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IIT JEE 2009 Chemistry Paper2 Code 1 Solutions


1.     In the following carbocation. H/CH3 that is most likely to migrate to the positively charged carbon is


        (A)    CH3 at C-4

        (B)    H at C-4

        (C)    CH3 at C-2

        (D)    H at C-2


Sol.   (D)


          driving force is conjugation from oxygen ,i.e, resonance stabilisation


2.     The correct stability order of the following resonance structures is

resonance structure

(A)    (I) > (II) > (IV) > (III)

(B)    (I) > (III) > (II) > (IV)

(C)    (II) > (I) > (III) > (IV)

(D)    (III) > (I) > (IV) > (II)


Sol.   (B)

        On the basis of stability of resonating structures.Since +ve and -ve charges when they are at some separation then you need to do work to maintain it at some distance.Hence their energy will be high and so they are lesser stable.

3.     The spin only magnetic moment value (in Bohr magneton unit) of Cr(CO)6 is

        (A)    0

        (B)    2.84

        (C)    4.90

        (D)    5.92


Sol.   (A)


        Cr(24) = [Ar]3d5 4s1

        So (CO) is strong ligand, in Cr(CO)6 no unpaired electron is present. Hence 'spin only' magnetic moment is zero.


4.     For a first order reaction A -> P, the temperature (T) dependant rate constant (k) was found to follow the equation log k = -(2000) 1/T + 6.0. The pre-exponential factor A and the activation energy Ea, respectively, are

        (A)    1.0 × 106 s-1 and 9.2 kJ mol-1

        (B)    6.0 s-1 and 16.6 kJ mol-1

        (C)    1.0 × 106 s-1 and 16.6 kJ mol-1

        (D)    1.0 × 106 s-1 and 38.3 kJ mol-1


Sol.   (D)

        Given, log K = 6 - 2000/T

         On comparing it with our standard equation , we have,

        Since, log K = log A - Ea/2.303RT So, A = 106 sec-1 and Ea = 38.3 kJ/mole


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