AIEEE 2007 Chemistry

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  • AIEEE 2007

    Chemistry

    1. The equivalent conductances of two strong electrolytes at infinite dilution in H2O (where ions move freely through a solution) at 25 C are given below

    CH C ONa2

    3Sc equivO

    0 191 0= . m

    HCl2S cm equiv0 1426 2= .

    What additional information/quantity one needs to cal-culate 0 of an aqueous solution of acetic acid?

    (a) 0 of chloroacetic acid (ClCH2COOH) (b) 0 of NaCl (c) 0 of CH3COOK (d) The limiting equivalent conductance of H

    H++ ( )0

    Solution(b) From Kohlrauschs law,

    CH COOH CH COONa HCl NaCl3 3 0 0 0 0

    = +

    2. Which one of the following is the strongest base in aque-ous solution?

    (a) Methylamine (b) Trimethylamine (c) Aniline (d) Diemethylamine

    Solution (d) In aqueous solution, the basicity order of 1 > 2 > 3

    amines and with methyl group, the basicity order is 2 > 1 > 3. In case of aniline, lone pair of nitrogen is involved in resonance, so it is weaker base than ali-phatic amines.

    3. The compound formed as a result of oxidation of ethyl benzene by KMnO4 is

    (a) benzyl alcohol. (b) benzophenone. (c) acetophenone. (d) benzoic acid.

    Solution(d) The reaction is

    CH2CH3

    Ethyl benzene Benzoic acid

    COOH

    H+KMNO4

    4. The IUPAC name of the following compound is

    (a) 3-ethyl-4,4-dimethylheptane. (b) 1,1-diethyl-2,2-dimethylpentane. (c) 4,4-dimethyl-5,5-diethylpentane. (d) 5,5-diethyl-4,4-dimethylpentane.

    Solution(a) The IUPAC name is

    3-Ethyl-4,4-dimethylheptane

    1234567

    H3C CH2 CH2 CH2 CH3

    CH3 CH2CH3

    CH3

    C CH

    5. Which of the following species exhibits diamagnetic behavior?

    (a) NO (b) O22 (c) O2

    + (d) O2

    Solution(b) The electronic configurations are as follows:

    O22 2 2 2 2 2

    2 2 2 2

    1 1 2 2 2

    2 2 2 2

    =

    = =

    s s s s s

    p p p p

    s s s s p

    p p p pz

    x y x y

    General Instructions

    (a) There are objective type questions with four options having single correct answer. (b) For each incorrect response, one fourth (1/4) of the total marks allotted to the question would be deducted.(c) No deduction from the total score will, however, be made if no response is indicated for an item in the answer sheet.(d) The candidates are advised not to attempt such item in the answer sheet if they are not sure of the correct response.(e) More than one answer indicated against a question will be deemed as incorrect response and will be negatively marked.

    AIEEE 2007 Chemistry.indd 1 12/19/2011 12:44:20 PM

  • 2 AIEEE 2007

    No unpaired electron, so it is diamagnetic.O2

    +

    =

    = =

    s s s s s

    p p p p

    1 1 2 2 2

    2 2 2 2

    2 2 2 2 2

    2 2 2 0

    s s s s p

    p p p pz

    x y x y*

    One unpaired electron, so it is paramagnetic.O2

    2 2 2 2 2

    2 2 1 1

    1 1 2 2 2

    2 2 2 2

    =

    = =

    s s s s s

    p p p p

    s s s s p

    p p p pz

    x y x y

    Two unpaired electrons, so it is paramagnetic.NO =

    = =

    s s s s s

    p p p p

    1 1 2 2 2

    2 2 2 2

    2 2 2 2 2

    2 2 1 0

    s s s s p

    p p p pz

    x y x y

    One unpaired electron, so it is paramagnetic. 6. The stability of dihalides of Si, Ge, Sn and Pb increases

    steadily in the sequence (a) PbX SnX GeX SiX2 2 2 2

  • AIEEE 2007 3

    100

    200

    180

    In absence of catalyst

    In presence of catalyst

    80

    Reaction coordinate

    Pote

    ntia

    l ene

    rgy

    Figure 1

    r (forward) backward kJmolH E E= = =

    ( ) 80 100 201

    14. The cell Zn Zn M Cu M Cu; ( V)cello| || |2 21 1 1 10+ + =( ) ( ) .E , was

    allowed to be completely discharged at 298 K. The rela-tive concentration of Zn2+ to Cu2+, [Zn2+]/[Cu2+] is

    (a) 9.65 104 (b) antilog (24.08) (c) 37.3 (d) 1037.3

    Solution(d) Using Nernst equation,

    E Encell cell

    o ZnCu

    =

    +

    +

    0 0591 22

    .log

    [ ][ ]

    For complete discharge, Ecell = 0, so

    Encell

    o2+

    2+

    ZnCu

    =

    0 05910

    .log

    [ ][ ]

    The reactions at anode and cathode areZn Zn ++ 2 2e

    Cu Cu2 2+ + eSo, n = 2,

    Ecello

    2+

    2+

    Zn

    Cu=

    0 05912

    .log

    Solving by substituting given value of Ecello V=1 10. ,

    we get

    1 10 20 0591..

    log[ ][ ]

    =

    ZnCu

    2+

    2+

    37 3 1037 3. log]

    [ ][ ][ ]

    .= =

    [ZnCu

    ZnCu

    2+

    2+

    2+

    2+

    15. The pKa of a weak acid (HA) is 4.5. The pOH of an aque-ous buffered solution of HA in which 50% of the acid is ionized is

    (a) 7.0 (b) 4.5 (c) 2.5 (d) 9.5

    Solution(d) For buffer solution (Henderson equation)

    pH p logSalt]Acid]

    4 5 logSalt]Acid]a

    = + = +K[[

    .[[

    As the acid is 50% ionized, which means that [salt] = [acid]

    pH 4 5 log 1 4.5= + =.

    Therefore,

    pOH 14 pH 14 4 5 9 5= = = . .

    16. Consider the reaction, 2A B Products+ . When con-centration of B alone was doubled, the half life did not change. When the concentration of A alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is

    (a) s1 (b) L mol1 s1 (c) It has no units. (d) mol1 s1

    Solution(b) For the reaction 2A B+ product, when concen-

    tration of B is doubled, the half life did not change, hence the reaction is of first order with respect to B because half life is independent of concentration of first-order reaction (t1/2 = 0.693/k). When the concen-tration of A is doubled, the reaction rate is doubled, hence the reaction is of first order with respect to A. So, the overall order is 1 + 1 = 2 and the units of rate constant for a second order reaction is L mol1 s1.

    17. Identify the incorrect statement among the following (a) 4f and 5f orbitals are equally shielded. (b) d-Block elements show irregular and erratic chemical

    properties among themselves. (c) La and Lu have partially filled d orbitals and no other

    partially filled orbitals. (d) The chemistry of various lanthanoids is very similar.

    Solution(a) 4f and 5f belongs to different energy levels, hence

    shielding effect is not the same for both of them. Shielding of 4f is more than 5f. Also, 5f is less deeply buried than 4f.

    18. Which one of the following has a square planar geometry? (a) [ ]PtCl4

    2 (b) [ ]CoCl42

    (c) [ ]FeCl42 (d) [ ]NiCl4

    2

    (Atomic numbers of Co = 27, Ni = 28, Fe = 26, Pt = 78)

    Solution(a) Most of the 4d and 5d series elements show square

    planar geometry.

    19. Which of the following molecules is expected to rotate the plane of plane-polarized light?

    (a) (b)

    CHO

    CH2OH

    HO H

    SH

    (c) (d)

    Ph Ph

    NH2H2N

    H H

    COOH

    H

    H2N H

    Solution(a) The compound has a chiral carbon atom.

    AIEEE 2007 Chemistry.indd 3 12/19/2011 12:44:30 PM

  • 4 AIEEE 2007

    (a) CHO

    CH2OH Chiral

    HO H

    Chiral carbon(A Carbon attachedto four dierentsubstituents)

    *

    (c)

    Ph Ph

    NH2H2N

    H H

    AchiralPlane of symmetry

    (b)

    SH Achiral

    Plane of symmetry(which divides themolecule into twoequal halves)

    (d) COOH

    H2N H

    H Achiral

    Two identicalgroups attached

    20. The secondary structure of protein refers to (a) fixed configuration of the polypeptide backbone. (b) a -helical backbone. (c) hydrophobic interactions. (d) sequence of a -amino acids.

    Solution(b) Secondary structure of proteins involves a -helical

    backbone and b -sheet structures. These are formed as a result of hydrogen bonding between different peptide groups.

    21. Which of the following reactions will yield 2,2-dibromo-propane?

    (a) CH3 CH2CH +HBr

    (b) CH3 C CH + 2HBr

    (c) CH3CH + HBrCHBr

    (d) + 2HBrCH CH

    Solution(b) The reaction is as follows:

    CH3 C CH HBr

    HBr

    CH2

    Br

    CH3 C+

    Br

    Br

    CH3 CH3C

    Markonikovsrule

    (Markonikovsrule)

    2,2-Dibromo-propane

    22. In the chemical reaction, CH CH NH CHCI KOH3 2 2 3 3+ + + +A B 3H O2 , the compounds (A) and (B) are, respectively,

    (a) C2H5NC and 3KCl (b) C2H5CN and 3KCl

    (c) CH3CH2CONH2 and 3KCl (d) C2H5NC and K2CO3Solution(a) This is an example of carbylamine reaction. So, the

    products will be C2H5NC and 3KCl.

    23. The reaction of toluene with Cl2 in presence of FeCl3 gives predominantly:

    (a) m-chlorotoluene. (b) benzoyl chloride. (c) benzyl chloride. (d) o- and p-chlorotoluene.

    Solution(d) The reaction involved is electrophilic addition reaction:

    CH3 CH3

    FeCl3

    Cl2Cl

    CH3

    Cl

    +

    Toluene(o, p-directing

    in nature)

    o-Chloro-toluene(Minor)

    p-Chloro-toluene(Major)

    24. Presence of a nitro group in a benzene ring (a) deactivates the ring towards electrophilic substitution. (b) activates the ring towards electrophilic substitution. (c) renders the ring basic. (d) deactivates the ring towards nucleophilic substitution.

    Solution(a) Nitro group reduces the electron density in the ben-

    zene ring due to strong I effect.

    NO O

    NOO

    +

    ONO

    +

    NO

    +

    NO OO

    Overall electron density on benzene ring decreases mak-ing electrophilic substitution difficult. Hence, it deacti-vates the ring towards electrophilic substitution.

    25. In which of the following ionization processes, the bond order has increased and the magnetic behavior has changed?

    (a) N N2 2+

    (b) C C2 2+

    AI