PART B - CHEMISTRY

 

36.       The ionization enthalpy of hydrogen atom is 1.312 ´ 106 J mol-1. The energy required to excite the electron in the atom from n = 1 to n = 2 is

            (1)  6.56 ´ 105 J mol-1                                      (2)  7.56 ´ 105 J mol-1

            (3)  9.84 ´ 105 J mol-1                                      (4)  8.51 ´ 105 J mol-1

36.       (3)

           

            E2 – E1= kz2 = 1.312 ×106

                        = 0.984 × 106 = 9.84 ×105 J mol-1

 

37.       Which one of the following pairs of species have the same bond order ?

            (1)  CN- and CN+                                            (2)  and CN-

            (3)  NO+ and CN+                                            (4)  CN- and NO+

37.       (4)

            CN = s1s2 s*1s2s2s2s*2s2 s2px2

                Bond order for CN- =

            NO+=  s1s2 s*1s2s2s2s*2s2 s2px2

            Bond order of NO+ =

 

38.       Which one of the following constitutes a group of the isoelectronic species ?

            (1)  NO+, , CN-, N2                                   (2)  CN-, N2, ,

            (3)  N2, , NO+, CO                                     (4)  , , CO, NO

38.       (1)

            Total electrons in NO+ = 14,       ,        = 14,       N2= 14

 

39.       For species are listed below:

            i.   

            ii.    H3O+

            iii.  

            iv.   HSO3F

            Which one of the following is the correct sequence of their acid strength ?

            (1)  ii < iii < i < iv                                              (2)  i < iii < ii < iv

            (3)  iii < i < iv < ii                                              (4)  iv < ii < iii < i


39.       (2)

            Correct order of acidic strength is

            (iv) > (ii) > (iii) > (i)

 

40.       The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be

            (1)  4.79                                                           (2)  7.01

            (3)  9.22                                                           (4)  9.58

40.       (2)

            pKa = 4.80  pKb = 4.78  therefore  pH of solution is

=

= 7 + 2.4  – 2.39

= 7.01

 

41.       The correct decreasing order of priority for the functional groups of organic compounds in the IUPAC system of nomenclature is

            (1)  -SO3H, -COOH, -CONH2, -CHO         (2)  -CHO, -COOH, -SO3H, -CONH2

            (3)  -CONH2, -CHO, -SO3H, -COOH         (4)  -COOH, -SO3H, -CONH2, -CHO

41.       (4)

            The correct decreasing order of priority is

            —COOH, — SO3H, —CONH2, —CHO

 

42.       The treatment of CH3MgX with CH3C º C - H produces

            (1)  CH3C º C - CH3                                       (2) 

            (3)  CH4                                                           (4)  CH3 - CH = CH2

42.       (3)

            CH3C º C — H + CH3MgX  ® CH4+ CH3C º C MgX

 

43.       The hydrocarbon which can react with sodium in liquid ammonia is

            (1)  CH3CH2C º CH                                        (2)  CH3CH = CHCH3

            (3)  CH3CH2C º CCH2CH3                             (4)  CH3CH2CH2C º CCH2CH2CH3

43.       (1)

            Only terminal alkynes have acidic hydrogen among different hydrocarbons.

            CH3CH2CºCHCH3CH2Cº

 

44.       The vapour pressure of water at 20oC is 17.5 mm Hg. If 18 g of glucose (C6H12O6) is added to 178.2 g of water at 20oC, the vapour pressure of the resulting solution will be

            (1)  15.750 mm Hg                                           (2)  16.500 mm Hg

            (3)  17.325 mm Hg                                           (4)  17.675 mm Hg

44.       (3)

            P0 = 17.5 mm of Hg

            w = 18 g,   molecular mass of glucose = 180,   W = 178.2 g

            P = ?

             

           

            Ps  = 17.325 mm of Hg

 

45.       Gold numbers of protective colloids A, B, C and D are 0.50, 0.01, 0.10 and 0.005, respectively. The correct order of their protective powers is

            (1)  C < B < D < A                                           (2)  A < C < B < D

            (3)  B < D < A < C                                           (4)  D < A < C < B

45.       (2)

            Smaller the value of gold number higher is the protective power

            Therefore D > B > C > A

 

46.       In a compound, atoms of element Y form ccp lattice and those of element X occupy 2/3rd of tetrahedral voids. The formula of the compound will be

            (1)  X2Y3                                                          (2)  X2Y

            (3)  X3Y4                                                          (4)  X4Y3

46.       (4)

            Effective number of atoms of X in the unit cell  =  of tetrahedral void

=

            Effective number of atoms of Y in the unit cell = 4

            Therefore formula of compound is 

 

47.       In context with the industrial preparation of hydrogen from water gas (CO + H2), which of the following is the correct statement?

            (1)  CO is removed by absorption in aqueous Cu2Cl2 solution

            (2)  H2 is removed through occlusion with Pd

            (3)  CO is oxidized to CO2 with steam in the presence of a catalyst followed by absorption of CO2 in alkali

            (4)  CO and H2 are fractionally separated using differences in their densities

47.       (3)

            CO is removed by its oxidation using steam to CO2 followed by absorption of CO2 in alkali.

            CO(g) + H2O(g)® CO2(g) + H2(g)

 

48.       Among the following substituted silanes the one which will give rise to cross linked silicone polymer on hydrolysis is

            (1)  RSiCl3                                                        (2)  R2SiCl2

            (3)  R3SiCl                                                        (4)  R4Si

 

 

 

 48.      (1)

      Hydrolysis followed by polymerization of RSiCl3 provides complex cross   linked

             polymer.

           

 

49.       Amount of oxalic acid present in a solution can be determined by its titration with KMnO4 solution in the presence of H2SO4. The titration gives unsatisfactory result when carried out in the presence of HCl, because HCl

            (1)  furnishes H+ ions in addition to those from oxalic acid

            (2)  reduces permanganate to Mn2+

            (3)  oxidizes oxalic acid to carbon dioxide and water

            (4)  gets oxidized by oxalic acid to chlorine

49.       (2)

            HCl reduces MnO4 to Mn2+

 

50.       Given V.

            The potential for the cell

            Cr |Cr3+ (0.1 M) | |Fe2+ (0.01 M)|Fe is

            (1)  0.339 V                                                     (2)  -0.339 V

            (3)  -0.26 V                                                     (4)  0.26 V

50.       (4)

            Net cell reaction is

           

           

            = [0.72 – 0.42 ] –

            = 0.30 –

            = 0.2606 V

 

51.       Which one of the following is the correct statement ?

            (1)  Beryllium exhibits coordination number of six

            (2)  Chlorides of both beryllium and aluminium have bridged chloride structure in solid phase

            (3)  B2H6 . 2NH3 is known as ‘inorganic benzene’

            (4)  Boric acid is a protonic acid

51.       (2)

.           Chlorides of both beryllium and aluminium have bridged chloride structures in solid phase.

52.       Identify the wrong statement in the following:

            (1)     Greenhouse effect is responsible for global warming

            (2)     Ozone layer does not permit infrared radiation from the sun to reach the earth

            (3)     Acid rain is mostly because of oxides of nitrogen and sulphur

            (4)     Chlorofluorocarbons are responsible for ozone layer depletion

52.       (2)

            Ozone layer does not permit ultraviolet radiation from the sun to reach the earth.

 

53.       The coordination number and the oxidation state of the element ‘E’ in the complex [E(en)2(C2O4)]NO2 (where (en) is ethylene diamine) are, respectively,

            (1)     4 and 2                                        (2)     4 and 3

            (3)     6 and 3                                        (4)     6 and 2

53.       (3)

            In [E(en)2 (C2O4)] NO2,   en and C2O42– are bidentate ligand.

Coordination number = 2 ×(no. of bidentate ligand)

           = 2 × 3= 6

            Let oxidation number of E be x.

            x + 2(0) + (–2)= + 1

            x = + 3

 

54.       In which of the following octahedral complexes of Co(at. no. 27), will the magnitude of D0 be the highest?

            (1)     [Co(C2O4)3]3-                             (2)     [Co(H2O)6]3+

            (3)     [Co(NH3)6]3+                              (4)     [Co(CN)6]3-

54.       (4)

            Since  is strongest legand in given legands therefore the magnitude of D0 will be highest.

 

55.       Large number of oxidation states are exhibited by the actinoids than those by the lanthanoids, the main reason being

            (1)     lesser energy difference between 5f and 6d than between 4f and 5d orbitals

            (2)     more energy difference between 5f and 6d than between 4f and 5d orbitals

            (3)     more reactive nature of the actinoids than the lanthanoids

            (4)     4f orbitals more diffused than the 5f orbitals

55.       (1)

            Actinoids exhibit larger number of oxidation state than lanthanoids because of lesser energy difference between 5f and 6d than between 4f and 5d orbitals.

 

56.       Which of the following factors is of no significance for roasting sulphide ores to the oxides and not subjecting the sulphide ores to carbon reduction directly?

            (1)     CO2 is thermodynamically more stable than CS2

            (2)     Metal sulphides are less stable than the corresponding oxides

            (3)     CO2 is more volatile than CS2

            (4)     Metal sulphides are thermodynamically more stable than CS2

56.       (3)

            All steps involved in metallurgical reduction process are based on favourable value of free energy change for the process.

 

57.       Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below:

           

            The energy involved in the conversion of to

            (using the data,  

             

            will be

            (1)     - 610 kJ mol-1                             (2)     - 850 kJ mol-1

            (3)     +120 kJ mol-1                              (4)     + 152 kJ mol-1

57.       (1)

             kJ mol-1

             kJ mol-1

             = – 381 kJ mol-1

           

            DH = DH1 + DH2 + DH3

            = 120 –349 –381

            = – 610 kJ mol-1

 

58.       In the following sequence of reactions, the alkene affords the compound ‘B’

                    

            The compound B is

            (1)     CH3COCH3                                (2)     CH3CH2COCH3

            (3)     CH3CHO                                    (4)     CH3CH­2CHO

58.       (3)

           

 

59.       Phenol, when it first reacts with concentrated sulphuric acid and then with concentrated nitric acid, gives

            (1)     o- nitrophenol                              (2)     p-nitrophenol

            (3)     nitrobenzene                                (4)     2, 4, 6-trinitrobenzene


59.       (2)

            During nitration of p-hydroxybenzene sulphonic acid, desulphonation as well as nitration takes place.

           

 

60.       Toluene is nitrated and the resulting product is reduced with tin and hydrochloric acid. The product so obtained is diazotised  and then heated with cuprous bromide. The reaction mixture so formed contains

            (1)     mixture of o- and p-dibromobenzenes

            (2)     mixture of o- and p-bromanilines

            (3)     mixture of o- and m-bromotoluenes

            (4)     mixture of o- and p-bromotoluenes

60.       (4)

           

           

 

61.       The organic chloro compound, which shows complete stereochemical inversion during a SN2 reaction, is

            (1)     (CH3)3CCl                                  (2)     (CH3)2CHCl

            (3)     CH3Cl                                         (4)     (C2H5)2CHCl

61.       (3)

            In the following options no chiral carbon is present therefore stereochemical inversion is not possible but CH3Cl will give SN2 reaction (backside attack)

62.       The absolute configuration of is

 

            (1)     R, R                                            (2)     R, S

            (3)     S, R                                            (4)     S, S

62.       (1)

             The absolute configuration of

is R , R for both the chiral centre.

 

63.       a- D-(+)-glucose and b-D-(+)-glucose are

            (1)     epimers                                       (2)     anomers

            (3)     enantiomers                                 (4)     conformers

63.       (2)

            a-D-(+) – glucose and b-D-(+) glucose are anomers.

 

64.       The electrophile,  EÅ attacks the benzene ring to generate the intermediate s-complex. Of the following, which s-complex is lowest energy?

           

64.       (1)

           

Nitrobenzene is deactivated towards electrophilic aromatic substitution at any position, but deactivation is strongest at the ortho and para positions. Reaction occurs at the meta position, but it is slower than the reaction of benzene.

 therefore   has lowest energy.

 

65.       Standard entropy of X2, Y2 and  XY3 are 60, 40 and 50 J K-1 mol-1, respectively. For the reaction, , to be at equilibrium, the temperature will be

            (1)     500 K                                         (2)     750 K

            (3)     1000 K                                       (4)     1250 K

65.       (2)

            DS for the reaction , is

 DS=50 – = – 40 J K–1

            Since at equilibrium DG = 0 so

 = 750 K

 

66.       For the following three reactions a, b and c, equilibrium constants are given:

           

           

           

            Which of the following relations is correct ?

            (1)     K2K3 = K1                                  (2)     K3 = K1K2

            (3)     K3. K23 = K12                              (4)    

66.       (2)