UNIT- 1.The Solid State

UNIT- 2: SOLUTIONMole fraction Ratio of number of moles of a component to the total number of moles of solutions, Molarity

Number of moles of solute dissolved in one litre of solution.Molality Number of moles of the solute dissolved in 1 kg of the solvent., relative lowering of vapour pressureNumericals based onelevation in boiling point

depression in freezing point

osmotic pressure of solution.

1. State Henry’s lawAns.The partial pressure of the gas in vapour phase is directly proportional to the molefraction of the gas in the solution.P = KH xApplication To increase the solubility of CO2 in soft drinks , the bottle is sealed under high pressure.2. State Raoutlt’s LawAns.For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to itsmole fraction.P1 = p01 x13. What are Ideal solutions?

Amorphous and Crystalline Solids Amorphous- short range order, Irregular shape eg-glassCrystalline Solids- long range order, regular shape eg : NaCl

No of atoms per unit cell (z ) Simple cubic -1, BCC- 2, FCC – 4 , End-Centred- 2Coordination Number FCC- 6:6 BCC- 8:8Calculation of number of voids Let the number of close packed spheres be N, then:

The number of octahedral voids generated = NThe number of tetrahedral voids generated = 2N

Relation between r and a Simple Cubica = 2r , BCC4r = a√3 FCC4r = a√2Packing Efficiency = ( )

Simple Cubic52.4% , BCC68% , FCC74%Calculations Involving Unit CellDimensions

Frenkel Defect: cation is dislocated to an interstitial site. It does not change the density ofthe solid. Frenkel defect is shown by ionic substance in which there is alarge difference in the size of ions, for example, ZnS, AgCl, AgBr and AgIdue to small size of Zn2+ and Ag+ ions.

Schottky Defect A vacancy defect. The number of missing cations and anions are equal.Density decreases. For example, NaCl, KCl, CsCl and AgBr.

Metal excess defect due to anionicvacancies (F-centres )

The anionic sites occupied by unpaired electrons are called F-centres Theyimpart yellow colour to the crystals of NaCl. Similarly, excess of lithiummakes LiCl crystals pink and excess of potassiummakes KCl crystals violet (or lilac).

Doping The conductivity of intrinsic semiconductors is increased by adding anappropriate amount of suitable impurity. This process is called doping

MINIMUM LEARNING PACKAGE IN CHEMISTRY FOR CBSE CLASS 12TH

PREPARED BY – Mr. O. P. PANDEY, Mr. B. K. SHARMA, Mr. A. P. SINGH, Mr. S. S. OJHA,DEPT. OF CHEMISTRY, KV ALIGANJ LUCKNOW

Ans.Solutions which obey Raoult‟s law . solute –solvent interaction is nearly equal to solute –solute or solvent – solvent interaction.Δmix H = 0 and Δmix V = 0Eg: n- hexane and n-heptane4. What are non-ideal solutions?Ans. Solutions which do not obey Raoult‟s law . Inter molecular force of attraction between solute –solvent is Weaker or strongerthan those between solute –solute or solvent – solvent interaction.Δmix H ≠ 0 Δmix V ≠ 04. colligative propertyAns.Properties which depend on the number of solute particle irrespective of their nature relative to the totalnumber of particle present in the solution.5. Ebullioscopic constantAns. Elevation of boiling point of one molal solution.6. Cryoscopic constantAns. Depression of freezing point of one molal solution.7. Define OsmosisAns. Flow of solvent molecule through semi permeable membrane from pure solvent to the solution. This processof flow of the solvent is called osmosis.8. Define Osmotic PressureAns. Excess pressure that must be applied to a solution to prevent osmosis.9. What is reverse osmosisAns. Pure solvent flows out of the solution through the semi permeable membrane when we apply pressure largerthan the osmotic pressure. Application of reverse osmosis is desalination of seawater.10. What are isotonic solutionAns.Two solutions having same osmotic pressure at a given temperature.11. Define van’t Hoff factorAns. i = Normal molar mass / abnormal molar massi= observed colligative property / calculated colligative property.

Unit 3- ELECTROCHEMISTRYKohlrausch’s law of independent migration of ions: The law states that molar conductivity of an electrolyte atinfinite dilution can be represented as the sum of the individual contribution of the anion and cation of theelectrolyte at infinite dilution λm∞ = λ∞( cation) + λ∞( anion)

Q.1.How does the molar conductivity of strong electrolyte vary with increase in concentration?Ans 1. For strong electrolyte molar conductivity decreases with increase in concentration.Q.2.Suggest a metal which can be used for cathodic protection of iron ? Ans 2. ZincQ.3.What is meant by cell constant ? Ans 4 .Cell constant = l/a Where, l = distance between two electrodea = cross-section area of electrodeQ.4.How is the cell potential related to free energy change?Ans 5. ΔG = -nFEcell

Q.5.Can you store CuSO4 solution in a zinc pot ?

Ans 7. No, because zinc is more reactive than copper & it will displace copper from CuSO4 solution.Q.6.How does concentration of sulphuric acid changes in Lead Storage Battery when current is drawn fromit ? Ans 8. Concentration of sulphuric acid decreases.Q.7.Suggest two materials other than hydrogen that can be used as fuels in Fuel Cell ? Ans 9. CO & CH4

Q.8.HCl does not give an acidic solution in Benzene ?Ans10. It does not give H+ in benzene(a non-polar solvent)Q9. How can you increase the reduction potential of an electrode.?Ans For the reaction It can be increased bya. Increase in concentration of Mn+ions in solutionb. By increasing the temperature

Q10.. Define corrosion. Write chemical formula of rust.Ans Corrosion is a process of destruction of metal as a result of its reaction with air and water, surrounding it. It isdue to formation of sulphides, oxides, carbonates, hydroxides, etc.Formula of rust-Fe2O3..XH2OQ11.. What are the products of electrolysis of molten and aqueous sodium chloride?Ans Molten sodium chloride:- Na, Cl2, and aqueous sodium chloride:-H2 and Cl2.Q12. Why is it not possible to determine the molar conductivity at infinite dilution for weak electrolytes byextrapolation?Ans Because the molar conductivity at infinite dilution for weak electrolytes doesnot increase linearly with dilution as for strong electrolytesSECONDARY CELL, Fuel CellsUNIT - 4 CHEMICAL KINETICSRATE LAW : Representation of rate of the reaction in terms of concentration of the reactants .For a general reaction of the type aA+bB→ cC+dD, rate law can be given as Rate = k [A]x[B]yRATE CONSTANT ( K ) : It is the rate of the reaction when concentration of the reactants is unity.ORDER OF A REACTION: It is the sum of the powers of all concentration terms in the rate law expression. Itcan be zero, one, two etc or fractional. Should be found out experimentally.

ELEMENTARY REACTIONS: Reactions which get completed in one step.COMPLEX REACTIONS: Reactions which take place in two or more steps.MOLECULARITY: The number of reacting species which collide simultaneously to bring about a chemicalreaction.PSEUDO FIRST ORDER REACTION: Reactions of higher order which follow first order kinetics.CH3COOC2H5 + H2O in presence of H+→ CH3COOH +C2H5OHActivation energy: The energy required by the reactant molecules to change to products.Activation Energy = Threshold Energy– Energy of reacting speciesQ 1 What is the order of reaction whose rate constant has the same units as the rate of reaction?Ans. . Zero orderQ 2. Why boiling of an egg or cooking of rice in an open vessel take more time at a hill station?Ans- At a hill station , due to higher altitude atmospheric pressure is low. Hence , water boils at aLower temperature . At a lower temperature , the rate of reaction decreases. Hence it takesLonger time.Q3.. Why are reactions of higher order less in number ?Ans- A reaction takes place because the molecules collide.The chances for a large number of molecules or ions tocollides simultaneously are less. Hence, the reactions of higher order . . are lessQ4. Give units of specific rate constant for a zero order reaction. Ans mol L-1 s-1.

Q5. Why the Rate of reaction does not remain constant throughout the course of reaction?Ans- Rate of reaction depands on concentration of reactants and since concentration becomesLess and less as the reaction progresses, the rate also goes on decreasing.Q6. Among the elementary processes in the proposed mechanism of reaction which is rateDetermining step? Ans- The slowest step.Q7. State any one condition under which a biomolecular reaction may be kinetically of firstOrder? Ans- A biomolecular reaction may become kinetically of first order if one of the reactantsIs present in excessQ.8.Write the difference between order and molecularity of reaction

Unit-5-Surface chemistryImportant features of solid catalystActivity-The activity of a catalyst depend on the strength of chemisorptions.

Selectivity–The selectivity of a catalyst is its ability to direct a reaction to yield a particular product.Shape selective catalysis: The catalytic reaction that depends upon the pore structure of the catalyst and thesize of reactant and product molecules is called shape selective catalysis. e.g. Zeolites are good shape selectivecatalysts.Enzyme catalysis Enzymes are protein molecules of high molecular mass which catalyse the biochemicalreaction. e.g. Inversion of cane sugar by invertase enzyme.

Lyophobic colloid (Solvent hating colloid) – these colloids can not be prepared by simply mixing ofdispersed phase into dispersion medium. e.g. metallic sols.Lyophilic colloid (Solvent loving colloid)- these colloids can be prepared by simply mixing of dispersionphase into dispersion medium. e.g. Starch sol.Multi-molecular colloid- on dissolution, a large number of atoms or smaller molecules of a substanceaggregate together to form species having size in colloidal range. The species thus formed are called Multimolecular colloids.e.g.Sulphur sol.Macromolecular colloids- macromolecules in suitable solvent form colloid, in which size of the particles arein range of colloidal range. e.g. starch sol.Associated colloids(micelles)- some substances in low concentration behave as normal strong electrolyte butat higher concentration exhibit colloidal behavior due to formation of aggregates. The aggregated particles arecalled micelles and also known as associated colloids.Peptization- Process of converting a precipitate into colloidal sol, by shaking it with suitable dispersionmedium in the presence of a small amount of electrolyte.Brownian movement-it is the zig-zag motion of colloidal particlesTyndall effect- Scattering of light by colloidal particles by which path of beam becomes clearly visible. Thiseffect is known as tyndall effect.Charge on colloidal particles– Colloidal particles which carry anelectric charge and nature of charge is sameon all particles.Electrophoresis- Movement of Colloidal particles towards opposite charge electrode in presence of externalelectric field.Coagulation– The process of settling of colloidal particles is called coagulation of the sol.Hardy-Schulze Law– Coagulating value of a coagulating ion is directly proportional to the charge on the ion.eg: Na+ < Ca 2+< Al3+ for negatively charged sol.

Cl- < CO32- < PO43- < [Fe (CN)6]4– for positive sol.Emulsion – Liquid – liquid colloidal system is known as Emulsion. There are two types of Emulsion.a)O/W type - Oil dispersed in water. Eg: milk, vanishing cream.b)W/O type – Water dispersed in oil. Eg: Butter & Cream.UNIT 6 GENERAL PRINCIPLES & PROCESS OF ISOLATION OF ELEMENTSProcess Froth Flotation:- Concentration is based on the facts that sulphide ore is wetted by oil & gangue particlesare wetted by water.Mond Process for Refining Nickel: In this process, nickel is heated in a stream of carbon monoxide forming avolatile complex, nickel tetracarbonyl The carbonyl is subjected to higher temperature so that it is decomposedgiving the pure metalNi+4CO----Ni(CO)4 At 330-350K & Strong heat Ni(CO)4 -----Ni +4COvanArkel Method for Refining Zirconium or Titanium: This method is very useful for removing all the oxygenand nitrogen present in the form of impurity in certain metals like Zr and Ti. The crude metal is heated in anevacuated vessel with iodine. The metal iodide being more covalent,volatilizesWhat are depressants? It is possible to separate two sulphide ore by adjusting proportion of oil to water in frothflotation process by using a substance known as depressant. e.g. NaCN is used to separate ZnS and PbS.principle of electro-refining. In this method of purification impure metal is made Anode and pure metal is made

the cathode. On passing electricity, pure metal is deposited at the cathode while the impurity dissolves in solution asanode mud At Cathode: Cu2++2e---- Cu

At Anode: Cu ----Cu2++2e

Explain the role of (i) Cryolite in the electrolytic reduction of alumina(ii) Carbon monoxide in the purification of nickel.Ans. (i) Cryolite lowers the melting point of mixture. Or it act as electrolyte.(ii)Carbon monoxide forms a volatile complex with nickel which on heating decomposesto give pure nickel.Q.Which is better reducing agent at 983K, carbon or CO during the extraction of iron?CO, (above 983K CO being more stable & does not act as a good reducing agent but carbon does.Q At which temperature carbon can be used as a reducing agent forFeO ?

Above 1123K, carbon can reduce FeO to Fe.

UNIT 7 P-BLOCK ELEMENTS1.Nitrogen does not form pentahalide although it exhibit +5 oxidation state.Due to absence d-orbitals N can not extend its valency beyond four2. NH3 acts as Lewis base. Because N has a lone pair electron so NH3 acts as a Lewis base

3. NH3 is stronger base than PH3 Ans Due to smaller size of nitrogen there is high electron density on nitrogen soelectron pair is easily available.4. All the five P-Cl bonds are not equal in PCl5 .The two axial bonds suffer more repulsion from equatorial bonds and hence are elongated.5. H3PO3 is dibasic (diprotic) but H3PO4 is tribasic.In H3PO3only two H atoms are linked to O which are ionisable the third H is attached to P and not ionisablebecause P is less electronegative. In H3PO4 all the three H atoms are with O and ionisable6. PCl5 is ionic in solid stateIt is due to the following conversion: 2PCl5 ---- [PCl4]+

+[PCl6]--

7. Nitrogen shows little catenation but phosphorous distinctly shows catenation propertyDue to smaller size of N there is repulsion between the lone pairs and N-N single bond is weaker than P-P8. +5 oxidation state of Bi is less stable than +3Because inert pair effect is very prominent in Bi , so +5 oxidation state is not stable9. N2 is less reactive at room temperatureDue to having triple bond and hence high bond dissociation energy(946 kJ/mol10. N exists as N2 and gas form but P exists as P4 and solid formDue to smaller size N can form pπ-dπ multiple bonding and exists as discrete N2 molecule but P can not form pπ-pπmultiple bonding.11. H2S is acidic while H2O is neutral.H-S bond is weaker due to larger size of S so proton release easier in H2S12. Compound of F & O is fluoride of oxygen not oxide of fluorine .F is more electronegative than O.13. SCl6 is not known but SF6 is knownF is strongest oxidizing agent so it can oxidizes S to its maximum oxidation state +6 . Cl can not. Again Cl haslarger size so steric repulsion is there in SCl614. H2O is liquid but H2S is gasO is electronegative so there is intermolecular H-bonding in water so it is liquid.15. O2 is gas but sulphur is solidDue to smaller size O can form pπ-pπ multiple bond and exists as discrete diatomic molecule.16. Group 16 elements are called chalcogensChalcogen means ore forming elements. They form several ores17. Halogens have maximum negative electron gain enthalpy(ΔegH)Because they have smallest size in their respective periods18. F has less electron gain enthalpy than that of Cl but fluorine is stronger oxidizing agent than chlorine.F has very small size so there is inter-electronic repulsion. F is stronger oxidizing agent due to its low bonddissociation energy and high hydration energy19. F exhibits only -1 oxidation state , other halogen shows +1, +3, +5, +7 oxidation statesF is most electronegative element and due to absence of d-orbitals it can not expand its octet so it does not exhibitpositive oxidation state.20. Bond dissociation energy of F2 is less than Cl2

Due to very small size of F there is interelectronic repulsion in F2 so it has low bond dissociation energy21. HF has lower acid strength than HIDue to larger size of I the H-I bond is weaker than H-F bond so HI is stronger22. He, Ne do not form compound with F. ANS--- Due to high IE23. Noble gases have very low b.ptBecause there is only weak dispersion force between their atoms.24. Ne used as warning signalBecause Ne – light has high fog penetration power25. Noble gases form compounds only with fluorine and oxygen

Because F & O are the most electronegative elements26. Out of noble gases only Xe forms compoundsBecause Xe has comparatively low IE and vacant orbitals for promotion of electrons27. Noble gases are mostly inertBecause they have completely filled valence orbitals i.e octet configuration28. He is used as diving apparatus .Because it is less soluble in blood with compare to nitrogen29. It is difficult to study the chemistry of Rn.Because Rn is radioactive and hence very unstable30. Noble gases have comparatively large atomic size.They are mono atomic so their van der Walls radii measured which is longer than covalent/ionic or metallic radii.Learn the increasing order of the property mentioned(i)Acidic strength: HOCl<HClO2<HCIO3<HCIO4

(ii)Acidity: Ga2O3<GeO2<AsO3<CIO2

(iii)Bond angle: SbH3<AsH3<PH3<NH3

(iv)Acidic strength: HF<HCl<HBr<HI(v)Ionic character: MI<MBr<MCl<MFDraw the structure of followings and name the geometryXeO3, PCl5 , NH3, ClF3 , H2S2O8, H3PO4 H3PO2 ,H3PO3, H2SO4 , XeOF4, N2O5 , XeF2, XeF4 etc

Unit 8 : d & f-Block elements1.Zn, Cd & Hg are not treated as true transition elementsBecause they have completely filled d-orbitals in their atomic as well as stable ionic state2.Cu & Ag are transition metals although they have completely filled d-orbitalsCu2+ & Ag+ have (n-1)d9 4s0 configuration.3.Some d-block elements have irregular(exceptional) electronic configurationDue to very small energy difference between (n-1)d & ns sub-shell4.Atomic size does not change appreciably in a row of transition metalsAlong the rows nuclear charge increases but the penultimate d-sub shell has poor shieldingeffect so atomic and ionic size remain almost same .5.Transition elements have variable oxidation statesDue to very small energy difference between (n-1)d & ns sub-shell electrons from both thesub-shell take part in bonding6.Transition metals have high melting and boiling pointsA large number of unpaired electrons take part in bonding so they have very strong metallicbonds and hence high m.pt & b.pt7.Transition metals have high enthalpy of atomizationA large number of unpaired electrons take part in bonding so they have very strong metallicbonds and hence high enthalpy of atomization8.Transition metals show catalytic propertiesBecause they have variable oxidation states and hence can form different intermediates. Theyalso provide large surface area.9.Transition metals and their salts are generally coloredBecause they have partially filled d-sub shell and hence d-d electron transition takes placewhen they absorb radiations from visible region and transmit complementary colors.10.Transition metals form coordination compoundsBecause they have large number of vacant orbitals in (n-1)d, ns, np & ns sub shells so they canaccept electron pairs from ligands11.Transition metals form alloysThey have comparable atomic size and hence can be mixed uniformly12.Transition metals form interstial compoundsBecause small atoms like H, C, N etc can be entrapped in their metallic crystals13.Zn, Cd & Hg have low boiling points and Hg is liquid.They have full filled 3d-orbitals and no electorns from d-orbitals are taking part in metallicbonding so they have weak metallic bonding. Due to larger atomic size Hg is liquid14.Transition metals and many of their compounds show paramagnetic behaviourBecause they have unpaired electrons

15.Mn2+ compounds are more stable than Fe2+Mn2+ has half- filled d-orbitals i.e 3d5 4s0 configuration16.Zr has similar size to that of Hf Ans:-Due to lanthanoid contraction.17.Actinoid contraction is greater from element to element than lanthanoid contraction.The 5f electrons are more effectively shielded from nuclear charge. In other words the 5f electrons themselvesprovide poor shielding from element to element in the series.18.Ce4+ is a strong oxidantIt can be converted into the most common oxidation state Ce3+

19.Write the steps involved in the preparation of K2Cr2O7 from chromite ore4FeCr2O4+8 Na2CO3+7O2 8Na2CrO4+2Fe2O3+8CO2

2Na2CrO4+2 H+ Na2 Cr2 O7 +2 Na++H2ONa2Cr2O7+ 2 KCl K2Cr2O7+ 2 NaClUnit 9- Coordination Compounds

IUPAC nameambidentate ligand? ligand which has two donor atoms but attach only with one donor site is calledambidentate ligand. e.g. NO2 - & ONO-

Which complex is used in the treatment of cancer? cis-platin or cis- [Pt (NH3)2 Cl2]Spectrochemical series? It is a series in which ligands can be arranged in the order of increasing field strength orin the order of increasing magnitude CFSE.Inner orbital complex? The complex which uses its inner i.e. (n-1) d orbital for complex formation is known as

inner orbital complex. Such complex is formed by strong field ligands.crystal field splitting energy? The energy difference between t2g & eg orbitals is known as Crystal Field SplittingEnergy (CFSE)Q.1- How would account for followings:-(i) [Ti(H2O)6]3+ is coloured while [Sc(H2O)6]3+ is colourless.(ii) [Fe(CN)6] 3- is weakly paramagnetic while is [Fe(CN)6] 4- diamagnetic.(iii) [Ni(CO)4] possesses tetrahedral geometry while [Ni(CN)4]2- is square planar.Q.2-Describe the type of hybridization, shape and magnetic property of followings:-(i)[Fe(H2O)6]2+ (II) [Co(NH3)]3+ (iii) [NiCN4]2-

Q.3.[NiCl4]2− is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why?Q.4- Draw a sketch to show the splitting of d-orbitals in an octahedral crystal field.

CHAPTER-14 BIOMOLECULESOligosaccharidesThose carbohydrates which give two to ten monosaccharides on hydrolysis e.g. sucrose on

hydrolysis gives glucose and fructose.Glycosidic Linkage: The linkage between two monosaccharide units through oxygen is called the glycosidic linkDenaturaion of Protein: when subjected to a physical change like temperature, pH etc undergoes uncoiling andlooses it’s biological activity.Renaturation of Protein: Some proteins regain their biological activity by reversible process it is calledrenaturation of proteins. In such a cases, when temperature and pH of a denatured proteins is brought back toconditions in which the native protein is stable, secondary and tertiary structures of proteins are restored to whichleads to recovery of biological activity.(i) Fat SolubleVitamins: Vitamin A,D,E and K. They are stored in liver and adipose tissues.(ii) Water Soluble Vitamins: vitamin B & C. They need to supplied regularly in diet as they are excreted in urineand cannot be stored (except vitamin B12) in our body. Their deficiency causes diseases.Nucleoside: The unit formed by the attachment of a base to 1’Position of sugar (Base+Sugar).Nucleotide: Nucleoside and phosphoric acid at 5’‐position. Nucleotides are bonded by phosphodiester linkagesbetween 5’ and 3’ carbon atoms of pentose sugar(Base + Sugar + Phosphoric Acid).

Q.1.Which types of bonds are present in a protein molecule?Peptide bonds, hydrogen bonds, sulphide bonds, ionic bonds etc.Q.2.What is animal starch? Glycogen.Q.3.Which types of bonds are present in a protein molecule?Peptide bonds, hydrogen bonds, sulphide bonds, ionic bonds etc.

Q.4.Which α‐helix or β‐helix is more stable?α‐helix is right handed and is more stable due to intermolecular H bonding between first and fourth amino acid.Q.5.Name a protein which is insoluble in water. KeratinQ.6.What are anomers? roup at C‐1.e.g,α‐glucose and β‐glucose.Q.7.What is invert sugar?An equimolar aqueous solution of glucose and fructose is called invert sugar.Q.8.Why are carbohydrates generally optically active?Ans. It is due to the presence of Chiral Carbon atoms in their moleculesQ.9.Why is cellulose not digested in human body?Ans. It is due to the fact that human beings do not have enzyme to digest cellulose.Q.10.Name the enzyme that is used to dissolve blood clots? Ans. StreptokinaseIsoelctric point–the pH at which there is no net migration of any ion towards electrode. e.g, amino acids have isoelectric point at pH= 5.5‐6.3Mutarotation‐it is spontaneous change in optical rotation when an optically active substance is dissolved inwater.e.g,α‐glucose when dissolved in water changes its optical rotation from 111o o 52.5o.Unit-15-POLYMERS1. Name a natural elastomer. Ans . Natural rubber.2. Write name of a synthetic polymer which is an ester. Ans. Nylon 6 or Nylon 6,6.3. Name the monomer of Nylon 6. Ans. Caprolactam4. Write the monomer units of Bakelite. Ans. Phenol and formaldehyde.5. Define a copolymer. Ans. The polymers made by addition polymerisation from two differentmonomers are termed as copolymers, e.g., Buna‐S, Buna‐N, etc.6. Write one use of PVC. Ans: In manufacture of rain coats & vinyl flooring.7.Give an example of thermoplastics. Ans: Thermoplastics are polythene, polystyrene, polyvinyls, etc.8. To which class of polymers does Nylon‐6,6 belong? Ans: PolyamidesBiodegradable polymer :- Polymers which disintegrate by themselves over a period of time due to

environmental degradation by bacteria etc. are called biodegradable polymers. e.g. PHBVDistinguish between the terms homopolymerAns: The addition polymers formed by the polymerisation of a single monomeric species are known ashomopolymers, e.g., polythene.Write the monomers of the following polymers: (i) Buna‐S(ii) Teflon (iii) Nylon‐6(do it)Ans. (i) 1, 3 – butadiene and styrene (ii) tetrafluoroethenefunction of sulphur in vulcanisation of rubber?Ans: Sulphur introduces sulphur bridges or cross links. So it is become more tensile strength, elasticity andresistance to abrasion etc.What is the main constituent of bubble gum? Ans‐ Styrene‐butadiene copolymer (SBR).

UNIT 16-CHEMISTRY IN EVERYDAY LIFE

1. Define the term chemotherapy? Treatment of diseases using chemicals is called chemotherapy.2. Why do we require artificial sweetening agents? To reduce calorie intake.3. What are main constituents of Dettol ? Choloro xylenol & Terpineol .4. Name the drugs that are used to control allergy? Antihistamines.5. Why is the use of aspartame limited to cold food and drinks? It is unstable and decompose at cookingtemperature and decompose.

6. What is a tranquilizer? Give an example This is the drug used in stress, mild mental disease.7. What type of drug chloramphenicol? It is a broad spectrum antibiotic.8. Why is biothional added to the toilet soap? It acts as antiseptics.9. What are food preservatives? The substances that prevent spoilage of food due to microbial growth. eg-sodium benzonate.10 Mention one important use of the following-(i) Equanil (ii)Sucrolose(i) Equanil - It is a tranquilizer. (ii) Sucrolose -It is an artificial sweetener.11.What is tincture of iodine? 2-3% iodine in alcohol-water mixture is called tincture of Iodine. It is apowerful antiseptics and is applied on wounds.ORGANIC CHEMISTRYNAME REACTIONS:-

1.

Friedel-CraftsAlkylation

+ CH3 ClAnhydrous AlCl3

CH3

2.

Friedel-CraftsAcylation

COCH 3

C H 3C O C l

A nhyd ro u s A lC l3

3.

Kolbe

OH

Na OH

ONa

i) CO2

ii) H+

OH

COOH

4.

Reimer-Tiemann

OH ONa

CHOCH3Cl + Na OHH+

OH

CHO

5.Williamson CH3-Br + CH3-ONa CH3-O- CH3 + NaBr

6.

Stephen CH3 CN+ SnCl2 + HCl CH3 CH NHH3O

+

CH3 CHO

7.

Etard

CH3

CrO2Cl2

H3O+

CHO

8.

Gatterman –Koch

CHO

CO / HCl

Anhydrous AlCl3

9.

Rosenmundreduction CH3

CCl

O H2

Pd / BaSO4CH3

CH

O

10.

Clemmensen

reduction CH3

CCH3

OZn - Hg

Conc. HClCH3 CH2 CH3

11.

Wolff-Kishnerreduction CH3

CCH3

O

CH3 CH2 CH3

i) NH2-NH2

ii) KOH / Ethylene glycol /

12.Tollens’ test R-CHO + 2 [Ag(NH3)2]+ + 3 OH- R-COO- + 2Ag + 2H2O + 4 NH3

13.Fehling’s test R-CHO + 2 Cu2+ + 5 OH- R-COO- + Cu2O + 3H2O

14.

IodoformCH3

CCH3

O I2 / NaOH

OR, NaOICHI3 + CH3COONa

15.

Aldolcondensation

CH3 CHOdil NaOH

CH3 CH CH2

OH

CHO

CH3 CH CH CHO2

16.

Cannizzaro HCHO + HCHOConc. NaOH

HCOONa + CH3 OH

17.

Hell-Volhard-Zelinsky(HVZ)

CH3 COOHi) Cl2 / Red Phosphorus

ii) H2OCH2 COOH

Cl

18.

Hoffmannbromamidedegradation

CH3 C NH2

OBr2

NaOHCH3 NH2

19.

CarbylamineR-NH2 + CHCl3 + 3 KOH

R-NC + 3 KCl + 3 H2O

20.

Diazo

N H 2

N a N O 2 + d il H C l

2 7 3 - 2 7 8 K

N 2+

Cl-

21.

Sandmeyer.

N 2+

Cl-

C u C l / H C l

C l

+ N 2

22.

Gatterman

N 2+

Cl-

C l

+ N 2C u / H C l

23.

Coupling N2+Cl

- + OHHOH-

N N OH

DISTINGUISH BY A SINGLE CHEMICAL TEST

1. All aldehydes ( R-CHO) give Tollens’ Test and produce silver mirror.RCHO + 2 [Ag(NH3)2]+ + 3 OH- RCOO- + 2 Ag + 2H2O + 4 NH3

Tollens’ Reagent silver pptNote: HCOOH(methanoic acid ) also gives this test, ketones(RCOR) do not give this test

2. .All aldehydes (R-CHO) and ketones(RCOR) give 2,4-DNP testRCOR + 2,4-DNP Orange pptR-CHO + 2,4-DNP Orange ppt

3. Aldehydes and ketones having CH3CO- (keto methyl) group give Iodoform Test. Alcohols having CH3CH- groupalso give Iodoform Test.

|OH

CH3CHO + 3I2 + 4 NaOH CHI3 + HCOONa + 3 NaI + 3H2OYellow ppt

The following compounds give Iodoform Test: ethanol (C2H5OH), propan-2-ol (CH3CH(OH)CH3),ethanal(CH3CHO), propanone(CH3COCH3), butanone ( CH3COCH2CH3) , pentan-2-one (CH3COCH2 CH2CH3) ,acetophenone ( PhCOCH3 )

4. All carboxylic acids ( R-COOH) give Bicarbonate TestRCOOH + NaHCO3 RCOONa + CO2 + H2O

effervescence

5. Phenol gives FeCl3 TestC6H5OH + FeCl3 (C6H5O)3Fe + 3 HCl

(neutral) (violet color)6. All primary amines (R/Ar -NH2) give Carbyl Amine Test

R-NH2 + CHCl3 + KOH(alc) R-NC + KCl + H2Ooffensive smell

7. Aniline gives Azo Dye Test ( Only for aromatic amines)C6H5NH2 + NaNO2 + HCl C6H5N2

+Cl- ; then add β-naphthol orange dye8. All alcohols (ROH) give Na-metal test

R-OH + Na R-ONa + H2

bubbles

9. For esters (RCOOR) : Hydrolyses first. Then see the products ( acid & alcohol) and give a test to identifythem

10. All alkenes (C=C) and alkynes (C≡C) decolorizes Br2 – water from red to colorless11. Lucas Test to distinguish primary, secondary and tertiary alcohols

Lucas reagent: ZnCl2/HCl30-alcohol + Lucas reagent immediate turbidity20-alcohol + Lucas reagent turbidity after sometime10-alcohol + Lucas reagent no turbidity

Organic Chemistry Explanations

1 Bond enthalpy decreases as : CH3-F > CH3-Cl > CH3-Br > CH3-I

Because C-X bond length increases from F to I due to greater size

2 Phenol can not be converted into chlorobenzene by using HCl

Because C-O bond in phenol has partial double bond character and OH gr is attached to sp2 carbon,

3 Thionyl chloride is preferred to convert ROH into RCl

Because the side products are gases and the product can be obtained as pure

4 30 alcohol easily can be converted into alkyl halide

Because 30 carbocations are stable

5 Benzene is converted into iodobenzene in presence of oxidizing agent like HNO3 or HIO3

C6H6 + I2/Fe C6H5I + HI Because HNO3 or HIO3 oxidises HI and prevent backward reaction

6 ROH is converted into RI by using KI in presence of H3PO4 not H2SO4

H2SO4 converts KI into HI and than into I2

7 B.pt. of alkyl halides (RX) is higher than hydrocarbons

Because R-X is polar so there is dipole dipole attraction

8 B.pt : R-I > R-Br > R-Cl > R-F

Because molecular mass decreases so van der Waals force decreases from RI to RF

9 B.pt : CH3-CH2-CH2-CH2-Br > CH3-CH2-CH(Br)-CH3 > (CH3)CBr

Because as the branching increases surface area decreases so van der Waals force of attraction decreases

10 p-dichloro benzene has higher m.pt than ortho and meta isomer

Because p-dichloro benzene has symmetrical structure so it fits well in the crystal lattice

11 Alkyl halide (RX) with KCN gives alkyl cyanide (RCN) where as with AgCN it gives isocyanide(RNC)

KCN is ionic and CN- is ambedent nucleophile but it link through C because C-C bond is more stable than C-N. In the other handAgCN is covalent and links through N only

12 SN2 reactivity : CH3-X > CH3CH2-X > CH3-CH(X)-CH3 > (CH3)3C-X ( 10 >20>30)

Due to steric hindrance nucleophile can not approach easily. In SN2 path release of X and linking of Nu- take placesimultaneously

13 SN1 reactivity : CH3-X < CH3CH2-X < CH3-CH(X)-CH3 < (CH3)3C-X ( 10 <20<30)

Because SN1 path involves formation of carbocation intermediate. And stability of carbocation is in the order : CH3+ < CH3CH2

+

< (CH3)2CH+ < (CH3)3C+ ( 10 <20<30)

14 SN reactivity : R-I > R-Br > R-Cl > R-F

Because as the size of halogen increases C-X bond becomes weaker

15 Aryl halides ( C6H5-X) are less reactive than alkyl halides(R-X) towards nucleophilic substitution

Because in aryl halide C-X bond has partial double bond character and X is attached to sp2 C. there is also the repulsionbetween Nu- and benzene ring which is electron reach

16 Chloroform is stored in dark coloured bottle as closed and completely filled

Because in air and light it converts into poisonous phosgene gas(COCl2)

17 In many countries DDT has been banned now

Because of its slow metabolism and it gas toxic effect on aquatic animals

18 Cyclohexyl chloride has greater dipole moment than chloro benzene

The lone electron pair of chlorine get delocalized towards benzene ring due to resonance decreasing the polarity of C-Cl bond.

19 Alkyl halides are immiscible in water although they are polar

In water there is intermolecular H-bonding but there is less attraction between R-X and H2O

20 Grignard reagent(RMgX) should be prepared in anhydrous condition

Because RMgX reacts with water and gives corresponding alkane

21 Alkyl halides undergo substitution when treated with aq KOH but in presence of alc KOH elimination takes place

Alcohol + KOH produces RO- which is a strong base so it extract H+ and elimination takes place

22 C-O-H bond angle in alcohol is less than regular tetrahedral angle

Due to lp-lp repulsion

23 In phenol the C-O bond length is less

Due to i) partial double bond character ii) O is attached to sp2 carbon

24 In ether the R-O-R bond angle is greater

Due to repulsion between two bulkier R group

25 To convert acid into alcohol LiAlH4 is not used

Because it is expensive so : RCOOH RCOOR’ then ester is reduced into RCH2OH by using H2/Pd

26 b.pt of alcohol(ROH) is higher than alkane(RH), ether(R-O-R), alkyl halide(R-X) and aryl halide(Ar-X)

Due to inter molecular H-bonding in R-O-H.

27 b.pt : n-butyl alcohol > sec. butyl alcohol > tert. Butyl alcohol

Because as the branching increases surface area decreases so van der Waals force of attraction decreases

28 Alcohols are highly miscible in water

Due to H-bonding with water

29 Acidity of alcohol : R-CH2-OH > R2CH-OH > R3C-OH

Because as the R gr increases +I effect increases so alkoxide ion becomes less stable

30 Alcohol is weaker acid than water

Because R-O- is less stable than HO-

31 Phenol (Ph-OH) is acidic in nature

Because phenoxide ion( Ph-O-) is resonance stabilized.

32 Acidity : nitro phenol > phenol > methyl phenol

Because –NO2 group is electron withdrawing it further increases the stability of phenoxide ion where as-CH3 group is electrondonating it destabilizes phenoxide ion

33 Esterification is carried out in presence of small amount of conc. H2SO4

Because it absorbs the water produced and accelerate the forward reaction.

34 R’COCl + R-OH R’COOR + HCl. Pyridine is used in this reaction

It is to remove HCl produced and to prevent the backward reaction

35 Tert. Alcohols are easier to dehydrate

Because the intermediate tert. carbo cation is stable.

36 -OH group in benzene ring is ortho and para directing for electrophilic substitution

Due to +R effect it increases the electron density at ortho and para positions

37 o-nitro phenol is steam volatile( low b.pt) but p-nitro phenol is not

In o-nitro phenol there is intra molecular H-bonding. But in p-nitro phenol there is inter molecular H-bonding so molecules getassociated and hence it has comparatively higher b.pt

38 Phenol with aq bromine gives 2,4,6-tribromo phenol but in non polar medium mono substitution takes place

In aq. Medium phenoxide ion is the main substrate which is not produced in non-polar medium. the benzene ring is moreactivated in phenoxide ion than in phenol itself.

39 CuSO4 and pyridine are mixed with ethanol used for industrial purpose

To prevent its misuse. CuSO4 gives colour pyridine gives smell

40 Ethers ( R-O-R) are polar

Due to angular structure of R-O-R there is net polarity.

41 Ethers are soluble in water

Due to H-bond with water and ether

42 Aldehydes (R-CHO) and ketones (R-CO-R) have higher b.pt than hydrocarbon and ether

Because they are polar so there is dipole-dipole attraction in aldehyde and ketones

43 Lower aldehydes and ketones are miscible with water

Because they form hydrogen bond with water

44 Aldehydes (R-CHO) are more reactive than ketones (R-CO-R) in nucleophilic addition

In ketone the two alkyl groups ( R ) have +I effect so they reduce the electrophilicity co carbonyl carbon. Also there is sterichindrance in ketone

45 Benzaldehyde ( C6H5CHO ) is less reactive than propanal ( CH3CH2CHO )

Due to resonance the electrophilicity of carbonyl carbon is less in benzaldehyde

46 NaHSO3 is used for separation of aldehydes

It forms a soluble compound with aldehyde which on hydrolysis give back the aldehyde

47 α-H of aldehyde and ketone is acidic in nature

Because the corresponding carbanion is resonance stabilized

48 Carboxylic acids ( R-COOH ) do not give nucleophilic addition reaction like RCHO & RCOR although it has >C=O

due to resonance the carbonyl carbon looses its electrophilicity

49 Carboxylic acids have higher b.pt than aldehyde, ketones and even than alcohols

There is extensive inter molecular H-bonding in carboxylic acid(RCOOH). Even in vapour phase it exists as dimer

50 Carboxylic acids are miscible in water

Due to H-bonding with water

51 R-COOH is acidic in nature

Because the conjugate base R-COO- (carboxylate ion) is stable due to resonance

52 Acidic Strength : Cl-CH2-COOH > CH3-COOH > CH3CH2-COOH

Because Cl has –I effect which stabilizes the conjugate base and ethyl gr has +I effect

53 In amines the C-N-H/C bond angle is less than 109.5o

Due to lp-bp repulsion

54 For reduction of nitro compounds into amines Fe/HCl is preferred instead of Sn/HCl

Because Fe + HCl FeCl2. on hydrolysis FeCl2 gives HCl. So just small quantity of HCl is required to initiate the reaction

55 To convert alkyl halide(R-X) into amines (R-NH2) ammonolysis is not suitable

Because on ammonolysis a mixture of pri, sec, tert and quaternary amine will be produced

56 Aniline on exposure to air and light turns into coloured

Due to atmospheric oxidation

57 Lower amines are soluble in water

Due to H-bonding with water. In case of higher amine alkyl group is larger which is hydrophobic

58 Amines (R-NH2) are less soluble than alcohols (R-OH)

In alcohol the H-bonding with water is stronger because O is more electronegative than N

59 Amines (R-NH2) are lower boiling than alcohols (R-OH)

In alcohol the inter molecular H-bonding is stronger because O is more electronegative than N

60 Order of b.pt : primary amine > sec. amine > tert. amine

The no of N-H bond decreases so extent of H-bonding also decreases

61 In gaseous phase the order os basic strength : 30-amine > 20-amine > 10-amine > NH3

Due to +I effect of alkyl groups the electron density on N increases. So 30 is strongest as it has 3 alkyl groups

62 In aqueous state the base strength order

:(C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3 ( 20 > 30 > 10 > NH3)

: (CH3)2NH > CH3NH2 > (CH3)3N > NH3 ( 20 > 10 > 30 > NH3)

In aqueous state +I effect, steric effect and salvation effect interplay. So the order is not regular

63 R-NH2 is stronger base than NH3

Due to +I effect of alkyl group electron density on N increases in R-NH2

64 Aniline ( C6H5-NH2) is weaker base than NH3 and R-NH2

In aniline the lone pair of electron of N is involved in resonance. So it is less available.

65 Base strength : p-methoxy aniline > aniline > p-nitro aniline

Methoxy group (-OCH3) has +R effect where as –NO2 group has –R effect so electron density in the first case increases but inthe second case it decreases.

66 Acylation of aniline is carried out in presence of pyridine

Pyridine removes HCl produced and favours forward reaction

67 -NH2 group in benzene ring is ortho –para directing for electrphilic substitution

Due to +R effect it increases the electron density at ortho and para position

68 Bromination of aniline gives 2,4,6-tribromo aniline

Because –NH2 group activates benzene ring by +R effect. So for mono substitution –NH2 group is acylated.

69 Nitration of aniline gives unusual meta-nitro aniline although –NH2 group is ortho-para directing

In presence of acid –NH2 is converted into –NH3+ which is meta directing

70 Aniline does not undergo Friedel Craft reaction

Aniline is base and reacts with anhydrous AlCl3 so N becomes positive which deactivates benzene ring.

CONVERSION FROM NCERT BOOK(1).Suggest a possible mechanism for the following reaction: (C.B.S.E. DELHI 2009)n-BuBr + KCN ⟶ n- BuCN(2) .

(3).