CHEMISTRY & SOLUTION
Final JEE - Main Exam Januar y, 2020/09-01-2020/Evening SessionFINAL JEE–MAIN EXAMINATION – JANUARY, 2020(Held On Thursday 09th JANUARY, 2020)TIME : 2 : 30 PM to 5 : 30 PMTEST PAPER WITH ANSWER & SOLUTIONCHEMISTRY1.The correct order of the spin-only magneticmoments of the following complexes is :(I) [Cr(H2O)6]Br2(II) Na4[Fe(CN)6](III) Na3[Fe(C2O4)3] (D0 P)(IV) (Et4N)2[CoCl4](1) (III) (I) (II) (IV)(2) (I) (IV) (III) (II)(3) (II) » (I) (IV) (III)(4) (III) (I) (IV) (II)NTA Ans. (2)ALLEN Ans. (2)[Cr(H2 O)6 ]2 Cr Þ [ Ar ] 3d 4 2H2O Weak field ligandegUnpaired e– 4Magnetic moment 24 BM 4.89 BMII[ Fe(CN) 6 ]4-AFe 2 Þ [ Ar ] 3d 6CN– Strong field ligandegt2gUnpaired e– 0Magnetic moment 0 BM 0 BMIII [ Fe(C2 O 4 )3 ]3-Fe 3 Þ [ Ar ] 3d 5As D 0 PMagnetic moment 3 BM 1.73 BMIV ( Et 4 N ) [ CoCl 4 ] 2-Co 2 Þ [ Ar ] 3d 7[CoCl4 ]-2 t2gegUnpaired electrons 3Magnetic moment 15 BM 3.87 BMHence order of magnetic moment isI IV III II2.The first and second ionisation enthalpies of ametal are 496 and 4560 kJ mol–1, respectively.How many moles of HCl and H 2 SO 4 ,respectively, will be needed to react completelywith 1 mole of the metal hydroxide ?(1) 1 and 0.5(2) 2 and 0.5(3) 1 and 1(4) 1 and 2NTA Ans. (1)ALLEN Ans. (1)Sol. IE values indicate, that the metal belongs to Istgroup since second IE is very high(Q only one valence electron)Metal hydroxide will be of type, MOH.MOH HCl MCl H2O(1mol) (1mol)LLt2gUnpaired e– 1ENSol. It2g1MOH 1 H2 SO 4 M2SO4 H2O221(1mol) ( mol)2So one mole of HCl required to react with onemole MOH.1mole of H2SO4 required to react with one2mole MOH.Soeg1
Final JEE - Main Exam Januar y, 2020/09-01-2020/Evening Session3.Which of the following reactions will notproduce a racemic product ?4.In the following reaction A is :(i) Br2, hn(ii) KOH (alc.)A ¾¾¾¾¾¾ (iii) O3(iv) (CH3)2S(v) NaOH (aq) DCH3(1)HClCH3–C–CH CH 2 ¾¾ HHOO(3)HCNCH3–CCH 2CH3 ¾¾ (1)(2)(3)(4)H3CHCl¾¾ CH3EN(2)NTA Ans. (3)HBr(4) CH3CH2CH CH2 ¾¾ NTA Ans. (1)HALLEN Ans. (1)CH3A(i) Br2(ii) KOH(iii) O3(iv) (CH3)2S(v) NaOH/DCH3CH3–CH–CH–CH3ACH3–CH–CH CH2 H – o chiral centre, so no racemisation �CH32OSol.LLSol.ALLEN Ans. (3)O3/(CH3)2SBr2Alc. KOHhvD
Final JEE - Main Exam Januar y, 2020/09-01-2020/Evening Session(1)(2)7.PressureA mixture of gases O2, H2 and CO are taken ina closed vessel containing charcoal. The graphthat represents the correct behaviour ofpressure with time is :Pressure5.TimeTimeBiochemical Oxygen Demand (BOD) is theamount of oxygen required (in ppm):(1) by anaerobic bacteria to breakdowninorganic waste present in a water body.(2) for the photochemical breakdown of wastepresent in 1 m3 volume of a water body.(3) by bacteria to break-down organic waste ina certain volume of a water sample.(4) for sustaining life in a water body.(4)ALLEN Ans. (3)Pressure(3)PressureNTA Ans. (3)TimeTimeNTA Ans. (4)Sol. Biochemical oxygen demand (BOD) is amountof oxygen required by bacteria to break downorganic waste in a certain volume of watersample.Among the statements (a)-(d) the correct onesare:EN8.ALLEN Ans. (4)(a) Lithium has the highest hydration enthalpyamong the alkali metals.6.Which polymer has 'chiral' monomer(s) ?(b) Lithium chloride is insoluble in pyridine.(1) Buna-N(2) Nylon 6,6(c) Lithium cannot form ethynide upon itsreaction with ethyne.(3) Neoprene(4) PHBVNTA Ans. (4)ALLEN Ans. (4)(d) Both lithium and magnesium react slowlywith H2O.(1) (a), (b) and (d) only(2) (b) and (c) only(3) (a), (c) and (d) onlyASol. PHBV :LLSol. Adsorption of Gases will decreasesNTA Ans. (3)(3-hydroxy butanoic acid)Sol. Lithium has highest hydration enthalpy amongalkali metals due to its small size.Poly b-hydroxy butyrate-co-b-hydroxy valerateCH3–CH–CH2–COOHOH OCH3–CH2–CH–CH2––C–OHOH(4) (a) and (d) onlyALLEN Ans. (3)LiCl is soluble in pyridine because LiCl havemore covalent character.Li does not form ethynide with ethyne.Both Li and Mg reacts slowly with H2O(3-hydroxy pentanoic acid)3
Final JEE - Main Exam Januar y, 2020/09-01-2020/Evening Session9.Amongst the following, the form of water withthe lowest ionic conductance at 298 K is:(1) distilled water(2) water from a well(3) saline water used for intravenous injection(4) sea waterNTA Ans. (1)ALLEN Ans. (1)Sol. Distilled water have lowest ionic conductance.10. Which of the following has the shortest C-Clbond?12.NH 2N(I)N(1) (I) (III) (IV) (II)(2) (III) (I) (II) (IV)(3) C1–CH CH2(3) (III) (II) (I) (IV)(4) C1–CH CH–NO2EN(4) (II) (III) (IV) (I)NTA Ans. (4)NTA Ans. (3)ALLEN Ans. (4)ALLEN Ans. (3)O13.Cl – CH CH—NOLLDue to –M effect of –NO2 and M effect of Clmore D.B. character between C – Cl bond. SoThe solubility product of Cr(OH)3 at 298 K is6.0 l0–31. The concentration of hydroxide ionsin a saturated solution of Cr(OH)3 will be :(1) (18 10–31)1/4(2) (2.22 10–31)1/4(3) (4.86 10–29)1/4(4) (18 10–31)1/2shortest bond length.NTA Ans. (1)In the figure shown below reactant A(represented by square) is in equilibrium withproduct B (represented by circle). Theequilibrium constant is :ALLEN Ans. (1)Sol. Cr(OH)3(s) Cr3 (aq.) 3OH–(aq.)A(1) 2H(IV)(III)(2) C1–CH CH–CH311.(II)NH 2(1) Cl–CH CH–OCH3Sol.The decreasing order of basicity of thefollowing amines is :(2) 1(3) 8NTA Ans. (1)(s)(3s)ksp 27(s)4 6 10–31Þ [3(s)]4 18 10 –31[OH–] 3(s) [18 10–31]1/4(4) 414.5 g of zinc is treated separately with an excessof(a) dilute hydrochloric acid andALLEN Ans. (1 or Bonus)(b) aqueous sodium hydroxide.Sol. Bonus (no reaction is given)The ratio of the volumes of H2 evolved in thesetwo reactions is : B (Assume reaction)A K 4[B] 11 ;2[A] 6(1) 1 : 4NTA Ans. (4)ALLEN Ans. (4)(2) 1 : 2(3) 2 : 1(4) 1 : 1
Final JEE - Main Exam Januar y, 2020/09-01-2020/Evening SessionSol.16.Zn 2HCl ¾¾ ZnCl 2 H 2Zn 2NaOH ¾¾ Na 2 ZnO2 H 215.The ratio of the volume of H2 is 1 : 1Consider the following reactions,(i) NaNO2/HCl, 0-5 C(ii) b-naphthol/NaOHMolisch's BarfoedTestTestColored gativePositive(4)ALLEN Ans. (4)Sol. Alanine does not show Biuret test becauseBiuret test is used for deduction of peptidelinkage & alanine is amino acid.NHCH3(3)NTA Ans. (4)EN(2)NH2Albumine is protein so have paptide linkageCH3LLCH3(i) NaNO2 / HCl, 0 - 5 C(ii) b-naphthol ColoredsolidABr2 / H2OC7H 6NBr 3bymonosaccharide but not disaccharide. PositiveMolisch's test is shown by glucose.The reaction of H3N3B3Cl3 (A) with LiBH4 intetrahydrofuran gives inorganic benzene (B).Further, the reaction of (A) with (C) leads toH 3 N 3 B 3 (Me) 3 . Compounds (B) and (C)respectively, are:(2) Borazine and MeMgBr(3) Borazine and MeBr(i) NaNO2/HCl0 – 5 C17.so it gives positive Biuret test.Positive Barfoed test is shown(1) Boron nitride and MeBrN2 ClNH2MeB(3) A Lactose, B Glucose, C AlanineCH3PNegative(4) A Lactose, B Glucose, C AlbuminCH3ALLEN Ans. (2)Negative(1) A Glucose, B Fructose, C AlbuminNH2NTA Ans. (2)A Positive(2) A Lactose, B Fructose, C AlanineNH2(1)BiuretTestA, B and C are respectively :C7H6NBr3The compound [P] is :Sol.A, B and C are three biomolecules. The resultsof the tests performed on them are givenbelow:(4) Diborane and MeMgBrCH3OHBr2/H2ONTA Ans. (2)ALLEN Ans. (2)MeBrNH2BrMeN NOHBr(colored dye)5
Final JEE - Main Exam Januar y, 2020/09-01-2020/Evening SessionSol.In tetrahydrofuraneH3 N 3 B3 Cl3 3LiBH 4 ¾¾¾¾¾¾ H3 N3 B3 H 3(T.H.F.)(B)Inorganic Benzene(Borazene)(A)20. 3LiCl(1) Both DS and S are functions of temperature. 3BH 3 .THF(2) S is not a function of temperature but DS isa function of temperature. H 3 N 3 B3 (CH 3) 3 3MgBrClH 3 N 3 B3Cl3 3CH 3 MgBr ¾¾(A)18.Each carbon atom is sp2 hybridizedTherefore each carbon has 3 sp 2 hybridorbitals.Hence total sp2 hybrid orbitals are 18.The true statement amongst the following is:(3) Both S and DS are not functions oftemperature.(C)The isomer(s) of [Co(NH3)4Cl2] that has/havea Cl–Co–Cl angle of 90 , is/are :(4) S is a function of temperature but DS is nota function of temperature.(1) meridional and trans(2) cis and transNTA Ans. (1)(3) trans onlyALLEN Ans. (1)(4) cis onlyALLEN Ans. (4)Sol.[ Co(NH3 )4 Cl2 ]òq rev.TENSol. ds NTA Ans. (4)21.has 2 geometrical isomersA cylinder containing an ideal gas (0.1 mol of1.0 dm3) is in thermal equilibrium with a largevolume of 0.5 molal aqueous solution ofCH3NClNH3H3NNH3H3NtransClCethylene glycol at its freezing point. If theClNH3LLClH3NNH3ciscis isomer has Cl–Co–Cl angle of 90 litres after equilibrium is achieved will be .(Given, Kf (water) 2.0 K kg mol–1,R 0.08 dm3 atm K–1 mol–1)The number of sp2 hybrid orbitals in a moleculeof benzene is :A19.stoppers S1 and S2 (as shown in the figure) aresuddenly withdrawn, the volume of the gas in(1) 24(2) 6NTA Ans. (4)ALLEN Ans. (4)Sol.HH(4) 18S1S2Ideal gasHH(3) 12Frictionlesspistonaq. ethylene glycolHHNTA Ans. (2.18 to 2.23)ALLEN Ans. (2.17 or 2.18)Sol. 0 – Tf' 2 0.5 1Tf' – 1ºC 272 K6
Final JEE - Main Exam Januar y, 2020/09-01-2020/Evening Session24.0.1 0.08 272for gas P 1The sum of the total number of bonds betweenchromium and oxygen atoms in chromate anddichromate ions is .P 2.176 atmP1V1 P2V2NTA Ans. (12)2.176 1 1 V2ALLEN Ans. (18.00)V2 2.176 litre22.10.30 mg of O2 is dissolved into a liter of seawater of density 1.03 g/mL. The concentrationSol.of O2 in ppm is .NTA Ans. (10)CrOALLEN Ans. (10.00)23.(4s 2p)A sample of milk splits after 60 min. at 300 Kof lactobacillus acidophilus in it doubles. Theactiva tion energy (in kJ/ mol) for this processæ t ö - Ea é 1 1 ùln ç 1 R ëê T2 T1 ûúè t2 øASol.1 ùæ 60 ö - Ea é 1ln ç êè 40 ø 8.3 ë 400 300 úûCrCrOOOO–Total Cr-O bonds 12(8s 4p)consider only linkages between Chromium andOxygen but in question total no. of bonds areLLALLEN Ans. (3.98 to 3.99 or –3.98 to –3.99)ONote :- But answer of NTA is 12. Theyæ2ö(Given, R 8.3 J mol–1 K–1, ln ç 0.4,è3øNTA Ans. (3.98 to 3.99 or –3.98 to –3.99)–Total number of bonds between chromium andoxygen in both structures are 18.and after 40 min. at 400 K when the populatione–3 4.0)–Total Cr-O bonds 610.3 10 -3 10 6 101030is closest to .O–OENSol. ppm Dichromate2–Cr2O 7OOChromate2–CrO4O25.asked so s and p bonds must be consideredseparately.Consider the following reactions(i)CH3MgBrCuA ¾¾¾¾¾ B ¾¾¾ 2-methyl573K(ii)H 3O 2-buteneThe mass percentage of carbon in A is .NTA Ans. (66.66 to 66.67)ALLEN Ans. (66.67)E 0.4 1200 8.3 3.984 kJ/mole7
Final JEE - Main Exam Januar y, 2020/09-01-2020/Evening SessionSol.CH3ACH3MgBrCuB H3OCH3–C CH–CH3573 K(2-methyl-2-butene)CH3–C–CH 2–CH3OCH3 MgBr H 3OCH3CH3–C–CH2–CH3OH(B)(A)Cu573 KCH3ENCH3–C CH–CH3(2-methyl-2-butene)C Þ 12 4 4848 100 66.66%72A% of C LLHÞ8 1 8O Þ 16 1 16Total728
A, B and C are three biomolecules. The results of the tests performed on them are given below: Molisch's Test Barfoed Test Biuret Test A Positive Negative Negative BPositive Positive Negative C Negative Negative Positive A, B and C are respectively : (1) A Glucose, B Fructos
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