IIT-JEE-2009 Question Paper 2

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IIT-JEE 2009

IIT-JEE-2009 Question Paper 2

Author: Sureshbala

IIT-JEE Solutions - 2009 Paper 1

IIT-JEE 2009:
Chemistry Paper 2

SECTION I

Q 1. The spin only magnetic moment value (in Bohr magneton units) of Cr(CO)_6 is

(A) 0
(B) 2.84
(C) 4.90
(D) 5.92

Ans.
$Cr=Ar3d^5 \, 3d^1$
CO being strong legand forms inner orbital complex. $\mu=Zero$

Q 2. In the following carbocation, H/CH_3 is most likely to migrate to the positively charged carbon is

$(A)\, CH_3 \, at\, C-4$
$(B)\, H \, at\, C-4$
$(C)\, CH_3 \, at\, C-2$
(D) H at C-2

Ans.

Q 3. For a first order reaction $A \Rightarrow P$ , the temperature (T) dependent rate constant (k) was found to follow the equation $\log K=-(2000)\cfrac {1}{T}+6.0$ The preexponential factor A and the activation energy E_a respectively, are

(A) $1.0 *10^6 s^{-1} \, and \, 9.2 kJ mol^{-1}$
(B) $6.0 s^{-1} \, and \, 16.6 kJ mol^{-1}$
(C) $1.0 *10^6 s^{-1} \, and \, 16.6 kJ mol^{-1}$
(D) $1.0 *10^6 s^{-1} \, and \, 38.3 kJ mol^{-1}$

Ans.

$K=Ae^{(-Ea/RT)}$
ln $k=in \, A-(Ea/RT)$
ln $k=ln \, A -(Ea/2.303 \, Rt)$
$\therefore A=10^6 \, and \, (Ea/2.303*R)=2000$
$Ea=2000*(25/3)*10^{-3}*2.303$ =(50*2.3)/3=5*7.66=38.30

Q 4. The correct stability order of the following resonance structures is

(A)(I) > (II) > (IV) > (III)
(B) (I) > (III) > (II) > (IV)
(C) (II) > (I) > (III) > (IV)
(D) (III) > (I) > (IV) > (II)

Ans.
Smaller the separation of opposite charge greater is the stability 1st structure is more stable because of the fact that -ve charge existing on the electronegative atom.

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IIT-JEE Solutions - 2009 Paper 1