CHEMISTRY TODAY | JUNE 157Vol. XXIVNo. 6June 2015Corporate Ofce : Plot 99, Sector 44 Institutional area, Gurgaon -122 003(HR).Tel : 0124-4951200e-mail : info@mtg.inwebsite : www.mtg.inRegd. Ofce406, Taj Apartment, Near Safdarjung Hospital, Ring Road, New Delhi - 110029.Managing Editor:Mahabir SinghEditor:Anil Ahlawat (BE, MBA)ContentsChemistry Musing Problem Set 238NCERT Corner11Learn Fast : 20Electron Displacements inOrganic CompoundsAdvanced Chemistry Bloc29AIPMT31Solved Paper 2015Ace Your Way CBSE XII37(Chapterwise Practice Paper : Series 1)Concept Map 46Some Basic Concepts of ChemistryKerala PMT48Solved Paper 201550 Challenging Problems58Concept Booster64Chemistry Musing Solution Set 2270WB-JEE71Solved Paper 2015You Ask, We Answer80Olympiad Problems82Crossword85rial editSend D.D/M.O in favour of MTG Learning Media (P) Ltd.Payments should be made directly to :MTG Learning Media (P) Ltd, Plot No. 99, Sector 44, Gurgaon - 122003 (Haryana)We have not appointed any subscription agent.Subscribe online at www.mtg.inOwned,PrintedandPublishedbyMahabirSinghfrom406,TajApartment, New Delhi - 29 and printed by Personal Graphics and Advertisers (P) Ltd., Okhla Industrial Area, Phase-II, New Delhi.Readers are adviced to make appropriate thorough enquiries beforeactinguponanyadvertisementspublishedinthismagazine.Focus/Infocus featuresaremarketingincentivesMTGdoesnotvouchorsubscribetotheclaimsand representations made by advertisers. All disputes are subject to Delhi jurisdiction only.Editor : Anil AhlawatCopyright MTG Learning Media (P) Ltd.All rights reserved. Reproduction in any form is prohibited.Individual Subscription Rates Combined Subscription Rates1 yr.2 yrs.3 yrs.1 yr.2 yrs.3 yrs.Mathematics Today330600775PCM90015001900Chemistry Today330600775PCB90015001900Physics For You330600775PCMB100018002300Biology Today330600775Science Teaching in Schools Below Danger MarkAlarm bells ringing!Science is not taught in all senior secondary schools under the Delhi Government. Out of 877 senior government schools, only 270teachscienceinclassesXIandXII. Thisgivesadoublebodyblow to an education system. First, this puts the lower income groups permanently in thelowerlevelasitisthestudentswhostudyingovernmentschools,whoare below the well off line. They cannot afford to pay the high fees and special fees demanded by private schools of the top classes. The private schools on their part cannotaffordtogivehighclasseducationwithrstclassfacilitiesforscience laboratories unless they charge the parents.Thegovernmentcannotsaythattheylackmoneytogiverstclassteaching facilitiesintheschoolsrunbythem. Thisistherstpriorityofthegovernment. We cannot afford to keep the lower middle classes in the lower level in the next generation also. The rst priority of the government is to give emphasis to give a rst class education to all. Every socialist country does it. I do not mean communist countries but countries like France and countries in North Africa. Students do not depend on their parents for education up to any level. The second body blow is tokeepthehavenotsawayfromhigherscienticeducation.Peoplechooseart subjects in colleges because of their choice. A minimum level science education up to the college level is a must for the development of the nation.Allourdreamsofindustrialisationandadvancementineveryeldshouldbe supportedbythestategovernments.Thesyllabusforeveryschool,particularly stategovernmentschoolsshouldbesameasfollowedbycentralschoolswith NCERT textbooks. If the schools do not have the ability to recruit good teachers, let NCERT recruit and train the teachers who alone should teach in high schools which cannot keep science subjects for lack of teachers. Anil AhlawatEditor8CHEMISTRY TODAY |JUNE 15JEE MAIN/PMTs1.Pickuptherearrangementinvolvingmigrationto electron defcient oxygen.(a)Pinacol pinacolone rearrangement(b)Hofmann rearrangement(c)Baeyer Villiger rearrangement(d)Noneof these2.Whichamongthefollowingstatementsiscorrect regardingthebondingofPtandC2H4inZeises salt?I.Itinvolvess-donationfromthep-orbitalof alkene into vacant metal hybrid orbital.II.Itinvolvesp-donationfromthep-orbitalof alkene into vacant metal hybrid orbital.III.Itinvolvesp-backdonationfromtheflled metald-orbital(orhybrid)intothevacant antibonding orbital of alkene.(a)I, II and III(b)I and III(c)II and III(d)III only3.Tevapourpressuresofbenzene,tolueneand xyleneare75torr,22torrand10torrrespectively at20C.Whichofthefollowingisnota possiblevalueofthevapourpressureofan equimolarbinary/ternarysolutionoftheseat 20C?Assumeallformidealsolutionswitheach other.(a) 4812(b)16(c) 3523(d)53124.Ethylene can be prepared in good yield by(a)CHCHN(CH) I3 2+33HeatCH2CH2 + (CH3)3N + HI(b)CH3CH2N+(CH3)3OH CH2CH2 + (CH3)3N + H2O(c)CH3CH2NH2CH2CH2 + NH3 (d)both (a) and (b).5.TeIP1,IP2,IP3,IP4andIP5ofanelementare 7.1,14.3,34.5,46.8,162.2eVrespectively.Te element is likely to be(a)Na(b)Si(c)F(d)CaChemistryMusingwasstartedfromAugust'13issueofChemistryTodaywiththesuggestionof ShriMahabirSingh.TheaimofChemistryMusingistoaugmentthechancesofbrightstudentspreparingfor JEE(Mainand Advanced)/ AIPMT/ AIIMS/OtherPMTs&PETswithadditionalstudymaterial.IneveryissueofChemistryToday,10challengingproblemsareproposedinvarioustopicsofJEE(Main andAdvanced)/AIPMT.ThedetailedsolutionsoftheseproblemswillbepublishedinnextissueofChemistry Today.Thereaderswhohavesolvedfveormoreproblemsmaysendtheirsolutions.Thenamesofthosewhosend atleastfvecorrectsolutionswillbepublishedinthenextissue.Wehopethatourreaderswillenrichtheirproblemsolvingskillsthrough"ChemistryMusing"andstandinbetter steadwhilefacingthecompetitiveexams.CHEMISTRYCHEMISTRY MUSINGMUSINGSet 23Solution Senders of Chemistry MusingSET 221.Krunal N. Jariwala (Ahmedabad)2.Pramila Kar, Cuttack (Odisha)3.Manas Rajput, Udaipur (Rajasthan)10CHEMISTRY TODAY |JUNE 15JEE ADVANCED6.In the reaction,2SO2 + O2 2SO3, starting with 2 moles of SO2 and 1 mole of O2 in 1 L fask, mixture required 0.4 moles of MnO4 in acidic medium. Hence, Kc is(a)2.0(b)0.4(c)1.6(d)2.6COMPREHENSIONBoilingpointofacovalentcompounddependson theintermolecularforces.Intermolecularforcesare theforcesofattractionandrepulsionbetweenthe interactingparticles(atomsandmolecules).Tisterm doesnotincludetheelectrostaticforcesthatexist between the two oppositely charged ions and the forces thatholdatomsofamoleculetogetheri.e.,covalent bonds.7.Liquifcationofwhichgaseoussubstancewillarise asaresultofmomentaryimbalanceinelectronic distribution? (a)NH3(g) (b)CO(g)(c)Na+(g)Cl(g)(d)Xe(g)8.Inwhichofthefollowingtriads,frstonehasthe highest boiling/melting point?(a)PH3, AsH3, SbH3(b)HBr, HCl, HF(c)CH3OCH3, CH3SCH3, CH3SeCH3(d)AlF3, SiF4, PF5 INTEGER VALUE9.Tenumberofstereogeniccentrespresentin following compound is10.2 g of a gas X areintroduced into an evacuated fask keptat25C.Tepressureisfoundtobe1atm. If 3 g of another gas Y are added to the same fask, thetotalpressurebecomes1.5atm.Assuming ideal behaviour, the molecular mass ratio of Y and X isMTG o f f e r s C l a s s r o o m S t u d y Ma t e r i a l f o r J EE( Mai n&Advanc ed) ,AI PMTand FOUNDATIONMATERIALforClass7,8,9, 10,11&12with YOURBRANDNAME& COVER DESIGN.This study material will save you lots of money spentonteachers,typing,proof-readingand printing.Also,youwillsaveenormoustime. Normally, a good study material takes 2 years to develop. But you can have the material printed with your logo delivered at your doorstep.Prot from associating with MTG Brand the most popular name in educational publishing for JEE (Main & Advanced)/AIPMT/PMT ....Order sample chapters on Phone/Fax/e-mail.Phone : 0124-495120009312680856, 09717933372e-mail: sales@mtg.in | www.mtg.inATTENTIONCOACHINGINSTITUTES:a great offer from MTGCONTENTPAPERPRINTINGEXCELLENTQUALITYCLASSROOM STUDY MATERIALYour logo hereCHEMISTRY TODAY |JUNE 15111.0.5 mole of each of H2, SO2 and CH4 are kept in a container. A hole was made in the container. Afer 3hours,theorderofpartialpressuresinthe container will be(a)pSO2 > pCH4 > pH2(b)pH2 > pSO2 > pCH4(c)pCH4 > pSO2 > pH2(d)pH2 > pCH4 > pSO22.(R) in the given reaction sequence is+ CH COCl3( ) P( ) Q ( ) Ranhyd. AlCl3HCNH O/dil. acid2(Optically active)(a) CHCH3OH(b) HOOCCCH3OH(c) ClCCH3OH(d) HOOCCCH3NH23.Te atomic weights of two elements A and B are 40 and80respectively.IfxgofAcontainsyatoms, how many atoms are present in 2x g of B?(a) y2(b) y4(c)y(d)2y4.Phenol associates in benzene to a certain extent to form dimer. A solution containing 2.0 102 kg of phenolin1.0kgofbenzenehasitsfreezingpoint decreasedby0.69K.Tedegreeofassociationof phenol is(Kf for benzene = 5.12 K kg mol1)(a)73.4%(b)50.1%(c)42.3%(d)25.1%5.Dettol is used as an antiseptic. It is a mixture of(a)furacine, terpineol(b)bithionol, iodoform(c)chloroxylenol, terpineol(d)veronal, serotonin.6.Te IUPAC name of the given compound isOH(a)2,4,4-trimethylhex-5-ene-5-ol(b)3,3,4,4-tetramethylbut-1-en-2-ol(c)3,3,5-trimethylhex-1-en-2-ol(d)none of the above.7.AssumetworadioactivesubstancesAandB disintegrating by d Adt[ ] = kA[A], t1/2(A) = 0 693 .kA =d Bdtk BB[ ][ ] ,2 t1/2(B) = 10k BB[ ]Ifbothhalf-lifeperiodsandconcentrationsare equal,ratesatthestartofreactionwillbeinthe ratio(a)0.693(b)kA/kB(c)kB(d)none of these.8.An example of a condensation homopolymer is(a)bakelite(b)melamine-formaldehyde resin(c)alkyd resin(d)perlon or nylon-6.9.TeHinsbergstestofacompound,C5H14N2 producesasolidthatisinsolublein10% aq.NaOH.Tissolidderivativedissolvesin10% aqueoussulphuricacid.Whichofthefollowing would best describe these facts?CLASSXI-XIINCERTCORNERThe questions given in this column have been prepared strictly on the basis of NCERT Chemistry.PapersofJEE(Main& Advanced)/AIPMT/AIIMS/otherPMTsaredrawn heavily from NCERT books. Practice Hard! All the best!!12CHEMISTRY TODAY |JUNE 15(a)NH2CH2CH2N(CH3)2(b)(CH3)2NCH2CH2NHCH3(c)NH2CH2C(CH3)2CH2NH2(d)(CH3)2NCH2N(CH3)210.Correctarrangementofthefollowingacidsin decreasing order of pKa values isCH3COOH,Cl2CHCOOH,F3CCOOH,IIIIIIClCH2COOHFCH2COOHIVV(a)I > III > II > IV > V(b)I > IV > V > II > III(c)III > II > I > V > IV(d)II > III > I > IV > V11.At273K,thedensityofacertaingaseousoxide at2atmosphereissameasthatofdioxygenat 5atmosphere.Temolecularmassoftheoxide (in g mol1) is(a)80(b)64(c)32(d)16012.If 0.189 g of a chlorine containing organic compound gives 0.287 g of silver chloride, then the percentage of chlorine in the organic compound is(a)35.47(b)47.57(c)37.57(d)45.3713.Amongthefollowingcomplexestheonewhich shows zero crystal feld stabilization energy (CFSE) is(a)[Mn(H2O)6]3+(b)[Fe(H2O)6]3+(c)[Co(H2O)6]2+(d)[Co(H2O)6]3+14.An organic compound P has 76.6% C and 6.38% H. Itsvapourdensityis47.Itgivesacharacteristic colour with aq. FeCl3. P when treated with CO2 and NaOHat140CunderpressuregivesQwhichon acidifcationgivesR.Rreactswithacetylchloride to give S, which is(a) OCOCH3OCOCH3 (b) OHOCOCH3 (c) OHOH (d) COOHOCOCH315.Which of the following descriptions about [FeCl6]4 is correct?(a)dsp3, inner orbital complex, diamagnetic(b)sp3d2, outer orbital complex, paramagnetic(c)d2sp3, inner orbital complex, paramagnetic(d)sp3d2, outer orbital complex, diamagnetic16.Which of the following on thermal decomposition yields a basic as well as an acidic oxide?(a)KClO3(b)Na2CO3(c)NaNO3(d)CaCO317.Whichofthefollowingorderabouttheirbasic strengths is correct?(a)NH3 < NH2OH < NH2NH2(b)NH2OH < NH2NH2 < NH3(c)NH3 > NH2OH > NH2NH2 (d)NH2OH > NH2NH2 > NH318.In acidic solution, MnO4 is(a)oxidised by 3 electrons (b)reduced by 3 electrons (c)oxidised by 5 electrons(d)reduced by 5 electrons.19.Teatomicnumbersofvanadium(V), chromium (Cr),manganese(Mn)andiron(Fe)are23,24, 25and26respectively.Whichoneofthesemay beexpectedtohavethehighestsecondionisation enthalpy?(a)V(b)Cr(c)Mn(d)Fe20.In the equilibrium N2O4(g) 2NO2(g)when 5 moles of each is taken and the temperature is kept at 298 K, the total pressure was found to be 20 bar. [Given: DGf (N2O4) = 100 kJ; DGf (NO2) = 50 kJ]Te reaction will be in(a)forward direction(b)reverse direction(c)equilibrium(d)cannot say.21.Te data for the reaction A + B C, isExp. [A]0[B]0Initial rate1 0.012 0.035 0.102 0.024 0.070 0.803 0.024 0.035 0.104 0.012 0.070 0.80Te rate law corresponds to the above data is(a)Rate = k[A][B]3(b)Rate = k[A]2[B]2(c)Rate = k[B]3(d)Rate = k[B]422.Teuncertaintyinthevelocityofaparticleofmass 6.626 1031 kg is 1 106 m s1. What is the uncertainty in its position (in nm)? (h = 6.626 1034 J s) (a) 12 (b) 2 5 .(c) 4 (d) 14CHEMISTRY TODAY |JUNE 151323.Which of the following orders is notin accordance with the property stated against it?(a)F2 > Cl2 > Br2 > I2 : Bond dissociation energy(b)F2 > Cl2 > Br2 > I2 : Oxidising power(c)HI>HBr>HCl>HF:Acidicpropertyin water(d)F > Cl > Br > I : Electronegativity24.CalculatetheamountofCa(OH)2requiredto removethehardnessofwaterfrom60,000litres containing 16.2 g of Ca(HCO3)2 per 100 litre.(a)1.11 kg(b)2.22 kg(c)3.33 kg(d)4.44 kg25.Whichoneofthefollowingisleastcovalentin nature?(a)NF3(b)BiF3(c)PF3(d)SbF326TevanderWaalsconstantsforfourgasesP,Q, RandSare4.17,3.59,6.71and3.8atmL2mol2. Terefore, the ascending order of their liquefaction is(a)R < P < S < Q(b)Q < S < R < P(c)Q < S < P < R(d)R < P < Q < S27.Which of the following haloalkanes would undergo SN2 reaction faster?I. CHCl2II. ClIII.IIV. Cl(a)I(b)II(c)III(d)IV28.For the properties mentioned, the correct trend for the diferent species is in(a)strength as Lewis acid BCl3 > AlCl3 > GaCl3(b)inert pair efect Al > Ga > In(c)oxidising property Al3+ > In3+ > Tl3+(d)frst ionization enthalpy B > Al > Tl29.Totalnumberoftertiarycarbonatomspresentin the following compound is (a)6(b)2(c)4(d)530.Forareaction,2K(g)+L(g) 2M(g);DU=10.5kJ andDS=44.1JK1.CalculateDGforthe reactionandpredictwhetherthereactionwillbe spontaneous or non-spontaneous?(a)+ 0.16 kJ, non-spontaneous(b) 0.16 kJ, spontaneous(c)+ 26.12 kJ, non-spontaneous(d)26.12 kJ, spontaneous31.Inwhichofthefollowingpairsofmolecules/ions both the species are paramagnetic?(a)C2, N2+(b)O22, F2(c)NO, CN(d)N22, NO+32.Te isoelectronic pairs are(a)Na+, Mg2+(b)Mg2+, F(c)NO+, N2(d)all of these.33.ForCr2O72+14H++6e2Cr3++7H2O, E=1.33Vat[Cr2O72]=4.5millimole, [Cr3+] = 15 millimole, E is 1.067 V. Te pH of the solution is nearly equal to(a)2(b)3(c)5(d)434. YcanbeobtainedbyEtardsreaction. Zundergoesdisproportionationreactionwith concentrated alkali. X could be(a) CCH(b) (c) CHCHCH3(d) CHCCH3CH335.4-methylpent-3-en-2-one.P is(a)propanone(b)ethanamine(c)ethanenitrile(d)ethanal.36.RbO2 is(a)peroxide and paramagnetic(b)peroxide and diamagnetic(c)superoxide and paramagnetic(d)superoxide and diamagnetic.37.Tetruestatementfortheacidsofphosphorus, H3PO2, H3PO3 and H3PO4 is (a)the order of their acidity is H3PO4 > H3PO3 >H3PO214CHEMISTRY TODAY |JUNE 15(b)all of them are reducing in nature(c)all of them are tribasic acids(d)thegeometryofphosphorusistetrahedralin all the three.38.Initial rate data at 25C are listed in the table for the reaction : NH4+ + NO2N2 + 2H2OExp. no.Initial [NH4+]Initial [NO2 ]Initial rate of consumption1. 0.24 0.10 7.2 1062. 0.12 0.10 3.6 1063. 0.12 0.15 5.4 106Te rate law for the reaction will be(a)k[NH4+ ] (b)k[NH4+ ][NO2](c)k[NH4+ ]2 (d)k[NH4+ ][NO2]139. A crystalline solid X reacts with dil. HCl to liberate agasY.YdecolourisesacidifedKMnO4.Whena gasZisslowlypassedintoanaqueoussolutionof Y, colloidal sulphur is obtained. X and Z could be, respectively(a)Na2S, SO3(b)Na2SO4, H2S(c)Na2SO3, H2S(d)Na2SO4, SO240.Te number of peptide bond(s) in the compound HC3ONHCH3NNHNH2OHOHC3CH3 is/are(a)1(b)2(c)3(d)441.Inasolid,oxideionsarearrangedinccp. One-sixth of the tetrahedral voids are occupied by cationsXwhileone-thirdoftheoctahedralvoids areoccupiedbythecationsY.Teformulaofthe compound is(a)X2YO3(b)XYO3(c)XY2O3(d)X2Y2O342.In the following reactions sequenceCH3CH2Br aq. KOH A KMnO/H4+ B NH3 C Br2alkali D;D is (a)CH3Br(b)CH3CONH2(c)CH3NH2(d)CHBr343.Whichofthefollowingcompoundsundergo electrophilic substitution most easily?(a)(b) (c)(d) 44.Temajorproductformedwhen2-bromo-2-methylbutane is refuxed with ethanolic KOH is(a)2-methylbut-2-ene(b) 2-methylbutan-1-o1(c)3-methylbutan-2-ol(d) 2-methylbutan-2-ol.45.SilicaisreactedwithNa2CO3.Whichgasis liberated?(a)CO(b)O2(c)CO2(d)O346.Which of the following structures for a nucleotide is not correct?(a)Cytosine-Ribose-Phosphate(b)Uracil-2-Deoxyribose-Phosphate(c)Uracil-Ribose-Phosphate(d)Tymine-2-Deoxyribose-Phosphate47.Teproductionofdihydrogengasviawater-gas shif reaction : CO + H O( ) ( ) g g 2CO + H2 2 ( ) ( ) g gCatalystTe CO2 gas is removed by scrubbing with solution of(a)sodium arsenite(b)calcium oxide(c)sodium phosphite(d)aluminium oxide.48. In a mixture of PbS, ZnS and FeS, each component is separated from other by using the reagents in the following sequence in froth foatation process(a)potassium ethyl xanthate, KCN(b)potassium ethyl xanthate, KCN, NaOH, CuSO4, acid(c)KCN, CuSO4, acid (d)none of the above.49. Biochemical Oxygen Demand (BOD) is a measure of organic material present in water. BOD value less than 5 ppm indicates a water sample to be(a)rich in dissolved oxygen(b)poor in dissolved oxygen(c)highly polluted(d)not suitable for aquatic life.50.Te correct order of frst ionisation potential among the following elements Be, B, C, N, O is(a)B < Be < C < O < N(b)B < Be < C < N < O(c)Be < B < C < N < O(d)Be < B < C < O < NCHEMISTRY TODAY |JUNE 151516CHEMISTRY TODAY |JUNE 15SOLUTIONS1.(a) :Initially,partialpressureswereequal.Now r1/d or1/M, thereforeamountsdifused out in the same time will be H2 > CH4 > SO2.Amounts lef will be H2 < CH4 < SO2\pSO2 > pCH4 > pH22.(b) : + CH COCl3( ) P( ) QAnhyd. AlCl3HCNH O/dil. acid2( ) R(Optically active)COCH3HOOCCCH3OHNCCCH3OH3. (c) :No. of moles of A = x40No. of atoms of A = x40 NA = yNo. of moles of B = 280 40x x=Now, no. of atoms of B = x40 NA = y4.(a) :MK wW Tff2( )( )obs.in kg= = = 5 12 2 0 101 0 0 6914 84 1022 1. .. .. kg mol= 148.4 g mol1Calculated molecular mass of phenol = 94 g mol1 iMM= = =2294148 40 633(( ) ..cal.)obs. 21 2C H OHC H OH)6 5 6 5 2 (/Total species = + = ( ) 1212 i i == 1 21 21 /oror = = = 2 1 2 1 0 633 0 734 ( ) ( . ) . i = 73.4%5.(c)6.(c) : 3,3,5-Trimethylhex-1-en-2-olOH1234567.(a) : t1/2(A) = t1/2(B)0.693=1[ ]0k k BA B; kAkB = 0.693 [B]0 = 0.693 [A]0Rate, rA = kA[A]0 ; rB = kB[B]02 = kB[A]02r kk AA ABrB =[ ]0 = 0.6938.(d) :Nylon-6 is a condensation homopolymer since itisobtainedbycondensationpolymerizationof only one type of monomer, i.e., caprolactum.9.(b) :3aminedoesnotreactwithHinsbergs reagent,2aminereactsbutsaltformedisnot soluble in alkali.1 amine reacts and salt formed is soluble in alkali. (CH ) NCH CH NHCH3 2 2 2 3C H SO Cl6 5 2Not soluble in aq. NaOH but dissolvesin sulphuric acid(CH ) NCH CH NSO C H3 2 2 2 2 6 5CH310.(b) :Moretheelectronwithdrawingefect,the compoundwillbemoreacidic.Morethenumber ofhalogenatoms,greaterwouldbethedispersion ofthenegativecharge.Hence,morewillbe stabilisationofanionandthecompoundwillbe more acidic. More the value of pKa, less acidic will bethecompound.SmallerthevalueofpKa,the compound will be more acidic.11.(a) :From ideal gas equation, PV = nRT PVwM RT = nwM=so,PM1 \PPMM1221= 2532 5 3228011= ==MM org mol1,12.(c) :According to Carius method,% of chlorine in organic compound

= 35 5143 5100 ..Mass of AgCl formedMass of substance taken = =35 5 0 287143 5 0 189100 37 57. .. .. %13.(b) :H2O is a weak feld ligand, hence Do < pairing energy.CFSE = (0.4x + 0.6y)Dowhere,xandyareno.ofelectronsoccupyingt2g and eg orbitals respectively.For [Fe(H2O)6]3+ complex ion, Fe3+ (3d5) = t32g eg2 = 0.4 3 + 0.6 2 =0.0or0 DqCHEMISTRY TODAY |JUNE 151714.(d) :To determine empirical formula from data :Element %Relative no. of atomsSimple ratioC 76.676 6126 38.. = 6H 6.386 3816 38.. = 6O 17.0217 02161 06.. = 1\Empirical formula (P) = C6H6OEmpirical formula weight = 94Molecular weight = 2 VD = 2 47 = 94\Molecular formula of P is C6H6O.SincePgivescolourwithaq.FeCl3ithasa phenolic group.Compound P should be C6H5OH (phenol).( ) PPhenolOHNaOH + CO2OHCOONaH+HOH( ) QOHCOOHCH COCl3( ) RSalicylic acidOCOCH3COOH( ) SAspirin (pain killer)15.(b) :In [FeCl6]4, Fe2+(3d6), Paramagneticduetopresenceoffourunpaired electrons.16.(d) :CaCO3 CaO + CO2(Basic)(Acidic)17.(b) :Basicityinnitrogencompoundsisattributed to the availability of lone pair of electrons.PuttingmoreelectronegativegroupsonNwill decrease its basicity.Terefore, the order of basicity of these compounds is :NH3 > NH2NH2 > NH2OH18.(d) :In acidic medium,MnO4 + 8H+ + 5e Mn2+ + 4H2OTe reaction shows gain of 5e i.e. reduction.19.(b) :TesecondionisationpotentialvalueofCr issufcientlyhigherthanthoseofneighbouring elements.Tisisbecauseoftheelectronic confgurationofCr+whichis3d5(half-flled), i.e., for the second ionisation, the electron is to be removed from very stable confguration.20.(b) :StandardGibbsfreeenergychangeforthe reaction, N2O4(g) 2NO2(g)DG = DGproducts DGreactants = 2 50 100 = 0DG = 2.303 RT log Kp = 0; Kp = 1Initially, pN2O4 = pNO2 = 10 barSo,Qppp(initial)22 4NON O= =( )210Initial Gibbs free energy of the above reaction,DG= DG + 2.303 RT log QpDG= 0 + 2.303 8.314 298 log10 = 5.705 103 kJ mol1Since initial Gibbs free energy change of the reaction is positive, so the reverse reaction will take place.21.(c) :A + B CLetrate = k[A]x [B]ywhere order of reaction is (x + y)Putting the values of exp. 1, 2, and 3, we get following equations.0.10 = k [0.012]x [0.035]y...(i)0.80 = k [0.024]x [0.070]y...(ii)0.10 = k [0.024]x [0.035]y...(iii)Dividing eq. (ii) by eq. (iii), we get 0 800 100 0700 035....= y 2y = 8y = 3Keeping [A] constant, [B] is doubled, rate becomes 8 times.Dividing eq. (iii) by eq. (i), we get 0 100 100 0240 012....= x 2x = 1x = 0 Keeping [B] constant, [A] is doubled, rate remains unafected. Hence rate is independent of [A].Rate [B]322.(d) :According to Heisenberg uncertainty principle, x ph 4 xhmv= 4[ Dp = mDv] = 6 626 104 6 626 10 1 103431 6.. =1 104149 m or nm18CHEMISTRY TODAY |JUNE 1523.(a) :Te bond dissociation energy order is incorrect, the correct order is Cl2 > Br2 > F2 > I2. Te lower value ofbonddissociationenergyoffuorineisdueto thehighinter-electronicrepulsionsbetweennon-bondingelectronsinthe2p-orbitalsoffuorine. As a result, FF bond is weaker in comparison to ClCl and BrBr bonds.24.(d) :Amount of Ca(HCO3)2 in 60,000 litres of water = 16 2 60 000100. ,= 9720 g = 9720162= 60 moles [ mol. wt. of Ca(HCO3)2 = 162]Ca(OH)2 + Ca(HCO3)2 2CaCO3 + 2H2O 1 mole1moleAmount of Ca(OH)2 required = 60 moles= 60 74 = 4440 g = 4.44 kg[ mol. wt. of Ca(OH)2 = 74]25.(b) :Onmovingdownthegroup,thecovalent character of halides decreases.26.(c) :Easilyliquefablegaseshavegreaterinter-molecularforceswhichisrepresentedbyhigh valueofa.Tegreaterthevalueofamorewillbe liquefability. So, the order is Q < S < P < R.27.(c) :AmongIandII,IwillundergoSN2reaction fasterasitisaprimaryalkylhalidewhileIIisa secondary alkyl halide. Among I, III and IV, III will undergoSN2reactionfastestasiodineisabetter leaving group because of large size. Hence, it will be released at a faster rate in the presence of incoming nucleophile.28.(a) :Lewisacidstrengthdecreasesdownthe group.29.(d) : Encircledcarbonatomsinthiscompoundare tertiarycarbonatoms(carbonatomsdirectly attached to three other carbon atoms).30.(a) :2K(g) + L(g) 2M(g)Dng= 2 3 = 1DH = DU + DngRT= 10.5 103 + (1 8.314 298)= 10500 + (2477.572) = 12977.57 J = 12.98 kJDG = DH TDS= 12.98 298 ( 44.1 103)= 12.98 + 13.14 = 0.16 kJSinceDGis+vehence,thereactionis non-spontaneous.31.(c) :Molecular orbital confguration ofC2 : s1s2 s*1s2 s2s2 s*2s2 p2px2 p2py2 (diamagnetic)N2+ : s1s2 s*1s2 s2s2 s*2s2 p2px2 p2py2 s2pz1(paramagnetic)O22 : s1s2 s*1s2 s2s2 s*2s2 s2pz2 p2px2 p2py2 p*2px2 p*2py2 (diamagnetic)F2 : s1s2 s*1s2 s2s2 s*2s2 s2pz2 p2px2 p2py2 p*2px2 p*2py2 (diamagnetic)NO : s1s2 s*1s2 s2s2 s*2s2 s2pz2 p2px2 p2py2 p*2px1(paramagnetic)CN : s1s2 s*1s2 s2s2 s*2s2 p2px2 p2py2 s2pz1(paramagnetic)N22 : s1s2 s*1s2 s2s2 s*2s2 p2px2 p2py2 s2pz2 p*2px1p*2py1 (paramagnetic)NO+ : s1s2 s*1s2 s2s2 s*2s2 s2pz2 p2px2 p2py2(diamagnetic)32.(d) : All are isoelectronic pairs.Na+ and Mg2+ have 10 electrons.Mg2+ and F have 10 electrons.NO+ and N2 have 14 electrons.33.(a) :For the given reaction, Cr2O72 + 14H+ + 6e 2Cr3+ + 7H2OE ERTnF= 2 303 .log[Products][Reactants]1 067 1 330 05916. ..log = [Cr ] [H O][Cr O ][H ]3+ 2272 72 + 141 067 1 33 9 85 1015 10 14 5 1033 2 73 14. . . log[ ] [ ][ . ][ ]= +H 0 2639 85 103..=2log[15103]log[4.5103] 14 log[H+]26.7 = 2 1.82 + 2.34 + 14 pH pH= =2814234.(b) : Y is obtained by Etards reaction,CHEMISTRY TODAY |JUNE 1519Z reacts with conc. NaOH,2HCHO+ NaOH Formaldehyde (Z) HCOONa + CH3OHSodium formate35.(c) : 36.(c)37.(d) :Tere is very little diference in their acidity. PHOHOHHPO3 2PHOHOOHHPO3 3PHOOHOOHHPO3 4Reducingnaturedependsonno.ofP Hbonds.Moretheno.ofP Hbonds,morewillbethe reducing nature. Tus, H3PO2 is stronger reducing agentthanH3PO3whileH3PO4doesnotactas reducing agent at all.H3PO2,H3PO3andH3PO4containone,twoand threeionisablehydrogenatoms(P OHbonds) respectively.As sp3 hybridised, therefore all are tetrahedral.38.(b)39.(c) :Na2SO3(s) + 2HCl(dil.) 2NaCl + SO2(g) + H2OX YH2O + SO2(g) H2SO3(aq)Sulphurous acid(An aq. solution of Y)H2SO3 + 2H2S(g) 3S+ 3H2OZ(Colloidal)40.(a) :HC3ONHCH3NNHNH2OHOHC3CH341.(b) :Suppose number of O2 ions = n. Ten number of octahedral voids = n and number of tetrahedral voids = 2nNo. of cations X present in tetrahedral voids= =1623nn No. of cations Y present in octahedral voids = =13 3nn\Ratio X : Y : O2 = =n nn3 31 1 3 : : : :Hence, formula is XYO3.42.(c): 43.(c) : Activating due to +I andLone pairs of O are inhyperconjugationconjugation with C O(weak efects)hence moderate activating group Lone pair on N leads toDeactivating group +M efect (a powerful efect)(Most activating)44.(a) :AccordingtoSaytzefsrule,thecompound undergoeseliminationreactiontoformmore substituted alkene as the major product.45.(c) :SiO2 + Na2CO3 Na2SiO3 + CO246.(b) :Uracil is present only in RNA which contains ribose as the sugar. Tus, nucleotide given in option (b) is incorrect.47.(a) 48.(b)49.(a) :Cleanwaterrichindissolvedoxygenwould have BOD value of less than 5 ppm whereas highly polluted water could have a BOD value of 17 ppm or more.50.(a)20CHEMISTRY TODAY |JUNE 15inductiveefectisalmostnegligiblebeyondtwo carbon atoms from the active atom or group.Inductiveefectdoesnotinvolveactualtransferofelectronsfromoneatomtoanotherbutmerely helps in displacing them permanently.Groups with I efect :NO2 > CN > COOH > F > Cl > Br > I > OCH3 > C6H5Decreasing order of I efect Groups with + I efect :(CH3)3C > (CH3)2CH > C2H5 > CH3 > HDecreasing order of +I efect Tephenomenonofinductiveefectisveryimportantinorganicchemistryasitishelpfulin explaining a number of facts.Reactivityofalkylhalides: Tepresenceof halogenatomsinthemoleculeofalkylhalide createsacentreoflowelectrondensitywhich isreadilyattackedbythenegativelycharged reagents. Dipolemoment: Astheinductiveefect increases, the dipole moment increases. Relativeacidstrengthofchloroaceticacidsand acetic acid : Cl CH C O H2H C C O H3O OChloroacetic acidp = 2.86 KaAcetic acidp = 4.76 Ka CH C O H C C O HOClClOClClClDichloroacetic acidp = 1.25 KaTrichloroacetic acidp = 0.65 KaTebehaviourofanorganiccompoundisinfuenced toalargeextentbytheelectrondisplacementstaking placeinitscovalentbonds.Tesedisplacementsmay beofpermanentnatureortemporarywhichtake placeinpresenceofanotherspeciesinthemolecule. Teacidityandbasicityoforganiccompounds,their stability,theirreactivitytowardsothersubstances,etc. caneasilybepredictedbyconsideringsuchelectronic displacements.Inductive EfectTisisapermanentefectoperatinginpolarcovalentbonds.Teinductionofapermanent dipole in a covalent bond bearing two unlike atoms ofdiferentelectronegativitiesisreferredtoasthe inductive efect.Tedevelopmentofpartial+veandvechargesisduetothedrifofthesharedpairofelectrons towardsthemoreelectronegativeatomresulting intheoriginofsmallfractionalchargesonthe constituent atoms.Whenacarbonatomisbondedtoahydrogen(CH)oranothercarbon(CC)atombya covalentbondsuchasinalkanes,thesharingof electron pair is symmetrical between them.C H or C:H(Symmetrical sharingof electron pair) or C:X C X+ (Origin of fractionalcharges due to greaterelectronegativity of ) XTe direction of displacement is shown by placingan arrow head midway along the line representing the sigma bond.C C C C Cl4 3 2 1+++Teinductiveefectofanatomoragroupofatoms diminishes rapidly with distance. In fact, the Electron Displacements in Organic Compounds Electron Displacements in Organic CompoundsCHEMISTRY TODAY |JUNE 1521Te decreasing order of acid strength :Cl3CCOOH > Cl2CHCOOH > ClCH2COOH> CH3COOHRelativeacidstrengthofformicacidandacetic acid : H C C O H3H C O HO Op = 4.76 Kap = 3.77 Ka Methylgrouphasanelectronreleasing inductiveefect(+Iefect).Tereforeacetic acid is a weaker acid than formic acid.Relativeacidstrengthoffuoroaceticacid,chloroaceticacid,bromoaceticacidand iodoaceticacid:Halogenatedacidsaremuch strongeracidsthantheparentacidandthe acidityincreasesalmostproportionatelywith the increase in electronegativity of the halogen presentwhichhelpsinrepellingtheproton from the hydroxy group of acid.FCH2COOH > ClCH2COOH > BrCH2COOH > ICH2COOH Sincetheinductiveefectdecreaseswith increaseindistanceofhalogenatomfrom thecarboxylicgroup,thestrengthofthe acid is proportionally decreased.CH3CH2CH(Cl)COOH > CH3CH(Cl)CH2COOHa-Chlorobutyric acidb-Chlorobutyric acid> CH2(Cl)CH2CH2COOH > CH3CH2CH2COOHg-Chlorobutyric acid n-Butyric acidRelative reactivity of toluene (methylbenzene)andbenzeneinaromaticsubstitution reactions : Aromatic substitution reactions are generally electrophilic in nature. Methyl has an electronreleasinginductiveefect(+Iefect). Terefore, toluene with higher electron density ismorereactivethanbenzeneinelectrophilic substitution reactions. CH3Toluene(High electron density)(More reactive towardselectrophilic substitution reactions)BenzeneRelativereactivityofnitrobenzeneandbenzene in electrophilic aromatic substitution reactions : +NO2Nitrobenzene ( effect due to NO ) I2(Lower electron density)(Less reactive towards electrophilicsubstitution reactions)BenzeneRelativeacidstrengthofwater,phenolandmethylalcohol:Ascomparedtowater, phenolismoreacidic(Iefect)butmethyl alcohol is less acidic (+I efect). Strengthofbase: Acompoundissaidtobe basicinnature,ifitiscapableofacceptinga proton. Base strength is defned as the tendency todonateanelectronpairforsharing.Te diference in base strength can be explained on the basis of inductive efect. Ascomparedtoammonia,methylamineis morebasic(+Iefect)whileanilineisless basicanddiphenylamineisastillweaker base (I efect).CH3NH2 > NH3 > C6H5NH2 > (C6H5)2NHTedecreasingorderofbasestrengthin alcohols is due to +I efect of alkyl groups.(CH3)3COH > (CH3)2CHOH > CH3CH2OH > CH3OH321methylGreater the tendency to donate electron pair forcoordinationwithproton,moreisthe basicnature,i.e.,morethenegativecharge onnitrogenatom(dueto+Iefectofalkyl group), higher is the basic strength.Tus,thebasicnaturedecreasesinthe order,Alkyl groupRelative basic strengthCH3R2NH > RNH2 > R3N > NH3C2H5 R2NH > R3N > RNH2 > NH3Relative stabilities of carbocations : (CH3)3C + > (CH3)2 +CH > CH3 +CH2 > +CH3Reactivityofcarbonylcompoundsinnucleophilicadditionreactions:+Igroup increaseselectronavailabilityoncarbonyl carbon.Tisthereforedecreasestherateof nucleophilicaddition.Ontheotherhand, electronwithdrawingIefectdecreases electronavailabilityoncarbonylcarbonand therebyincreasestherateofnucleophilic addition.22CHEMISTRY TODAY |JUNE 15CCl C H > H C H > H C C H3 3> CH C CH3 3O O OODecreasing order of nucleophilic additionInductive efect is also dependent on the diferencein the state of hybridisation of the atoms linked by covalent bond.Relative acidity of hydrocarbons may be givenas : Relativebasicityofcorrespondingcarbanionsis as : Teacidstrengthofacrylicacid(CH2 CHCOOH)isconsiderablyhigher thanthatofpropionicacid(CH3CH2COOH) duetotheelectronwithdrawinginductive efect of the sp2-hybridised carbon atom of the a,b-doublebondeventhoughtheresonance efectofthea,b-unsaturatedcarbonsystem would tend to decrease the acid strength.Field EfectInductive efect is a permanent efect in the groundstate of the molecule and usually operates through single bonds. However, when the inductive efect is transmittedthroughspaceorsolventmolecules,it is known as feld efect.Tetwochlorineatomsin(I)and(II)exertthesame inductive efect with respect to the position of electronsassociatedwiththeCOOHgroupand consequentlythetwocompoundsshouldexhibit the same acid strength. HHClClCOOHCOOHCl HCl Hp = 5.67 Ka(I)p = 6.07 Ka(II)However,thetwochlorineatomsinthemolecule (I),beingclosertotheacidgroupascomparedto (II),willexertgreaterelectronwithdrawingefect in (I) as compared to (II). Hence the isomer (I) is a stronger acid than (II).Electromeric EfectTis is a temporary efect operating in unsaturatedcompounds only at the demand of a nearby reagent andassoonasthisattackingreagentisremoved, the original condition is restored.Itinvolvesthecompletetransferof p-electronsof multiplebond,becausep-bondsarelooselyheld and easily polarisable.C OC O::::: + Tusthecompletetransferofsharedpairof p-electronsofamultiplebondtothemore electronegativeatomofthebondedatomsdueto therequirementofanattackingreagentiscalled electromeric efect (E-efect).Whenthetransferof p-electronstakesplace towardstheattackingreagent(electrophile),the efect is called +E-efect.C C+ H C C++HCH CH CH + H CH CH CH3 2+3 3+PropeneWhenthetransferofelectronstakesplaceawayfromtheattackingreagent,theefectiscalled E-efect.C O + CN C O CNC O+CN C O : ::CNstep (i)slow:::H+step (ii)fastC OHCNWhenthe I-andE-efectoccurtogetherina molecule,theymaybeassistingoropposing eachother.Whentheyareopposing,theE-efect generally dominates over I-efect.Applications : Electrophilic addition reactions of unsaturatedcompoundsinvolvethepolarisationofthe carbon- carbon double bond in the presence of attacking electrophiles like H+.CHEMISTRY TODAY |JUNE 1523Nucleophilicadditionreactionsofcarbonylcompoundsinvolvepolarisationthrough electromeric efect of the carbon-oxygen double bond in the presence of a nucleophile.Electrophilicsubstitutionreactionsofbenzenoidsinvolvepolarisationthrough electromeric efect of the benzene ring when an electrophile (E+) approaches them. E+E++ +EH: OCH3::OCH3:+OCH3:+E+E+H EH: OCH3:+HEE+E+OCH3:+OCH3:::HHyperconjugationHyperconjugation is the stabilizing interaction thatresultsfromtheinteractionoftheelectronsina sigma bond (usually C H or C C) with an adjacent emptyorpartiallyfllednon-bondingp-orbitalor antibonding p-orbital to give an extended molecular orbital that increases the stability of the system.Conjugated dienes like 1,3-butadiene have been foundto be more stable than simple alkenes like 1-butene. Tis has been explained in terms of delocalization of p-electrons.Besidesconjugationevenalkylgroupsbearinghydrogen on the carbon that is attached to doubly bonded carbon atoms tend to increase the stability of alkenes.Propene (CH 3 CH CH2) for example has been found to be more stable than ethene (CH2 CH2) by about 11 kJ/mole.Tiscanbeexplainedintermsofdelocalization ofelectronswhichtakesplacebytheoverlapping between a p-orbital of carbon and a s-orbital of the H of methyl group. As a result of this overlapping, each pair of electrons does not just bind together two atoms i.e., the doubly bonded carbons or the carbon and hydrogen but all the four atoms. Tis delocalization whichinvolvess-bondorbitalsalso,isreferredto as hyperconjugation or s, p-conjugation.Teconceptofhyperconjugationwasdevelopedonthebasisofdiscoveryofanomalouselectron releasing pattern of alkyl groups. Te inductive (+I) efectofalkylgroupsisnormallyinthefollowing order :Baker and Nathan observed that when alkyl groupsareattachedtoanunsaturatedsystem,theorder ofinductiveefectisdisturbedandinsomecases actually reversed. For example, the rate of reaction ofp-alkylbenzylbromidewithpyridinewas contrary to what was expected from the order of the inductiveefectofthesubstituentalkylgroupi.e., rate of the above reaction follows the order :methyl > ethyl > isopropyl > tert butyl.RRCH Br + C H N2 5 5CH NC H + Br2 5 5+TisefectisknownasBaker-Nathanefect.Itis apermanentefect.Infacthyperconjugationisan extension of resonance.Resonanceefectinvolvesdelocalizationofp-electrons of two or more conjugated double bonds whilehyperconjugationinvolvesdelocalizationof s-electrons. Hyperconjugation can be described as double bond - no bond resonance.Conjugationbetweenthe s-electronsofsingle bondandp-electronsofmultiplebondi.e., s,p-conjugation is known as hyperconjugation.Hyperconjugation is of two types : Sacrifcialhyperconjugation: Teessential conditionistheattachmentofalkylgroupto double bond or triple bond. Carbon atom of alkyl group attached to double bondmustcontainatleastonehydrogenatom in hyperconjugation.24CHEMISTRY TODAY |JUNE 15 H C C C H C C CHH HH+H C C C H+C C CHH+HHIt involves a sort of sacrifce of bond.Isovalenthyperconjugation: Tiskindof hyperconjugationinvolvesnosacrifceof bonds. Ethyl radicals have the same number of real bonds as the classical structure.H C C+H C C H+C CHHHHH+HHHHHHHH C CHH+HHSignifcance of hyperconjugation : Heat of hydrogenation : C C + H CH CH + (kcal)2E Lessertheheatofhydrogenation,lesseristhe internal energy and more is the stability of the system. Hyperconjugation decreases the heat of hydrogenation. C CCH3CH3HHC CCH3CH3HHtrans-2-Butene= 27.6 kcal/mol Hcis-2-Butene= 28.6 kcal/mol HCH CH CH3 2CH CH2 2Propene= 30.1 kcal/mol HEthylene= 32.8 kcal/mol H cis-2-Buteneislessstablethantrans-2-butene duetorepulsionbetweentwobulkiergroups close to each other.Stabilityofcarboniumions: Greaterthe numberofH-atomspresentonthecarbon atoms a to unsaturation, more are the resonating formspossibleduetohyperconjugationand thus greater is the stability of carbonium ion.Bondlength: Hyperconjugation,like conjugationandresonance,alsoafectsbond length. H C CH CH2H C CH CH2HHHH+1.46 1.353 Bondlengthinpropeneis1.46incontrast to normal 1.54 (in propane). It is due to the partialdoublebondcharacteracquiredand hence a little shorter.Dipolemoment: Sincehyperconjugation causesthedevelopmentofcharges,italso afects the dipole moment in the molecule.H C OH C CH OH C CH OH HHH+H = 2.27 D = 2.72 DOrtho -para directing property of methyl group intolueneispartlydueto+Iefectandpartly due to hyperconjugation. H H H HHC H HCH+HCH+HC H+Resonance EfectTeconceptofresonancepertainstothefactthatthere are a number of compounds which could be assignedtwoormoreLewisstructures,difering only in the relative position of electrons. However, theactualpropertiesofthesubstancearenot representedbyanyofthesestructuresbutbya structurewhichisablendorhybridofvarious contributing structures. For example, the following threeLewisstructurescanbewrittenforthe carbonate anion.COO OCOOOCOOOTesestructuresrevealthatthecarbonateanion contains two carbon-oxygen single bonds and one carbon-oxygendoublebondbutneitherofthese predictionsisconsistentwiththeobservationthat CHEMISTRY TODAY |JUNE 1525all the C O bonds in carbonate anion are of equal length (1.30 ).If single and double bonds are present alternativelyinamolecule(incaseofconjugatedsystem)then p-electronsaredelocalizedi.e.,electronscanfow fromoneparttoanotherpartofthesystem.Tis fow of electrons is due to resonance and it results in polarity of the system.CH2CH CH CH2CH2CH CH CH2+Tisefectiscalledmesomericefectandis transmittedthroughoutthechain.Tisisalsoa permanent efect like inductive efect.Resonance efect is of two types : + M or +R efect : Te groups which donate the electrons to the double bond or to a conjugated system are said to have +M efect or +R efect. e.g., OH, OR, NHR, NR2, Cl, Br, I, etc. M or R efect : Te groups which withdraw theelectronsfromthedoublebondorfroma conjugated system towards themselves are said tohaveMefectorRefect.e.g.,, CHO, CN, NO2, COOR, etc.Te term mesomerism is used synonymously withresonance.Moleculeshowingresonanceinvolves overlap of p-orbitals in both directions and there is participationofeachp-electroninmorethanone bond. In other words resonance and delocalization are used ofen in the same sense.Resonanceenergy: Resonancehybridismore stablethananysinglediscretestructure.Tiscan beexplainedintermsofenergyofstabilization known as resonance energy. Te resonance energy iscalculatedfromthediferencebetweenthe theoretical and experimental heats of hydrogenation of the compound. Te main postulates of resonance theory are :Resonatingstructuresshoulddiferonlyinthe position of electrons, not in the position of atoms.Resonatingstructuresshouldhavethesamenumber of unpaired electrons.Greater the stability of a contributing structure,greater is its contribution to the hybrid.Smallerthediferenceintheenergycontents(orstabilities)ofthecontributingstructures, greater is the resonance energy of that hybrid.Tepolarcontributingstructuresinvolvingdistinct charges, are less stable (and hence less important)thanthosewhichdonotinvolve any charges.Greater the number of contributing structures,greater is the stability.Largerthenumberofbondsinacontributingstructure,greateristhestabilityofthat structure.Alltheatomsshouldhaveoctetofelectronsexcepthydrogenwhichhasduplet.Te resonancestructureswhichviolateoctetrule, should not be considered.Te skeleton of the molecule should be planar inconjugated system which is necessary in order to achieve maximum overlap of p-orbitals. Any structuralfeaturethatdestroyscoplanarity oftheconjugatedsysteminhibitsresonance. Tis inhibition is known as steric inhibition of resonance.Inresonancehybrid,thebondlengthsarediferentfromthoseinthecontributing structures.No. of p-bonds No. of contributing structures Resonance energy StabilityApplications : Byknowingtheresonatingstructures,wecanget bond order in a given ion as :Bond order =Total no. of bondsTotal no. of resonating structuuresResonating structures of dienes : CH CH CH CH2 2CH CH CH CH2 2CH CH CH CH2 2::++(I)(II)(III)26CHEMISTRY TODAY |JUNE 15 Tecontributingstructures(II)and(III) provideasatisfactoryexplanationfor1,2and 1,4additionandstabilityofconjugateddienes over simple alkenes.Allthecarbon carbonbondsinbenzeneare equivalent, (there are no true single and double bondsasexpectedfromthecontributing structures)havingexactlythesamelength (139 pm). (I) (II) (III) Besidesthis,benzenehaslargeresonance energy(151kJ/mole)andhencestabilityis expected for this system.Telowerp Kavaluesofnitromethaneand acetyl acetone are simply because of resonance stabilisation of their conjugate bases. CH NCH N2 2:+OO+OOResonance stabilised conjugate base of nitromethaneTeacidityofcarboxylgroupisprimarilybecause of ease of proton release and later, the conjugate base carboxylate ion, is stabilised by forming two equivalent contributing forms.R C R C ; R COOOO H+OO H::::::::ResonancehybridR R C COH:::O:OO H+::::H+:Acidityof a-hydrogenincarbonyl compounds:Tisisduetotheresonance stabilisation of resulting carbanions. C CC CC COH+O OH:::: ::Carbanion(Resonance stabilised)Stabilityoftriphenylmethylcation: Its stability is due to the extensive delocalisation of the positive charge.+CC C etc.Acidiccharacterofphenols: Phenolisa weakacid.Teacidicnatureofphenolisdue totheformationofstablephenoxideionin solution. Phenoxide ion is stable due to resonance.OO O OOOHTenegativechargeisspreadthroughoutthe benzenering.Tischargedelocalizationisa stabilizingfactorinthephenoxideionand increasedacidityofphenol.Noresonanceis possibleinalkoxideions(RO)derivedfrom alcohols.Tenegativechargeislocalizedon oxygen atom. Tus, alcohols are less acidic than phenols.Efectofsubstituentsontheacidityofphenols:Presenceofelectronattracting group(NO2,NR+3,CN,CHO,COOH) onthebenzeneringincreasestheacidityof phenol.Nitrogroupinnitrophenolifpresent in ortho and para positions would stabilise the phenoxideionbydispersalofnegativecharge throughresonanceormesomericefect,to agreaterextentandwillbemoreacidicthan phenol.Baseweakeningefectofamines: Incaseof arylamines,resonanceefectcomesintoplay. Aniline,forexampleisaresonancehybridof structures I to V.CHEMISTRY TODAY |JUNE 1527 NH2NH2+NH2+NH2+NH2(I) (II) (III)(IV) (V) Anilineisfarlessweakerbasethanaliphatic amines,primarilyduetoitsresonance stabilizationandsecondarily,theanilinium cationlosesthestabilizationbecauseof protonation.Electronwithdrawinggroups likeNO2whenpresentato-orp-position with respect to aniline will further weaken the basiccharacterofaniline.e.g.,p-nitroaniline isaweakerbasethananiline.Teextrabase weakening efect when nitro is in the o-position, isprimarilyduetoshortdistance(Iefect assistingtheresonance),andsecondarily, becauseofdirectinteraction(bothstericand hydrogenbonding).Tus,o-nitroanilineis suchaweakbasethatitssaltsarelargely hydrolysed in aqueous solution. Incontrasttoamines,theamidesandimides aremuchlessbasicbecauseoftheresonance stabilisationofthemolecules.Itispractically difcult to protonate an imide.H N C OH N C O C2 2R::::::+ ONH2+R RResonance hybrid(acid amide)OCCOphthalimideNHOHOCCON + H O2OO ON NNOO OResonance Efect vs Inductive EfectBotharepermanentefects,butthereare signifcantdiferencesbetweenthetwowhichare outlined below :Resonance efect operates in unsaturated (preferablyonewithaconjugatedsystem)compoundswhile inductiveefectoperatesincompoundscontaining s-bonds. In other words, p-electrons are involved in resonance efect but only s-electrons are involved in inductive efect.Inductiveefectisdistancedependent(itsintensitydecreasessharplywithdistancefromthe crucial atom in the chain) while resonance efect is not.Resonance efect involves delocalisation of electrons,butthereisnosuchdelocalisationincompounds showing inductive efect.Directive or Orientation EfectTesubstituentalreadypresentonthebenzenering directstheincomingsubstituenttooccupy,(2or6) ortho, meta (3 or 5) or para (4) position. Tis direction dependsonthenatureofthefrstsubstituentandis called directive or the orientation efect.Class I : ( o, p-directing groups) : R(alkyl), OH, SH,NH2,OR,NHR,NR2,NHCOR,Cl, Br,I,CH2Cl,CH2OH,CH2NH2,CH2CN, CH2COOH,CHCH2,CHCHCOOH, C6H5, N N, NC, etc.ClassII:( m-directinggroups):SO3H,NO2, CHO,COOH,CN,NH3Cl,SO2Cl,COCl, COOR,COR,CCl3, NH NH ,N+3+2+3, , R R, etc.Allortho,para-directingsubstituentspossess atleastonenon-bondingelectronpaironthekey atomTe only exception to the above rule is the methyl or alkyl group.Teory of directive efects : Te resonance theory clearly explains why certain substituents are ortho-para directing, while others are meta directing.28CHEMISTRY TODAY |JUNE 15ortho -paraDirecting:Tenon-bondingelectron pair of the key atom of the substituent is delocalized ontheringbyinteractionwiththep-systemdue towhichorthoandparapositionsattaingreater electrondensityandtheelectrophile(E+)would naturallyattackattheseelectronrichcentres formingorthoandparaisomers.ortho-para Directing groups activate the benzene ring towards electrophilicsubstitutionwhilemetadirecting groupsdeactivatethebenzenenucleustowards electrophilic substitution.F, Cl, Br and I (halogens) are exceptions to the aboverule.Tesegroupsareo-,p-directingbut deactivate the ring. +NH2 NH2NH2+E+EEo-productp-productmeta -Directing:Tesubstituentwithdraws electronsfromorthoandpara-positions.Tus, m-positionbecomesapointofrelativelyhigh electrondensityandfurthersubstitutionby electrophile occurs at meta-position.Any substituent or group which releases (donates) electrons into the ring (i.e., o, p-directing) activates thebenzeneringforfurthersubstitution.Te substituent which withdraws electrons (m-directing) fromtheringdeactivatesthebenzeneringfor further substitution.In case of halogen I efect predominates resonance efect.NO2groupismeta-directing(electron withdrawing), its mechanism can be explained asO O ON NO O ON+ON+O ON++OAllmeta-directinggroupshaveeitherapartial positive charge or a full positive charge on the atom directly attached to the ring.IntroductionofaTirdSubstituentintoBenzene RingThepositionoccupiedbyathirdsubstituentgroup enteringthebenzeneringismainlydecidedbythe nature of the two groups already present on the benzene ring.Whenbothgroupsare( o-,p-directing)thenthe directive infuence of each group is in the following order :O > NR2, > NH2 > OH > OMe > NHCOCH3 > CH3 > XTenewgroupentersp-positionpreferablywith respect to more powerful group.Incasethep-positionisblocked,thethirdgroup entersthepositionorthotothemorepowerful group.Whenbothgroupsare meta-directingthethird groupisaccomodatedaccordingtothefollowing order :Me3N+ > NO2 > CN > SO3H > CHO > COMe > COOHTenewsubstituentoccupiesmeta-positionwith respect to the more powerful group.When two groups exert diferent directive infuencethen o, p-directors take precedence.Incasetheinfuenceoftwogroupsreinforceeachother,thethirdgroupisattachedtooneposition, only.CHEMISTRY TODAY |JUNE 1529Ideal SolutionAnidealsolutionisdefnedasamixtureinwhichthe moleculesofdiferentspeciesareindistinguishable. Unliketheidealgas,theyexertforcesoneach other.Whentheseforcesaresameforallmolecules, independentofspecies,thenasolutionissaidtobe ideal. In an ideal solution, the equilibrium between the solutionconcentrationandthevapourphasepartial pressure is described by Raoults law for each component. Tis means each component follows Raoults law in the entire range of concentrations. Note that ideally dilute solutionisdiferentthanidealsolution.Intheideally dilutesolutionthesolutefollowsHenryslawand solventtheRaoultslawbutintheidealsolutionboth solvent and the solute follow Raoults law. Ideally dilute solutionsareeasilyfoundbutidealsolutionsareonly nearlyachieved.Substanceswithsimilarmolecules usuallyformnearlyidealsolutions.Toluenebenzene, ethylenedibromidepropylenedibromide,carbon tetrachloridesilicon tetrachloride, ethyl chlorideethyl bromide are such examples.Mathematically,PT = pAxA+ pBxBwhere the terms pA and pB are vapour pressures of the purecomponentsAandBrespectively.xAandxB are the mole fractions of components A and B in the liquid phase.Consideraspecifcexampleofasolutionofbenzene and toluene.Total vapourpressureVapour pressureof pure benzeneVapour pressureof puretolueneMole fraction oftoluene is 1 here atthe originMole fractionMole fraction ofbenzene is 1 hereTe partial vapour pressure of benzene and toluene in a solution. Te graph is not to scale but showing at least benzene is more volatile than toluene.Thermodynamics of Ideal SolutionForanidealsolution,statevariablessuchasV,U, HandGareobtainedbyaddingthecorresponding contributions of solvent A and solute B.For example, the volumeVAB = nA Vm, A + nB Vm,BwhereVm, AandVm, BarethemolarvolumesofA andBrespectively.Foranidealsolutionatconstant T and P, the mixing quantities behave as follows : mixG RT n xi ii= ln , mixS R n xi ii= lnAnd,mixV = 0, mixH = 0TevalueofmixGisactuallythevalueofGforthe solutionrelativetothepurecomponent.Physically mixH = 0 means no heat is lost or gained by the system when two components of the solution interact. In other words,thestrengthofforcesbetweenAandBis almost of same strength as A-A and B-B.Total Vapour Pressure versus CompositionSince total mole fraction of both the components taken togetherisone,theexpressionforthetotalvapour pressure can be written as :PT = pB + (pA pB)xASotheplotofthetotalvapourpressureagainstmole fractionofcomponentAinliquidphasegivesa straightlinegraphwithinterceptasthevapour pressure of pure component B and slope of the line as (pA pB).Itisverynaturalthatduringtheevaporation, themorevolatilecomponentwillpassintothe vapourphasepreferentially.Terefore,thevapour phasecompositionisdiferentthantheliquidphase compositionandthevapourphasecompositioncan beobtainedwiththehelpofDaltonslawofpartial pressure.Mukul C. Ray, Odisha(IDEAL SOLUTIONS)30CHEMISTRY TODAY |JUNE 15 ypPx pp p p xAATA AAB A B A= =+ ( )Toexpressthetotalvapourpressureinterms ofcompositionofvapourphase,theaboveequation isrearrangedtogetanexpressionofxAinterms of yA. xy pp p p yAA BA B A A=+ ( )SubstitutingthisvalueofxAintheexpressionoftotal pressure,PT = p B+ (p A p B )xAor, P p p py pp p p yT B A BA BA B A A= + + ( )( )or Pp pp p p yTB AA B A A=+ ( )So assuming pA < pB,yB11yApBpADiagramof total vapourpressure versus vapourphase compositionpApBxB11xADiagramof total vapourpressure versus liquidphase compositionTese plots straightaway follow from the mathematical equationsshowingthedependenceoftotalvapour pressureonvapourphasecomposition(lefdiagram) and on liquid phase composition (right one). Whenthetwodiagramsaresuperimposed,diagram asshownisobtained.Herethex-axisisshowing amountfraction.Forlinearlineitistheliquidphase composition (line is called liquidus or the bubble point line)andfortheotherone-itisthevapourphase composition(lineiscalledvapourousorthedew point line). Note that all these diagrams are at constant temperature.PureA PureBpApBPmaxPminLiquid-vapourequilibriumLiquid compositionVapour compositionAmount fractionFrom the diagram, the conclusions are :If the system pressure (may be the pressure createdby the piston of the cylinder in which the liquid is kept) is more than maximum pressure, only liquid phase exists.Ifthesystempressureislessthanminimumpressure, only vapour phase exists.If the system pressure is in between, then a liquid- vapour equilibrium is observed.Boiling Point versus CompositionFor an ideal solution, the graph of boiling point versus amountfractionisshown.Teliquidhavinghigher vapourpressurehasthelowerboilingpoint.SinceB is more volatile than A, the boiling point of the former isless.Onemorenotablepointisthepositionoflines and their curve like nature. None of the line is a straight line and it immediately follows from their mathematical equations. Also the vapourous curve called as the dew pointlineisonthetopandtheliquiduslinecalled the bubble point line is at the bottom. TBTMM LTLTAThe dew point lineThe bubblepoint lineTesignifcanceofthediagramcanbeeasilyrealized. Forasolutionwhosecompositionisshownbypoint L, the boiling point would be TL. Te vapours coming out would have composition shown by point M. When these vapours are condensed, the resulting liquid would have boiling point TM. By repeating this process again andagain,themorevolatilecomponentBcanbe obtainedfromvapoursandlessonefromtheresidual liquid. CHEMISTRY TODAY |JUNE 15311.Which of the following pairs of ions are isoelectronic and isostructural?(a)SO32, NO3(b)ClO3 , SO32(c)CO32, SO32(d)ClO3, CO322.Te reaction,CH CONa + CH CH Cl3 3 2CH3CH3NaClCH COCH CH3 2 3CH3CH3is called(a)Etard reaction (b)Gattermann-Koch reaction(c)Williamson synthesis(d)Williamson continuous etherifcation process.3.Teboilingpointof0.2molkg1solutionofXin waterisgreaterthanequimolalsolutionofYin water. Which one of the following statements is true in this case?(a)Molecular mass of X is less than the molecular mass of Y.(b)Yisundergoingdissociationinwaterwhile X undergoes no change.(c)X is undergoing dissociation in water.(d)MolecularmassofXisgreaterthanthe molecular mass of Y.4.Teactivationenergyofareactioncanbe determined from the slope of which of the following graphs?(a)ln . k vsT1(b) Tk vsT ln.1(c)ln k vs. T(d) ln.kTvs T5.Biodegradablepolymerwhichcanbeproduced from glycine and aminocaproic acid is(a)buna-N(b)nylon 6,6(c)nylon 2-nylon 6(d)PHBV.6.A single compound of the structure, isobtainablefromozonolysisofwhichofthe following cyclic compounds?(a) CH3HC3(b)CH3CH3(c) HC3CH3(d) HC3HC3SOLVED PAPER 2015AIPMTAIPMTWe are happy to inform our readers that out of the 45 questions asked in AIPMT 2015,many questions were either exactly same or of similar type from the MTG Books. Hurray!!Hurray!!Exam Q. No. MTG Book Q. No. P. No. Exam Q. No. MTG Book Q. No. P. No.1 27 years AIPMT 48 26 18 27 years AIPMT 4 1132 27 years AIPMT 20 184 19 NCERT Fingertips 42 1264 27 years AIPMT 28 140 30 AIPMT Guide 195 2785 NCERT Fingertips 56 337 31 27 years AIPMT 32 259 27 years AIPMT 2 162 32 27 years AIPMT 15 8811 Objective Chemistry 2 148 37 Objective Chemistry 1 87112 27 years AIPMT 43 123 43 27 years AIPMT 11 9Here, the references of few are given :and more such questions 32CHEMISTRY TODAY |JUNE 157.TeKspofAg2CrO4,AgCl,AgBrandAgIare respectively,1.11012,1.81010,5.01013, 8.31017.Whichoneofthefollowingsaltswill precipitatelastifAgNO3solutionisaddedtothe solution containing equal moles of NaCl, NaBr, NaI and Na2CrO4?(a)AgBr(b)Ag2CrO4(c)AgI(d)AgCl8.Te correct bond order in the following species is(a)O2+ < O2 < O22+(b)O2 < O2+ O2 < O2+(b)O2 < O2 > O2+(c)O2 > O2 > O2+(d)O2 < O2 < O2+12.Which one of the following electrolytes has the same valueofvantHoffactor(i)asthatofAl2(SO4)3 (if all are 100% ionised)?(a)Al(NO3)3(b)K4[Fe(CN)6](c)K2SO4(d)K3[Fe(CN)6]13.Metalsareusuallynotfoundasnitratesintheir ores.Out of the following two (I and II) reasons which is/are true for the above observation?I.Metal nitrates are highly unstable.II.Metal nitrates are highly soluble in water.(a)I is false but II is true.(b)I is true but II is false.(c)I and II are true.(d)I and II are false.14.Treatment of cyclopentanone O with methyl lithium gives which of the following species?(a)Cyclopentanonyl radical(b)Cyclopentanonyl biradical(c)Cyclopentanonyl anion(d)Cyclopentanonyl cation15.TefunctionofSodiumpumpisabiological processoperatingineachandeverycellofall animals.Whichofthefollowingbiologically important ions is also a constituent of this pump?(a)K+(b)Fe2+(c)Ca2+(d)Mg2+16.Tetotalnumberofp-bondelectronsinthe following structure is HC3HC3H HHC2HHCH3CH3(a)12(b)16(c)4(d)817.When initial concentration of a reactant is doubled in a reaction, its half-life period is not afected. Te order of the reaction is(a)second(b)more than zero but less than frst(c)zero(d)frst.18.A given metal crystallises out with a cubic structure havingedgelengthof361pm.Iftherearefour metalatomsinoneunitcell,whatistheradiusof one atom?(a)80 pm(b)108 pm (c) 40 pm(d)127 pm19.Whichofthefollowingstatementsiscorrectfora reversible process in a state of equilibrium?(a)DG = 2.30 RT log K(b)DG = 2.30 RT log K(c)DG = 2.30 RT log K(d)DG = 2.30 RT log K20.A mixture of gases contains H2 and O2 gases in the ratio of 1 : 4 (w/w). What is the molar ratio of the two gases in the mixture?(a)16 : 1(b)2 : 1(c)1 : 4(d)4 : 121.Cobalt(III)chlorideformsseveraloctahedral complexeswithammonia.Whichofthefollowing will not give test for chloride ions with silver nitrate at 25C?(a)CoCl35NH3(b)CoCl36NH3(c)CoCl33NH3(d)CoCl34NH322.If the value of equilibrium constant for a particular reaction is 1.6 1012, then at equilibrium the system will contain(a)mostly products(b)similar amounts of reactants and products(c)all reactants(d)mostly reactants.CHEMISTRY TODAY |JUNE 153323.Whichofthefollowingspeciescontainsequal number of s- and p-bonds?(a)(CN)2(b)CH2(CN)2(c)HCO3(d)XeO424.Whichofthesestatementsabout[Co(CN)6]3is true?(a)[Co(CN)6]3hasfourunpairedelectronsand will be in a high-spin confguration.(b)[Co(CN)6]3 has no unpaired electrons and will be in a high-spin confguration.(c)[Co(CN)6]3 has no unpaired electrons and will be in a low-spin confguration.(d)[Co(CN)6]3hasfourunpairedelectronsand will be in a low-spin confguration.25.An organic compound X having molecular formula C5H10O yields phenylhydrazone and gives negative responsetotheiodoformtestandTollenstest.It produces n-pentane on reduction. X could be(a)3-pentanone(b)n-amyl alcohol(c)pentanal(d)2-pentanone.26.Given : CH3OCH3CH3OCH3CH3OCH3(I) (II) (III)Whichofthegivencompoundscanexhibit tautomerism?(a)II and III(b)I, II and III(c)I and II(d)I and III27.Bithionalisgenerallyaddedtothesoapsasan additive to function as a/an(a)bufering agent(b)antiseptic(c)sofener(d)dryer.28.Given :HC3CH3CH3HC3 CH2CH3HC2CH2CH2(I) (II) (III)Te enthalpy of hydrogenation of these compounds will be in the order as(a)II > III > I(b)II > I > III(c)I > II > III(d)III > II > I29.Which of the following is the most correct electron displacementforanucleophilicreactiontotake place?(a) HH CCCCCl3HH2(b) HH CCCCCl3HH2(c) HH CCCCCl3HH2(d) HH CCCCCl3HH230.Solubilityofthealkalineearthmetalsulphatesin water decreases in the sequence(a)Sr > Ca > Mg > Ba (b)Ba > Mg > Sr > Ca(c)Mg > Ca > Sr > Ba (d)Ca > Sr > Ba > Mg31.Maximumbondangleatnitrogenispresentin which of the following?(a)NO2+(b)NO3(c)NO2(d)NO232.In Dumas method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at300Ktemperatureand725mmpressure.Ifthe aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is(a)16.76(b)15.76(c)17.36(d)18.2033.Nitrogendioxideandsulphurdioxidehavesome propertiesincommon.Whichpropertyisshown by one of these compounds, but not by the other?(a)Is soluble in water.(b)Is used as a food preservative.(c)Forms acid-rain.(d)Is a reducing agent.34.TereactionofC6H5CH CHCH3withHBr produces(a)C6H5CH2CH2CH2Br(b) CHCHCH3Br(c)CHCHCHCH6 5 2 3Br(d)CHCHCHCH6 5 2 3Br35.Which of the following processes does not involve oxidation of iron?(a)Formation of Fe(CO)5 from Fe.(b)LiberationofH2fromsteambyironathigh temperature.(c)Rusting of iron sheets.(d)DecolourisationofblueCuSO4solutionby iron.34CHEMISTRY TODAY |JUNE 1536.Whichofthemisnotequaltozeroforanideal solution?(a)DVmix(b)DP = Pobserved PRaoult(c)DHmix (d)DSmix37.Becauseoflanthanoidcontraction,whichofthe following pairs of elements have nearly same atomic radii?(Numbersintheparenthesisareatomic numbers)(a)Zr(40) and Hf(72) (b)Zr(40) and Ta(73)(c)Ti(22) and Zr(40)(d)Zr(40) and Nb(41)38.Which property of colloidal solution is independent of charge on the colloidal particles?(a)Electro-osmosis(b)Tyndall efect(c)Coagulation(d)Electrophoresis39.Teelectrolyticreductionofnitrobenzenein strongly acidic medium produces(a)azobenzene(b)aniline(c)p-aminophenol(d)azoxybenzene.40.Consider the following compounds :CH CCH3CH3CH3CH3(I) (II) (III)Hyperconjugation occurs in(a)III only(b)I and III(c)I only(d)II only.41.Teenolicformofethylacetoacetateasshown below has(a)9 sigma bonds and 2 pi-bonds(b)9 sigma bonds and 1 pi-bond(c)18 sigma bonds and 2 pi-bonds(d)16 sigma bonds and 1 pi-bond.42.TespeciesAr,K+andCa2+containthesame number of electrons. In which order do their radii increase?(a)Ca2+ < K+ < Ar(b)K+ < Ar < Ca2+(c)Ar < K+ < Ca2+(d)Ca2+ < Ar < K+43.Te angular momentum of electron in d orbital is equal to(a) 2 3 (b) 0 (c) 6 (d)2 44.A device that converts energy of combustion of fuels like hydrogen and methane, directly into electrical energy is known as(a)dynamo(b)Ni-Cd cell(c)fuel cell(d)electrolytic cell.45.In which of the following compounds, the C Cl bond ionisation shall give most stable carbonium ion?(a) HCHCl(b) (c) HCClH C3H C3(d) CClH C3H C3CH3SOLUTIONS1.(b) :Species Hybridisation Shape No. of esSO32sp3Pyramidal 42ClO3sp3Pyramidal 42CO32sp2Triangular planar32NO3sp2Triangular planar322.(c) :Williamsonsynthesisisthebestmethodfor the preparation of ethers.3.(c) :DTb = iKb mFor equimolal solutions, elevation in boiling point willbehigherifsolutionundergoesdissociation i.e., i > 1.4.(a) :According to Arrhenius equation,k = AeEa/RT ln ln k AERTa= Slope =lnkERa1/THence,iflnkis plottedagainst1/T, slope of the line will beERa.5.(c) :nH2N CH2COOH + nH2N (CH2)5COOH GlycineAminocaproic acid( HN CH2CO NH (CH2)5CO)nNylon 2-nylon 66.(c) :HC3CH3CH2CHCH2COCH3CH3CHEMISTRY TODAY |JUNE 15357.(b) :Salt KspSolubilityAg2CrO4 1.1 1012 = 4s3 sKsp= = 40 65 1034.AgCl1.8 1010 = s2s Ksp= = 1 34 105.AgBr5 1013 = s2s Ksp= = 0 71 106.AgI 8.3 1017 = s2s Ksp= = 0 9 108. Solubility of Ag2CrO4 is highest thus, it willbe precipitated at last.8.(b) :O2 O2 O2+ O22+B.O. : 1.5 2.0 2.5 3.09.(c) :Magnetic moment (m) = n n ( ) +22.84 B.M. corresponds to 2 unpaired electrons.Cr2+3d4, 4 unpaired electronsCo2+3d7, 3 unpaired electronsNi2+3d8, 2 unpaired electronsTi3+3d1, 1 unpaired electron10.(d) :Number of d-electrons is Fe2+ = 6Number of p-electrons inCl = 1111.(d) : O2 < O2 < O2+B.O. : 1.52.0 2.512.(b) :Al2(SO4)3 2Al3+ + 3SO42, i = 5Al(NO3)3 Al3+ + 3NO3, i = 4K4[Fe(CN)6] 4K+ + [Fe(CN)6]4, i = 5K2SO4 2K+ + SO42, i = 3K3[Fe(CN)6] 3K+ + [Fe(CN)6]3, i = 413.(a) :Allnitratesaresolubleinwaterandarequite stable as they do not decompose easily on heating.14.(c) : OLi+CH3O+CH3CyclopentanonylanionH OHHO CH315.(a)16.(d) :Terearefourdoublebonds.Hence,no.of p-electrons = 2 4 = 8.17.(d) :Halflifeperiodofafrstorderreactionis independent of initial concentration, tk1 20 693/.=18.(d) :Z = 4, i.e., structure is fcc.Hence,ra= = = 2 23612 2127 65 .pm 127 pm19.(a)20.(d) :Number of moles of H2 = 12Number of moles of O2 = 432Hence, molar ratio = 12432:= 4 : 121.(c) :Foroctahedralcomplexes,coordination number is 6.Hence,CoCl33NH3i.e.,[Co(NH3)3Cl3]willnot ionise and will not give test for Cl ion with silver nitrate.22.(a) :TevalueofKishighwhichmeansreaction proceeds almost to completion i.e., the system will contain mostly products.23.(d) :CH2(CN)2,

NCCCNHH (6 s + 4 p)HCO3 , OCOHO(4 s + 1 p)XeO4, (4 s + 4 p)(CN)2, N C C N(3 s + 4 p)24.(c) :[Co(CN)6]3, oxidation no. of Co = +3Co3+ = 3d6As CN is a strong feld ligand, so all electrons will be paired up and complex will be low spin complex.25.(a) :Compound X yields phenylhydrazone OCgroup.Negative iodoform test OCH C3 group is absent.Negative Tollens test ketoneHence, the compound is 3-pentanone.CH CH CCH CH3 2 2 3OReductionCH CH CH CH CH3 2 2 2 3(C H O)5 1036CHEMISTRY TODAY |JUNE 1526.(b) :In keto-enol tautomerism,(I) OHC3CH3HC3CH3OHhere, a-H participates.(II) OCH3OHCH3CH3CH3here, a-H participates.(III) here, g-H participates (p-tautomerism).27.(b)28.(d) :Enthalpyofhydrogenationisinversely proportional to the stability of alkenes.Stability of alkenes : I > II > IIIEnthalpy of hydrogenation : I < II < III29.(a) :Nucleophilewillattackastablecarbocation (SN1 reaction).H C CHCHCH Cl3 2H C CHCHCH3 2+(Stable due to + effectof CH group)I330.(c) :Solubilityofalkalineearthmetalsulphates decreases down the group because hydration energy decreases.31.(a) :Species NO3NO2NO2NO2+Hybridisation sp2sp2sp2sp (linear)Bond angle 120 134 115 180So, NO2+ has maximum bond angle.32.(a) :Mass of organic compound = 0.25 gExperimental values,At STP, V1 = 40 mLV2 = ?T1 = 300 KT2 = 273 KP1 = 725 25 = 700 mmP2 = 760 mmPVTP VT1 112 22=VPV TT P21 1 21 2700 40 273300 76033 52 = = = .mL22400 mL of N2 at STP weighs = 28 g\33.52 mL of N2 at STP weighs = 28 33 5222400 .= 0.0419 g% of NMass of nitrogen at STPMass of organic compound taken= 100 0 = =0 04190 25100 16 76... %33.(b) :NO2 is not used as a food preservative.34.(c) : CHCHCH3HBrCHCH CH2 3BrCHCH CH2 3More stable(Benzyl carbocation)35.(a) :Oxidation number of Fe in Fe(CO)5 is zero.36.(d) :Foranidealsolution,DSmix>0whileDHmix; DVmix and DP = 0.37.(a) :ZrandHfhavenearlysameradiidueto lanthanoid contraction.38.(b) :Tyndall efect is scattering of light by colloidal particles which is independent of charge on them.39.(c) : NO2NH2OHp-AminophenolPhenylhydroxylamine40.(a) :Hyperconjugationcanoccuronlyin compound III as it has a-hydrogen atoms.41.(c)42.(a) :In case of isoelectronic species, radius decreases with increase in nuclear charge.43.(c) :Angular momentum= + l l ( ) 1 For d orbital, l = 2 Angular momentum= + = 2 2 1 6 ( ) 44.(c)45.(d) :HC3CCH3CH3ismoststableduetohyper-conjugation. CHEMISTRY TODAY |JUNE 1537ORInacompoundAX,theradiusofA+ionis 95pmandthatofXionis181pm.Predictthe crystal structure of AX and write the coordination number of each of the ion.10.4%NaOHsolution(mass/volume)and6%urea solution(mass/volume)areequimolarbutnot isotonic. Why?11.Assign reasons for the following :(i)Phosphorus doped silicon is a semiconductor.(ii)Schottky defect lowers the density of a solid.(iii)Someoftheveryoldglassobjectsappear slightly milky instead of being transparent.12.(i)What is meant by vant Hof factor?(ii)Teosmoticpressureofa0.0103molar solutionofanelectrolyteisfoundtobe 0.70atmat27C.CalculatethevantHoffactor.(R = 0.082 L atm K1 mol1)Whatconclusiondoyoudrawabout themolecularstateofthesoluteinthe solution?1.Which type of solids are called giant molecules?2.Whyaresodawaterandsofdrinkbottlessealed under high pressure?3.Why is Frenkel defect not found in pure alkali metal halides?4.Whattypeofsolutionisformedwhenethanolis mixed with water?5.What happens when ferrimagnetic Fe3O4 is heated at 850 K?6.Whenlime(CaO)isdissolvedinwater,solution becomes quite hot. What is the efect of temperature on the solubility of CaO in water?7.WhyisFeO(s)notformedinstoichiometric composition?8.Defnevapourpressure.Whathappenstothe vapour pressure when(i)a volatile solute is dissolved in the liquid?(ii)a non-volatile solute is dissolved in the liquid?9.TecompoundCuClhasZnSstructureandthe edge length of the unit cell is 500 pm. Calculate its density.[Atomic mass of Cu = 63 u, Cl = 35.5 u]GENERAL INSTRUCTIONS(i)All questions are compulsory.(ii)Q. no. 1 to 5 are very short answer questions and carry 1 mark each.(iii)Q. no. 6 to 10 are short answer questions and carry 2 marks each.(iv)Q. no. 11 to 22 are also short answer questions and carry 3 marks each.(v)Q. no. 23 is a value based question and carries 4 marks.(vi)Q. no. 24 to 26 are long answer questions and carry 5 marks each.(vii)Use log tables if necessary, use of calculators is not allowed.Time Allowed : 3 hoursMaximum Marks : 70CHAPTERWISE PRACTICE PAPER : THE SOLID STATE | SOLUTIONSSeries 138CHEMISTRY TODAY |JUNE 1513.Explain the following terms with suitable examples :(i) Antiferromagnetism(ii)Frenkel defect(iii)Forbidden zone14.Whatweightofthenon-volatileurea (NH2CONH2)needstobedissolvedin100g of water in order to decrease the vapour pressure of waterby25%?Whatwillbethemolalityofthe solution?15.(i)Whatisthediferencebetween13-15and 12-16 compounds?(ii)Incorrundum,oxideionsarearrangedin hexagonalclosepackingandaluminiumions occupytwo-thirdoftheoctahedralvoids. What is the formula of corrundum?16.Basedonsolute-solventinteractions,arrange thefollowinginorderofincreasingsolubilityin n-octane and explain.Cyclohexane, KCl, CH3OH, CH3CN17.Whatisasemiconductor?Describetwomain typesofsemiconductorsgivingexampleandtheir distinctive features.18.(i)Why is the elevation in boiling point of water diferent in the following solutions?(a)0.1 molar NaCl solution(b)0.1 molar sugar solution.(ii)What advantage does osmotic pressure method have over the elevation in boiling point method for determining molecular masses?19.(i)What general name is given to binary mixtures whichshowdeviationsfromRaoultslawand whosecomponentscannotbeseparatedby fractional distillation?(ii)Why is it not possible to obtain pure ethanol by fractional distillation?(iii)How many types of such mixtures are there?OR(i)(a) State Henrys law for solubility of a gas in liquid and explain its signifcance. (b)Atthesametemperaturehydrogenis more soluble in water than helium. Which of them will have a higher value of KH and why?(ii)Mention two applications of Henrys law.20.Distinguish between the following pairs of terms :(i)Hexagonalclosepackingandcubicclose packing(ii)Crystal lattice and unit cell(iii)Tetrahedral void and octahedral void.21.(i)Non-stoichiometriccuprousoxide,Cu2Ocan bepreparedinthelaboratory.Inthisoxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor?(ii) Classifyeachofthefollowingasap-typeor n-type semiconductor :(a)Ge doped with In(b)B doped with Si22.Temolalfreezingpointdepressionconstantof benzene(C6H6)is4.90Kkgmol1.Selenium existsasapolymerofthetypeSex.When3.26g ofseleniumisdissolvedin226gofbenzene,the observedfreezingpointis0.112Clowerthanfor purebenzene.Deducethemolecularformulaof selenium. (At. mass of Se = 78.8 g mol1).23.Ms. Abha, a public health worker, always tells people in the villages not to cook food in open kadahi. She insists on using pressure cooker or covered pans for cookingasitwillpreservethenutritionalvalueof the food and cook food faster.(i)What values are displayed by Ms. Abha?(ii)What is the reason for fast cooking in pressure cooker?(iii)What is molal elevation constant?(iv)How is it related to elevation in boiling point?24.(i)Aluminiumcrystallisesinacubicclosepacked structure. Radius of atom in the metal is 125 pm.(a)What is the length of the side of the unit cell?(b)How many unit cells are there in 1 cm3 of aluminium?(c)How many atoms are there in 1 unit cell?(ii)Defnethefollowingtermsinrelationto crystalline solids :(a)Lattice point(b)Coordination numberGive one example in each case.OR(i)Explain anisotropic and isotropic properties of solids.CHEMISTRY TODAY |JUNE 1539(ii)Calculate the packing efciency of a crystal for a face centred cubic lattice.25.(i)TwoelementsAandBformcompounds havingmolecularformulaAB2andAB4. Whendissolvedin20gofbenzene(C6H6), 1.0 gofAB2lowersthefreezingpointby 2.3K whereas1.0gofAB4lowersitby1.3K.Te molaldepressionconstantforbenzeneis 5.1Kkgmol1.Calculateatomicmassesof A and B.(ii)What happens when RBCs are placed in(a)1% NaCl solution?(b)Pure water?OR(i)Illustrateelevationinboilingpointwiththe helpofvapourpressure-temperaturecurveof a solution. Show that elevation in boiling point is a colligative property.(ii)Calculatethemassofanon-volatilesolute (molarmass40gmol1)whichshouldbe dissolvedin114gofoctanetoreduceits vapour pressure to 80%.26.(i)Calculate the number of atoms per unit cell in a(a)face centred cubic structure(b)body centred cubic structure.(ii)Iftheradiusoftheoctahedralvoidisrand radius of the atoms in close packing is R, derive relation between r and R.OR(i)Explain(a)Tebasisofsimilaritiesanddiferences between metallic and ionic crystals.(b)Ionic crystals are hard and brittle.(ii)What are F-centres?(iii)Zinc oxide on heating becomes yellow. Why?SOLUTIONS1.Covalent(network)solidsarecalledgiant molecules.2.Soda water and sof drink bottles are sealed under highpressurebecausesolubilityofCO2ishighat higher pressure.3.Frenkeldefectisnotfoundinpurealkalimetal halides because alkali metal ions cannot ft into the interstitial sites.4.Whenethanolismixedwithwater,non-ideal solutionshowingpositivedeviationsfromideal behaviour is formed.5.WhenferrimagneticFe3O4isheatedat850Kit losesferrimagnetismandbecomesparamagnetic duetogreateralignmentofspinsisonedirection on heating.6.DissolutionofCaOinwaterisanexothermic processi.e.,heatisevolved.Accordingto LeChateliersprinciple,increaseintemperature decreasessolubilitywhiledecreaseintemperature increases solubility.7.InFeOcrystals,someFe2+ionsarereplacedby Fe3+ions.Tobalancenetcharge,threeFe2+ions arereplacedbytwoFe3+ions.Eventually,there wouldbelessamountofmetalascomparedto stoichiometric composition.8.Te pressure exerted by gas molecules on its liquid layer at equilibrium is called vapour pressure.(i)Whenavolatileliquidisaddedtoanother liquidthentheresultantvapourpressure mayincrease,decreaseormayremainsame dependingonthestrengthofsolute-solvent interaction.(ii)Netvapourpressuredecreaseswhenanon-volatile solute is added to the liquid.9.Given, structure = fcc, a = 500 pm For CuCl, M = 63 + 35.5 = 98.5 g mol1, d = ?For fcc, Z = 4Using formuladZ MN aA=3ord = 4 98 56 022 10 500 10123 1 3 30 3. g mol. ( ) m cm ol= 5.234 g cm3ORGiven, rA+ = 95 pm, rX = 181 pm, structure = ?C.No. = ? Radius ratio = rrAX+= =95pm181 pm. 0 524Radiusratioliesbetween0.414to0.732hence, cations are in the octahedral voids of ccp of anions.Hence, structure is fcc, C.No. = 6.10.Both the solutions, 4% NaOH (W/V) and 6% urea (W/V) have same concentration (1 M) but these are not isotonic because NaOH undergoes dissociation in solution. Terefore, number of particles in NaOH solution is more than that in urea solution.40CHEMISTRY TODAY |JUNE 1511.(i)Whensilicon(havingfourvalenceelectrons) is doped with phosphorus (having fve valence electrons),phosphorusoccupiessomeofthe latticesitesinthecrystal.Fouroutoffve electrons are used in the formation of covalent bonds with neighbouring atoms. Te ffh extra electron becomes delocalised and moves in the crystal lattice making it a semiconductor.(ii)InSchottkydefect,equalnumberofcations andanionsaremissingfromtheirnormal positioninthecrystallatticeso,itdecreases the density of the crystal.(iii)Tisisbecauseduetoheatingduringtheday andcoolingatnighti.e.,annealingovera number of years, glass acquires some crystalline character.12.(i)vantHoffactoristheratioofthenormal molecularmasstotheobservedmolecular massortheratiooftheobservedcolligative property to the normal colligative property.(ii)p = iCRTor, 0.70 = i 0.0103 0.082 (27 + 273)or,i = =0 700 0103 0 082 3002 76.. ..Since i >1, solute molecules are dissociated in the solution.13.(i)Substanceswhichareexpectedtopossess paramagnetismorferromagnetismonthe basisofunpairedelectronsbutactuallythey possesszeronetmagneticmomentarecalled antiferromagneticsubstances,e.g.,MnO. Itisduetothepresenceofequalnumberof magnetic moments in the opposite directions.(ii)Frenkel defect arises when smaller ions (usually cations)inthelatticeoccupyinterstitialsites leavinglatticesitesvacant.Tisdefectis generallyfoundinioniccrystalswhereanion ismuchlargerinsizethanthecation,e.g., AgBr, ZnS, etc. Due to this defect density does not change, electrical conductivity increases to a small extent and there is no change in overall chemical composition of the crystal.(iii)Te energy gap between the fully flled valence band and the empty conduction band is called forbidden zone.14.Given :p = 100wA = 100 gps = 75wB = ?For urea, MB = 60 g mol1For water, MA = 18 g mol1p ppnnw Mw MssBAB BA A = =// 257560100 18= wB//orwB = 111 gMolality of the solution= 111601100100011gg molgg kg = 18.5 mol kg1 or 18.5 m15.(i)Group13-15compoundsviz.AlP,GaAs,etc. havelargecovalentcharacterwhereasGroup 12-16 compounds viz. ZnS, CdS, HgTe, etc. do not possess covalent bonds but have sufcient ionic character.(ii)Let the number of oxide ions in the packing be N.Te number of octahedral voids = N Terefore, Al3+ ions = 2323 = NNRatio ofAl3+ :O2232 3= =NN : :Hence, formula of corrundum is Al2O3.16.Cyclohexaneandn-octanebotharenon-polar. Hence, they mix completely in all proportions.KClisanioniccompoundwhilen-octaneis non-polar.Hence,KClwillnotdissolveatallin n-octane.CH3OHandCH3CN,botharepolarbutCH3CN islesspolarthanCH3OH.Asthesolventisnon-polar,CH3CNwilldissolvemorethanCH3OHin n-octane.Tus, the order of solubility will be :KCl < CH3OH < CH3CN < Cyclohexane17.Tesolidswhichhaveconductivitiesbetween 106 to 104 ohm1 m1 are called semiconductors.e.g., germanium and silicon. 42CHEMISTRY TODAY |JUNE 15Tetwomaintypesofsemiconductorsareas follows :(i)n-type semiconductor : When a silicon crystal isdopedwithatomsofgroup-15elements, suchasP,As,SborBithenonlyfourofthe fvevalenceelectronsofeachimpurityatom participate in forming covalent bonds and ffh electronisalmostfreetoconductelectricity. Such type of semiconductors are called n-type semiconductorsasincreaseinconductivityis due to negatively charged electrons.+ +n-type semiconductor p-type semiconductor+++ +++(ii)p-type semiconductor : When a silicon crystal is doped with atoms of group-13 elements, such asB,Al,GaorIn,eachimpurityatomforms only three covalent bonds with the host atom. Te place where the fourth electron is missing is called a hole which moves through the crystal likeapositivechargeandhenceincreasesits conductivity. Such type of semiconductors are called p-type semiconductors.18(i)DTb is diferent in the two cases because NaCl dissociates into Na+ and Cl ions in the solution whilesugardoesnotdissociate.Terefore, numberofparticlesincaseofNaClishigher than that in sugar.(ii)Advantagesofosmoticpressuremethod overelevationinboilingpointmethodareas follows :(a)Osmotic pressure measurement occurs at room temperature.(b)Tismethodusesmolaritiesinsteadof molalities.(c)Value of osmotic pressure is large even for dilute solutions.(d)It is more suitable for biomolecules which are not stable at higher temperature.19.(i)Azeotropic mixtures. (ii)Itisnotpossibletoobtainpureethanolby fractionaldistillationbecauseamixtureof 95.6%ethanolwith4.4%waterformsan azeotrope i.e., a constant boiling mixture.(iii)Tere are two types of such mixtures :(a)Minimumboilingazeotropes:Tey showlargepositivedeviationsfromRaoults law (as ethanol-water system).(b)Maximumboilingazeotropes:Tey showlargenegativedeviationsfromRaoults law (as nitric acid-water system).OR(i)(a)Henryslawstatesthatthesolubilityof agasinliquidataconstanttemperatureis directly proportional to the partial pressure of the gas in equilibrium with the liquid.Mathematically, p = KH.xwherepisthepressureofthegasabovethe solution and x is the mole fraction of the gas in the solution and KH is Henrys constant.(b)KH is a function of the nature of the gas. HigherthevalueofKHatagivenpressure, lower is the solubility of the gas in the liquid.As helium is less soluble in water, so it will have a higher value of KH than hydrogen.(ii)(a)In the production of beverages. (b)In deep sea diving 20.(i)HexagonalclosepackinghasABAB.....type arrangement.i.e.,eachthirdlayerisparallel tothefrstwhereasincubicclosepacking, arrangement is ABCABCABC....type in which each fourth layer is parallel to the frst.(ii)Crystallatticeisthearrangementoflattice pointsinthreedimensionalspacewhereas unitcellisthesmallestrepeatingunitwhich representsarrangementoflatticepointsina crystal lattice.(iii)Avoidsurroundedbyfouratomswhose centres when joined form tetrahedron is called tetrahedral void whereas a void surrounded by six atoms whose centres when joined form an octahedron is called octahedral void.Number of tetrahedral voids = 2 Number ofatoms forming ccpNumber of octahedral voids = Number of atoms forming ccpCHEMISTRY TODAY |JUNE 154321.(i)Coppertooxygenratioislessthan2:1in Cu2O.TisshowsthatsomeCu+ionshave beenreplacedbyCu2+ions.Tomaintain electrical neutrality, every two Cu+ ions will be replacedbyoneCu2+ionandhenceahole willbecreatedwhichwillberesponsiblefor conduction.Tusthissubstanceisap-type semiconductor.(ii)(a)Geisgroup-14element(4valence electrons)andInisgroup-13element (3valenceelectrons).Henceaholeiscreated and it is a p-type semiconductor.(b)B is group-13 element (3 valence electrons) and Si is group-14 element (4 valence electrons). Hence, due to presence of a free electron it is a n-type semiconductor.22.Given Kf = 4.90 K kg mol1wB = 3.26 gwA = 226 gDTf = 0.112CAtomic mass of Se = 78.8 g mol1MK ww TBf BA f= = 10004 90 3 26 1000226 0 112 . ..orMB = 631 g mol1Number of atoms of Se in a molecule = =63178 88.Molecular formula of selenium is Se8.23.(i)Ms. Abha showed social values of saving energy and fuel as well as concern towards the health of villagers.(ii)In pressure cooker, pressure is high and water boils at higher temperature. Hence more heat is available for cooking the material.(iii)Molal elevation constant may be defned as the elevation in boiling point when molality of the solution is unity. (iv)DTb = Kb m24.(i)(a)For fcc (or ccp), a = 2 2 2 1 414 125 r = . pm = 354 pm(b)a = 354 pm = 3.54 108 cmVolume of one unit cell = a3 = (3.54 108 cm)3= 4.44 1023 cm3Number of unit cells = Total volumeVolume of one unit cell== 14 44 102 25 10323 322cm. cm.(c)1 unit cell has four atoms of aluminium.(ii)(a)Teconstituentparticlesofacrystalline solidsaredenotedbypoints.Tesearecalled latticepoints.Latticepointsmaybeatoms, molecules or ions. (b)Te number of nearest neighbours of any constituent particle in a crystal lattice is called itscoordinationnumber.Tecoordination number of an atomin the bcc structure is 8.OR(i)Someofthephysicalpropertiesofcrystalline solidslikeelectricalconductivity,refractive index etc. show diferent values when measured alongdiferentdirectionsinthesamecrystal. Tisiscalledanisotropicpropertyofthe crystal.Since, the arrangement of particles is diferent alongdiferentdirections,thevalueofsame physical property is found to be diferent along each direction.Incaseofamorphoussubstances,these propertiesareidenticalinalldirections.Tis property is called isotropy.(ii)Edge length = a bABCDEFGHaFace diagonal= b = ACIn DABCAC2 = BC2 + AB2b2 = a2 + a2 = 2a2b =2aIf radius of the spherical atom is r then b a r = = 2 4 ra=2 2Volume of the four spherical atoms of radius r in the unit cell is 4343r =432 243a= a33 2Volume of the cube = a3Packing efciency = Volume of one spherical atom in unit cellVolume of the cubee44CHEMISTRY TODAY |JUNE 15Packing efciency = aa333 2 = 3 20 74 = .25.(i)Given, w2 = 1 gw1 = 20 gDTf(AB2) = 2.3 KDTf(AB4) = 1.3 KKf = 5.1 K kg mol1Applying the formula, MK ww Tff2211000= MAB ( )... g mol21000 5 1 120 2 3110 871= = MAB ( )... g mol41000 5 1 120 1 3196 151= =SupposeatomicmassesofAandBareaandb respectively thenMolar mass of AB2 = a + 2b = 110.87 g mol1 ...(i)Molar mass of AB4 = a + 4b = 196.15 g mol1...(ii)Eqn. (ii) - Eqn. (i) gives2b = 85.28 or b = 42.64Substituting in eqn. (i) we geta + 2 42.64 = 110.87or a = 25.59Tus atomic mass of A = 25.59 uAtomic mass of B = 42.64 u(ii)(a)1%NaClsolutionishypertonic,hence RBCs will shrink due to plasmolysis.(b)Pure water is hypotonic hence RBCs will burst due to osmosis.OR(i) 1.0 atmTTemperatureb TbTbVapour pressureSolventSolutionofnon-volatilesoluteChange of vapour pressure of a solvent and solution of non-volatile solute with temperature.Elevationinboilingpointisacolligative propertyasitdependsonlyonnumberof molesofsoluteparticlesdissolvedin1000g ofthesolventandnotuponthenatureofthe solute i.e.DTb m(ii)Number of moles of solute=w40molLet p = 100 then ps = 80Dp = 20Number of moles of solvent (octane) = =1141141gg molmol1[Molar mass of C8H18 = 114 g mol1]Now, ppx =2 =+201004040 1ww// 0 240140.w w+ =w = 10 g26.(i)(a)Numberofatomsinaunitcelloffcc structure : Tis type of unit cell has 8 atoms on the corners and 6 atoms on the face centres.Contribution by atoms on the corners = 18 8 = 1Contribution by atoms on the face centres= 126 3 =Total number of atoms per unit cell = 3 + 1 = 4 (b)Numberofatomsinaunitcellofbcc structure : Tis type of unit cell has 8 atoms on the corners and one atom within the body.Contribution by 8 atoms present on the corners= 18 8 = 1Contributionbyatomspresentwithinthe body = 1Number of atoms present per unit cell = 1 + 1 = 2(ii)Derivationoftherelationshipbetweenthe radius (r) of the octahedral void and the radius (R) of the atoms in close packing :Aspherefttingintotheoctahedralvoidis shown by shaded circle. A sphere above and a spherebelowthissmall(shaded)spherehave not been shown in the fgure.CHEMISTRY TODAY |JUNE 1545 RABCR RRr rABCisarightangledtriangle.Applying Pythagoras theorem,BC2 = AB2 + AC2(2R)2 = (R + r)2 + (R + r)2 = 2(R + r)2 ( )( )2222RR r = +or ( ) ( ) 22 2R R r = +2 R R r = +orr R R R = = 2 2 1 ( ) r = R(1.414 1) = 0.414 ROR(i)(a)Similaritiesbetweenmetallicandionic crystals : 1. In both ionic and metallic crystals constituent particlesareheldbyelectrostaticforceof attraction.2.Inboththesolids,bondsarenon-directional.3. Both the solids have high melting points.Diferencesbetweenmetallicandionic crystals :1. Ionic crystals are made of cations and anions whereas metallic solids are made of metal ions and sea of electrons.2.Ioniccrystalsareinsulatorsinsolid statebecausetheirionsarenotfreeto movewhereasmetalliccrystalsconduct electricity due to the presence of sea of mobile electrons.3. Ionic crystals are hard and brittle but metallic crystals are hard, malleable and ductile. Some metallic crystals are sof also.(b)Ionic crystals are hard due to the presence ofstronginterionicelectrostaticforcesof attraction.However,whenanionicsolidissubjectedto stress, ions of same charge come close together andtherepulsiveforcesbetweenthemcause the crystal to break into pieces.Tus, ionic crystals are hard but brittle.(ii)Electronstrappedinanionicvacanciesare calledF-centresastheyimpartcolourtothe crystal.(iii)When ZnO is heated, it loses oxygen and turns yellow due to the following reaction :ZnO Zn2+ + 12O2 + 2eTeexcessZn2+ionsmovetointerstitialsites andelectronstoneighbouringinterstitial sites.Teseelectronsimpartcolourbyabsorbinga portion of energy from visible light.47CHEMISTRYTODAY|JUNE '1546CHEMISTRYTODAY|JUNE '15Introductionof most fundamental andimportant tools of chemistrywhichhelpinvarious calculations.CONCEPTSOME BASIC CONCEPTS OF CHEMISTRY SOME BASIC CONCEPTS OF CHEMISTRY SI system has seven base units pertaining tosevenfundamental scientific quantities :Physical quantity SI unitLength ( ) l metre (m)Mass ( ) m kilogram (kg)Time ( ) t second (s)Electric current ( ) I ampere (A)Thermodynamictemperature ( ) Tkelvin (K)Amount of substance ( ) n mole (mol)Luminous intensity ( ) Ivcandela (cd)SI Units SI systemallows the use of prefixes toindicate the multiples or submultiples of aunit :deci - 10 deka - 101 1centi - 10 hecto - 102 2milli - 10 kilo - 103 3micro - 10 mega - 106 6nano - 10 giga - 109 9pico - 10 tera - 1012 12 Anumber is representedas 10 xn is ve if decimal is moved towards right nand is +ve if it is movedtowards left. nScientific Notation Theseareall certaindigitswithlastdigituncertain. All non-zerodigits are significant . Zeros preceding to first non-zero digit arenot significant. Zeros between two non-zero digits aresignificant. Zeros on the right side of the decimal aresignificant.Significant Figures Requiredunit = Givenvalue conversionfactor Some useful conversionfactors :Length1=10 cm=10 m8 101nm=10 m, 1pm=10 m9 12Volume 1L =1000mL=1000cm=1dm=10 m3 3 3 3Pressure 1atm=760mmor torr=101325Pa1bar =10 Nm =10 Pa5 2 5Energy1calorie =4.184J1eV=1.602210 J191J =10 ergs7Dimensional AnalysisMass is the quantity of matter containedinthe substance andis constant whereas weight variesfrom place toplace.Exact numbers have aninfinite number of