility of thermal reduction of ore.

→ The criterion of feasibility is that at a given temperature, Gibbs energy of the reaction must be negative.

→ Iron obtained from Blast furnace contains 4% carbon and many impurities in smaller amounts (eg : P, S, Si, Mn). This is known as pig iron.

→ Copper is extracted from CuFeS₂ FeS is oxidised to FeO and is removed as FeSiO₃.

→ Bauxite is leached with NaOH which removes impurities.

→ A metal extracted by any method is usually contaminated with some impurity. For obtaining metals of high purity, the follow¬ing methods are in vogue.

→ Distillation : Useful for low boiling metals like zinc and mercury containing high boiling metals as impurities.

The extracted metal in the molten state is distilled to obtain the pure metal as distillate.

→ Liquation: Low melting metal like tin can be made to flow on a sloping surface leaving behind high melting inpurities.

→ Zone refining: Impurities are more soluble in the melt than in the solid state of the metal. As the heater moves forward the pure metal crystallises out of the melt and the impurities pass on to the adjacent molten zone. Impurities get concentrated at one end. This end is cut off.

→ Aluminium foils are used as wrappers for Chacolates.

→ Copper alloys are tougher than the metal itself. Ex : Brass (with zinc, and bronze (with tin).

→ Zinc is used for galvanising iron.

→ Cast iron is used for casting stoves, railway sleepers etc.

Students can go through TS Inter 2nd Year Chemistry Notes 4th Lesson Surface Chemistry will help students in revising the entire concepts quickly.

TS Inter 2nd Year Chemistry Notes 4th Lesson Surface Chemistry

→ The phenomenon of attraction and conse-quent accumulation or adherence of mole-cules of a substance on the surface of a liquid or solid is called adsorption.

→ The phenomenon of concentration of mole-cules of a gas or a liquid on a surface of solid or liquid is called absorption.

→ The substance adsorbed on the surface of a liquid or solid is called adsorbate.

→ The substance on whose surface the adsorption occurs is called adsorbent. Eg. in the adsorption of acetic acid on charcoal adsor-bate is acetic acid and charcoal is adsorbent.

→ Adsorption is two types Physical adsorption and Chemical adsorption.

→ The adsorption due to physical forces or – van der Waals’ forces is known as physical adsorption. This is also known as van der Waals’ adsorption.

→ If the adsorption is due to chemical attractive forces, it is known as chemical adsorption.

→ The amount of a gas adsorbed on metal surface or a solid depends on

• surface area of the adsorbent
• nature of gas
• pressure of gas
• temperature.

→ The graph showing relation to pressure (p) and the ratio of masses of adsorbate (x) and adsorbent (m) at a constant temperature is called adsorption isotherm.

→ Freundlich adsorption isotherm is \(\frac{x}{m}\) = Kp^1/n which indicates that at any given temperature the amount of gas (x) adsorbed by unit mass of the adsorbent (m) is related to this adsorption equilibrium pressure (p).

→ The graph showing the relation between x/m and temperature at constant pressure is called as adsorption isobar.

→ The substances which accelerates the rate of a reaction is known as positive catalyst. Positive catalyst increase the rate of reac-tion by lowering the activation energy.

→ Substances which decreases the rate of a reaction or retard the reaction are called negative catalyst or inhibitors.

→ Substance which itself has no catalytic acti-vity but will increase the activity of catalyst is called promoter or activator. A promotor thus be regarded as a catalyst for catalyst e.g. Al₂O₃ K₂O, MO powder act as promoter to iron catalyst in Haber process for the manufacture of ammonia.

→ The substances which reduces or even com-pletely destroy the activity of a catalyst are known as catalytic poisons or anticatalysts.

→ If one of the products or intermediate formed in a reaction acts as catalyst for the same reaction it is known as autocatalyst and the phenomenon is known as autocatalysis.

→ If the catalyst and reactants are not in the same phase, then it is known as heterogen-eous catalysis. Generally the catalyst is solid while the reactants are liquids or gases. Most of the solid catalysts are either transition metals or their compounds.

→ Zeolites can act as molecular seives which allow certain molecules which are smaller than the pore size to pass through them while the molecules bigger than the pore size cannot pass through them.

→ Enzymes are protein molecules which are complex nitogeneous organic compounds which are present in the living plants and animals.

→ Metal ions such as Na+, Mn²⁺, Co²⁺, Cu²⁺ etc which activate the enzyme catalysis are called activators.

→ If the particle size of the solute in the binary system is in the range of 1 μ to 1 mμ a colloidal solution is formed.

→ In the colloids the phase constituting the colloidal particles is called dispersion phase while the medium in which the colloidal particles are distributed is called dispersion medium. The dispersion medium or disper¬sion phase may be solid or liquid.

→ The dispersion medium in a colloid is named as lyo. Colloidal solutions in which the parti¬cles of the dispersed phase have great affinity for the dispersion medium are called lyophilic colloids. The colloidal solutions in which particles of the dispersed phase have no affinity for the dispersion medium are called lyophobic colloids.

→ If water is dispersion medium lyophilic colloids are called hydrophilic colloids while lyophobic colloids are called hydrophobic colloids.

→ The process of transferring back a preci-pitate into colloidal form is called peptisation. It is the reverse of coagulation.

→ When a beam of light is made to fall on a colloidal solution, the path of the beam of light is illuminated by a bluish light. The luminosity by the path of beam is known Tyndall effect. It is due to the scattering of light from the surface of the colloidal par¬ticles. The scattering of light due to the absorption of light energy and then scatter the light of shorter wavelength.

→ The constant rapid zig-zag of colloidal particles in a collodial solution is known as Brownian movement named after its discoverer.

→ The migration of electrically charged sol particles under an applied electric field is called electrophoresis strictly cataphoresis or anaphoresis according to the electrode to which the particles move.

→ The potential difference between the fixed layer and the diffused layer having oppo-site charge is called the electro kinetic potenital or zeta potential and the double layer is called electrical double layer.

→ At a particular pH the sol particles become neutral and exhibit no movement in an ele-ctric field and this pH is called isoelectric point.

→ When electrophoresis of dispersed particles is prevented by suitable means the medium can be made to move under the influence of an applied potential. This phenomenon is referred to as electro osmosis.

→ Coagulation or flocculation is the process of breaking up of a colloidal solution by which colloidal particle come close and result in the precipitation of the dispersed phase.

→ If a colloidal solution is stored for longer periods smaller particles dissolved and crystallise out on the larger particles causing the coagulation of colloid. This is known as ageing.

→ The minimum concentration of an electrolyte which is able to cause coagulation or flocculation of sol is termed as flocculation value.

→ The coagulation of lyophilic colloid by adding salt solutions of high concentrations is called ‘salting out’. This may be due to the removal of solvent layer around the lyophilic colloid by the salt.

→ The process by which the sol particles are prevented from coagulation by electrolyte due to the previous addition of some lyophilic sol is called protectioii of colloid.

→ The number of milligrams of protective colloid which just prevents the coagulation of 10 ml of given gold sol when 1 mL of 10% solution of NaCl is added to it is called gold number.

→ A colloidal system involving one liquid dispersed in another is known as emulsion.

→ The substances that stabilises an emulsion are called emulsifier or emulsifying agent.

→ Emulsions are two types

• oil in water (0/w)
• water in oil (w/o) depending on the dispersion medium and dispersion phase.

Students can go through TS Inter 2nd Year Chemistry Notes 3rd Lesson Electrochemistry and Chemical Kinetics will help students in revising the entire concepts quickly.

TS Inter 2nd Year Chemistry Notes 3rd Lesson Electrochemistry and Chemical Kinetics

→ In an electrolytic cell electrical energy is converted into chemical energy while in Galvanic cell or Voltaic cell chemical energy is converted into electrical energy.

→ In both galvanic and electrolytic cells oxidation take place at anode and reduction take place at cathode.

→ Nernst equation gives the dependence of the electrode. Potential on the concentration of ions with which the electrode is reversible.

→ For a metal electrode the reduction reaction is Mnⁿ⁺ + ne^– ⇌ M. The Nernst equation to calculate the electrode potential at different concentration is
E = E° – \(\frac{2.303 \mathrm{RT}}{\mathrm{nF}}\) log \(\frac{[\mathrm{M}]}{\left[\mathrm{M}^{\mathrm{n}+}\right]}\)
E = E° – \(\frac{2.303 \mathrm{RT}}{\mathrm{nF}}\)log C
Since M is solid, its activity will be unity.

→ For a non – metal electrode the reduction reaction is A + ne^– ⇌ A^n-. The Nernst equation to calculate electrode potential at different concentration is
E = E° – \(\frac{2.303 \mathrm{RT}}{\mathrm{nF}}\) log\(\frac{\left[\mathrm{A}^{\mathrm{n}-}\right]}{\left[\mathrm{A}^{-}\right]}\) or
E = E – \(\frac{2.303 \mathrm{RT}}{\mathrm{nF}}\) log [A^n-]

→ General equation for electrode potential of any electrode is
E = E° – \(\frac{2.303 \mathrm{RT}}{\mathrm{nF}}\) log\(\frac{\text { [Products }]}{[\text { Reactan } t s]}\)

→ For a cell reaction of the type aA + bB ⇌ cC + dD
The Nernst equation
E = E° – \(\frac{0.0591}{n}\) log \(\frac{[\mathrm{C}]^{\mathrm{c}}[\mathrm{D}]^{\mathrm{d}}}{[\mathrm{A}]^{\mathrm{a}}[\mathrm{B}]^{\mathrm{b}}}\)

→ From the standard Gibbs energy the equilibrium constant by the equation
ΔG° = – RT /n K.

→ Substances whose melts or aqueous solutions conduct electric current are called electrolytes. All salts, acids and based are referred to as electrolytes.

→ Substances whose melts or aqueous solutions do not conduct are called non – electrolytes e.g. non – polar covalent substances like urea, glucose, sugar etc., are non – electrolyte.

→ The flow of electrons across the boundary is accompanied by chemical reaction i.e., oxidation-reduction. Such a reaction is called electrolysis. Electrolysis take place only at electrodes.

→ If a substance ionise completely, it is known as strong electrolyte while the substances which show very little ionisation in solution are called weak electrolytes.

→ Kohlrausch’s law states that the equivalent conductivity of an electrolyte at infinite dilution (λ₀) is the sum of the equivalent conductivities of the cation and anions.
λ₀ = λ₀⁺ + λ₀^–
where λ₀⁺ is the ionic conductance of
the cation and λ₀^– is the ionic conductance of the anion.

→ Kohlrauch law is also useful to calculate As for any electrolyte from the X° of individual ions.

→ Equivalent conductance of weak electrolytes can be calculated from the conductances of completely dissociated strong electrolytes
e.g. Λ°_CH3COOH = Λ°_CH3COONa + Λ°_HCl – Λ°_NaCl

→ First law : The amount of the substance liberated or deposited or dissolved at an electrode during electrolysis of an electro¬lyte is directly proportional to the quantity of electricity passing through the solution of electrolyte or the melt.

→ Electrochemical equivalent of a substance is the amount of substance deposited or liberated or dissolved or underwent electrode reaction at an electrode by passing one ampere current for one second i.e., one coulomb.

→ One Faraday i.e., 96,500 coloumbs is equal to the charge present on one mole
(6.023 × 10²³) of electrons or protons.
m = \(\frac{E \times c \times t}{96500}\)
∴ e = \(\frac{E}{96500}\)

→ Second law of Faraday states that, if the same quantity of electricity is passed through different electrolyte solutions or melts the amount of the different substances liberated or deposited or dissolved or had undergone reaction at the electrode are directly pro¬portional to their chemical equivalents.
\(\frac{W_1}{E_1}=\frac{W_2}{E_2}=\frac{W_3}{E_3}\)

→ A primary cell is a cell in which the electrical energy is obtained at the expense of chemical reactions. A primary cell works as long as the active chemicals reacting. The dry cell which generates a voltage of ~ 1.25 – 1.50V is an example of the primary cell.

→ A secondary cell is a cell in which the electrode reactions are reversed by the appli¬cation of an external current. Hence such a cell once used can be recharged.

→ Corrosion is a process of deterioration and consequent loss of solid metallic material through an unwanted chemical or electro-chemical attack by its environment, starting at the surface.

→ Corrosion is prevented by coating the metal surface with a thin film of paint, grease, metal (eg : Zn, Sn, Ni, Cu, Cr), metal oxide (eg: Fe₃O₄).

→ The rate of reaction is defined as the change in molar concentration of reactant or product per unit time.

→ At anytime the rate of reaction depends on the concentration of the reactants at that instant.

→ The units of rate of reaction are moles litre^-1 sec^-1 or moles litre^-1 min^-1 or moles litre^-1 hr^-1.

→ The rate of reaction is directly proportional to the (concentration of the reactants)11 or Cn.

→ For gaseous reactions the rate of reaction is directly proportional to (pressure of the reactants)”.

→ Generally rise of 10°C in temperature doubles the specific rate of reaction.

→ Catalyst increases the rate of reaction by changing the path of the reaction i.e., mak-ing the reaction to proceed in the path pf lower activation energy.

→ The mathematical expression of the rate of reaction on concentration terms of reac¬tants is known as rate expression or rate equation or rate law.

→ Units of rate constant are mole^1-n litre^n-1 sec^-1.

→ The total number of reactant molecules tak-ing part in the slowest step or rate limiting step or in the formation of intermediate species is known as molecularity of the reaction.

→ The order of a reaction is the total number of molecules whose concentrations changes during the chemical reaction or the sum of the powers of the concentration terms in the rate equation.

→ Reactions in which the rate of reaction is independent of the concentration of the reacting substances is called zero order reactions.

→ The reactions which appear to be higher order actually follow lower order Kinetics are called Pseudo Chemical reactions e.g. acid catalysed ester hydrolysis and hydrolysis of cane sugar to give glucose are pseudo first order reactions.

→ The reactions in which the products formed in the first stage may react with each other or with the original reactant to give new products are known as consecutive reactions.

→ The reactions proceeding in a series of successive reactions initiated by a suitable primary process are called chain reactions.
e.g. Formation of HCl from H₂ and Cl₂ and Chlorination of alkanes.

→ The minimum energy which must be associated with reactant molecules so that their mutual collision result in a chemical reaction is called Threshold energy.

→ Collisions which yield the product are called effective collisions or fruitful collisions or activated collisions.

→ The difference in energy between the threshold energy and the energy of the normal colliding molecules is known as activation energy.
Activation energy = Threshold energy – Energy of normal colliding molecules

→ The ratio of the rate constants at two different temperatures (preferably 35°C and 25°C) is known as temperature coefficient.

Students can go through TS Inter 2nd Year Chemistry Notes 2nd Lesson Solutions will help students in revising the entire concepts quickly.

TS Inter 2nd Year Chemistry Notes 2nd Lesson Solutions

→ A homogeneous mixture of two or more substances whose composition can be varied with certain limits is known as true solution e.g. salt in water, sugar in water, air, alloys etc.

→ Gaseous solutions are those in which solvent is gas, solute may be gas or liquid or solid. Eg:

• Gas in gas : air, mixture of O₂ and N₂
• Liquid in gas: moisture in air
• Solid in gas : camphor in air.

→ Liquid solutions are those in which solvent is liquid while solute may be gas or liquid or solid. Eg:

• Gas in liquid: aerated water, soda water
• Liquid in liquid : alcohol in water
• Solid in liquid: salt in water, sugar in water.

→ Solid solutions are those in which solvent is solid and solute may be gas or liquid or solid. Eg:

• Gas in liquid: occlusion of H₂ in palladium
• Liquid in solid : amalgams (liquid Hg in Zn)
• Solid in solid: Alloys like brass, bronze etc.

→ Mole fraction is the ratio of the number of moles of one component to the total number of moles (solute and solvent) in a binary solution.

→ In a solution, the sum of mole fraction of all components = 1

→ Molarity (M) is the number of moles of solute present in one litre of solution.

→ Normality is the number of gram equivalents of solute present in one litre of solution, represented by ‘N’.

→ Relation between molarity (M) and normality (N) of any solution are related as
Molarity × \(\frac{\text { Molecular Weight }}{\text { Equivalent Weight }}\) = Normality
Molarity = \(\frac{\text { Normality } \times \text { Equivalent Weight }}{\text { Molecular Weight }}\)

→ Henry’s law states that the mass of gas dissolving in a given amount of liquid is directly proportional to the pressure of the gas above the liquid at equilibrium.
m ∝ p or m = k.p (k is constant) k value depends on the nature of the gas, nature of the solvent,.temperature and the units of pressure.

→ Raoult’s states that partial pressure of a component (say liquid A) in solution is proportional to the mole fraction.

→ The decrease in the vapour pressure of a liquid when a non – volatile solute is dissolved in it is called the lowering of vapour pressure.

→ The ratio of the lowering of vapour pressure showing (p – pg) to the vapour pressure of the pure solvent (p) is known as the relative lowering of vapour pressure.

→ Raoult’s law states that the relative lowering of vapour pressure of a dilute solution of a non-volatile solute is equal to the mole fraction of the solute.
\(\frac{p-p_s}{p}=\frac{n}{n+N}\) where n and N are the moles of solute and solvent in a solution.

→ The properties of dilute solutions which depend on the number of particles (ions or molecules) of the solute dissolved in the solution are called colligative properties.

→ Lowering of vapour pressure, elevation of boiling point, depression in freezing point and osmotic pressure are colligative properties.

→ Since vapour pressure of a solution is less than the vapour pressure of pure solvent, the solution boils at a higher temperature than the boiling point of pure solvent and is known as elevation of boiling point.

→ Since the vapour pressure of a solution is less than the vapour pressure of pure solvent, the solution freezes at a lesser temperature than the freezing point of pure solvent and is known as depression in freezing point.

→ The inflow of solvent from a dilute solution into the concentrated solution of the solute when the two solutions are separated by a semipermeable membrane is also called as osmosis.

→ According to vant Hoff, all laws that applicable to gases are also applicable to dilute solutions and these laws are called vant Hoffs laws.

→ Solutions of same osmotic pressure at a given temperature are called isotonic solutions.

→ The molecular weights determined by using colligative properties of substances which associate or dissociate will be abnormal and are called abnormal molecular weights.

→ The ratio of the observed colligative property and calculated colligative property is called vant Hoff factor ‘i’.

Students can go through TS Inter 2nd Year Chemistry Notes 1st Lesson Solid State will help students in revising the entire concepts quickly.

TS Inter 2nd Year Chemistry Notes 1st Lesson Solid State

→ Solids are of two types. They are crystal-line solids and amorphous solids.

→ Crystalline solids are of four types. They are :

• Covalent solid
• Ionic solid
• Molecular solid and
• Metallic solid.

→ Depending on the type of attractive forces between molecules the molecular solids are again categorised into three types :

• Non – polar molecular solids: In which atoms such as Ar, Kr, Xe etc., or molecules of covalent compounds are held together by weak van der Waal’s forces. These solids have low m.pts and the particles are widely separated than in close packed ionic or metallic lattices.
• Polar molecular solids: In which molecules are held together by relatively stronger dipole-dipole interactions. These solids are soft and non – conductors of electricity. Their m.pts are higher than those of non – polar molecular solids e.g., solid CO₂.
• Hydrogen bonded molecular solids : are those in which molecules participate in hydrogen bonding. In crystals of benzoic acid, hydrogen bonds cause the association into dimers which are then held together van der Waal’s forces. These are also non – conductor of electricity.

→ The smallest repetitive unit of a crystal lattice which is used to describe the lattice is called the unit cell. Crystals possess the same symmery as their constituent unit cells.

→ In a simple or primitive cubic lattice the lattice points are located at the corners of each unit cell and can contribute only 1/8 of each particle at the corner to the unitj cell shared by unit cells in space lattice, So a simple cubic unit cell has particle per unit cell.

→ In a body centred cubic unit cell particles are located at the centre of the ceil as well as at the corners.
8 (at corners) × \(\frac{1}{8}\) + 1 (at body centre) × 1
= 2 particles.

→ In a face centred cubic unit cell atoms are found at the centre of the six faces of the cell as well as at each of the eight corners. The number of particles per unit cell in fee is
6 (at centre of each face) × \(\frac{1}{2}\) + 8 (at corners) × \(\frac{1}{8}\) (at comers) = 4 particles.

→ In hep and ccp structures the coordination number i.e., the number of surrounding atoms in contact with atom is 12.

→ The void created when six spherical particles are contact with each other is called octahedral void or octahedral hole.

→ The void created when four spherical particles are in contact with each other called tetrahedral void or tetrahedral hole.

→ In a close packed structure of N atoms there are 2N tetrahedral voids and N octahedral voids because the octahedral voids are larger than tetrahedral voids.

→ When X – rays are incident on a crystal face, they are reflected by the atoms in different planes.

→ Bragg’s equation is useful to calculate the distance between the repeating planes of particles in crystals from the reflected x – rays. It is, nλ = 2d sin θ where n is an integer like 1, 2, 3 and represents order of reflection, λ is the wavelength of the x – rays used and d is the distance between the repeating places.

→ A Schottky defect consists of a pair of holes in the crystalline lattice due to the absence of one positive ion and one negative ion.

→ Frenkel defect is created when an ion occupies an interstitial site instead occupying its correct lattice site.

→ The metal excess defect is due to the absence of a negative ion from its lattice site leaving a hole which is occupied by an electron, there by maintaining the electrical balance.

→ The solids having F – centres have colour and the intensity of the colour increases with increase in the number of F – centres.

→ Metal excess defects also occurs when an extra positive ion occupies an interstitial position in the lattice and to maintain ele-ctrical neutrality one electron is included in an interstitial position e.g. ZnO, CaO, Cr₂O₃ and Fe₂O₃.

→ The solids with metal excess defect contain free electrons and behave as n – type semi-conductor.

→ Metal deficiency defect is due to the absence of a positive ion from its lattice point and the charge can be balanced by an adjacent metal ion having an extra positive charge e.g. FeO, NiO, FeS and Cul.

→ Crystals with metal deficiency defects are p – type semiconductors.

→ Doping is a process of mixing pure silicon or germanium with an impurity.

→ n-type semiconductors (n – stands for negative) are obtained due to metal excess defect or by adding trace amounts of V or 15th group elements (P, As) to pure silicon or germanium.

→ p-type semiconductors (p – stands for positive) are obtained due to metal defi-ciency defect or by doping with impurity atoms containing less electrons (i.e., atoms of III or 13th group).

→ Diamagnetic solids contain paired electrons (↑↓) and repel the external magnetic field.

→ Paramagnetic solids contain unpaired electrons and are attracted into the applied magnetic field.

→ In ferromagnetic solids there occurs mag-netic interaction between the neighbouring centres (domains) and the electrons in these centres interact in parallel direction (↑↑↑↑↑). This interaction leads to an increase in magnetic moment. Iron, cobalt and nickel are ferromagnetic substances.

→ In antiferromagnetic solids there occurs magnetic interaction between the neigh-bouring centres and the electrons in these centres interact in antiparallel (↑↓ ↑↓ ↑↓) direction which leads to a decrease in magnetic moment e.g., [Cu(CH₃COO)^– H₂O]₂;
VO (CH₃COO)₂, MnO, MnO₂, Mn₂O₃.

→ In ferrimagnetic solids there occurs mag-netic interactions between the neighbou-ring centres and the electrons in these cen-tres interact in such a way which leads to the presence of uncompensated spins (↑↑↓↑↑↓) in the opposite direction resulting some magnetic moment e.g. magnetite (Fe₃O₄); ferrite M FeO₄ (where M = Mg²⁺, Cu , Zn²⁺ etc.)

Students must practice these Maths 2B Important Questions TS Inter Second Year Maths 2B System of Circles Important Questions Very Short Answer Type to help strengthen their preparations for exams.

TS Inter Second Year Maths 2B System of Circles Important Questions Very Short Answer Type

Question 1.
Show that the circles given by the equations x² + y² + 4x – 2y – 11 = 0, x² + y² – 4x – 8y + 11 = 0 intersect each other orthogonally.
Solution:
Given equations of the circles are
x² + y² + 4x – 2y – 11 = 0 ……(1)
x² + y² – 4x – 8y + 11 = 0 ………(2)
Comparing (1) with x² + y² + 2gx + 2fy + c = 0,
we get g = 2, f = -1, c = -11
Comparing (2) with x² + y² + 2g’x + 2f’y + c’ = 0,
we get g’ = -2, f = -4, c’ = 11
Now, 2gg’ + 2ff’ = 2(2)(-2) + 2(-1)(-4)
= -8 + 8
= 0
c + c’ = -11 + 11 = 0
∴ 2gg’ + 2ff’ = c + c’
∴ The two circles cut each other orthogonally.

Question 2.
Show that the circles given by the equations x² + y² – 2lx + g = 0, x² + y² + 2my – g = 0 intersect each other orthogonally.
Solution:
Given equations of the circles are
x² + y² – 2lx + g = 0 ……..(1)
x² + y² + 2my – g = 0 …………(2)
Comparing (1) with x² + y² + 2gx + 2fy + c = 0,
we get g = -l, f = 0, c = g
Comparing (2) with x² + y² + 2g’x + 2f’y + c’ = 0,
we get g’ = 0, f’ = m, c’ = -g
Now, 2gg’+ 2ff’ = 2(-l)(0) + 2(m)(0)
= 0 + 0
= 0
c + c’ = g – g = 0
∴ 2gg’+ 2ff’ = c + c’
∴ The two circles cut each other orthogonally.

Question 3.
Show that the circles given by the equations x² + y² – 2x – 2y – 7 = 0, 3x² + 3y² – 8x + 29y = 0, intersect each other orthogonally. [May ’15 (AP)]
Solution:
Given equations of the circles are
x² + y² – 2x – 2y – 7 = 0 ……….(1)
3x² + 3y² – 8x + 29y = 0
\(x^2+y^2-\frac{8}{3} x+\frac{29}{3} y=0\) ………(2)
Comparing (1) with
x² + y² + 2gx + 2fy + c = 0,
we get g = -1, f = -1, c = -7
Comparing (2) with
x² + y² + 2g’x + 2f’y + c’ = 0, we get

c + c’ = -7 + 0 = -7
∴ 2gg’ + 2ff’ = c + c’
∴ The two circles cut each other orthogonally.

Question 4.
Find ‘k’ if the pair of circles x² + y² + 4x + 8 = 0, x² + y² – 16y + k = 0 are orthogonal. [Mar.’16 (AP)]
Solution:
Given equations of the circles are
x² + y² + 4x + 8 = 0 ………(1)
x² + y² – 16y + k = 0 ……….(2)
Comparing (1) with x² + y² + 2gx + 2fy + c = 0,
we get g = 2, f = 0, c = 8
Comparing (2) with x² + y² + 2g’x + 2f’y + c’ = 0,
we get g’ = 0, f = -8, c’ = k
Since the given circles are orthogonal then
2gg’ + 2ff’ = c + c’
⇒ 2(2)(0) + 2(0)(-8) = 8 + k
⇒ 0 + 0 = 8 + k
⇒ k = -8

Question 5.
Find ‘k’ if the pair of circles x² + y² + 2by – k = 0, x² + y² + 2ax + 8 = 0 are orthogonal.
Solution:
Given equations of the circles are
x² + y² + 2by – k = 0 ……….(1)
x² + y² + 2ax + 8 = 0 ……..(2)
Comparing (1) with x² + y² + 2gx + 2fy + c = 0,
we get g = 0, f = b, c = -k
Comparing (2) with x² + y² + 2g’x + 2f’y + c’ = 0,
we get g’ = a, f’ = 0, c’ = 8
Since the given circles are orthogonal then
2gg’ + 2ff’ = c + c’
⇒ 2(0)(a) + 2(b)(0) = -k + 8
⇒ 0 + 0 = -k + 8
⇒ k = 8

Question 6.
Find ‘k’ if the pair of circles x² + y² – 6x – 8y + 12 = 0, x² + y² – 4x + 6y + k = 0 are orthogonal.
Solution:
Given equations of the circles are
x² + y² – 6x – 8y + 12 = 0 ……….(1)
x² + y² – 4x + 6y + k = 0 ………(2)
Comparing (1) with x² + y² + 2gx + 2fy + c = 0,
we get g = -3, f = -4, c = 12
Comparing (2) with x² + y² + 2g’x + 2f’y + c’ = 0,
we get g’ = -2, f’ = 3, c’ = k
Since the given circles are orthogonal then
2gg’+ 2ff’ = c + c’
⇒ 2(-3)(-2) + 2(-4)(3) = 12 + k
⇒ 12 – 24 = 12 + k
⇒ k = -24

Question 7.
Find the angle between the circles x² + y² – 12x – 6y + 41 = 0, x² + y² + 4x + 6y – 59 = 0. [(AP) May ’19, ’18; (TS) ’15, Mar. ’17, ’15]
Solution:
Given equations of the circles are
x² + y² – 12x – 6y + 41 = 0 ……….(1)
x² + y² + 4x + 6y – 59 = 0 ……….(2)
Comparing (1) with x² + y² + 2gx + 2fy + c = 0,
we get g = -6, f = -3, c = 41
∴ C₁ = (-g, -f) = (6, 3)

Comparing (2) with x² + y² + 2g’x + 2f’y + c’ = 0,
we get g’ = 2, f = 3, c’ = -59
∴ C₂ = (-g’, -f’) = (-2, -3)

Question 8.
Find the angle between the circles x² + y² + 4x – 14y + 28 = 0, x² + y² + 4x – 5 = 0. [(AP) May ’17]
Solution:
Given equations of the circles are
x² + y² + 4x – 14y + 28 = 0 ………(1)
x² + y² + 4x – 5 = 0 ……….(2)
Comparing (1) with x² + y² + 2gx + 2fy + c = 0, we get
g = 2, f = -7, c = 28
∴ C₁ = (-g, -f) = (-2, 7)

Comparing (2) with x² + y² + 2g’x + 2f’y + c’ = 0,
we get g’ = 2, f’ = 0, c’ = -5
∴ C₂ = (-g’, -f’) = (-2, 0)

If ‘θ’ is the angle between the circles then

Question 9.
Find the angle between the circles x² + y² + 6x – 10y – 135 = 0, x² + y² – 4x + 14y – 116 = 0.
Solution:
Given equations of the circles are
x² + y² + 6x – 10y – 135 = 0 ………(1)
x² + y² -4x + 14y – 116 = 0 ………..(2)
Comparing (1) with x² + y² + 2gx + 2fy + c = 0,
we get g = 3, f = -5, c = -135
∴ C₁ = (-g, -f) = (-3, 5)

Comparing (2) with x² + y² + 2g’x + 2f’y + c’ = 0,
we get g’ = -2, f’ = 7, c’ = -116
∴ C₂ = (-g’, -f’) = (2, -7)

Question 10.
Show that the angle between the circles x² + y² = a² , x² + y² = ax + ay is \(\frac{3 \pi}{4}\). [(TS) Mar. ’20, ’16, May ’16 (TS) Mar. ’14]
Solution:
Given equations of the circles are
x² + y² – a² = 0 ………..(1)
x² + y² – ax – ay = 0 ……….(2)
Comparing (1) with x² + y² + 2gx + 2fy + c = 0,
we get g = 0, f = 0, c = -a²
C₁ = (-g, -f) = (0, 0)

Question 11.
Find the equation of the radical axis of the circles x² + y² – 2x – 4y – 1 = 0, x² + y² – 4x – 6y + 5 = 0. [(AP) May ’16]
Solution:
Given equations of the circles are
S = x² + y² – 2x – 4y – 1 = 0
S’ = x² + y² – 4x – 6y + 5 = 0
∴ The equation of the radical axis of the given circle is S – S’ = 0
⇒ (x² + y² – 2x – 4y – 1) – (x² + y² – 4x – 6y + 5) = 0
⇒ x² + y² – 2x – 4y – 1 – x² – y² + 4x + 6y – 5 = 0
⇒ -2x – 4y – 1 + 4x + 6y – 5 = 0
⇒ 2x + 2y – 6 = 0
⇒ x + y – 3 = 0

Question 12.
Find the equation of the radial axis of the circles x² + y² – 5x + 6y + 12 = 0 and x² + y² + 6x – 4y – 14 = 0.
Solution:
Given equations of the circles are
S = x² + y² – 5x + 6y + 12 = 0
S’ = x² + y² + 6x – 4y – 14 = 0
The equation of the radical axis of the circles S’ = 0 and S = 0 is S – S’ = 0
⇒ x² + y² – 5x + 6y + 12 – (x² + y² + 6x – 4y – 14) = 0
⇒ x² + y² – 5x + 6y + 12 – x² – y² – 6x + 4y + 14 = 0
⇒ -11x + 10y + 26 = 0
⇒ 11x – 10y – 26 = 0

Question 13.
Find the equation of the radical axis of the circles 2x² + 2y² + 3x + 6y – 5 = 0 and 3x² + 3y² – 7x + 8y – 11 = 0. [(AP) Mar. ’17]
Solution:
Given equations of the circles are
S = 2x² + 2y² + 3x + 6y – 5 = 0

Question 14.
Find the equation of the common chord of x² + y² – 4x – 4y + 3 = 0, x² + y² – 5x – 6y + 4 = 0. [(TS) Mar. ’20]
Solution:
Given equations of the circles are
S = x² + y² – 4x – 4y + 3 = 0
S’ = x² + y² – 5x – 6y + 4 = 0
The equation of the common chord (radical axis) of the given circles is S – S’ = 0
⇒ (x² + y² – 4x – 4y + 3) – (x² + y² – 5x – 6y + 4) = 0
⇒ x² + y² – 4x – 4y + 3 – x² – y² + 5x + 6y – 4 = 0
⇒ x + 2y – 1 = 0

Question 15.
Find the equation of the common chord of (x – a)² + (y – b)² = c², (x – b)² + (y – a)² = c² (a ≠ b).
Solution:
Given equations of the circles are
S = (x – a)² + (y – b)² = c²
⇒ x² + a² – 2ax + y² + b² – 2by – c² = 0
⇒ x² + y² – 2ax – 2by + a² + b² – c² = 0
S’ = (x – b)² + (y – a)² = c²
⇒ x² + b² – 2bx + y² + a² – 2ay – c² = 0
⇒ x² + y² – 2bx – 2ay + a² + b² – c² = 0
The equation of the common chord (radical axis) of the given circles is S – S’ = 0
⇒ x² + y² – 2ax – 2by + a² + b² – c² – (x² + y² – 2bx – 2ay + a² + b² – c²) = 0
⇒ x² + y² – 2ax – 2by + a² + b² – c² – x² – y² + 2bx + 2ay – a² – b² + c² = 0
⇒ -2ax – 2by + 2bx + 2ay = 0
⇒ 2x(b – a) – 2y(b – a) = 0
⇒ x – y = 0

Question 16.
Find the equation of the common tangent of x² + y² + 10x – 2y + 22 = 0, x² + y² + 2x – 8y + 8 = 0.
Solution:
Given equations of the circles are
S = x² + y² + 10x – 2y + 22 = 0
S’ = x² + y² + 2x – 8y + 8 = 0
The equation of the common tangent at the point of contact (radical axis) of the circles S = 0, S’ = 0 is S – S’ = 0
⇒ x² + y² + 10x – 2y + 22 – (x² + y² + 2x – 8y + 8) = 0
⇒ x² + y² + 10x – 2y + 22 – x² – y² – 2x + 8y – 8 = 0
⇒ 8x + 6y + 14 = 0
⇒ 4x + 3y + 7 = 0

Question 17.
If the angle between the circles x² + y² – 12x – 6y + 41 = 0 and x² + y² + kx + 6y – 59 = 0 is 45°, find k. [(TS) Mar. ’18]
Solution:
Given equations of the circles are
x² + y² – 12x – 6y + 41 = 0 ………(1)
x² + y² + kx + 6y – 59 = 0 ………..(2)
Comparing (1) with x² + y² + 2gx + 2fy + c = 0,
we get g = -6, f = -3, c = 41
C₁ = (-g, -f) = (6, 3)

Given that, θ = 45°
If ‘θ’ is the angle between the circles, then

Question 18.
Show that the circles given by the equations x² + y² – 2x + 4y + 4 = 0, x² + y² + 3x + 4y + 1 = 0 intersect each other orthogonally.
Solution:
Given equations of the circles are
x² + y² – 2x + 4y + 4 = 0 ……..(1)
x² + y² + 3x + 4y + 1 = 0 ……….(2)
Comparing (1) with x² + y² + 2gx + 2fy + c = 0,
we get g = -1, f = 2, c = 4
Comparing (2) with x² + y² + 2g’x + 2f’y + c’ = 0,
we get g’ = \(\frac{3}{2}\), f’ = 2, c’ = 1
Now, 2gg’+ 2ff’ = 2(-1)(\(\frac{3}{2}\)) + 2(2)(2)
= -3 + 8
= 5
c + c’ = 4 + 1 = 5
∴ 2gg’+ 2ff’ = c + c’
∴ The two circles cut each other orthogonally.

Question 19.
Find the equation of the radical axis of the circles x² + y² – 3x – 4y + 5 = 0, 3(x² + y²) – 7x + 8y – 11 = 0. [(TS) May ’17]
Solution:
Given equations of the circles are
S = x² + y² – 3x – 4y + 5 = 0
S’ = 3[x² + y²] – 7x + 8y – 11 = 0
\(x^2+y^2-\frac{7}{3} x+\frac{8}{3} y-\frac{11}{3}=0\)
∴ The equation of the radical axis of the given circles is S – S’ = 0

Question 20.
Find the equation of the radical axis of the circles x² + y² + 4x + 6y – 7 = 0, 4(x² + y²) + 8x + 12y – 9 = 0. [(TS) May ’19. Mar. ’19. (AP) ’19]
Solution:
Given equations of the circles are
S = x² + y² + 4x + 6y – 7 = 0
S’ = 4(x² + y²) + 8x + 12y – 9 = 0
⇒ x² + y² + 2x + 3y – \(\frac{9}{4}\) = 0
The equation of the radical axis of the given circles is S – S’ = 0
⇒ x² + y² + 4x + 6y – 7 – (x² + y² + 2x + 3y – \(\frac{9}{4}\)) = 0
⇒ x² + y² + 4x + 6y – 7 – x² – y² – 2x – 3y + \(\frac{9}{4}\) = 0
⇒ 2x + 3y – 7 + \(\frac{9}{4}\) = 0
⇒ 8x + 12y – 28 + 9 = 0
⇒ 8x + 12y – 19 = 0

Question 21.
Find the equation of the common chord of the pair of circles x² + y² + 2x + 3y + 1 = 0, x² + y² + 4x + 3y + 2 = 0.
Solution:
Given equations of the circles are
S = x² + y² + 2x + 3y + 1 = 0
S’ = x² + y² + 4x + 3y + 2 = 0
The equation of the common chord (radical axis) of the given circles is S – S’ = 0
⇒ x² + y² + 2x + 3y + 1 – x² – y² – 4x – 3y – 2 = 0
⇒ -2x – 1 = 0
⇒ 2x + 1 = 0