Detailed Solution Csir Net Dec 2015 Final

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CSIR Chemistry

Transcript of Detailed Solution Csir Net Dec 2015 Final

  • PART B (21) Using crystal field theory, identify from the

    following complex ions that shows the same eff

    (spin only) values

    (A) [CoF6]3- (B) [IrCl6]3- (C) [Fe(H2O)6]2+

    (1.) A and B (2.) B and C

    (3.) A and C (4.) A, B and C

    Soln (3)

    All the complex are d6 and contain weak field ligand

    but except Ir all form high spin complex

    *[4d and 5d metals never form high spin complex]

    Hence spin only magnetic moment will be same for

    Co(III) and Fe(II)

    (22) The order of increasing bronsted acidity for

    boron hydrides is

    (1.) B5H9 < B6H10 < B10H14

    (2.) B10H14 < B5H9 < B6H10

    (3.) B6H10 < B10H14 < B5H9

    (4.) B10H14 < B6H10< B5H9

    Soln (1)

    Acidity of boron hydride depends on the size of

    borane greater the sizes higher will be the acidity.

    This is because the negative charge, formed upon

    deprotonation, can be better delocalized over a large

    anion with many boron atoms than over a small one.

    (23) The molecule C3O2 has a linear structure. This

    compound has

    (1.) 4 and 4 bonds (2.) 3 and 2


    (3.) 2 and 3 bonds (4.) 3 and 4


    Soln (1)

    Structure of carbon sub oxide C3O2

    (24) Identify the complex ions in the sequential order

    when ferroin is used as an indicator in the titration of

    iron(II) with potassium dichromate. (phen = 1, 10-


    (1.) [Fe(phen)3]2+ and [Fe(phen)3]3+

    (2.) [Fe(phen)3]3+ and [Fe(phen)3]2+

    (3.) [Fe(CN)6]4- and [Fe(CN)6]3-

    (4.) [Fe(CN)6]3- and [Fe(CN)6]4-

    Soln (1)

    The color changes observed with redox indicators at

    the equivalence point is due to structural changes in

    the indicator molecules. The redox indicators used

    are organic molecules; they undergo structural

    changes upon being oxidized or reduced. This can he

    illustrated with ferroin indicator which is the iron(II)

    complex of the organic compound 1,10-

    phananthroline (Fig. 4.8}.

    Each of the two nitrogen atoms in 1, 10-

    phenanthroline has an unshared pair of electrons that

    can be shared with the Fe(II) ion. Three such ligand

    molecules get attached to one Fe(II) ion to form a

    blood red complex ion. This complex can he

    oxidized to the corresponding light blue

    Fe(III) complex with the same structure. Therefore, a

    sharp colour change from red to blue occurs on

    oxidation of the complex.

    [Ph3Fe]2+ [Ph3Fe]3+ + e- E0 = -1.06 V

    The indicator is prepared by mixing equivalent

    quantities of iron(II) sulphate and 1, 10-

    phananthroline. The resulting Fe(II) complex

    sulphate is called Ferroin. The Fe(III) complex

    sulphate is called ferrin. The colour changes occur at

    about -1.11 V (not at -1.06), since the color of ferroin

    is so much more intense than that of ferrin.

    (25) The ring size and the numbers of linked

    tetrahedral present in [Si6O18]12- are, respectively,

    (1.) 6 and 6 (2.) 12 and 6

    (3.) 12 and 12 (4.) 6 and 12

    Soln (1)

    Cyclic metasilicates [SiO3)n]2n- having 3, 4, 6 or S

    linked tetrahedra are known, though 3 and 6 are the

    most common. These anions are shown

    schematically in Fig. and are exemplied by the

    mineral benitoite [BaTi{Si3O9}], the synthetic

    compound [K4{Si4O8(OH)4}], and beryl

    [Be3Al2{Si6O8}] and murite [Ba,0(Ca, Mn,

    Ti]4{Si8O24}(Cl, OH, O)2].4H2O.

  • (26) The W-W bond order in [W(5-C5H5)(-

    Cl)(CO)2]2 is

    (1.) three (2.) two (3.) one (4.) zero

    Soln (4)

    To calculate bond order

    (a) Calculate TVE : a

    (b) Calculate n x 18 : b [n = number of metal]

    (c) Calculate

    2 = [number of M-M bond]

    [W(5 C5H5)(y Cl)(CO)2]2

    (a) TVE = 36 [12 + 10 + 6 + 8] = a

    (b) n x 18 = 2 x 8 = 36 = b


    2 = 3636

    2 = 0 [Ans]

    (27) The correct statement for Mn-O bond lengths in

    [Mn(H2O)6]2+ is

    (1.) All bonds are equal

    (2.) Four bonds are longer than two others

    (3.) Two bonds are longer than four others

    (4.) They are shorter than the Mn-O bond in [MnO4]-

    Soln (1)

    Mn aqua complex do not allow any distortion

    because of symmetrical d5 distribution.

    All Mn-O bond are equal.

    (28) Among the following, species expected to show

    fluxional behavior are

    (A) [NiCl4]2- (tetrahedral),

    (B) IF7 (pentagonal bipyramidal),

    (C) [CoF6]3- (octahedral),

    (D) Fe(CO)5 (trigonal, bipyramidal)

    (1.) B and C (2.) B and D

    (3.) C and D (4.) A and D

    Soln (2)

    Fe(CO)5 show fluxional behavior by process called

    berry pseudo rotation.

    And molecule IF7 shows similar mechanism called

    Bartell mechanism which exchanges the axial atoms

    with one pair of the equatorial atoms with an energy

    requirement of about 2.7 kcal/mol

    Fig : Mechanism of Berry Pseudorotation

    (29) The oxidizing power of [CrO4]2-, [MnO4]2-, and

    [FeO4]2- follows the order

    (1.) [CrO4]2- < [MnO4]2- < [FeO4]2-

    (2.) [FeO4]2- < [MnO4]2- < [CrO4]2-

    (3.) [MnO4]2- < [FeO4]2- < [CrO4]2-

    (4.) [CrO4]2- < [FeO4]2- < [MnO4]2-

    Soln (1)

    Since oxidation number is same (VI) in all three

    metals. Hence, in going from Cr to Mn to Fe, size

    decrease due to which oxidizing power increase.

    (31) The structures of XeF2 and XeO2F2 respectively


    (1.) bent, tetrahedral (2.) linear, square planar

    (3.) linear, see-saw (4.) bent, see-saw

    Soln (3)

    XeF2 : linear molecule in vapor state

    Structure of XeF2

    The three 5p electrons are promoted to higher energy

    5d sub-level.

    The 5s three 5p and one 5d orbitals hybridize to give

    ve sp3d hybrid orbitals. The four singly occupied

    orbitals are used for bond formation to two uorine

    and two oxygen atoms. The fth hybrid orbital

    contains the lone pair. The other 5d electrons of

    xenon which do not take part in the hybridization

    scheme are involved in -bond formation to two

    oxygen atoms. The structure of XeO2F2 is

    represented in Fig.

  • (32)The biological functions of the cytochrome P450

    and myoglobin are, respectively

    (1.) Oxidation of alkene and O2 storage

    (2.) O2 transport and O2 storage

    (3.) O2 storage and electron carrier

    (4.) Electron carrier and O2 transport

    Soln (1)

    The biological function of cytochrome P450 is to act

    in a mono oxygenase reaction.e.g insertion of one

    atom of oxygen into aliphatic position of organic


    RH +O2 R-O-H

    The function of Mb is more subtle than Hb. Besides

    being a simple repository for dioxygen, it also serve

    as dioxygen reserve against which organism can

    draw during increased metabolism (O2 deprivation).

    (Reference : Inorganic Chemistry , Huheey)

    (33) Spin motion of which of the following gives

    magnetic moment

    (A) Electron (B) Proton (C) Neutron

    Correct answer is

    (1.) A and B (2.) B and C

    (3.) A and C (4.) A, B and C

    Soln (4)

    Magnetic moment is induced by the spin of

    elementary particles (electron and quarks in the

    proton and neutron of atomic nuclei).

    * Neutron is electrically neutral but a non-zero

    magnetic moment because of its internal quark


    (Reference : P.W. Atkins, Quanta: A handbook of

    concepts. Oxford University)

    (34) The metallic radii are abnormally high for which

    of the following pairs?

    (1.) Eu, Yb (2.) Sm, Tm

    (3.) Gd, Lu (4.) Nd, Ho

    Soln (1)

    The lanthanide and actinide elements are all metallic

    and are formally members of Group 3 of the Periodic

    Table. They all form 3+ ions in their compounds and

    in aqueous solution, with few exceptions. The

    actinide elements form a larger range of oxidation

    states than the lanthanides, but the 3+ ions are used

    in this section for comparison purposes. Figure

    shows the lanthanides metallic and 3+ ionic radii.

    There is an almost regular decrease in metallic radius

    along the lanthanide series, with discontinuities at Eu

    and Yb. Most of the Lanthanide metals contribute

    three electrons to bands of molecular orbitals

    constructed from the 5d and 6s atomic orbitals, their

    other valency shell

    electrons remaining as 4fn congurations that do not

    interact with those on neighboring atoms. In the

    cases of Eu and Yb, the atoms only contribute two

    electrons to the 5d/6s bands and retain the

    congurations 4f7and 4F14, respectively, and so

    maximize their exchange energies of stabilization.

    The contributions of only two electrons each to the

    bonding bands makes the Eu and Yb metallic radii

    larger than those of their neighbors that contribute

    three electrons.

    The 3+ ions have regular 4f orbital llings, and there

    is an almost linear decrease in their ionic radii along

    the series. This is known as the Lanthanide

    contraction, and has consequences for the transition


    (35) Correct statement for coulometry is

    (1.) it is based on faradays law of electrolysis

    (2.) it is a type of voltammetry

    (3.) it is based on Ohms law

    (4.) it uses ion selective electrode

    Soln (1)

    Coulometry is an analytical in which we measure

    amount of amount of water transferred during an

    electrolysis reaction by measuring the amount of

    electricity consumed/produced.

    (36) Deoxy Hemocyanin is

    (1.) heme protein and paramagnetic

    (2.) Colorless and diamagnetic

    (3.) O2 transport and paramagnetic

    (4.) Blue colored and diamagnetic

    Soln (2)

  • Hemocyanin is blue pigment which contains Cu in (I)

    oxidation state.

    Since it is in deoxy form it is colorless and also

    diamagnetic (d10) configuration.

    (37) The major product formed in the following

    reaction is

    Soln (2)

    Step 1 : Removal of nitrogen to form carbene

    Step 2 : attack of Oxygen lone pair to form 5

    membered cyclic adduct

    Step 3 : 1,2 alkyl shift to form rearranged product

    (38) In the mass spectrum of 1, 2-dichloroeth-ane,

    approximate ratio of peaks at m/z values 98, 100, 102

    will be

    (1.) 3 : 1 : 1 (2.) 9 : 6 : 1

    (3.) 1 : 1 :2 (4.) 1 : 2 : 1

    Soln (2)

    When chlorine is present, the M + 2 peak becomes

    very significant. The heavy isotope these elements is

    two mass units heavier than the lighter isotope. The

    natural abundance of 37Cl is 32.5% that of 35Cl.

    When Chlorine is present, the M + 2 peak becomes

    quite intense. If a compound contains two chlorine

    atoms, a distinct M + 4 peak, as well as an intense M

    + 2 peak, should be observed.

    The M : M+2 : M+4 peaks for two chlorine atom is

    in the ratio of 100 : 65.3 : 10.6 which on

    simplification gives 9 : 6 : 1

    (39) The correct statement about the following

    reaction is that

    (1.) A = and the reaction proceeds

    through carbine intermediate

    (2.) A = and the reaction proceeds

    through nitrene intermediate

    (3.) A = and the reaction proceeds

    through Norrish type II path

    (4.) A = and the reaction proceeds through

    Norrish type I path

    Soln (1)

    This reaction is an example of wolff rearrangement

    in which cyclic diazoketone , on rearrangement leads

    to ring contraction.

    The R group migrates with retention of configuration

    and the resulting ketene can be trapped by methanol

    to give the corresponding ester.

    (40) The major product formed in the following

    reaction is

    Soln (3)

  • Step 1 : The rst step is an acetal exchange in which

    the allylic alcohol displaces methanol.

    Step 2 : . A second molecule of methanol is now lost

    in an acid-catalysed elimination reaction to give the

    vinyl group.

    Step 3 : It is simple example of [3,3] sigmatropic

    known as Claisen rearrangement

    (41) The major products A and B in the following

    reactions are

    Soln (3)

    Heating dimethylmenthyl amine oxide gave a

    mixture of 2- and 3menthene, whereas isomeric neo

    menthylamine oxide gave only 2menthene. This is

    an example of pyrolytic syn elimination

    (42) The number of Chemical shift non equivalent

    protons expected 1H NMR spectrum of -pinene is


    1. 7 2. 8 3. 9 4. 10

    Soln (4)

    All the protons are different and hence give 10

    distinct signals in 1H NMR.

    (43) Correctly matched structure and carbonyl

    stretching frequency set is

    Column A Column B


    X. 1750 cm-1


    Y. 1770 cm-1


    Z. 1800 cm-1

    (1.) P-Y, Q-Z, R-X (2.) P-Y, Q-X, R-Z

    (3.) P-Z, Q-Y, R-X (4.) P-X, Q-Z, R-Y

    Soln (1)

    -lactone has very peculiar IR stretching frequency at

    around ~ 1770 cm-1.

    Next, Conjugation lowers IR stretching frequency.

    Therefore structure R must correspond to ~ 1750 cm-


    Hence , the correct option is (1)

    (44) the major product formed by photochemical

    reaction of (2E, 4Z, 6E)-decatriene is

    Soln (1)

  • This is an example of electrocyclic ring closing

    reaction of 4n + 2 electron system by Conrotatory

    photochemical mode.

    (45) The major product formed in the following

    reaction is

    Soln (3)

    This is an example of favorskii rearrangement of


    (46) The major product formed in the following

    reaction is

    Soln (2)

    The formation of eight-membered rings by radical

    cyclization has been found to occur using -acyl

    radical species. The reaction occurs at a rate that is

    faster remarkably even than 5-exo-trig cyclization.

    Hence, the eight-membered lactone 92 was formed,

    rather than a ve-, six- or seven-membered lactone,

    on treatment

    Of the bromide with tributyltin hydride

    (47) D-Mannose upon fluxing in acetone with CuSO4

    and H2SO4 gives


    Soln (3)

    This is an example of protection of 1,2 diol by

    acetone in the presence of acid. For the protection of

    1,2 diol the two hydroxyl group must be in cis-

    orientation. Hence only anomeric OH group remain

    unaffected rest all OH will form 5-membered cyclic

    acetal as thermodynamic product.

    (48) The structure of the compound that matches the 1H NMR data given below is 1H NMR (DMSO-D6): 7.75 (dd, J = 8.8, 2.4 Hz,

    1H), 7.58 (d J = 2.4 Hz, 1H), 6.70 (d J = 8.8 Hz, 1H),

    6.50 (broad s, 2H), 3.80 (s, 3H).

    Soln (2)

  • Steps to Analyze NMR Spectra

    a. First we know that delta value of aromatic protons

    comes at 7.2 . So the first delta value 7.75 indicates

    the proton is deshielded . so it must be adjacent to

    EWG. Further since it is doublet of doublet hence

    there must be meta and ortho coupling based on their

    J values.

    So 1 and 4th option are straight away ruled out since

    the above condition is not followed.

    Further in 3rd , all the three aromatic protons will give

    doublet of doublet which is clearly not the case.

    (49) The major product formed in the following

    reaction is

    Soln (1)

    This is an example of intramolecular Diels alder


    (50) The major product formed in the following

    reaction is

    Soln (4)

    (51) The major product formed in the following

    reaction is

    Soln (4)

    This is an example of Swern Oxidation in which last

    step is the intermolecular SN2 displacement by

    chloride ion in the absence of base

    (52) Identify two enantiomers among the following


    (1.) A and B (2.) A and C

  • (3.) B and D (4.) C and D

    Soln (4)

    A and B are homomers ; A and C are diastereomers

    B and D are diastereomers ; C and D are enantiomer

    (53) The correct expression for the product

    ((). ()) [Mn and Mw are the number-average

    and weight average molar masses, respectively, of a

    polymer] is

    (1.) N-1 i Ni Mi (2.) N-1 i Ni Mi2

    (3.) N / NM (4.) N / NM2

    Soln (2)

    Number average and weight average molar mass is

    given by the expression





    x 2

    = 12

    (54) The concentration of a reactant K varies with

    time for two different reactions as shown in the

    following plots:

    The order of these two reactions I and II,

    respectively, are

    (1.) zero and one (2.) one and zero

    (3.) zero and two (4.) two and zero

    Soln (3)

    For the zero order reaction

    x = kT

    (a x) = -RT (linear with negative slope)

    For 2nd order reaction 1



    [0]+ (linear with intercept



    (55) .) sp hybrid orbitals are of the form C12s + C22pz

    (2s and 2pz are normalized individually). The

    coefficients of the normalized form of the above sp

    hybrid orbitals are

    (1.) C1 = 1

    2, C2 =


    2 (2.) C1 =


    2, C2 =



    (3.) C1 = 1

    2, C2 =


    2 (4.) C1 =


    2, C2 =



    Soln (1)

    Condition for normalization = 1

    = C12s + C22pz

    12(2)2 + 2

    2(2)2 + 212 2 2 = 1

    The last two terms cancel out each other because

    they are orthogonal

    Also |2|2 = 1 |2|

    2= 1 [condition of


    C12 + C22 = 1

    Since Sp orbital have equal contribution toward

    atomic orbital C1 = C2

    2C12 = 1 C1 = 1


    C2 = 1


    (56) For a single cubic crystal lattice, the angle

    between the [2 0 1] plane and the xy plane is

    (1.) less than 300 (2.) between 300 and


    (3.) between 450 and 600 (4.) greater than 600

    Soln (4)

    The expression for finding the angle between the two

    plane is given by

    =12 + 12 + 12

    12 + 1

    2 + 12 . 2

    2 + 22 + 2


    For the XY Plane , miller indices are [001]

    1 = 2 1 = 0 1 = 1 ; 2 = 0 , 2 = 0 , 2 = 1

    On solving

    =2.0 + 0.0 + 1.1

    22 + 02 + 12 . 02 + 02 + 12




    2.23= 0.44

    Therefore = cos-1 (0.44) = 63.89

    (57) The concentration of a MgSO4 solution having

    the same ionic strength as that of a 0.1 M Na2SO4

    solution is

    (1.) 0.05 M (2.) 0.067 M

    (3.) 0.075 M (4.) 0.133 M

    Soln (3)

    Ionic strength of 0.1M Na2SO4




  • =1

    2[0.2 x 12 + 0.1 x 22]

    = 0.3

    MgSO4 will have ionic strength equal to 0.3

    0.3 = 1

    2[m x 22 + m x 22]

    0.3 = 4m

    m = 0.3/4 = 0.075

    (58) H of a reaction is equal to slope of the plot of

    (1.) G versus (1/T) (2.) G versus T

    (3.) (G/T) versus T (4.) (G/T) versus (1/T)

    Soln (1)

    H = U + PV .. (1st Law of thermodynamics)

    H = U + ngRT

    U - H = -ngRT

    Fe2O3(s) + 3C(s) 2Fe(s) + 3(CO)

    Ans = -(3-0)= -3RT

    (59) Two different non-zero operators and (A

    B) satisfy the relation

    (A + B) (A - B) = A2 - B2, when

    (1.) AB = A2 and BA = B2

    (2.) AB + BA = 0

    (3.) A and B are arbitrary

    (4.) AB - BA = 0

    Soln (4)

    Expanding and simplifying the expression gives the

    option (4) as answer

    (A + B) (A B) = A2 AB + BA B2

    = A2 B2 (only when AB BA = 0)

    (60) For the following reaction,


    is given by

    (1.)1[] 1[]2 22[]

    (2.) 21[] 21[]2 2[]

    (3.) 1




    2 2[]

    (4.) 21[] 21[]1 2 2[]

    Soln (2)

    Rate of formation of B

    is given by



    = 1[] 1[]



    = 21[] 21[]



    = 2[]

    overall rate is given by

    = 22[] 21[]

    2 1[]

    (61) The standard deviation of speed (c) for

    Maxwells distribution satisfies the relation

    (1.) c T (2.) c

    (3.) c 1/T (4.) c 1/

    Soln (2)

    Standard deviation for the velocity of Maxwell

    distribution is given by

    = 3

    (62) In Kohlrausch law m = 0m - , 0m and

    (1.) depends only on stoichiometry

    (2.) depends only on specific identity of the


    (3.) are independent on specific identity of the


    (4.) are mainly dependent on specific identity of the

    electrolyte and stoichiometry, respectively

    Soln (1)

    The justification of Kohlrauschs law on theoretical

    grounds cannot be obtained within the framework of

    a macroscopic description of conduction. It requires

    an intimate view of ions in motion. A clue to the type

    of theory required emerges from the empirical

    findings by Kohlrausch: (l) the c1/2 dependence and

    (2) the intercepts 0 and slopes A of the versus c1/2

    curves depend not so much on the particular

    electrolyte (whether it is KCI or NaCl) as on the type

    of electrolyte (whether it is a l:l or 2:2 electrolyte)

    (63) If the reduced mass of the diatomic molecule is

    doubled without changing its force constant, the

    vibrational frequency of the molecule will be

    (1.) 2 times the original frequency

    (2.) 2 times the original frequency

    (3.) twice the original frequency

    (4.) unchanged

    Soln (2)

  • Vibrational frequency of molecule is given by




    if reduced mass is doubled, then will become 1


    times the original frequency.

    (64) The degeneracy of an excited state of a particle

    in 3-dimensional cubic box with energy three times

    its ground state energy is

    (1.) 3 (2.) 2 (3.) 1 (4.)


    Soln (1)

    For particle in 3D box


    22[12 + 2

    2 + 32]

    In the ground state n1 = n2 = n3 = 1



    Energy when it wil be 3 times to its ground state will

    be 3 x32

    22 =



    It is possible when m1 = n2 = 2, n3 = 1

    Total degeneracy is 3 (221), (122) and (212)

    (65) If the pressure p system is greater than the p

    (surroundings), then

    (1.) work is done on the system by the surroundings.

    (2.) work is done on the surroundings by the system.

    (3.) work done on the system by the surroundings is

    equal to the work done on the surroundings by the


    (4.) internal energy of the system increases

    Soln (2)

    When the pressure of the system is greater than

    pressure of surrounding, expansion will take place.

    Hence, during expansion work is done by the system

    on the surrounding

    (66) H of a reaction is equal to slope of the plot of

    (1.) G versus (1/T) (2.) G versus T

    (3.) (G/T) versus T (4.) (G/T) versus (1/T)

    Soln (4)

    G = H TS

    Dividing the whole equation be T brought out


    Equation of straight line

    y = mx + c

    y =

    x =


    Slope = H

    Intercept = -S

    (67) The correct statement among the following is

    (1.) N2 has higher bond other than N2+ and hence has

    larger bond length compared to N2+

    (2.) N2+ has higher bond other than N2 and hence has

    larger bond length compared to N2

    (3.) N2 has higher bond other than N2+ and hence has

    higher dissociation energy compared to N2+

    (4.) N2 has lower bond other than N2+ and hence has

    lower dissociation energy compared to N2+ energy

    Soln (3)

    N2 = KK (2S)2 *(2S)2 (2px)2 (2py)2 6(2pz)2

    B.O = 82

    2 = 3 (most stable, diamagnetic)

    N2+ = KK 22 2

    2 22 2

    2 621

    B.O = 72

    2 = 2.5 (less stable, paramagnetic)

    (68) The correct form for the simple Langmuir

    isotherm is

    (1.) = Kp (2.) = (Kp)1/2

    (3.) = Kp / (1 + Kp) (4.) = (1 + Kp) / Kp

    Soln (3)

    The concentration of vacant sites on the surface of an

    adsorbent will be directly proportional to the fraction

    of the surface that remains uncovered and hence will

    be directly proportional to the factor (1 - ). Thus we

    can write,

    Rate of absorption p(l )

    i.e. Rate of adsorption = ka p(l - )

    Now since at equilibrium

    Rate of absorption = rate of desorption,

    it follows that

    ka (1 - )p = kd

    i.e. =


    ( )

    1+( )=



    (69) The correct match of the compound in column A

    with the description in column B is

    Column A Column B



    Oil of








    (1.) P-Y, Q-Z, R-X (2.) P-Z, Q-X, R-Y

    (3.) P-Z, Q-Y, R-X (4.) P-X, Q-Z, R-Y

    Soln (2)

    Aspirin : acetyl salicylic acid

  • Ibuprofen : ISO butyl phenyl propanoic acid

    Oil of winter green : methyl salicylate

    (70) The formation constant for the complexation of

    M+ (M = Li, Na, K and Cs) with cryptant, C222

    follows the order

    (1.) Li+ < Cs+ < Na+ < K+

    (2.) Li+ < Na+ < K+ < Cs+

    (3.) K+ < Cs+ < Li+ < Na+

    (4.) Cs+ < K+ < Li+ < Na+

    Soln (2)

    The ability of cryptant to trap an alkali metal cation

    depends on size of both cage and metal ion, the better

    the math between these; the more effectively the ions

    can be trapped.

    PART C

    (71) Identify radioactive capture from the following

    nuclear reactions

    (1.) 9Be(, n) 8Be (2.) 23Na(n, ) 24Na

    (3.) 63Cu (p, p 3n 9) 24Na (4.) 107Ag (n, n) 107Ag

    Soln (2)

    Radioactive capture reactions are described as

    follows Parent Nucleus, Projectile Daughter

    nucleus, ejectile 23Na11 + 0n6 11Na24 + 0n6 (y rays)

    Rest all examples do not follow this trend.

    (72) Match the metalloproteins in column A with its

    biological function and metal center in column B

    Column A Column B

    (a.) Hemoglobin (i.) Electron carrier

    and iron

    (b.) Cytochrome b (ii.) Electron carrier

    and copper

    (c.) Vitamin B12 (iii.) O2 transport and


    (d.) Hemocyanin (iv.) Group transfer

    reactions and


    (v.) O2 storage and


    (vi.) O2 transport and


    The correct match is

    (1.) (a.)-(vi); (b.)-(i); (c.)-(iv) and (d.)-(iii)

    (2.) (a.)-(v); (b.)-(i); (c.)-(iv) and (d.)-(iii)

    (3.) (a.)-(iv); (b.)-(v); (c.)-(i) and (d.)-(ii)

    (4.) (b.)-(v); (b.)-(vi); (c.)-(ii) and (d.)-(iv

    Soln (1) Self explanatory

    (73) [(3-C3H5)Mn(CO)4] shows fluxional behaviour.

    The 1H NMR spectrum of this compound when it is

    in the non-fluxional state shows

    (1.) one signal

    (2.) two signals in the intensity ratio of 4 : 1

    (3.) three signals in the intensity ratio of 2 : 2 : 1

    (4.) five signals of equal intensity

    Soln (3)

    Non fluxional behavior can be observed if NMR is

    carried out at low temperature. For this complex, 3

    signals are observed

    one for H3

    two for H1 and H4

    third for H2 and H5

    In the intensity ratio 2 : 2 : 1

    (74) Reaction of [Mn2(CO)10] with I2 results in A

    without loss of CO. Compound A, on heating to

    1200C loses a CO ligand to give B, which does not

    have a Mn-Mn bond. Compound B reacts with

    pyridine to give 2 equivalents of C. Compounds A, B

    and C from the following respectively, are

  • (1.) II, V and IV (2.) II, III and IV

    (3.) V, III and IV (4.) II, V and III

    Soln (1)

    (a) Dimanganese decacarbonyl reacts with halogens

    to form carbonyl halides.

    Mn2CO10 + l2 2Mn(Co)5 I

    (75) The final product of the reaction of carbonyl

    metalates [V(CO)6]- and [Co(CO)4]- with H3PO4,

    respectively, are

    (1.) V(CO)6 and HCo(CO)4

    (2.) HV(CO)6 and Co2(CO)8

    (3.) [H2V(CO)6]+ and HCo(CO)4

    (4.) V(CO)6 and Co2(CO)8

    Soln (4)

    [V(CO)6]- + H3PO4 [H V(CO)6] + V(CO)6

    2[Co(CO)4]- 2+ 2[HCo(CO)4]

    2 Co2(CO)8

    (76) For, uranocene, the correct statement(s) is/are:

    (1.) Oxidation use of uranium is +4.

    (2.) it has cyclocotatetraenide ligands

    (3.) it is a bend sandwich compound

    (4.) it has -2 charge.

    Correct answer is

    (1.) A and B (2.) B and C

    (3.) A and D (4.) B only

    Soln (1)

    Uranium compound was the 1st metallocene of Cot2-

    to be synthesized

    UCl2- + 2Cot2- U(Cot)2

    From the given reaction, it is concluded that among

    the following option (A) and (B) are correct.

    (77) The number of lone pair(s) of electrons on the

    central atom in [BrF4]-, XeF6 and [SbCl6]3- are,


    (1.) 2, 0 and 1 (2.) 1, 0 and 0

    (3.) 2, 1 and 1 (4.) 2, 1 and 0

    Soln (3)

    Formula to calculate Bond Pair and Lone Pair

    A. calculate TVE (total valence electron)

    B. Divide TVE by 8 if 8 < TVE < 56

    C. The quotient will give number of bond pair

    D. while dividing the remainder by 2 will give lone

    pair on central atom

    BrF4- : TVE = 5 x 7 +1 = 36

    8 = 4 +


    2 = 4 + 2 (lp)

    XeF6- : 4 + 42 = 50

    8 = 6 +


    2 = 6 + 1 (lp)

    SbCl63- : 5 + 6 x 7 + 3 = 50

    8 = 6 +


    2 = 6 + 1 (lp)

    (78) Consider the following reaction:

    N3P3Cl6 + 6 HNMe2 N3P3Cl3(NMe2)3 +



    The number of possible isomers for [A] is

    (1.) 4 (2.) 3 (3.) 2 (4.)


    Soln (2)

    The cyclophosphazenes, first discovered by Liebig in

    1834, have aroused great interest and, especially

    within the last twenty years, have been intensively

    investigated. Much has been discovered about their

    chemical and physical properties.

    Two opposite points of view were put forward, one

    based on a completely delocalized molecular orbital

    covering the whole of the cycle , the other based on

    the three-centered P-N-P islands with little or no

    interaction between them.

    (79) Three electronic transitions at 14900, 22700 and

    34400 cm-1 are observed in the absorption spectrum

    of [CrF6]3- and the corresponding transition are

    (1.) 7800 and 4A2g 4T2g

    (2.) 14900 and 4A2g 4T2g

    (3.) 14900 and 4T2g 4T1g(F)

    (4.) 7800 and 4T2g 4T1g(F)

    Soln (2)

    C6H63-: d3 electronic configuration orgel diagram for

    d3 electronic transition are-

  • 4A2g 4A2g corresponds to the transition of lowest


    (80) The calibration curve in spectrofluorimetric

    analysis becomes non-linear when

    (1.) Molecular weight of analyte is high

    (2.) Intensity of light source is high

    (3.) Concentration of analyte is high

    (4.) Molar absorptivity of analyte is high

    Soln (3)

    According to Lambert-Beer law

    A = C. l

    When A is plotted v/s conc. Curve should nearly be

    of straight line. However, only limitation is that conc.

    Of analytes should be low

    (81) The reaction of BCl3 awith NH4Cl gives product

    A which upon reduction by NaBH4 gives product B.

    product B upon reacting with HCl affords compound

    C, which is

    (1.) Cl3B3N3H9 (2.) [ClBNH]3

    (3.) [HBNH]3 (4.) (ClH)3B3N3(ClH)3

    Soln (1)

    Ammonium chloride reacts with Boron chloride to

    produce trichloridoborazine and hydrogen chloride.

    This reaction takes place at temperature 140 150 0C. catalyst C6H6, C6H5Cl.

    (82) The approximate positions of vco band (cm-1) in

    the solid-state infrared spectrum and the Fe-Fe bond

    order in [Fe(5 - C5H5)( CO)(CO)]2 (non-centro-

    symmetric) respectively, are

    (1.) (2020, 1980, 1800) and one

    (2.) (2020, 1980, 1800) and two

    (3.) (2020, 1980) and one

    (4.) (2143) and one

    Soln (1)

    Three bands are expected in CO in the complex

    [Fe(5-C5H5)(-Co)(CO)]2 since it is non Centro-


    One band for bridging CO around 1800

    Two bands (one symm, one asymm) around 2020

    (asymm) and 1980 (symm)

    To calculate bond order

    TV E = 34

    nx18 36 (there are 2 Fe atoms)

    M.M bond = 36 34

    2 = 1

    (83) Protonated form of ZSM-5 catalyzes the

    reaction of ethene with benzene to produce

    ethylbenzene. The correct statement for this catalyst

    process is

    (1.) alkyl carbocation is formed

    (2.) carbanion is formed

    (3.) benzene is converted to (C6H5)+ group

    (4.) vinyl radical is formed

    Soln (1)

    ZSM-5 is an acid catalyst and is typically used for

    Alkylation of aromatic He

    (84) [MnO4]- is deep purple in color whereas [ReO4]

    is colorless. This is due to greater energy require for

    (1.) d-d transition in the Re compound compared to

    the Mn compound

    (2.) d-d transition in the Mn compound compared to

    the Re compound

    (3.) charge transfer from O to Re compared to O to


    (4.) charge transfer from O to Mn compared to O to


    Soln (3)

    The 5d orbitals of Re are higher in energy than the 3d

    orbitals of Mn, so an LMCT excitation requires more

    energy for ReO-. In addition, since the molecular

    orbitals derived primarily from the 3d orbitals of

    MnO4- are lower in energy than the corresponding

    MOs of ReO4-, MnO4- is better able to accept

    electrons; it is a better oxidizing agent.

    (85) The resonance Raman stretching frequency (vo-o,

    in cm-1) of O2 is 1580. The vo-o for O2 in bound oxy-

    hemoglobin is close to

    (1.) 1600 (2.) 1900

    (3.) 800 (4.) 1100

    Soln (4)

    It is known fact that stretching frequency of any

    molecule in the Free State is always higher that the

    same in bound state

  • Here it is given that o-o is 1580 of free O2. Hence in

    bound Hb it must be less than that, so correct answer

    is 1100 (more precisely 1180 cm-1)

    (86) Pick the correct statements about Atomic

    Absorption Spectrometry (AAS) from the following

    (1.) Hg lamp is not suitable source for AAS

    (2.) Graphite furnace is the best atomizer for AAS

    (3.) Non-metals cannot be determined with AAS

    (4.) AAS is better than ICP-AES for simultaneous

    determination of metal ions

    Correct answer is

    (1.) A, B and C (2.) B, C and D

    (3.) C, D and A (4.) D, A and B

    Soln (1)

    Common source used in AAS is HCl (Hallow

    Cathode Lamp) and EDL (Electrodes Discharge


    Best atomizer is graphite tube

    Since it is based on the principle of flame

    detection, generally non metal cannot be detected

    ICP-AES is better technique that trivial AAS.

    (87) Aqueous Cr2+ effects one electron reduction of

    [Co(NH3)5Cl]2+ giving compound Y. compound Y

    undergoes rapid hydrolysis. Y is,

    (1.) [Co(NH3)5]2+ (2.)


    (3.) [Co(NH3)4(OH)2] (4.) [Cr(H2O)5Cl]2+

    Soln (1)

    The reduction of cobalt(III) (in [Co(NH3)5Cl]2+) by

    chromium(II) (in [Cr(H2O)6]2+) and was specically

    chosen because (I) both CO(III) and Cr(III) form

    inert complexes and (2) the complexes of Co(II) and

    Cr(III) are labile.

    Under these circumstances the chlorine atom, while

    remaining rmly attached to the inert Co(III) ion, can

    displace a water molecule from the labile Cr(III)

    complex to form a bridged intermediate:

    [Co(NH3)3Cl]2+ + [Cr(H2O)6]2+ [(H3N)5Co-Cl-

    Cr(OH2)5]4+ + H2O

    The redox reaction now takes place within this

    dinuclear complex with formation of reduced Co(II)

    and oxidized Cr(III). The latter species forms an inert

    chloroaqua complex, but the cobalt(II) is labile, so

    the intermediate dissociates with the chlorine atom

    remaining with the chromium:

    [(H3N)5Co-Cl-Cr(OH2)5]4+ [(H3N)5Co]2+ +


    (88) Which of the following statements are TRUE for

    the lanthanides?

    (1.) The observed magnetic moment of Eu3+ at room

    temperature is higher than that calculated from spin-

    orbit coupling

    (2.) Lanthanide oxides are predominantly acidic in


    (3.) The stability of Sm(II) is due to its half-filled


    (4.) Lanthanide(III) ions can be separated by ion

    exchange chromatography

    Correct answer is

    (1.) A and D (2.) A and B

    (3.) A and C (4.) B and C

    Soln (1)

    Oxides of lanthanides are generally basic in nature.

    Sm: [Xe]6s24f6

    Sm: [Xe]6s24f6 Sm(II) is not stable because of half-

    filled subshell

    Rest all statement are correct above Lanthanides (for

    detailed description on above statement of

    lanthanides, refer, inorganic chemistry, Shriver &


    (89) The correct statement about the substitution

    reaction of [Co(CN)5(Cl)]3- with OH- to give

    [Co(CN)5(OH)]3- is,

    (1.) it obeys first order kinetics

    (2.) its rate is proportional to the concentration of

    both the reactants

    (3.) it follows the SN1CB mechanism

    (4.) its rate is dependent only on the concentration of


    Soln (1)

    Complexes such as [Co(CN)5Cl]3- [Co(py)4Cl2]+

    would not be expected typical base hydrolysis and

    reaction proceeds slowly without dependence of OH-


    (Reference : Inorganic Chemistry , Huheey)

    (90) Mossbauer spectrum of a metal complex gives

    information about

    (A) oxidation state and spin state of metal

    (B) types of ligands coordinated of metal

    (C) nuclear spin state of metal

    (D) geometry of metal

    Correct answer is

    (1.) A and C (2.) B and C

    (3.) A, B and D (4.) B and D

    Soln (3)

    Mssbauer spectroscopy is unique in its sensitivity to

    subtle changes in the chemical environment of the

    nucleus including oxidation state changes, the effect

  • of different ligands on a particular atom, and the

    magnetic environment of the sample.

    (91) Using wades rules predict the structure type of


    (1.) nido (2.) closo

    (3.) arachno (4.) hypho

    Soln (2)

    Wades rule

    1 carbon is equal to BH fragment as both are isolobal

    C2B5H11 = B7H9 B7H72- (hence closo)

    (92) Among the following complexes

    A. [Co(ox)3]3-, B. trans-[CoCl2(en)2]+, C.

    [Cr(EDTA)]- the chiral one(s) is/are

    (1.) A and B (2.) C and B

    (3.) C only (4.) A and C

    Soln (4)

    The essential condition to show the chirality is the

    lack of any symmetry element and it show have


    Chiral complex is able to show handedness

    properly (right or left handedness) by and


    However the complex trans-[CoCl2(en)2]+

    contains plane of symmetry and hence are achiral

    (93) Choose the correct statements about Tanabe-

    Sugano diagrams:

    (1.) E/B is plotted against 0/B.

    (2.) The zero energy is taken as that of the lowest


    (3.) Terms of the same symmetry cross each other.

    (4.) Two terms of the same symmetry upon increase

    of ligand field strength bend apart from each other.

    Correct answer is

    (1.) A and B (2.) A and C

    (3.) A, B and D (4.) A, B, C and D

    Soln (3)


    Reference: Inorganic Chemistry, James Huheey

    (94) The intermediate and final major product of

    photolysis of Z

    From the following:


    (1.) A and D (2.) B and D

    (3.) B and C (4.) A and C

    Soln (2)

    In the following photolysis reaction, first the removal

    of CO takes place to form 16e- complex and then the

    change in hapticity from 1 to 3. Hence correct

    option is (2)

    (95) The number of valence electrons provided by

    [Ru(CO)3] fragment towards cluster bonding is

    (1.) 1 (2.) 14 (3.) 6 (4.)


    Soln (4)

    For transition metal, number of electrons available

    per building block for cluster bonding

    L = d + y2 12

    d = number of valence shell electron

    y = number of ligands

    L = number of e- contributed by each ligand


    Ru(d8) = 8 + 6 12

    = 2

    (96) The major products A and B in the following

    reaction sequence are

    Soln (1)

    A. The first step is the synthesis of 9-

    Borabicyclo[3.3.1] nonane

  • B. Second is the Boronic Suzuki coupling which is

    highly stereospecific in which geometry of the

    reactant is faithfully reproduced in the product.

    (97) The major product A and B in the following

    reaction sequence are

    Soln (3)

    Mechanism :

    (98) The major product A and B in the following

    reaction sequence are

    Soln (3)


    (99) The major product formed in the following

    reaction is

    Soln (2)

    It is an example of light induced photo-isomerization

    of benzene via Dewar benzene intermediate

    (100) The correct reagent combination to effect the

    following transformation is

    (1.) A = NaBH4, BF3.OEt2; B = MeMgBr (2.5

    equiv.), THF then H3O+

    (2.) A = BH3.THF; B = MeLi (2.5 equiv.), THF then


    (3.) A = BH3.THF; B = (i) aq. NaOH then H3O+ ,

    (ii) MeLi (2.5 equiv.), THF then H3O+

    (4.) A = (i) Me3Al, MeNHOMe, (ii) MeMgBr, THF

    then H3O+; B = LiAlH4, THF

  • Soln (3)

    Method of Analysis :

    A. NaBH4 cannot be the reagent of choice since it

    generally do not reduces amides.

    B. reagent 2 and 4 results in the formation of tertiary

    alchohol as resulting ketone is more reactive than

    ester itself.

    C. So the proper strategy should be (i) first reduce

    the amide with borane (ii) then , convert ester to acid

    which is usually less reactive and then perform

    alkylation which will stop at ketone oxidation state


    (101) The major product A and B in the following

    reaction sequence are

    Soln (2)

    This is an example of Mannich reaction which will

    take place prefentially at 2 and 5 position.

    (102) The major product A and B in the following

    reaction sequence are

    Soln (3)

    a. First step is the metathesis ring closing reaction by

    grubbs catalyst.

    b. Second is the reduction of amide be lithium

    aluminium hydride.

    (103) A concerted [1,3]-sigmatropic rearrangement

    took place in the reaction shown below. The structure

    of the resulting product is

    Soln (3)

    This is an example of 1,3 carbon shift and according

    to woodward Hoffman rule it should be inversion.

    (104) The major product formed in the following

    reaction is

    Soln (3)

    The ease of the anionic oxy-Cope rearrangement and

    its high level of stereo-control make this reaction a

    popular and valuable synthetic method. For example,

    stereo-controlled rearrangement of the potassium salt

    of the 3-hydroxy-1, 5-diene to give the

    cyclohexanone. The diastereoselectivity across the

    new carbon-carbon single bond reects the

    preference for a chair-shaped transition state with the

    methoxy group in the pseudo equatorial position.

  • (105) The major product A and B in the following

    reaction sequence are

    Soln (1)

    The diol 1 must first be protected/activated and this

    can be achieved by selective acylation (or

    mesylation) of the less hindered primary alcohol

    group. For example, treatment with a carboxylic acid,

    dicyclohexylcarbodiimide (DCC) and 4-

    dimethylaminopyridine gives the required ester.

    Oxidation of the secondary alcohol can be achieved

    by using one of a number of oxidizing agents such as

    pyridinium dichromate (PDC). Formation of the

    ketone was followed by _-elimination on alumina

    to give the unsaturated ketone

    (106) The major product formed in the following

    reaction sequence is

    Soln (4)

    Boronic esters RB(OR')2 react with

    methoxy(phenylthio)methyllithium LiCH(OMe)SPh

    to give salts, which, after treatment with HgCl2 and

    then H2O2, yield aldehydes. This synthesis has been

    made enantioselective, with high ee values (> 99%),

    by the use of an optically pure boronic ester.

    (107) The major product of the following reaction is

    Soln (1)

    This is an example of acid-catalyzed diels Alder


    (108) The major products A and B in the following

    reaction sequence are

    Soln (1)

    (109) The major product A and B in the following

    reaction sequences are

  • (1.) A = D-threose; B = D-glucose

    (2.) A = D erythrose; B = D-glucose + D-mannose

    (3.) A = D-threose; B = D-glucose + D-mannose

    (4.) A = D-tartaric acid; B = D-glucose

    Soln (2)

    (110) The major product A and B in the following

    reaction sequence are

    Soln (2)

    This is an example of wolf-kischner reduction

    Reference : Organic Mechanisms Reactions,

    Stereochemistry and Synthesis , Reinhard Bruckner

    (111) The major product of the following reaction is

    Soln (3)

    This is an example of chelation control felkin-Ahn


    (112) The following transformation involves


    (1.) Claisen rearrangement cope rearrangement

    ene reaction

    (2.) Cope rearrangement - Claisen rearrangement

    ene reaction

    (3.) Cope rearrangement ene reaction Claisen


    (4.) ene reaction Claisen rearrangement - Cope


    Soln (2)

    Mechanism :

    (113) The mechanism and the product formed in the

    following reaction, respectively, are

  • Soln (2)

    Mechanism :

    Since Alkyl halide is tertiary in nature hence only

    possibility of SN1 mechanism.

    Moreover , Since in SN1 there is formation of

    carbocation hence racemization take place hence

    stereochemistry is not defined.

    (Reference : Organic Chemistry , clayden)

    (114)The major product A and B in the following

    reaction sequence are

    Soln (4)

    Mechanism :

    (115) The major product A and B in the following

    reaction sequence are

    Soln (4)

    This is an example of wittig reaction

    (116) The major product formed in the following

    reaction is

    Soln (4)

  • This is an example of prins-pinacol cyclization

    Subjection of ketal 1 to a Lewis acid gives the

    corresponding oxacarbenium 2, which is poised to

    undergo a Prins cyclization to furnish intermediate 3.

    The empty p orbital in 3 (at C10) is aligned

    periplanar to the C1-C2 bond thus allowing this bond

    to migrate to form bicyclic ketone bearing a

    quaternary center at C10.

    (117) The major product formed in the following

    reaction is

    Soln (2)

    This is an example of intermolecular heck coupling

    Palladium is very sensitive to steric effect and

    generally forms less hindered complex

    further migration of alkene by hydro palladation

    are prevented by Ag2CO3 which rapidly removes

    oxide re-addition of Pd-H to alkene

    Overall elimination must be Syn.

    (118) The major product formed in the following

    reaction is

    Soln (2)


    (119) The major product A and B in the following

    reaction sequence are

    Soln (3)

    (a) 1st step in the chlorination of anime with


    (b) 2nd step is the HLF reaction with retention in


  • (120) The major product formed in the following

    reaction is

    Soln (4)

    Mechanism :

    (121) The spectroscopic technique that can

    distinguish unambiguously between trans-1, 2-

    dichloroethynele and cis-1, 2-dichloroethynele

    without any numerical calculation is

    (1.) Microwave spectroscopy

    (2.) UV Visible spectroscopy

    (3.) X-ray photoelectron spectroscopy

    (4.) -ray spectroscopy

    Soln (1)

    In microwave, only molecule possessing permanent

    dipole moment can be distinguished

    cis and trans have marked different in their dipole


    UV is for conjugated system hence cannot be the


    (122) 10 mL aliquots of a mixture of HCl and HNO3

    are titrated conductometrically using a 0.1M NaOH

    and 0.1M AgNO3 separately. The titer volumes are

    V1 and V2 mL, respectively. The concentration of

    HNO3 in the mixture is obtained from the


    (1.) V1 - V2 (2.) 2V1 - V2

    (3.) V2 - V1 (4.) 2V2 V1

    Soln (3)

    (123) A reversible expansion of 1.0 mol of an ideal

    gas is carried out from 1.0 L to 4.0 L under

    isothermal condition at 300 K. G for this process is

    (1.) 300 R . ln2 (2.) 600 R .ln2

    (3.) -600 R. ln2 (4.) -300 R . ln2

    Soln (3)

    G = Wmax

    WD = -RT In 2

    = -R x 300 x ln22

    = -600 R ln 2

    (Reference : Physical chemistry , Atkins)

    (124) The non-spontaneous process among the

    following is

    (1.) Vaporization of superheated water at 1050C and

    1 atm pressure

    (2.) Expansion of gas into vacuum

    (3.) Freezing of super cooled water at -100C and 1

    atm pressure

    (4.) Freezing of water at 00C and 1 atm pressure

    Soln (3)

    Expansion of gas into vacuum and freezing of water

    at 0 C is always a spontaneous process.

    However , it is sometimes possible to cool liquid

    water at low temperature without solidification. The

    liquid below the freezing point is in the supercooled

    state.This state is not quite stable and is known as

    metastable state. The metastable state is

    spontaneously converted into stable state by addition

    of small amount of ice to the supercooled liquid ,

    which will result in the solidification of water. This

    indicates that former process is non-spontaneous at

    the same temperature.

    (125) The 1H NMR frequency at 1.0 T is 42.4 MHz.

    If the gyromagnetic ratios of 1H and 13C are 27 x 107

    and 6.75 x 107 T-1 s-1, respectively, what will be the 13C frequency at 1.0 T?

    (1.) 10.6 MHz (2.) 169.6 MHz

    (3.) 42.6 MHz (4.) 21.3 MHz

    Soln (1)

    since we know that gyromagnetic ratio of C13 is

    almost time to that of 1H hence even without

    calculation, it can be inferred that

    13 =1


    if 1H NMR frequency at 1.0T is 42.4 Mhz so the

    C13 frequency at the same field strength would be

    42.4/4 = 10.6 MHz

    (126) The first order rate constant for a unimolecular

    gas phase reaction A products that follows

    Lindemann mechanism is 2.0 s-1 at pA = 1 atm and

    4.0 s-1 at pA =2 atm. The rate constant for the

    activation step is

    (1.) 1.0 atm-1 s-1 (2.) 2.0 atm-1 s-1

  • (3.) 4.0 atm-1 s-1 (4.) 8.0 atm-1 s-1

    Soln (4)





    Therefore, two different pressures we have










    So, = (1



    2) (





    = ( 1 1

    2 ) x (




    = 8 atm-1s-1

    (127) Stability of lyophobic dispersions is

    determined by

    (1.) inner-particle electric double layer repulsion and

    intra-particle van der Waals attraction

    (2.) inner-particle electric double layer attraction and

    intra-particle van der Waals repulsion

    (3.) inner-particle excluded volume repulsion and

    intra-particle van der Waals attraction

    (4.) inner-particle excluded volume attraction and

    intra-particle van der Waals repulsion

    Soln (1)

    The theory of the stability of lyophobic dispersions

    was developed by B. Derjaguin and L. Landau and

    independently by E. Verwey and J.T.G. Overbeek,

    and is known as the DLVO theory. It assumes that

    there is a balance between the repulsive interaction

    between the charges of the electrical double layers on

    neighboring particles and the attractive interactions

    arising from van der Waals interactions between the

    molecules in the particles. The potential energy

    arising from the repulsion of double layers on

    particles of radius a has the form

    = +22

    (128) According to transition state theory, one of the

    vibrations in the activated complex is a loose

    vibration. The partition function for this loose

    vibration is equal to (kB is the Boltzmanns constant

    and h is the Plancks constant)



    (3.) kBT (4.)

    Soln (4)

    Reactants A and B have 3N-6 vibrational degrees of

    freedom if non-linear, 3N-5 if linear. The same is

    true of the activated complex, which has 3(NA + NB)-

    6 vibrational modes if it is non-linear. One of these

    modes is of a different character from the rest

    corresponding to a very loose vibration that allows

    the complex to dissociate into products. For this

    degree of freedom we can use a vibrational partition

    function q* in which the vibrational frequency v

    tends to zero. i.e.

    = lim0


    1 =


    1 (1 /)=

    (129) The average end to end distance of a random

    coil polymer of 106 monomers (in units of segment

    length) is

    (1.) 106 (2.) 105 (3.) 104 (4.)


    Soln (4)

    Average end to end distance of random coil polymer

    is given by

    R2 = l.N1/2, where N = no of monomer l = segment


    R2 = 106 = 103

    (130) The radial part of the hydrogenic wave

    function is given as r( r)-r (, are constants).

    This function is then identifiable as

    (1.) 2s (2.) 3p (3.) 4d (4.)


    Soln (2)

    Radial part of hydrogenic wave function is given as

    (L )-g. The corresponds to one radial node.

    Number of radial nodes = n l 1 =1

    n = 3 l = 1

    Orbital wave function must be 3p

    (131) A normalized state is constructed as the

    linear combination of the ground state (0) and the

    first excited state (1) of some harmonic oscillator

    with energies and 3/2 units, respectively. If the

    average energy of the state is 7/6, the probability of

    finding 0 in will be

    (1.) 1/2 (2.) 1/3 (3.) 1/4 (4.)


    Soln (2)

    According to Schrodinger Equation

    H = E

    i.e. Hamiltonian operated on a wave function will

    give a same wave function back with an eigen value.

    Here the eigen value for the 1st two state is given

    and 3/2

    Linear combination will give

    = C10 + C21, where e1 and c2 are normalizable


    = 1+


    2 + 221

    2 +21210 = 0

  • The last term in the integral vehicles due to ortho


    C12 + C12 = 1

    Now its easy to find the expectation value of energy

    H = ||2En = ||

    2(n + ) h

    H = C12 (hw) + C12 (3/2hw)

    C12 + C12 = 1.. (1) 1

    2C12 +


    2C12 =


    6 (given) (2)


    212 +


    221 =




    212 +


    221 =



    C12 = 2/6 =1/3

    (132) The symmetry-allowed atomic transition

    among the following is

    (1.) 3F 1D (2.) 3F 3D

    (3.) 3F 1P (4.) 3F 3P

    Soln (2)

    Solution rule for the atomic transition is

    l = 1 (parity should be changed by unity)

    s = 0 (spin should not change)

    Hence option 2 is correct

    (133) Given that E0 (Cl2/Cl-) = 1.35 V and Ksp

    (AgCl) = 10-10 at 250C, E0 corresponding to the

    electrode reaction 1

    22() +

    +(. ) +

    () is

    (1.) 0.75 V (2.) 1.05 V

    (3.) 1.65 V (4.) 1.95 V


    = 0.06 ]

    Soln (1)

    According to the Nernst equation

    Ecell = E0cell + 2.303




    Since Ksp is given which is the condition for

    equilibrium hence ECL will be zero

    E0cell = 2.303


    = 0.06 log 10-10 since log10 = 1

    E0cell = 0.6

    E0cell = Eox + Ered

    Eox = E0cell Ered = 0.6 1.35 = 0.75 V

    (134) The ground state electronic configuration of C2

    using all electrons is

    (1.) 12 1

    222 2

    222 2


    (2.) 12 1

    222 2

    222 2


    (3.) 12 1

    222 2

    222 2

    1 21

    (4.) 12 1

    222 2


    Soln (4)

    C 1s22s2p2

    MOT diagram for C2

    1s2 2s2 has been omitted for clarity

    C2 = 61s261s262s261s22p4

    (135) The temperature dependence of the vapor

    pressure of solid A can be represented by log p =

    10.0 - 1800

    , and that of liquid A by log p = 8.0 -



    The temperature of the triple point of A is

    (1.) 200 K (2.) 300 K

    (3.) 400 K (4.) 500 K

    Soln (1)

    Triple point correspondence of unique value of

    pressure hence two equation for the pressure must

    equate to the same value

    log = 10 1800


    log = 8 1400


    (1) and (2) must be same


    10 1800

    = 8


    10 8 =1400


    2 =400

    2T = 400

    T = 200 K

    (136) The molecule with the smallest rotational

    constant(in the microwave spectrum) among the

    following is

    (1.) N CH (2.) HC CCl

    (3.) ClC CF (4.) B CCl

    Soln (3)

    Expression for the rational constant is given by

    = 2/82

    1/I ( is inversely proportional to reduced


    i.e. heavier will be the molecule, lesser will be the

    rotational constant. Hence 3 is the correct option

    (137) vmax and Km for an enzyme catalyzed reaction

    are 2.0 x 10-3 Ms-1 and 1.0 x 10-6 M, respectively.

    The rate of the reaction when the substrate

    concentration is 1.0 x 10-6 M is

    (1.) 3.0 x 10-3 s-1 (2.) 1.0 x 10-3 s-1

    (3.) 2.0 x 10-3 s-1 (4.) 0.5 s-1

  • Soln (2)

    According to Michaelis Menten equation


    = =



    =2 103[106]

    106 + 106=

    2 109

    [1 + 1]106=2 109

    2 106


    (138)A certain 2 level system has stationary state

    energies E1 and E2 (E1 < E2) with normalized wave

    functions 1 and 2 respectively. In the presence of a

    perturbation V, the second order correction to the

    energy for the first state (1) will be

    (1.) 1||212

    (2.) 1||221

    (3.) |1||2|


    12 (4.)



    Soln (3)

    Since we know that second order correction is always

    hence it follows that expression should contain E1

    E2 (since E1 < E2). Therefore correct option is 3

    (139) The vibrational frequency of a homonuclear

    diatomic molecule is v. The temperature at which the

    population of the first excited state will be half that

    of the ground state is given by

    (1.) hv.In2/kB (2.) hv/(In2.kB)

    (3.) In2/(hv.kB) (4.) hv.log2/kB

    Soln (2)

    Expression for the Boltzmann population is given by 21= E/RT

    According to given condition 2 =1



    2= hv/kT

    ln 1

    2= hv/kT or ln2 = hv/kT

    T = hv/ln2.kB

    (140) For a simple cubic crystal, X-ray diffraction

    how intense reflections for angles 1 and 2 which are

    assigned to [1 0 1] and [1 1 1] planes , respectively.

    The ratio sin1/sin2 is

    (1.) 1.5 (2.) 1.22 (3.) 0.82 (4.)


    Soln (3)

    According to Braggs law,

    = 2dsin

    sin =

    2 sin 1/d where d =









    = 0.85

    (141) The standard EMF of the cell

    Pt, H2(g)HCl (soln.) AgCl(s), Ag(s)

    (1.) Increase with T (2.) decrease with T

    (3.) remains unchanged with T

    (4.) decreases with HCl

    Soln (3)

    This cell is an example chemical cell without

    transference. It consist of hydrogen electrode on the

    left hand side and Ag|AgCl electrode on the right ,

    dipping directly in hydrochloric acid giving cell with

    no liquid junction. In the cell, the hydrogen electrode

    is negative and reversible with respect to hydrogen

    ion whereas the Ag|AgCl electrode is positive and

    reversible with respect to chloride ions.

    EMF of the cell is given by (without derivation)

    Ecell = ||




    The cell reaction indicates that cell EMF results from

    chemical reaction , namely , the reduction of silver

    chloride by hydrogen gas to solid silver and

    hydrochloric acid.

    Furthermore , the EMF of the cell as given by above

    equation depends on the activity of hydrochloric acid

    and the pressure of hydrogen gas. If the pressure of

    the gas is 1 atm equation becomes

    = ||


    (142) The irreducible representations of C2h are Ag,

    Bg, Au, and Bu. The Raman active modes of trans-1,

    3-butadiene belongs to the irreducible representations

    (1) Ag and Bg (3) Ag and Au

    (2) Au and Bg (4) Bg and Bu

    Soln (1)

    The character table for C2h point group is given by

    E C2


    i h Linear Rotation


    Ag 1 1 1 1 Rz 2 , 2, 2 , Bg 1 -1 1 -1 Rx , Ry xz , yz

    Au 1 1 -1 -1 z

    Bu 1 -1 1 1 x , y

    From above it is evident that Raman active modes

    belong to Ag and Bg representation

    (143) The molecule diborane belongs to the

    symmetry point group

    (1.) C2v (2.) C2h (3.) D2d (4.)


    Soln (4)

    Diborane has three perpendicular C2 axes are 3

    perpendicular mirror plane

  • (144) Possible term symbol(s) of the excited states of

    Na atom with the electronic configuration

    [1s22s22p63p1] is/are

    (1.) 2S1/2 (2.) 2P3/2 and 2P1/2

    (3.) 1S0 and 1P1 (4.) 3P0 and 3P1

    Soln (2)

    Excited state configuration of Na11 1s22s2p63p1

    Term symbol: 2s + 1LJ

    S = , 2s + 1 = 2, L = 1

    J = |L - S|to|L + S|

    1 + = 3/2 or 1 =

    Possible term symbol are: 2P3/2 and 2P1/2

    (145) Though a constant shift of energy levels of a

    system changes the partition function, the properties

    that do not change are

    (1.) average energy, entropy and heat capacity

    (2.) average energy and entropy

    (3.) average energy and heat capacity

    (4.) entropy and heat capacity

    Soln (2)

    The expression for the average energy U in the

    partition fn is given by the


    ) =



    )exp () =

    Similarly, for entropy we have expression (


    where F = -RT[In Z]

    Therefore both the terms depends only on

    temperature but not on energy levels