NEET UG 2013, Chemistry Question Paper (Date of Examination: ) Class Unit No. Unit name Question No. Total No. of

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1 TARGT Publications NT UG 13 Question Paper NT UG 13, Question Paper (Date of xamination: ) Class Unit No. Unit name Question No. Total No. of XI NT XII NT Questions I Some Basic Concepts of II Structure of Atom 6, 7, 5 3 III Classification of lements and Periodicity in Properties IV Chemical Bonding and Molecular Structure 58, 6, 77, 85 V States of Matter: Gases and Liquids 56 1 VI Thermodynamics VII quilibrium 68, 71 VIII Redox Reactions IX Hydrogen 73 1 X s-block lements (Alkali and Alkaline earth metals) XI Some p-block elements 63, 6, 75 3 XII Organic -Some Basic Principles and 7, 81 Techniques XIII Hydrocarbons: Alkanes Alkenes Alkynes Aromatic Hydrocarbons 8, 83, 88, 9 XIV nvironmental Total = I Solid State 55, 57 II Solutions 5, 8 III lectrochemistry 9, 51, 53 3 IV Chemical Kinetics 8, 5 V Surface VI General Principles and Processes of Isolation of lements VII p-block lements 6, 61, 66, 69 VIII d and f -Block lements 67, 7, 76 3 IX Co-ordination Compounds 59, 7 X Haloalkanes and Haloarenes XI Alcohols Phenols and thers Alcohols XII Aldehydes, Ketones and Carboxylic Acids Phenols thers 79 1 Aldehydes Ketones 65, 89 Carboxylic Acids XIII Organic Compounds Containing Nitrogen 87 1 XIV Biomolecules XV Polymers 8, 86 XVI in veryday life 78 1 Total = 5 Total = 5 1

2 W Code NT UG 13 Question Paper TARGT Publications 6. The value of Planck s constant is Js. The speed of light is nm s 1. Which value is closest to the wavelength in nanometer of a quantum of light with frequency of s 1? (A) 1 (B) 5 (C) 5 (D) 75 Given: Speed of light, c = nm s 1 Frequency, ν = s 1 Planck s constant = Js. To find: Wavelength, λ =? nm Formula: ν = c λ Calculation: From formula, λ = 3 1 nms s 17 1 = 5 nm 7. What is the maximum numbers of electrons that can be associated with the following set of quantum numbers? n = 3, l = 1 and m = 1 (A) 1 (B) 6 (C) (D) An orbital can accommodate a maximum of electrons. 8. What is the activation energy for a reaction if its rate doubles when the temperature is raised from C to 35 C? (R = 8.31 J mol 1 K 1 ) (A) 3 kj mol 1 (B) 69 kj mol 1 (C) 3.7 kj mol 1 (D) 15.1 kj mol 1 Given: Initial temperature, T 1 = C = ( + 73)K = 93 K Final temperature, T = 35 C = ( )K = 38 K Final rate of the reaction (K ) = Initial rate of the reaction (K 1 ) Gas constant R = 8.31 Jmol 1 K 1 To find: Activation energy, a =? Formula: K log K = a T T R T 1.T Calculation: From formula, a log = a = a = 15 = 3.67 kj mol kj mol 1

3 TARGT Publications NT UG 13 Question Paper 9. A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of ph = 1 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be (A).59 V (B).59 V (C).118 V (D) 1.18 V Given: ph of the HCl solution = 1 To find: oxidation potential of electrode, cell =? Calculation: cell =.59 1 log = +.59 V 5. A reaction having equal energies of activation for forward and reverse reactions has (A) S = (B) G = (C) H = (D) H = G = S = When nergy of activation for forward reaction = nergy of activation for reverse reaction G =, H = and S = 51. At 5 C molar conductance of.1 molar aqueous solution of ammonium hydroxide is 9.5 ohm 1 cm mol 1 and at infinite dilution its molar conductance is 38 ohm 1 cm mol 1. The degree of ionisation of ammonium hydroxide at the same concentration and temperature is (A).8% (B).8% (C).8% (D).8% Given: Molar conductance of.1 molar aqueous solution of ammonium hydroxide at 5 C, λ m = 9.5 ohm 1 cm mol 1 Molar conductance of.1 molar aqueous solution of ammonium hydroxide at infinite dilution, λ = 38 ohm 1 cm mol 1 m To find: Degree of ionization of ammonium hydroxide at same concentration and temperature =? λm Formula: Degree of ionization = λ 1 m Calculation: From formula, λm Degree of ionization = λ 1 m = 38 =.8% 5. Based on equation = Z J certain conclusions are written. Which of them is NOT n CORRCT? (A) The negative sign in equation simply means that the energy of electron bound to the nucleus is lower than it would be if the electrons were at the infinite distance from the nucleus (B) Larger the value of n, the larger is the orbit radius (C) quation can be used to calculate the change in energy when the electron changes orbit (D) For n = 1, the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit In the smallest allowed orbit, electron is strongly attracted by the nucleus. 3

4 W Code NT UG 13 Question Paper 53. A button cell used in watches functions as following Zn (s) + Ag O (s) + H O (l) Ag (s) + If half cell potentials are + e Zn (s) ; =.76 V Zn + (aq) Zn + (aq) + (aq) Ag O (s) + H O (l) + e Ag (s) + (aq), =.3 V The cell potential will be (A) 1.1 V (B). V (C).8 V (D) 1.3 V Given: cathode =.3 V anode =.76 V To find: cell =? Formula: cell = cathode Calculation: From formula, cell = cathode anode anode =.3 (.76) = 1.1 V TARGT Publications 5. How many grams of concentrated nitric acid solution should be used to prepare 5 ml of. M HNO 3? The concentrated acid is 7% HNO 3. (A) 5. g conc. HNO 3 (B) 9. g conc. HNO 3 (C) 7. g conc. HNO 3 (D) 5. g conc. HNO 3 Given: M of conc. HNO 3 =. V of conc. HNO 3 = 5 ml. M V = No. of Moles of HNO 3 = 5 1 =.5 To find: Grams of concentrated nitric acid =? Calculation: HNO 3 required = = 5 g 55. The number of carbon atoms per unit cell of diamond unit cell is (A) (B) 8 (C) 6 (D) 1 Diamond has ZnS type structure i.e. 8 carbon atoms per unit cell. 56. Maximum deviation from ideal gas is expected from (A) H (g) (B) N (g) (C) CH (g) () NH 3(g) Among the given gases, NH 3 has the highest T c value and thereby shows maximum deviation from ideal gas behaviour.

5 TARGT Publications NT UG 13 Question Paper 57. A metal has a fcc lattice. The edge length of the unit cell is pm. The density of the metal is.7 g cm 3. The molar mass of the metal is (N A Avogadro's constant = mol 1 ) (A) g mol 1 (B) 3 g mol 1 (C) 7 g mol 1 (D) g mol 1 Given: Metal has fcc lattice, Z = The density of the metal, d =.7 g cm 3 The edge length of the unit cell, = pm. Volume, v = l 3 = (. 1 8 ) 3 m Avogadro s constant, N A = mol 1 To find: The molar mass of the metal, M =? Formula: Z M d = V NA Calculation: From formula, Z M d = V NA M.7 = 3 ( 8 ) M =.7 (.) = 7 g/mol. 58. Dipole-induced dipole interactions are present in which of the following pairs? (A) H O and alcohol (B) Cl and CCl (C) HCl and He atoms (D) SiF and He atoms H δ+ δ Cl... He HCl is polar and He is non-polar They show dipole induced dipole interaction. 59. A magnetic moment of 1.73 BM will be shown by one among the following (A) [Cu(NH 3 ) ] + (B) [Ni(CN) ] (C) TiCl (D) [CoCl 6 ] Magnetic moment (µ) = n(n+ ) 1.73 = n(n+ ) n = 1 So, compound must contain one unpaired electron. The compound is [Cu(NH 3 ) ] Roasting of sulphides gives the gas X as a byproduct. This is a colorless gas with choking smell of burnt sulphur and causes great damage to respiratory organs as a result of acid rain. Its aqueous solution is acidic acts as a reducing agent and its acid has never been isolated. The gas X is (A) H S (B) SO (C) CO (D) SO 3 The gas is SO. 5

6 W Code NT UG 13 Question Paper TARGT Publications 61. Which is the strongest acid in the following? (A) H SO (B) HClO 3 (C) HClO (D) H SO 3 Among the oxo acids, higher the oxidation state of central atom, stronger is the acid. In case of HClO, Cl has the maximum oxidation state of +7. Thus HClO is the strongest acid. 6. Which of the following is paramagnetic? (A) CO (B) O (C) CN (D) NO + O has one unpaired electron, in the antibonding molecular orbital. It is paramagnetic. 63. Which of the following structure is similar to graphite? (A) BN (B) B (C) B C (D) B H 6 Boron nitride has structure same as that of graphite consisting of sheets made up of hexagonal rings of alternate B and N atoms joined together. 6. The basic structural unit of silicates is (A) SiO (B) SiO Basic structural unit of silicates is SiO (C) 65. Reaction by which Benzaldehyde CANNOT be prepared? SiO 3 (D) SiO (A) + CrO Cl in CS followed by H 3 O + (B) COCl + H in presence of Pd-BaSO (C) + CO + HCl in presence of anhydrous AlCl 3 (D) + Zn/Hg and conc. HCl Zn/Hg and con. HCl cannot reduce group to CHO 66. Which of the following does NOT give oxygen on heating? (A) KClO 3 (B) Zn(ClO 3 ) (C) K Cr O 7 (D) (NH ) Cr O 7 On heating (NH ) Cr O 7, the products formed are N, Cr O 3 and H O. Oxygen is not formed in this reaction. (NH ) Cr O 7 N ( ) + Cr O 3 + H O 6

7 TARGT Publications NT UG 13 Question Paper 67. Which of the following lanthanoid ions is diamagnetic? (At. nos. Ce = 58, Sm = 6, u = 63, Yb = 7) (A) Ce + (B) Sm + (C) u + (D) Yb + Yb + has an electronic configuration of f 1. Since there is absence of unpaired electron, it is diamagnetic in nature. 68. Identify the CORRCT order of solubility in aqueous medium (A) CuS > ZnS > Na S (B) ZnS > Na S > CuS (C) Na S > CuS > ZnS (D) Na S > ZnS > CuS Solubility product values decreases in the order Na S > ZnS > CuS. 69. XeF is isostructural with (A) TeF (B) ICl (C) SbCl 3 (D) BaCl XeF is isostrucutral with ICl as both of them possess linear geometry. 7. An excess of AgNO 3 is added to 1 ml of a.1 M solution of dichlorotetraaquachromium(iii) chloride. The number of moles of AgCl precipitated would be (A).1 (B). (C).3 (D).1 [Cr(H O) Cl ]Cl has one ionizable Cl. 1 3 moles of the complex precipitates the same amount of AgCl. 71. Which of these is least likely to act as a Lewis base? (A) CO (B) F (C) BF 3 (D) PF 3 BF 3 is a Lewis acid as it is electron deficient. 7. KMnO can be prepared from K MnO as per the reaction: 3MnO + H O MnO + MnO + The reaction can go to completion by removing ions by adding (A) HCl (B) K (C) CO (D) SO 3K MnO + CO KMnO + MnO K CO Which of the following is electron-deficient? (A) ( ) (B) (SiH 3 ) (C) (BH 3 ) (D) PH 3 (BH 3 ), Diborane is electron deficient. 7

8 W TARGT Publications Code NT UG 13 Question Paper 7. Structure of the compound whose IUPAC name is 3-thyl--hydroxy--methylhex-3-en-5-ynoic acid is (A) (B) (C) (D) thyl--hydroxy--methylhex-3-en-5ynoic acid 75. Which of these is NOT a monomer for a high molecular mass silicone polymer? (A) MeSiCl 3 (B) Me SiCl (C) Me 3 SiCl (D) PhSiCl 3 (Me) 3 SiCl is not a monomer for a high molecular mass silicone polymer 76. Which of the following statements about the interstitial compounds is INCORRCT? (A) They retain metallic conductivity (B) They are chemically reactive (C) They are much harder than the pure metal (D) They have higher melting points than the pure metal Interstitial compounds are chemically inert. 77. Which one of the following molecules contains no π bond? (A) CO (B) H O (C) SO (D) NO H O molecule consists of two O H σ bonds and two lone pair of electrons. There is no π bond present. H H O 78. Antiseptics and disinfectants either kill or prevent growth of microorganisms. Identify which of the following statements is NOT TRU. (A) A.% solution of phenol is an antiseptic while 1% solution acts as a disinfectant (B) Chlorine and Iodine are used as strong disinfectants (C) Dilute solutions of Boric acid and Hydrogen Peroxide are strong antiseptics (D) Disinfectants harm the living tissues Dilute solutions of boric acid and H O are mild antiseptics. 8

9 TARGT Publications NT UG 13 Question Paper 79. Among the following ethers, which one will produce methyl alcohol on treatment with hot concentrated HI? (A) CH CH CH O (B) CH CH O (C) C O (D) CH CH O C O + hot conc. HI 8. Nylon is an example of (A) Polyester (B) Polysaccharide (C) Polyamide (D) Polythene Nylons are polyamides + C I Methanol 81. The structure of isobutyl group in an organic compound is (A) CH CH (B) CH CH (C) CH CH CH (D) C ( ) CH CH group is known as isobutyl group 8. Nitrobenzene on reaction with conc. HNO 3 /H SO at 8-1 C forms which one of the following products? (A) 1, -Dinitrobenzene (B) 1, 3-Dinitrobenzene (C) 1, -Dinitrobenzene (D) 1,, -Trinitrobenzene 9

10 W Code NT UG 13 Question Paper NO group is a meta directing group. TARGT Publications Nitrobenzene on reaction with conc.hno 3 /H SO (i.e. Nitrating mixture) at 8-1 C forms 1,3-Dinitrobenzene. NO NO cons.hno 3/HSO NO Nitrobenzene 1, 3-Dinitrobenzene 83. Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating? (A) C N (B) SO 3 H (C) (D) NO NO group is more deactivating than the other groups given molecules of urea are present in 1 ml of its solution. The concentration of solution is (A). M (B).1 M (C).1 M (D).1 M Given: 6. 1 molecules of urea are present in 1 ml of its solution To find: The concentration of solution =? Calculation: 1 mole of urea contains molecules molecules of urea contains = 3 moles moles of urea are present in 1 ml of solution Concentration of the solution is measured in terms of 1dm 3 i.e. 1 ml Concentration = =.1 M Which of the following is a polar molecule? (A) BF 3 (B) SF (C) SiF (D) XeF Due to the presence of a lone pair of electron on "S", SF has a distorted geometry. It is see-saw shaped and has a resultant dipole moment. Therefore SF is a polar molecule. 86. Which is the monomer of Neoprene in the following? (A) CH = CH C CH (B) CH = C CH = CH (C) CH = C CH = CH Cl (D) CH = CH CH = CH Chloroprene (CH = C CH = CH ) Cl 1

11 TARGT Publications NT UG 13 Question Paper NO NO 87. In the reaction A A is + N Cl Br Br (A) HgSO /H SO (B) Cu Cl (C) H 3 PO and H O (D) H + /H O NO NO H3PO /HO + N Cl Br Br H 3 PO and H O replaces N Cl group by hydrogen. 88. The radical CH is aromatic because it has (A) (B) (C) (D) 6 p-orbitals and 6 unpaired electrons 7 p-orbitals and 6 unpaired electrons 7 p-orbitals and 7 unpaired electrons 6 p-orbitals and 7 unpaired electrons 6p orbitals and 6 unpaired electrons contributes to aromaticity. 89. The order of stability of the following tautomeric compounds is O CH = C CH C O C CH C I O II O C = CH C III (A) I > II > III (B) III > II > I (C) II > I > III (D) II > III > I III is stabilised by intramolecular hydrogen bonding. 9. Which of the following compounds will NOT undergo Friedal-Craft's reaction easily? (A) Cumene (B) Xylene (C) Nitrobenzene (D) Toluene Nitrobenzene does not undergo Friedel-Crafts reaction due to the strong deactivating effect of NO group. 11