HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 1. For the reaction A(g) + B(g) C(g) + D(g); ∆ H = - Q KJ The equilibrium constant cannot be disturbed by (A) Addition of A (B) Addition of D (C) Increasing of pressure (D) Increasing of temperature Ans : C Since number of moles of gaseous products and reactants remain same, change in presence cannot disturb the equilibrium. 2. An organic compound 'X' on treatment with PCC in dichloromethane gives the compound Y. Compound ‘Y' reacts with I 2 and alkali to form yellow precipitate of triiodomethane. The compound X is (A) CH 3 CHO (B) CH 3 COCH 3 (C) CH 3 CH 2 OH (D) CH 3 COOH Ans : C CH 3 - CH 2 - OH (X) PCC CH 3 CHO I 2 / Alkali CHI 3 (Y) Yellow ppt 3. A com pound ‘ A ’ (C 7 H 8 O) is insoluble in NaHCO 3 solution but dissolve in NaOH and gives a characteristic colour with neutral FeCl 3 solution. When treated with Bromine water compound ‘A’ forms the compound B with the formula C 7 H 5 OBr 3 ‘ A ’ is A) B) OH CH 3 C) OH CH 3 CH 3 OH OH D) Ans : B Compound A (molecular Formula C 7 H 8 O) is insoluble in water and dil. NaHCO 3 , but dissolves in dil. NaOH and gives colour with aq. FeCl 3 . Hence it is phenol. On treatment with Br 2 it readily gives precipitate of C 7 H 5 OBr 3 Therefore, compound is m – cresol. CREATIVE LEARNING CLASSES, KARKALA K - CET DETAILED SOLUTIONS - 2021 DEPARTMENT OF CHEMISTRY SET – C 3 HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 4. In set of reactions, Identify D CH 3 COOH 2 SOCl   A 3 Benzene Anh AlCl  B HCN   C 2 H O   D A) COOH CH 3 OH B) CN CH 3 OH OH CH 3 CN C) COOH CH 3 OH D) Ans : D CH 3 COOH SOCl 2 CH 3 COCl (A) C 6 H 6 Anh. AlCl 3 COCH 3 HCN C H 3 OH CN H 2 O C H 3 OH COOH (B) (C) (D) 5. K a values for acids H 2 SO 3 , HNO 2 , CH 3 COOH and HCN are respectively 1.3 x 10 -2 , 4 x 10 -4 , 1.8 x 10 -5 and 4 x 10 -10 , which of the above acids produces stronger conjugate base in aqueous solution? (A) H 2 SO 3 (B) HNO 2 (C) CH 3 COOH (D) HCN Ans : D As Ka decreases the acidic character also decreases and its conjugate base become more stronger. 6. Acetaldehyde HgSO 4 / H 2 SO 4 PCC (i) SnCl 2 / HCl (ii) H 3 O + A, B and C respectively are (A) ethanol, ethane nitrile and ethyne (B) ethane nitrile, ethanol and ethyne (C) ethyne, ethanol and ethane nitrile (D) ethyne, ethane nitrile and ethanol HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA Ans : C C H CH CH 3 - CH 2 - OH CH 3 CN HgSO 4 /H 2 SO 4 PCC SnCl 2 H 3 O + CH 3 CHO 7. The reagent which can do the conversion CH 3 COOH   CH 3 – CH 2 – OH is (A) LiAlH 4 / ether (B) H 2 , Pt (C) NaBH 4 (D) Na and C 2 H 5 OH Ans : A CH 3 COOH LAH/Ether CH 3 CH 2 OH 8. (i) CH MgBr (i) B H Conc H SO 3 2 6 2 4 CH CHO A B C 3 + - Δ (ii) H O (ii) H O, OH 3 2       A and C are (A) Identical (B) Position isomers (C) Functional isomers (D) Optical isomers Ans : B CH 3 CHO CH 3 MgBr H 3 O + C H 3 OH CH 3 H Conc. H 2 SO 4 CH 2 = CH - CH 3 B 2 H 6 H 2 O, OH - CH 3 CH 2 CH 2 OH (A) (B) (C) Hence (A) and (C) are positional isomers. 9. Which of the following is not true for oxidation? (A) addition of oxygen (B) addition of electronegative element (C) removal of hydrogen (D) removal of electronegative element Ans : D Removal of electromagnetic element is reduction. 10. Which is the most suitable reagent for the following conversion? O O || || CH - CH = CH - CH - C - CH CH - CH = CH - CH - C- OH 3 2 3 3 2   (A) Tollen's reagent (B) Benzoyl peroxide (C) I 2 and NaOH solution with subsequent acidification (D) Sn and NaOH solution Ans : C CH 3 - CH = CH - CH 2 - CO - CH 3 I 2 / NaOH / H + CH 3 - CH = CH - CH 2 - COOH HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 11. alc.NH 2CH Cl 3 3 C H CH Cl A B 6 5 2     The product B is (A) N, N- Dimethyl phenyl methanamine (B) N, N- Dimethyl benzenamine (C) N- Benzyl – N – methyl methanamine (D) phenyl - N, N - dimethyl methanamine Ans : A C 6 H 5 CH 2 Cl Alc NH 3 C 6 H 5 CH 2 NH 2 2CH 3 Cl C 6 H 5 - CH 2 - N (CH 3 ) 2 12. The method by which aniline cannot be prepared is (A) Nitration of benzene followed by reduction with Sn and con HCI (B) Degradation of benzamide with bromine in alkaline solution (C) Reduction of nitrobenzene with H 2 / Pd is ethanol (D) Potassium salt of phthalimide treated with chlorobenzene followed by the hydrolysis with aqueous NaOH solution Ans : D Aniline cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide. 13. Permanent hardness cannot be removed by (A) Using washing soda (B) Calgon's method (C) Clark's method (D) lon exchange method Ans : C Clark’s method is used to remove temporary hardness. 14. A hydrocarbon A(C 4 H 8 ) on reaction with HCI gives a compound B(C 4 H 9 Cl) which on reaction with 1 mol of NH 3 gives compound C(C 4 H 10 N). On reacting with NaNO 2 and HCl followed by treatment with water, compound C yields an optically active compound D. The D is (A) | 3 | CH -CH 2 3 Cl CH C H   (B) | 3 | CH -CH 2 3 OH CH - C -H (C) | 3 | CH -CH 2 3 NH2 CH - C -H (D) | 3 | CH -CH 2 3 H CH - C -H Ans : B CH 3 - CH = CH - CH 3 HCl CH 3 - CH(Cl) - CH 2 - CH 3 1mol NH 3 CH 3 - CH(NH 2 ) - CH 2 - CH 3 NaNO 2 / HCl H 2 O CH 3 - CH(OH) CH 2 - CH 3 (A) (B) (C) (D) Optically Active HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 15. RNA and DNA are chiral molecules, their chirality is due to the presence of (A) D-Sugar component (B) L-Sugar component (C) Chiral bases (D) Chiral phosphate ester unit Ans : A Deoxy ribose and ribose sugars are D-chiral sugars in DNA and RNA. 16. The property of the alkaline earth metals that increases with their atomic number is (A) Ionisation enthalpy (B) Electronegativity (C) Solubility of their hydroxide in water (D) Solubility of their sulphate in water Ans : C Due to decrease in lattice energy of the hydroxide salt and size of cation increases down the group. 17. Primary structure in a nucleic acid chain contains bases as G A T G C ...... The chain which is complementary to this chain is (A) G G T G A .... (B) T G A A G..... (C) C T A C G ..... (D) T T T A G ..... Ans : C G A T G C T A C C G      18. In the detection of II group acid radical, the salt containing chloride is treated with concentrated Sulphuric acid, the colourless gas is liberated. The name of the gas is (A) Hydrogen chloride gas (B) Chlorine gas (C) Sulphur dioxide gas (D) Hydrogen gas Ans : A In II group acid radical Cl 2 gas liberated reacts with conc. H 2 SO 4 give HCl gas which is colourless. 19. The number of six membered and five membered rings in Buckminster Fullerene respectively Is (A) 20,12 (B) 12, 20 (C) 14, 18 (D) 14, 11 Ans: A In Buck minster fullerene there are 26 membered and 12 5-membered rings. 20. In Chrysoberyl, a compound containing Beryllium, Aluminium and oxygen, oxide ions form cubic close packed structure. Aluminium ions occupy 1/4th of tetrahedral voids and Beryllium ions occupy 1/4th of octahedral voids. The formula of the compound is (A) BeAlO 4 (B) BeAl 2 O 4 (C) Be 2 AlO 2 (D) BeAlO 2 Ans : B Be is at octahedral void. Therefore 1 4 Be Al is at tetrahedral voids HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA ∴ 1 1 ×2 4 2 Al =Al Oxide is at CCP 1 1 1 4 2 Be Al O 4         = Be Al 2 O 4 21. The correct statement regarding defects in solids is (A) Frenkel defect is a vacancy defect (B) Schottky defect is a dislocation defect (C) Trapping of an electron in the lattices leads to the formation of F-centre (D) Schottky defect has no effect on density. Ans : C Trapping of an electron in the anionic vacancy leads to the formation of F – center 22. A metal crystallises in BCC lattice with unit cell edge length of 300 pm and density 6.15 gcm -3 . The molar mass of the metal is (A) 50 gmol -1 (B) 60 gmol -1 (C) 40 gmol -1 (D) 70 gmol -1 Ans: A 3 × a × NA M = z  8 3 23 6.15 (3 10 ) 6.022 10 2       -1 -1 = 49.99gmol 50 gmol  23. Henry's law constant for the solubility of N 2 gas in water at 298 K is 1.0 x 10 5 atm. The mole fraction of N 2 in air is 0.8. The number of moles of N 2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure is (A) 4.0 x 10 -4 (B) 4.0 x 10 -5 (C) 5.0 x 10 -4 (D) 4.0 x 10 -6 Ans : A 2 2 N total N P = P ×  5 0.8   = 4 atm H W.K.T,P= K  2 2 N H H O n P = K n 2 N 5 n 4 =1×10 × 10 2 5 4 10 10 N n     -4 = 4×10 mol 24. A pure compound contains 2.4 g of C, 1.2 x 10 23 atoms of H, 0.2 moles of oxygen atoms. Its empirical formula is (A) C 2 HO (B) C 2 H 2 O 2 (C) CH 2 O (D) CHO HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA Ans : D 2.4 C = = 0.2 mol 12 O = 0.2mol 23 23 1.2×10 H = = 0.2 mol 6.022×10 ∴ By taking simple ratio of numbers of moles of elements 0.2 0.2 0.2 0.2 0.2 0.2 C H O C 1 H 1 O 1 = CHO 25. Choose the correct statement (A) K H value is same for a gas in any solvent (B) Higher the K H value more the solubility of gas (C) K H value increases on increasing the temperature of the solution (D) Easily liquefiable gases usually has lesser K H values Ans : C Henry’s constant is temperature dependent. 26. The K H value (K bar) of Argon (l), Carbondioxide (II) formuldehyde (III) and methane (IV) are respectively 40.3, 1.67, 1.83 × 10 -5 and 0.413 at 298 K. The increasing order of solubility of gas in liquid is (A) I < II < IV < III (B) III < IV < II < I (C) I < III < II < IV (D) I < IV < II < III Ans : A The relationship between K H and solubility is According to Henry’s law P = K χ B H B B B H A n P =K . n   A B For dilute solution n >>> n Solubility = B A B H P .n n = K i.e solubility H 1 K  ∴ (i) Argon K H = 40.3 (ii) carbon dioxide = 1.67 (iii) Formaldehyde = 1.83 x 10 -5 (iv) Methane = 0.413 ∴ III > IV > II > I HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 27. The vapour pressure of pure liquids A and B are 450 and 700 mm of Hg at 350 K respectively. If the total vapour pressure of the mixture is 600 mm of Hg, the composition of the mixture in the solution is (A) x A = 0.4, x B = 0.6 (B) x A = 0.6, x B = 0.4 (C) x A = 0.3, x B = 0.7 (D) x A = 0.7, x B = 0.3 Ans : C Given 0 A P 450 MM of Hg  0 B P 700 MM of Hg  Total P 600 MM of Hg  According to Raoult’s law, 0 0 0 Total A B A B P = (P - P ) χ + P A 600 (450 700) χ 700    A 250 χ = 100 100 0.4 250 A    B A χ =1 - χ = 0.6 o A A A P = P . χ = 450 × 0.4 = 180 mm o B B B P = P . χ = 700 × 0.6 = 420 mm Mole fraction of liquid A = 180 0.3 180 420   ∴ Mole fraction of liquid B = 1 – 0.3 = 0.7 28. Consider the following electrodes 2 P Zn (0.0001 M) / Zn   2 Q Zn (0.1 M) / Zn   2 R Zn (0.01 M) / Zn   2 S Zn (0.001 M) / Zn   o 2 E Zn / Zn 0.76 V    Electrode potentials of the above electrodes in volts are in the order (A) P S R Q    (B) S R Q P    (C) Q R S P    (D) P Q R S    Ans : C For concentration cells electrode potential ∝ concentration of solution 29. The number of angular and radial nodes in 3p orbital respectively are (A) 3,1 (B) 1,1 (C) 2,1 (D) 2,3 Ans : B Number of angular nodes for 3p = 1 Radial nodes = 0 – l – 1 = 3 – 1 – 1 = 1 ∴ 1, 1 HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 30. The resistance of 0.01 m KCI solution at 298 K is 1500  . If the conductivity of 0.01m KCl solution at 298 K is 0.146 3 1 10 S cm    The cell constant of the conductivity cell in 1 cm  is (A) 0.219 (B) 0.291 (C) 0.301 (D) 0.194 Ans: A G = K. R = 0.146 x 10 -3 x 1500 = 219 x 10 -3 = 0.219 cm -1 31. 2(g) (s) (s) (aq) H 2AgCI 2Ag 2HCI   0 0 cell E at 25 C for the cell is 0.22 V. The equilibrium constant at 25 o C is (A) 7 2.8 10  (B) 8 5.2 10  (C) 5 2.8 10  (D) 4 5.2 10  Ans : A 2(g) (s) (aq) H + 2AgCl 2Ag + 2HCl For the above reaction n = 2 o C nFE log K = 2.303 RT 2 × 96500 × 0.22 = 2.303 × 8.314 × 298 C log K = 7.44 K C = Antilog 7.44 = 2.8 x 10 7 32. For a reaction A+2B  Products, when concentration of B alone is increased half life remains the same. If concentration of A alone is doubled, rate remains the same. The unit of rate constant for the reaction is (A) 1 S  (B) 1 1 L mol S   (C) 1 1 mol L S   (D) 1 atm  Ans : A R = k[A] 0 [B] 1 -1 -1 -1 1 -1 R mol L ×S k = = =S [B] mol L 33. The third ionisation enthalpy is highest in (A) Alkali metals (B) Alkaline earth metals (C) Chalcogens (D) Pnictogens Ans : B Alkaline earth metals : GEC (Noble gas) ns2 On removing 2 electrons it attains stability. 34. If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of the reaction is given by (A) 4.606 t k  (B) 2.303 t k  (C) 0.693 t k  (D) 6.909 t k  HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA Ans : A 2.303 100 t = log k 1 4.606 t = k 35. The rate of a gaseous reaction is given by the expression k [A][B] 2 . If the volume of vessel is reduced to one half of the initial volume, the reaction rate as compared to original rate is (A) 1 16 (B) 1 8 (C) 8 (D) 16 Ans : C R 2 = K [2A] [2B] 2 R 1 = K [A] [B] 2 on comparing we get R 2 = 8R 1 36. The correct IUPAC name of ............. (A) 4-Ethyl-1-Fluoro-2-nitrobenzene (B) 1-Ethyl-4-Fluoro-3-nitrobenzene (C) 3-Ethyl-6-Fluoronitrobenzene (D) 5-Ethyl-2-Fluoronitrobenzene Ans : D NO 2 F CH 3 5 - ethyl – 2 – fluoronitrobenzene 37. Higher order (>3) reactions are rare due to (A) Shifting of equilibrium towards reactants due to elastic collisions (B) Loss of active species on collision (C) Low probability of simultaneous collision of all reacting species (D) Increase in entropy as more molecules are involved Ans : C 38. Arrange benzene-hexane and ethyne in decreasing order of their acidic behavior (A) Benzene > n-hexane > ethyne (B) n-hexane > Benzene > ethyne (C) ethyne > n-hexane > Benzene (D) ethyne > Benzene > n-hexane Ans : D More the % of S-charcter of C more will be acidic nature. ∴ Ethyne > Benzene > n – hexane sp sp 2 sp 3 39. A colloidal solution is subjected to an electric field than colloidal particles more towards anode. The amount of electrolytes of BaCl 2 , AlCl 3 and NaCl required to coagulate the given colloid is in the order (A) NaCl > BaCl 2 > AlCl 3 (B) BaCl 2 > AlCl 3 > NaCl (C) AlCl 3 = NaCl = BaCl 2 (D) AlCl 3 > BaCl 2 > NaCl HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA Ans : A Greater the valence of the flocculating ion added, greater is its power to cause coagulation. And coagulatory power follows order AlCl 3 > BaCl 2 > NaCl Higher the coagulating power smarter the quantity is need. ∴ NaCl > BaCl 2 > AlCl 3 40. Which of the following is an incorrect statement? (A) Hydrogen bonding is stronger than dispersion forces (B) Sigma bonds are stronger than π – bonds (C) Ionic bonding is non-directional (D) σ - electrons are referred to as mobile electrons Ans : D 41. Zeta potential is (A) Potential required to bring about coagulation of a colloidal sol. (B) Potential required to give the particle a speed of 1 cm -1 S (C) Potential difference between fixed charged layer and the diffused layer having opposite charges (D) Potential energy of the colloidal particles. Ans : C 42. Which of the following compound on heating gives N 2 O? (A) Pb(NO 3 ) 2 (B)NH 4 NO 3 (3) NH 4 NO 2 (D) NaNO 3 Ans : B Δ 4 3 2 2 NH NO N O + 2H O   43. Which of the following property is true for the given sequence NH 3 > PH 3 > AsH 3 > SbH 3 > BiH 3 ? (A) Reducing property (B) Thermal stability (C) Bond angle (D) Acidic character Ans : B & C As size of central atom increases thermal stability decreases down the group. Bond angle decreases down the group due to decrease in bond pair bond pair repulsion. 44. The correct order of boiling point in the following compound is (A) HF > H 2 O > NH 3 (B) H 2 O > HF > NH 3 (C) NH 3 > H 2 O > HF (D) NH 3 > HF > H 2 O Ans : B H 2 O > HF > NH 3 Move the electronegativity more will be boiling point and more the number of H-bonds per molecule more will be boiling point. H 2 O has surrounding 4 H 2 O molecules but HF has two molecules. 45. XeF 6 on partial hydrolysis gives a compound X, which has square p yramidal geometry ‘X’ is (A) XeO 3 (B) XeO 4 (C) XeOF 4 (D) XeO 2 F 2 HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA Ans : C 2 H O 6 4 XeF XeOF   46. A colorless, neutral, paramagnetic oxide of Nitrogen ‘P’ on oxidation gives reddish brown gas Q, Q on cooling gives colorless gas R, R on reaction with P gives blue solid S. Identify P, Q, R, S respectively. (A) N 2 O NO NO 2 N 2 O 5 (B) N 2 O NO 2 N 2 O 4 N 2 O 3 (C) NO NO 2 N 2 O 4 N 2 O 3 (D) NO NO N 2 O 4 N 2 O 5 Ans : C Oxidation Cooling NO NO NO N O N O 2 2 4 2 3       47. Which of the following does not represent property stated against it? (A) CO +2 < Fe +2 < Mn +2 - Ionic size (B) Ti < V < Mn – Number of oxidation states (C) Cr +2 < Mn +2 < Fe +2 – Paramagnetic behavior (D) Sc > Cr > Fe – Density Ans : D 48. Which one of the following is correct for all elements from Sc to Cu? (A) The lowest oxidation state shown by them is +2 (B) 4s orbital is completely filled in the ground state. (C) 3d orbital is not completely filled in the ground state (D) The ions in +2 oxidation states are paramagnetic Ans : D Ions in +2 oxidation state have unpaired electrons hence they are paramagnetic. 49. When the absolute temperature of ideal gas is doubled and pressure is halved, the volume of gas (A) Will be half of original volume (B) Will be 4 times the original volume (C) Will be 2 times the original volume (D) Will be 1/4 th times the original volume Ans : B 1 1 2 2 1 2 P V P V = T T 1 1 2 1 1 1 1 P 2T V =V × × = 4V 0.5P T 2 1 V = 4 V 50. Which of the following pairs has both the ions colored in aqueous solution? (Atomic numbers of Sc = 21, Ti = 22, Ni = 28, Cu = 29, Mn = 25) (A) Sc 3+ , Mn 2+ (B) Ni 2+ , Ti 4+ (C) Ti 3+ , Cu + (D) Mn 2+ , Ti 3+ Ans : D Mn 2+ and Ti +3 contains unpaired electrons. 3d 5 3d 1 . These are coloured. HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 51. For the crystal field splitting in octahedral complexes, (A) The energy of the e g orbitals will decrease by (3/5) ∆ o and that of the t 2g will increase by (2/5) ∆ o (B) The energy of the e g orbitals will increase by (3/5) ∆ o and that of the t 2g will decrease by (2/5) ∆ o (C) The energy of the e g orbitals will increase by (3/5) ∆ o and that of the t 2g will increase by (2/5) ∆ o (D) The energy of the e g orbitals will decrease by (3/5) ∆ o and that of the t 2g will decrease by (2/5) ∆ o Ans : B 52. Peroxide effect is observed with the addition of HBr but not with the addition of HI to unsymmetrical alkene because (A) H-I bond is stronger that H-Br is not cleaved by the free radical (B) H-I bond is weaker than H-Br bond so that Iodine free radicals combine to form Iodine molecules (C) Bond strength of HI and HBr are same but free radicals are formed in HBr (D) All of the these Ans : B H – Br undergoes homolysis to form free radical. But Iodine free radicals have greater tendency to combine. • homolysis R - O - O - R 2R - O   • • Peroxide H - I H + I  • R - O + H - Br R - O - H + Br   • • 2 I + I I   53. The IUPAC name of     3 3 5 Co NH CO Cl     is (A) Pentaamminecarbonatocobalt (III)Chloride (B) Carbonatopentamminecobalt (III) Chloride (C) Pentaamminecarbonatocobaltate (III) Chloride (D) Pentaammine cobalt (III) Carbonate Chloride Ans : A Pentaammine carbonato cobalt (III) chloride 54. Homoieptic complexes among the following are             3 2 4 2 2 3 2 4 A K Al C O , B CoCl en C K Zn OH              (A) A only (B) (A) and (B) only (C) (A) and (C) only (D) (C) only Ans : C K 3 [Al(C 2 O 4 ) 3 ] and K 2 [Zn(OH) 4 ] In these complexes central metal surrounded by one type of ligands. HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 55. The correct order for wavelengths of light observed in the complex ions       2+ 3+ 3- 3 3 5 6 6 CoCl NH , Co NH and Co CN             is (A)       2+ 3+ 3- 3 3 5 6 6 CoCl NH > Co NH > Co CN             (B)       3+ 3- 2+ 3 3 6 6 5 Co NH > Co CN > CoCl NH             (C)       3- 2+ 3- 3 6 5 6 Co CN > CoCl NH > Co CN             (D)       3+ 3- 3 3 6 5 6 Co NH > CoCl NH > Co CN             Ans : A [CoCl(NH 3 ) 5 ] 2+ > [Co(NH 3 ) 6 ] 3+ > [Co(CN) 6 ] 3- In [CoCl(NH 3 ) 5 ] less CFS and [Co(CN) 6 ] 3- shows maximum CFS. ∴ Wavelength absorbed by [Co(CN) 6 ] 3- is less and [CoCl(NH 3 ) 5 ] +2 is more. 56. O 2 N CH 3 Br 2 UV light A The compound A (major product) is A) O 2 N CH 3 B) O 2 N CH 3 Br C) O 2 N CH 3 Br O 2 N CH 3 Br D) Ans : B O 2 N CH 3 Br 2 / UV light O 2 N CH 3 Br 57. Bond enthalpies of A 2 , B 2 and AB are in the ratio 2 : 1 : 2. If bond enthalpy of formation of AB is -100 KJ mol -1 . The bond enthalpy of B 2 is (A) 100KJ mol -1 (B) 50KJ mol -1 (C) 200 KJ mol -1 (D) 150KJ mol -1 Ans : C A 2 + B 2 ⟶ 2AB Let B.E of A 2 = 2 x , B 2 = x , AB = 2 x ∆ H = (A 2 + B 2 ) – 2AB = 2 x + x – 2(2 x ) -100 x 2 = 3 x – 4 x -200 = - x x = 200 kJ mol -1 HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 58. The order of reactivity of the compounds C 6 H 5 CH 2 Br, C 6 H 5 CH(C 6 H 5 )Br, C 6 H 5 CH(CH 3 )Br and C 6 H 5 C(CH 3 )(C 6 H 5 )Br in 2 N S reaction is A) H 5 C 6 CH 3 C 6 H 5 Br < H 5 C 6 H C 6 H 5 Br H 5 C 6 H CH 3 Br H 5 C 6 H H Br < < B) H 5 C 6 H H Br H 5 C 6 H CH 3 Br H 5 C 6 H C 6 H 5 Br H 5 C 6 CH 3 C 6 H 5 Br < < < C) D) < < < < < < H 5 C 6 H CH 3 Br H 5 C 6 H CH 3 Br H 5 C 6 H H Br H 5 C 6 H H Br H 5 C 6 H C 6 H 5 Br H 5 C 6 H C 6 H 5 Br H 5 C 6 CH 3 C 6 H 5 Br H 5 C 6 CH 3 C 6 H 5 Br Ans : A For 2 N S reduction reactivity order of alkyl halides is 1 ° > 2 ° > 3 ° 59. The major product of the following reactions is CH 2 = CH - CH 2 - OH HBr Excess   product (A) CH 3 – CHBr - CH 2 Br (B) CH 2 = CH - CH 2 Br (C) CH 3 – CHBr - CH 2 - OH (D) CH 3 – CHOH - CH 2 OH Ans : A CH 2 = CH - CH 2 - OH HBr (Excess) CH 3 - CH(Br) - CH 2 - Br 60. CH 3 CH 3 + O 2 C - O - OH C H 3 CH 3 H + H 2 O A + B The product ‘ A ’ g ives white precipitate when treated with bromine water. The product ‘B’ is treated with Barium hydroxide to give the product C. The compound C is heated strongly form product D. The product D is (A) 4-Methylpent-3-en-2-one (B) But -2-enal (C) 3-Methylpent -3-en-2-one (D) 2-Methylbut-2-enal Ans : A HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA OH + CH 3 COCH 3 (A) (B) OH (A) + 3Br 2 OH Br Br Br + 3HBr CH 3 COCH 3 Ba(OH) 2 C H 3 OH CH 3 CH 2 COCH 3 heat C H 3 CH 3 CH - CO - CH 3 (B) (C) (D) The Product (D) is 4 – methyl pent – 3 – en – 2 – one Department of Chemistry  Mr. Adarsha M. K  Mr. Vishwanath Math  Mr. Yogisha Vasishta  Mr. Nagaraja H CREATIVE EDUCATION FOUNDATION MOODBIDRI (R), KARKALA Website : www.creativeedu.in Phone No. : +91 9019844492

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