Chemistry Model Practice Paper For AIEEE 2010 Q. 1. Arrange the carbanions , in order of their decreasing stability: 1. 2. 3. 4. Sol: Answer : (4) Q. 2. In Cannizzaro reaction given below the deprotonation of PhCH2OH 1. the attack of at the carboxyl group 2. the transfer of hydride of the carbonyl group 3. the abstraction of proton from the carboxylic group 4. Sol.: Cannizzaro reaction involves hydride transfer as the rate limiting step. Answer : (3) Q. 3. Two liquids X and Y form an ideal solution. At 300 K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mmHg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mmHg. Vapour pressure (in mmHg) of X and Y in their pure states will be, respectively: Downloaded From: http://www.iitportal.com Downloaded From: http://www.iitportal.com500 and 600 1. 200 and 300 2. 300 and 400 3. 400 and 600 4. Sol: Answer : (4) Q. 4. In a fuel cell, methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is ; At 298 K standard Gibb’s energies of formation for CH3OH(l), H2O(l) and CO2(g) are –166.2, –237.2 and –394.4 kJ mol–1 respectively. If standard enthalpy of combustion of methanol is –726 kJ mol–1, efficiency of the fuel cell will be 97% 1. 80% 2. 87% 3. 90% 4. Sol: Answer : (1) Q. 4. In a fuel cell, methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is ; At 298 K standard Gibb’s energies of Downloaded From: http://www.iitportal.com Downloaded From: http://www.iitportal.comformation for CH3OH(l), H2O(l) and CO2(g) are –166.2, –237.2 and –394.4 kJ mol–1 respectively. If standard enthalpy of combustion of methanol is –726 kJ mol–1, efficiency of the fuel cell will be 97% 1. 80% 2. 87% 3. 90% 4. Sol: Answer : (1) Q. 5. On the basis of the following thermochemical data: -343.52 kJ 1. -22.88 kJ 2. -228.88 kJ 3. +228.88 kJ 4. Q. 6. The half life period of a first order chemical reaction is 6.93 minutes. The time required for the completion of 99% of the chemical reaction will be (log 2 = 0.301): 460.6 minutes 1. 230.3 minutes 2. 23.03 minutes 3. 46.06 minutes 4. Sol: Answer : (4) Downloaded From: http://www.iitportal.com Downloaded From: http://www.iitportal.comQ. 7. Copper crystallises in fcc with a unit cell length of 361 pm. What is the radius of copper atom?181 pm 1. 108 pm 2. 127 pm 3. 157 pm 4. Sol: Answer : (3) Q. 8. Given The value of standard electrode potential for the change, -0.270 V 1. –0.072 V 2. (3) 0.385 V 3. (4) 0.770 V 4. Sol: Answer : (4) Q. 9. In which of the following arrangements, the sequence is not strictly according to the property written against it? B < C < O < N; increasing first ionization enthalpy 1. CO2 < SiO2 < SnO2 < PbO2: increasing oxidising power 2. Downloaded From: http://www.iitportal.com Downloaded From: http://www.iitportal.comHF < HCl < HBr < HI: increasing acid strength 3. NH3 < PH3 < AsH3 < SbH3: increasing basic strength 4. Sol.: On moving down the group, the basic tendency of hydrides of group 15 decreases. Answer : (4) Q. 10. Knowing that the chemistry of lanthanoids (Ln) is dominated by its +3 oxidation state, which of the following statements is incorrect? Ln (III) hydroxides are mainly basic in character. 1. Because of the large size of the Ln (III) ions the bonding in its compounds is predominantly ionic in character. 2. The ionic sizes of Ln (III) decrease in general with increasing atomic number. 3. Ln (III) compounds are generally colourless. 4. Answer : (4) Q. 11. The set representing the correct order of ionic radius is: 1. 2. 3. 4. Sol.: On moving down the group, the ionic radius increases. In case of isoelectronic ions, more the charge on the ion, smaller will be its size. Answer : (3) Q. 12. Buna–N synthetic rubber is a copolymer of: 1. Downloaded From: http://www.iitportal.com Downloaded From: http://www.iitportal.com2. 3. 4. Sol.: Buna–N is a copolymer of buta–1,3–diene and acrylonitrile. Answer : (4) Q. 12. Buna–N synthetic rubber is a copolymer of: 1. 2. 3. 4. Sol.: Buna–N is a copolymer of buta–1,3–diene and acrylonitrile. Answer : (4) Q. 13. Which of the following statements is incorrect regarding physiosorption? Enthalpy of adsorption is low and positive. 1. It occurs because of Van der Waal’s forces. 2. More easily liquefiable gases are adsorbed readily. 3. Under high pressure it results into multi molecular layer on adsorbent surface. 4. Downloaded From: http://www.iitportal.com Downloaded From: http://www.iitportal.comSol.: for adsorption is negative. Adsorption is a spontaneous process, so is negative and in case of adsorption is also negative. Answer : (1) Q. 14. The number of stereoisomers possible for a compound of the molecular formula Me is: 6 1. 3 2. 2 3. 4 4. Sol: Isomer 1: cis d ; Isomer 2: cis l ; Isomer 3: trans d ; Isomer 4: trans l Answer : (4) Q. 15. The two functional groups present in a typical carbohydrate are: –OH and –CHO 1. –OH and –COOH 2. –CHO and –COOH 3. 4. Sol.: Carbohydrate is a polyhydroxy aldehyde or polyhydroxy ketone or any compound which give these on hydrolysis. group includes aldehydic as well as ketonic group. Answer : (4) Q. 16. Calculate the wavelength (in nanometer) associated with a proton moving at 14.0 nm 1. Downloaded From: http://www.iitportal.com Downloaded From: http://www.iitportal.com0.032 nm 2. 0.40 nm 3. 2.5 nm 4. Sol: Answer: (3) Q. 17. The bond dissociation energy of B–F in BF3 is 646 kJ mol–1 whereas that of C–F in CF4 is 515 kJ mol–1. The correct reason for higher B–F bond dissociation energy as compared to that of C–F islower degree of interaction between B and F in BF3than that between C and F in CF4. 1. smaller size of B atoms as compared to that of C atom. 2. stronger bond between B and F in BF3 as compared to that between C and F in CF4 3. significant interaction between B and F in BF3 whereas there is no possibility of such interaction between C and F in CF4 . 4. Sol: Boron in BF3 has a vacant p–orbital, allowing back bonding while carbon in CF4 has no vacant orbital, so no back bonding is feasible. Thus, B–F bond is stronger than CF4. Answer: (4) Q. 18. The major product obtained on interaction of phenol with sodium hydroxide and carbon dioxide is: phthalic acid 1. benzoic acid 2. salicylaldehyde 3. salicylic acid 4. Sol: Downloaded From: http://www.iitportal.com Downloaded From: http://www.iitportal.com. Answer: (4) Q. 19. Solid Ba(NO3)2 is gradually dissolved in solution. At what concentration of Ba2+ will a precipitate begin to form? 1. 2. 3. 4. Sol: Concentration of Ba2+ needed to precipitate Answer: (3) Q. 20. Which one of the following reactions of xenon compounds is not feasible? 1. 2. 3. 4. Sol: XeF6 reacts violently with water and gets hydrolysed to give highly explosive XeO3. ; Reaction (2) is reverse of this, so it is not feasible. Answer: (2) Q. 21. The IUPAC name of neopentane is: 2,2–dimethylbutane 1. Downloaded From: http://www.iitportal.com Downloaded From: http://www.iitportal.com2–methylbutane 2. 2,2-dimethylpropane 3. 2–methylpropane 4. Sol.: Answer: (3) Q. 22. In context with the transition elements, which of the following statements is incorrect? Once the d5 configuration is exceeded, the tendency to involve all the 3d electrons in bonding decreases. 1. In addition to the normal oxidation states, the zero oxidation state is also shown by these elements in complexes. 2. In the highest oxidation states, the transition metal show basic character and form cationic complexes. 3. In the highest oxidation states of the first five transition elements (Sc to Mn), all the 4s and 3d electrons are used for bonding. 4. Sol.: In highest oxidation state, the transition metal will form complexes which will have large degree of covalent character, which shows the acidic behaviour. Answer: (3) Q. 23. Which of the following pairs represents linkage isomers? [PtCl2(NH3)4]Br2 and [PtBr2(NH3)4]Cl2 1. [Cu(NH3)4][PtCl4] and [Pt(NH3)4] [CuCl4] 2. [Pd(PPh3)2(NCS)2] and [Pd(PPh3)2(SCN)2] 3. Downloaded From: http://www.iitportal.com Downloaded From: http://www.iitportal.com[Co(NH3)5NO3]SO4 and [Co(NH3)5SO4]NO3 4. Sol: Linkage isomerism is shown by ambidentate ligands. (SCN– and NCS– are ambidentate groups). Answer: (3) Q. 24. In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is 1. 2. 3. 4. Sol: Answer: (4) Q. 25. Which of the following on heating with aqueous KOH, produces acetaldehyde? CH3CHCl2 1. CH3COCl 2. CH3CH2Cl 3. CH2ClCH2Cl 4. Sol: Answer: (1) Downloaded From: http://www.iitportal.com Downloaded From: http://www.iitportal.comQ. 26. A liquid was mixed with ethanol and a drop of concentrated H2SO4 was added. A compound with a fruity smell was formed. The liquid was: CH3COOH 1. CH3OH 2. HCHO 3. CH3COCHb3 4. Sol: Ethyl acetate (ester) has fruity smell. Answer: (1) Q. 27. Using MO theory predict which of the following species has the shortest bond length? 1. 2. 3. 4. Sol: The bond order of , , and is 1, 3, 2.5 and 1.5 respectively. Since, has highest bond order, so it has shortest bond length. Answer: (2) Q. 28. A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following statements is correct regarding the behaviour of the solution? n–heptane shows +ve deviation while ethanol shows –ve deviation from Raoult’s Law. 1. The solution formed is an ideal solution. 2. The solution is non-ideal, showing +ve deviation from Raoult’s law. 3. The solution is non-ideal, showing –ve deviation from Raoult’s Law. 4. Sol: When n–heptane is added to ethanol, the H-bonding interaction between ethanol molecules is loosened, so weaker interaction among molecules leads to higher vapour pressure. Thus, it forms a non-ideal solution with positive deviation from Raoult’s law. Answer: (3) Downloaded From: http://www.iitportal.com Downloaded From: http://www.iitportal.comQ. 29. The alkene that exhibits geometrical isomerism is: 2–methyl–2–butene 1. propene 2. 2–methyl propene 3. 2–butene 4. Sol: Terminal alkenes never show geometrical isomerism and non–terminal alkene with 2 different groups attached to each atom having restricted rotation is capable of showing geometrical isomerism. Answer: (4) Q. 30. Which of the following has an optical isomer? 1. 2. 3. 4. Sol: Square planar complexes can not show optical isomerism. So, option (2) and (3) are ruled out. Compound in option (4) i.e. has a plane of symmetry, so it is also optically inactive. Answer: (1) Downloaded From: http://www.iitportal.com Downloaded From: http://www.iitportal.com