AIEEE 2013 Exam Date (JEE MAIN)

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Prepare for IIT 2012 Exam with Solved IIT Question Papers

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IIT-JEE Solutions - 2011 Paper 2
IIT-JEE 2011 Solutions Paper 1

IIT-JEE Solutions - 2009 Paper 1

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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

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    manglesh patidar
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Sureshbala
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About: Worked for more than 6 years in renowned corporate institutes as their core faculty/lead content developer for C.A.T,G.R.E, G.M.A.T and Campus Recruitment Training Programs.

Last Updated At Jul 04, 2013
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