CHEMISTRY – A4 1. For the formation of which compound in Ellingham diagram 0 G becomes more and more negative with increase in temperature? (A) ZnO (B) Cu 2 O (C) CO (D) FeO Ans: C 2. Which of the following compound does not give dinitrogen on heating? (A) NH 4 NO 3 (B) (NH 4 ) 2 Cr 2 O 7 (C) Ba(N 3 ) 2 (D) NH 4 NO 2 Ans: A (A) 4 3 2 2 2 NH NO N O H O ⎯⎯ → + (B) ( ) 4 2 7 2 2 3 2 2 4 NH Cr O N Cr O H O ⎯⎯ → + + (C) 3 2 2 ( ) 3 Ba N Ba N ⎯⎯ → + (D) 4 2 2 2 2 NH NO N H O ⎯⎯ → + 3. Aqueous solution of raw sugar when passed over beds of animal charcoal, it becomes colourless. Pick the correct set of termino lo gies that can be used for the above example. Adsorbent Adsorbate Process A) Animal Charcoal Colouring substance Adsorption B) Colouring Substance Animal Charcoal Adsorption C) Solution of Sugar Animal Charcoal Sorption D) Animal Charcoal Solution of Sugar Absorption Ans: A 4. For Freundlich adsorption isotherm, a graph of a log (x/m) v/ s. log (P) gives a straight line. The slope of line and its Y - axis intercept respectively are (A) 1 log , log K n (B) 1 , K n (C) 1 log , K n (D) 1 , log K n Ans: D 1 1 log log log n x x k p k p m m n = = + , Slope = 1 n , intercept = log k 5. When FeCl 3 is added to excess of hot water gives a sol ‘X’. When FeCl 3 is added to NaOH (aq) solution, gives sol ‘Y’. X and Y formed in the above processes respectively are (A) 2 3 2 2 3 2 / / Fe O xH O Cl and Fe O xH O OH − − (B) 3 2 3 2 2 3 2 / / Fe O xH O Fe and Fe O xH O OH + − (C) 3 2 3 2 2 3 2 / / Fe O xH O OH and Fe O xH O Fe − + (D) 2 3 2 2 3 2 / / Fe O xH O H and Fe O xH O Na + + Ans: B 3 3 2 2 3 2 / FeCl hot H O Fe O xH O Fe + + → 3 ( ) 2 3 2 / aq FeCl NaOH Fe O xH O OH − + → 6. The reducing agent in the given equations: ( ) ( ) ( ) 2 ( ) 2 2 ( ) ( ) 4 8 2 4[ ( ) ] 4 g s aq aq aq aq Ag CN H O O Ag CN OH − − − + + + → + 2 2 ( ) ( ) 4 ( ) ( ) 2[ ( ) ] [ ( ) ] 2 aq s aq s Ag CN Zn Zn CN Ag − − + → + (A) H 2 O (B) CN - (C) Zn (D) O 2 Ans : C Zn acts as a reducing agent. 7. Which of the following is CORRECT with respect to melting point of a transition element? (A) M n > Fe (B) Ti > V (C) V > Cr (D) Cr > Mn Ans: D Due to more number of unpaired electrons 8. 2 2 4 2 3 2 2 4 aMnO bS O H O xMnO ySO zOH − − − − + + → + + , a and y respectively are (A) 3;6 (B) 8;8 (C) 8;3 (D) 8;6 Ans: D 2 2 4 2 3 2 2 4 8 3 8 6 2 MnO S O H O MnO SO OH − − − − + + → + + 9. Which formula and name combination in INCORRECT? (A) [CoCl 2 (en) 2 ]Cl – Dichloridodiethylenediammine cobalt (II) chloride (B) [Co(NH 3 ) 4 (H 2 O)Cl]Cl 2 – Tetraammnieaquachlordiocobalt (III) chloride (C) K 3 [Al(C 2 O 4 ) 3 ] – Potassium trioxalatoaluminate (III) (D) [Pt(NH 3 ) 2 Cl(NO 2 )] – Diamminechloridonitrito – N – platinum (II) Ans: A 2 2 [ ( ) ] CoCl en Cl Dichlor ido bis(ethylenediamine) cobalt (III) chloride 10. Which of the following system in an octahedral complex has maximum unpaired electrons? (A) d 4 (low spin) (B) d 7 (high spin) (C) d 9 (high spin) (D) d 6 (low spin) Ans: B A) d 4 (low spin) 2 unpaired electrons B) d 7 (high spin) 3 unpaired electrons C) d 9 (high spin) 1 unpaired electrons D) d 6 (low spin) 0 unpaired electrons 11. The correct dec reasing order of basicity of hyd rid e s of Group - 15 elements is (A) AsH 3 > SbH 3 > NH 3 > PH 3 (B) NH 3 > PH 3 > AsH 3 > SbH 3 (C) SbH 3 > AsH 3 > PH 3 > NH 3 (D) PH 3 > AsH 3 > SbH 3 > NH 3 Ans: B 12. Which one of the following oxoa c ids of phosphorus can reduce AgNO 3 to metallic silver? (A) H 4 P 2 O 6 (B) H 3 PO 4 (C) H 3 PO 2 (D) H 4 P 2 O 7 A ns: C H 3 PO 2 acts as reducing agent because it has two P - H bond 13. In solid state, PCl 5 is a/an (A) Ionic solid with [PCl 4 ] + and [PCl 6 ] - (B) Covalent solid present in the form of P 2 Cl 10 (C) Octahedral structure (D) Ionic solid [PCl 6 ] + and [PCl 4 ] - Ans: A Ionic solid with [PCl 4 ] + [PCl 6 ] − 14. In which one of the following pairs, both the elements does not have (n – 1)d 10 ns 2 configuration in its elementary state? (A) Hg, Cn (B) Cu, Zn (C) Zn, Cd (D) Cd, Hg Ans: B 15. 3 2 3 PCC CH CH CH CH OH CH CH CH CHO − = − ⎯⎯⎯ → − = − . Hybridization change involved at C – 1 in the above reaction (A) sp 2 to sp 3 (B) sp to sp 2 (C) sp 3 to sp (D) sp 3 to sp 2 Ans: D 16. If a didenta te ligand ethane - 1, 2 - diamine is progressively added in the molar ratio : ::1:1, 2 :1,3:1 en Ni to [Ni(H 2 O) 6 ] 2+ aq solution, following co - ordination entities are formed. I. 2 2 4 ( ) [ ( ) ] aq Ni H O en + - pale blue II. 2 2 2 2 ( ) [ ( ) ( ) ] aq Ni H O en + - blue/purple III. 2 3 ( ) [ ( ) ] aq Ni en + - violet The wavelength in nm of light adsorbed in case of I and III are respectively (A) 310 nm and 500 nm (B) 600 nm and 535 nm (C) 475 nm and 310 nm (D) 300 nm and 475 nm Ans: B Strength of ligand 1 absorbed Visible light range 400 nm to 750 nm. 17. Which of the following is an organometallic compound? (A) (CH 3 COO) 2 Ca (B) CH 3 O N a (C) CH 3 COONa (D) CH 3 CH 2 MgBr Ans: D Grignard reagent 18. A pair of compounds having the same boiling points are (A) Benzene and naphthalene (B) (+) butan - 2 – ol and ( - 1) butan – 2 – ol (C) cis but - 2 - ene and trans but - 2 - ene (D) n - hexane and neo - hexane A ns: B Boiling point depends on molar mass and optical isomers have same boiling point. 19. Identify A, B and C in the sequence 4 2 0 3 2 0 LiAlH HNO KCN alc C CH CH Br A B C ⎯⎯⎯ → ⎯⎯⎯ → ⎯⎯⎯ → (A) CH 3 CH 2 CN, C 2 H 5 OH, C 2 H 5 N 2 Cl (B) CH 3 CH 2 CN, CH 3 CH 2 NH 2 , C 2 H 5 OH (C) CH 3 CH 2 CN, CH 3 CH 2 CH 2 NH 2 , CH 3 CH 2 CH 2 OH (D) CH 3 CH 2 NC, CH 3 CH 2 OH, CH 3 CH 2 CH 2 NH 2 Ans: C 0 4 0 3 2 3 2 3 2 2 2 2 3 2 2 LiAlH KCN C alc CH CH Br CH CH CN CH CH CH NH HNO CH CH CH OH ⎯⎯⎯ → − ⎯⎯⎯ → + ⎯⎯ ⎯ → − − 20. Compounds P and R in the following reaction are 3 2 6 2 4 + - 3 2 2 (i) CH MgBr (i) B H Conc.H SO 3 heat (ii) H O ii) H O /OH CH CHO P Q R ⎯⎯⎯⎯ ⎯ → ⎯⎯⎯⎯→ ⎯⎯⎯⎯⎯ → (A) Metamers (B) Identical (C) position isomers (D) Functional isomers Ans: C 3 2 6 2 4 + - 3 2 2 (i) CH MgBr (i) B H Conc.H SO 3 3 3 3 2 3 2 2 heat (ii) H O ii) H O /OH CH CHO CH (CH )CHOH CH CH=CH CH CH CH OH ⎯⎯⎯⎯ ⎯ → ⎯⎯⎯⎯→ ⎯⎯⎯⎯⎯ → P & R are positional isomers. 21. Aniline does not undergo (A) Friedel - Craft reaction (B) Bromination (C) Nitration (D) Sulphonation Ans: A AlCl 3 coordinated with NH 2 group of aniline. 22. The heat of phenyl methyl ether with HI produces an aromatic compound A which on treatment with con. HNO 3 gives B. A and B respectively are, (A) Iodobenzene, 1 - Iodo - 4 - nitrobenzene (B) Phenol, Picric acid (C) Methanol, Ethanoic acid (D) Picric acid, Phenol Ans: B O CH 3 strong bond OH + CH 3 I Phenol(A) OH HI Conc. HNO 3 OH NO 2 NO 2 O 2 N 2,4,6-trinitrophenol or picric acid (B) 23. OH NaOH A (i) CO 2 (ii) H + B (i) (CH 3 CO) 2 O (ii) H + Y (Major product) Y in the above reaction is (A) Cumene (B) Picric acid (B) Salycylaldehyde (D) Aspirin Ans: D OH NaOH (i) CO 2 (ii) H + (i) (CH 3 CO) 2 O (ii) H + Aspirin ONa OH COOH OCOCH 3 COOH 24. A better reagent to oxidize primary alcohols into aldehyde is: (A) Acidified 2 2 7 K Cr O (B) 3 CrO (C) PCC (D) Alkaline KM n O 4 Ans: C PCC is a mild oxidising agent. 25. In the reaction : 2 3 ( ) 6 5 ( ) i SnCl HCl con KOH ii H O C H CN X Y Z + + ⎯⎯⎯⎯⎯ → ⎯⎯⎯⎯ → + , Formation of X, formation of Y and Z are known by (A) Wolff - Kishner reduction, Wurtz reaction. (B) Stephen reaction, Cannizaro reaction. (C) Rosenmund reduction, Cannizaro reaction. (D) Clemmensen reduction, Sandmeyer reaction. Ans: B 6 5 X C H CHO = − 6 5 2 Y C H CH OH = − 6 5 Z C H COOK = − 26. In the reaction: NH 2 N 2 + Cl - N 2 + BF 4 - NO 2 P Q R P,Q and R respectively are: (A) 2 3 2 , , NaNO dil HCl BF Cu NaNO + + (B) 3 2 3 , , NaNO dil HCl F Cu NaNO + + (C) 2 4 2 , , NaNO dil HCl HBF Cu NaNO + + (D) 2 2 3 , , NaNO con HCl F Cu NaNO + + A ns: C NH 2 N 2 + Cl - N 2 + BF 4 - NO 2 NaNO 2 + dil.HCl HBF 4 Cu + NaNO 2 27. Thyroxine produced in the thyroid gland is an iodinated derivative of _____ (A) Tyrosine (B) tryptophan (C) threonine (D) lysine Ans: A 28. Sucrose is dextrorotatory but after hydrolysis the mixture show laevorotation, this because of (A) Racemic mixture is formed (B) Laevorotation of fructose is more than dextrorotation of glucose (C) Laevorotation of glucose is more than dextrorotation of fructose (D) Sucrose is a non - reducing sugar Ans: B 29. The correct order of match between column X and column Y is: X Y I. Vitamin A i.Muscular weakness II. Vitamin D ii. Increased blood clotting time III. Vitamin E iii . Night - blindness IV. Vitamin K iv. Osteomalacia (A) I - iii, II - ii, III - iv, IV - i (B) I - iii, II - iv, III - i, IV - ii (C) I - iv, II - iiii, III - ii, IV - i (D) I - ii, II - I, III - iii, IV - iv Ans: B 30. Which of the following monomers form biodegradable polymers? (A) Phenol and formaldehyde (B) 3 - hydroxybutanoic acid and 3 - hydroxypentanoic acid (C) Ethylene glycol and p h thalic acid (D) Caprolactum and 1, 3 - Butadiene Ans: B 31. Match the list - I with List - II in the following: List - I List - II 1. Caprolactum (a) ( 2 | ) 3 n CH CH CH − 2. V inyl chloride (b) ( 2 | ) 6 5 n CH CH C H − 3. Styrene (c) ( 2 | ) n CH CH Cl − 4. Propene (d) ( || ( O C 2 ) CH 5 | ) N H n (A) 1 - d, 2 - c, 3 - a, 4 - b (B) 1 - d, 2 - c, 3 - b, 4 - a (C) 1 - c, 2 - d, 3 - a, 4 - b (D) 1 - a, 2 - d, 3 - c, 4 - b Ans: B 32. Which one of the following is a non - narcotic analgesic? (A) Aspirin (B) Morphine (C) Heroin (D) Codeine Ans: A 33. Receptors are proteins and crucial to body communication process. These receptors are embedded in (A) Endocrine gland (B) Chromosomes (C) Cell membrane (D) Protein Ans: C 34. A gas at a pressure of 2 atm is heated from 25 0 C to 323 0 C and simultaneously compressed to 2 3 rd of its original value. Then the final pressure is ( A) 2 atm ( B) 4 atm ( C) 1.33 atm ( D) 6 atm Ans : D 1 2 P = 1 V V = 1 25 273 298 T C K = + = 2 ? P = 2 323 273 596 T C K = + = , 2 2 3 V V = 1 1 2 2 1 2 PV PV T T = 1 1 2 2 1 2 PV T P T V = 2 = v 596 2 298 3 v 596 3 1788 298 298 = = = 6 atm 35. Lattice enthalpy for NaCl is + 788 kJ mol - 1 and 0 1 Hyd Δ H 784kJ mol − = − . Enthalpy of solution of NaCl is ( A) 1 572 kJ mol − − ( B) 1 4 kJ mol − − ( C) 1 572 kJ mol − + ( D) 1 4 kJ mol − + Ans : D 788 784 4 / sol L Hyd H H H kJ mol = + = − = + 36. At 500 K, for a reversible reaction ( ) ( ) ( ) 2 2 2 g g g A B AB + in a closed container. 5 2 10 C K − = . In the presence of catalyst, the equilibrium is attaining 10 times faster. The equilibrium constant C K in the presence of catalyst at the same temperature is ( A) 10 2 10 − ( B) 5 2 10 − ( C) 4 2 10 − ( D) 6 2 10 − Ans : B Equilibrium constant does not affect by adding catalyst. 37. A weak acid with pK a 5.9 and weak base with pK b 5.8 are mixed in equal proportions. pH of the resulting solution is ( A) 7 ( B) 7.05 (C) 7.005 (D) 7.5 Ans : B 1 7 2 a b pH pK pK = + − 1 7 5.9 5.8 2 = + − 1 7 0.1 2 = + = 7 + 0.05 = 7.05 38. Temperature of 25 0 C in Fahrenheit and Kelvin scale respectively are ( A) 45 0 F and 260.16 K ( B) 47 0 F and 312.15 K ( C) 77 0 F and 298.15 K ( D) 17 0 F and 298.15 K Ans : C 0 F = 0 9 32 5 C + K = 0 C + 273.15 39. The number of protons, neutrons and electrons in the ion 32 2 16 S − respectively are ( A) 18, 16, 16 ( B) 16, 16, 16 ( C) 16, 18, 16 ( D) 16, 16, 18 Ans : D 32 2 16 S − no. of protons = 16 no. of neutrons = 32 – 16 = 16 no. of electrons = 16 + 2 = 18 16, 16 , 18 40. The correct order of first ionisation enthalpy of given elements is (A) C < B < Be < Li ( B) Li < Be < B < C ( C) Li < B < Be < C ( D) Be < Li < B < C Ans : C Li < B < Be < C 520 k J < 801 k J < 899 J < 1086 kJ 41. Which of the f ollowing statement is INCORRECT ? ( A) Bond length of O 2 < Bond length of 2 2 O − ( B) Bond length of O 2 > Bond length of 2 2 O − ( C) Bond length of O 2 > Bond length of 2 2 O + ( D) Bond length of 2 O + < Bond length of 2 2 O − Ans : D Bond order of 2 2.5 O and + = Bond order of 2 2 1.0 O − = Bond order of 2 O + Bond order of 2 2 O − 42. IUPAC name of the compound is ( A) 1, 1, 2, 2 – tetra methylethene ( B) 2, 3 – dimethyl butene ( C) 2, 3 – dimethylbut - 2 - ene ( D) 2, 3 – dimethyl butyne Ans : C 43. Among the following : .. I II III IV V The set which represents aromatic species is ( A) II and III ( B) I, II and IV (C) I, II and III ( D) III, IV and V Ans : B According to Huckle’s Rule 44. Which one of the following gases converts haemoglobin into carboxy haemoglobin ? (A) NO ( B) CO 2 ( C) CO ( D) O 2 Ans : C CO gas strongly bonds to haemoglobin complex to form carboxy haemoglobin. 45. What is the oxidation of S in H 2 S 2 O 8 ? (A) +7 (B) +6 (C) +5 (D) +4 Ans : B S O O S O O O O OH O H 46. A 30% solution of hydrogen peroxide is ( A) ‘50 volume’ hydrogen peroxide ( B) ‘100 volume’ hydrogen peroxide ( C) ‘30 volume’ hydrogen peroxide ( D) ‘10 volume’ hydrogen peroxide Ans : B 3% H 2 O 2 is labelled as 10 V 30% H 2 O 2 will be 100 V 47. A pair of amphoteric oxides is (A) BeO, MgO ( B) BeO, ZnO ( C) Al 2 O 3 , Li 2 O ( D) BeO, BO 3 Ans : B BeO and ZnO are amphoteric. 48. The composition of water gas is ( A) ( ) ( ) 2 g g CO H O + ( B) ( ) ( ) 2 g g CO H + ( C) ( ) ( ) 2 g g CO N + ( D) ( ) 4 g CH Ans : B 49. The swelling in feet and ankles of an aged person due to sitting continuous ly for long hours during travel , is reduced by soaking the feet in warm salt water. This is because of : ( A) Edema ( B) Diffusion ( C) Reverse Osmosis ( D) Osmosis Ans : D 50. A sample of wate r is found to contain 5.85 % w w of AB (molecular mass 58.5) and 9.50 % w w 2 XY (molecular mass 95). Assuming 80 % ionisation of AB and 60 % ionisation of 2 XY , the freezing point of water sample is [Given : f K for water 1.86 K 1 kgmol − , Freezing point of pure water is 273 K and A, B and Y are monovalent ions] ( A) 280.44 K ( B) 281.75 K ( C) 264.25 K ( D) 265.56 K Ans : D 2 f f f 2 1 w ×1000 ΔT = iK m = iK M ×w f 2 f f For AB 5.85×1000 ΔT = 1.8 ×1.86× = 3.348K 58.5×100 For XY 9.5×1000 ΔT = 2.2 ×1.86× = 4.092K 95×100 Total ΔT 3.348K + 4.092K = 7.44K Freezing point of water = 273K - 7.44K = 265.56K = 51. Match the column A(type of crystalline solid) with the column B (example for each type) : A B P. Molecular Solid i. SiC Q. Ionic Solid ii. Mg R. Metallic Solid iii. 2 H O S. Network Solid iv. MgO ( A) P - ii, Q - iv, R - iii, S - i ( B) P - iii, Q - iv, R - ii, S - i ( C) P - iii, Q - I, R - ii, S - iv ( D) P - iv, Q - iii, R - ii, S – i Ans: B 52. A metal crystallises in a body c e ntered cubic lattice with the metallic radius 0 3 A . The volume of the unit cell in 3 m is (A) 29 6.4 10 − ( B) 10 4 10 − ( C) 29 64 10 − ( D) 29 4 10 − Ans: A 3 3 -10 3 29 3 For BCC 4r 4r 4 3×10 m a = a = = 6.4 10 m 3 3 3 − = 53. If ‘a’ stands for the edge length of the cubic systems – The ratio of radii in simple cubic, body centered cubic and face centered cubic unit cells is (A) 1 3 2 : : 2 2 2 a a a ( B) 1 1 : 3 : 2 2 a a a ( C) 1 : 3 : 2 a a a ( D) 1 3 1 : : 2 4 2 2 a a a Ans: D 54. Dimerisation of solute molecules in low dielectric constant solvent is due to : (A) Co - ordinate bond ( B) Ionic bond ( C) Hydrogen bond ( D) Covalent bond Ans: C 55. For a reaction, the value of rate constant at 300 K is 5 1 6.0 10 s − . The value of Arrhenius factor A at infinitely high temperature is : (A) 5 6 10 300 − ( B) 5 6 10 ( C) 5 / 300 6 10 Ea R e − ( D) / 300 Ea R e − Ans: B At temperature / Ea RT K Ae − = / Ea R K Ae − = 0 K Ae K A = = 56. The rate constants 1 k and 2 k for two different reactions are 16 2000/ 10 T e − and 15 1000/ 10 T e − respectively. The temperature at which 1 2 k k = is : (A) 1000 2.303 K ( B) 1000 K ( C) 2000 2.303 K ( D) 2000 K Ans: A 16 2000/ 1 10 T K e − = 15 1000/ 2 10 T K e − = According to Arrhenius equation If K 1 = K 2 16 2000/ 15 1000/ 10 10 T T e e − − = 16 1000/ 15 2000/ 10 10 T T e e − − = 1000/ 2000/ 10 T T e − + = 1000/ 10 T e = 1000 ln10 T = 1000 1000 1000 ln10 2.303log10 2.303 T k = = = 57. During the electrolysis of brine, by using inert electrodes, (A) Na deposits on cathode ( B) 2 Cl liberates at anode (C) 2 O liberates at anode ( D) 2 H liberates at anode Ans: B 58. Consider the following 4 electrodes A : ( ) ( ) 0.0001 / ; s Ag M Ag + B : ( ) ( ) 0.1 / s Ag M Ag + C : ( ) ( ) 0.01 / ; s Ag M Ag + D : ( ) / 0 ( ) 0.001 / ; 0.80 Ag Ag s Ag M Ag E V + + = + then reduction potential in volts of the electrodes in the order (A) A > D > C > B ( B) A > B > C > D ( C) B > C > D > A ( D) C > D > A > B Ans : C According to Nernst equation, o + E [Ag ] 59. The resistance of 0.1 M weak acid HA in a conductivity cell is 3 2 10 Ohm The cell constant of the cell is 1 0.78 cm − and 0 m of acid HA is 390 S cm 2 1 mol − The pH of the solution is ( A) 5 ( B) 3 ( C) 3.3 ( D) 4.2 Ans : B * 3 1 3 1 2 -1 m 3 3 3 2 -1 2 + 2 3 m 0 2 -1 m + 3 0.78 0.39 10 Scm 0.39 10 Scm , = 3.9 S cm mol 2 10 C 0.1 10 mol cm 3.9 S cm mol 1 10 [H ] C 0.1 10 10 390 S cm mol pH log[H ] = log[10 ] = 3 G R − − − − − − − − − − = = = = = = = = = = = = − − 60. In which one of the following reactions, rate constant has the unit mol L - 1 s - 1 ? ( A) ( ) ( ) ( ) 2 2 2 2 g g g NO O NO + → ( B) Decomposition of HI on the surface of Gold ( C) Acid catalysed hydrolysis of CH 3 COOCH 3 ( D) 3 2 4 CHCl Cl CCl HCl + → + Ans : B ******