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MEMORY BASED (IIT-JEE - 2006) TEST PAPER
CHEMISTRY
INSTRUCTIONS:
(i)
Question no. 1 to 12 has only one correct option. You will be awarded 3 marks for right answer and 1 mark for
wrong answer.
(ii)
Question no. 13 to 20 has one or more than one correct option(s). You will be awarded 5 marks if you answer all
correct options and only correct option(s) and 1 mark will be awarded for wrong answer.
(iii)
Question no. 21 to 32 are based on small write up first go through it then answer these questions. You will be
awarded 5 marks for right answer and 2 marks for wrong answer.
(iv)
Question no. 33 to 36 are subjective problems. Circle their correct answers. There is no negative marking for it.
Each question carries 6 marks.
(v)
Question no. 37 to 40 carry 6 marks each. These may have more than one correct options. There is no negative
marking for these.
PART - A :
(Question no. 1 to 12 has only one correct option. You will be awarded 3 marks for right answer and 1
mark for wrong answer.)
1.
The ratio of P to V at any instant is constant and is equal to 1, for a monoatomic ideal gas under going a process.
What is the molar heat capacity of the gas
(A)
Sol.
3R
2
(B)
4R
2
(C)
5R
2
(D) 0
(B)
From first law of Thermodynamics, E = q + w nCvdT = nCdT – PdV
.....(1)
Now according to process, P = V
and according to ideal gas equation, PV = nRT
We have ,
V2 = nRT
On differentiating, 2VdV = nRdT
So from first equation we have,
So,
Cv = C –
R
2
and
nCvdT = nCdT –
PdV = VdV =
n R dT
2
n R dT
2
Hence C =
4R
2
TM
RESONANCE
Page # 1
2.
Sol.
A white precipitate is obtained when a solution is diluted with H2O and boiled. On addition of excess NH4Cl or
NH4OH. The volume of precipitate decreases leaving behind a white gelatinous precipitate. Identify the precipitate
which dissolves in NH4OH or NH4Cl.
(A) Al(OH)3
(B) Zn(OH)2
(C) Mg(OH)2
(D) Ca(OH)2
(B)
Zn2+ + 2NH4OH  Zn(OH)2 + 2NH4+
Zn(OH)2 + 4NH4+  [Zn(NH3)4]2+ + 2H2O + 2H+
3.
Sol.
4.
B(OH)3 + NaOH  Na[B(OH)4] (aq)
Then addition of which of the following proceeds the reaction in the forward direction.
(A) cis-1, 2 diol
(B) Trans 1, 2 diol
(C) Borax
(D) Na2 HPO4
(A)
Boric acid is a very weak acid. It does not react with alkali explicitly. However, the weak orthoboric acid may be
converted into comparatively strong monobasic acid by addition of any cis 1,2 diol to pull the reaction in forward
direction significantly.
K = 4 × 106 at 298
K = 41 at 400 K
Which is correct statements is correct?
(A) If N2 is added at equlibrium condition, the equilibrium will shift to the forward direction because according to IInd
law of thermodynamics the entropy must increases in the direction of spontaneous reaction.
N2 + 3H2
2 NH3
(B) The condition for equilibrium is 2GNH3 = 3GH2 + GN2 where G is Gibbs free energy per mole of the gaseous
Sol.
species measured at that partial pressure.
(C) Addition of catalyst does not change Kp but changes H.
(D) At 400 K addition of catalyst will increase forward reaction by 2 times while reverse reaction rate will be changed
by 1.7 times.
(B)
5.
Ag + NH3
[Ag(NH3)]+
+
Sol.
K1 = 6.8 × 103
+
[Ag(NH3)] + NH3
[Ag(NH3)2]
K2 = 1.6 × 103
Calculate equilibrium constant for final complex formation
(A) 1.08 × 107
(B) 1.08 × 109
(C) 1.08 × 105
(A)
Equilibrium constant
K = K1 × K2
= 1.6 × 6.8 × 106
= 1.08 × 107
(D) 1.08 × 104
TM
RESONANCE
Page # 2
6.
The direct conversion of A to B is difficult, hence it is carried out by the following shown path:
S(A  C) = 50
S(C  D) = 30
S(B  D) = 20
The entropy change for the process A  B is
Sol.
(A) 100
(D)
(B) – 60
(C) – 100
(D) + 60
A  C ; S = 50
C  D; S = 30
D  B; S = 20
For A  B: S = 50 + 30 – 20 = 60
7.
Sol.
In blue solution of copper sulpahte excess of KCN is added then solution becomes colourless due to the formation
of
(A) [Cu(CN)4]2–
(B) Cu2+ get reduced to form [Cu(CN)4]3–
(C) Cu(CN)2
(D) CuCN
(B)
Cu+2 + 2CN–  Cu(CN)2
2Cu(CN)2  2CuCN + (CN)2
CuCN + 3KCN  K3[Cu(CN)4]
8.
IUPAC name of
Sol.
(A) Benzoyl chloride
(C) Chloroacetobenzene
(D)
(B) Chlorophenyl ketone
(D) Benzenecarbonyl chloride
TM
RESONANCE
Page # 3
9.
Arrange in the increasing order of boiling points
(A) I < IV < III < II
(B) IV < III < II < I
Sol.
(C)
10.
 X ,
CH2 = CH – CH3  
Sol.
11.
(C) I < II < III < IV
(D) I < II < IV<III
has highest boiling point due to maximum intermolecular H-bonding.
NOCl
X is :
CH2  CH  CH3
(A) |
|
Cl
NO
CH2  CH  CH3
|
(B) |
NO Cl
(C) ON – CH2 – CH2 – CH2 – Cl
(D) ON  CH  CH2  CH3
|
Cl
CH2  CH  CH3
|
|
NO Cl
(B)
NaHCO
3
  

NaHCO
3
  

A mixture of benzenesulphonic acid and paranitrophenol react with NaHCO3 separately. The gases produced are
respectively.
(A) SO2, CO2
(B) SO2, CO
(C) SO2, NO2
(D) CO2, CO2
TM
RESONANCE
Page # 4
Sol.
(D)
Since a strong acid displaces a weaker acid from its salt and forms its own salt therefore
+ NaHCO3 
+
H2CO 3
Weak acid

H2 O  CO 2 
+ NaHCO3 
+
H2CO 3
Weak acid

H2 O  CO 2 
12.
The open chain keotne with lowest molecular weight and its next homolog on reaction with NH2 – OH gives ...........
(A) 2 oxime as products
(B) 3 oximes as products
(C) only 1 product is optically active
(D) all optically active product is obtained
O
O
||
||
Sol.(B) Lowest molecular weight ketone is
and its next homolog is
CH3  C  CH3
CH3  C  CH2CH3
+ NH2OH 
+ NH2OH 
+
3 oximes (I, II, III) are produced.
PART (B) :
Question no. 13 to 20 has one or more than one correct option(s). You will be awarded 5 marks if you
answer all correct options and only correct option(s) and 1 mark will be awarded for wrong answer.
13.
Ans.
In CO, bond length is 1.128 Å then in Fe(CO)5 bond length of CO is :
(A) 1.128
(B) 1.15
(C) 1.72
(D) 1.118
(B)
The metal (Fe) makes the back bonding to CO. The metal to ligand back bonding creates a effect which strengthens
the bond between CO and the metal (Fe). Which results in the decrease in bond order of CO, hence increase in the
bond length.
TM
RESONANCE
Page # 5
14.
Sol.
When CO2 is passed over water then which of the following is obtained
(A) HCO3–, CO22–
(B) HCO3–, CO2 , H2 CO3
–
2–
(C) HCO3 , CO2
(D) HCO3–, CO32–, CO2 , H2 CO3
(D)
CO2 + H2O
H2CO3
H2CO3
H+ + HCO3–
HCO3–
H+ + CO3–2
15.
MgSO4 + NH4Cl + Na2 HPO4  White crystalline precipitate
Sol.
(A) MgCl2 . MgSO4
(C)
(B) MgSO4
(C) Mg(NH4)PO4 (D) MgSO4 . Mg(PO4)2
Mg+2 + NH4OH + Na2HPO4  Mg(NH4) PO4
16.
PV
versus P at a particular temperature, for real gases A,
nRT
B & C. Then which of the following statement is incorrect :
(A) For the gas A, a = 0 and its dependence on P is linear at all pressure.
(B) For the gas B, b = 0 and its dependence on P is linear at all pressure.
(C) For the gas C, which is typical real gas for which neither a nor b = 0. By knowing the minima and the point of
intersection, with Z = 1, a and b can be calculated.
(D) At high pressure, the slope is positive for all real gases.
(B)
Above sketch is the plot of compressibility factor (z) =
Sol.
If a = 0, then V =
n R dT
+ nb,
P
nRT
If b = 0, then V =
2
P
17.
Sol.
n a
V2
Now z =
1
PV
=P
nRT
nRT
Now z =
1
PV
=P
nRT
nRT
 n R dT

 nb  = 1 + Pb a linear relation on P

 P

RT
nRT
P
n 2a
V2
not a linear relation onm P
The product ‘P’ containing nitrogen is in the following reaction
CH3 – NH2 + KOH + CHCl3  P + KCl + H2O + 2HCl
(A)
(B)
(B)
(C) CH3 – NHCl
(D) CH3 – C  N
TM
RESONANCE
Page # 6
18.
P
Q + Phenol
What are P & Q.
(A)
O
||
+ CH3  C  CH3
(B)
O
||
+ CH3  CH2  C  H
(C)
+
O
||
CH

C
 CH3
+
3
(D)
Sol.
O
||
CH3  CH2  C  H
(D)

+
In the formation of product P the electrophile
rearranges to
for the electrophilic
substitution.
TM
RESONANCE
Page # 7
19.
Cl / h
Fractional
(CH3)2C – CH2CH3 2 [N]     [P]
distillati on
The number of possible isomers of [N] and number of fractions of [P] are
(A) (6, 6)
(B) (6, 4)
(C) (4, 4)
CH3
|
Sol.(B) CH3  C  CH2  CH3
|
H
CH3
|
CH2  C*  CH2 CH3
|
|
Cl
H
(d  )
(D) (3, 3)
Cl / h
2 

+
CH3
|
CH3  C CH2 CH3
|
Cl
CH3
CH3
|
|
*
CH
C
C
H
CH




 CH2  CH2
CH
C
+
3
3 +
3
|
|
|
|
H Cl
H
Cl
(d  )
Total 6 isomers will be formed. On distillation only 4 fraction will be found since (d + l) mixture will not be separated
by distillation.
20.
is produced on heating a compound with conc. alkali. Which of the compound will only give this
lactone?
(A)
(B)
Sol.(D)
(C)

(D)


TM
RESONANCE
Page # 8
PART (C) :
Question no. 21 to 32 are based on small write up first go through it then answer these questions. You will
be awarded 5 marks for right answer and 2 marks for wrong answer.
Passage - 1
Tollen reagent's reaction with glucose is as follow :
Ag+ + e–  Ag

= 0.800 V
Ereduction
C6H12O6 + H2O  C6H12O7 + 2H+ + 2e–
E oxidation = – 0.05 V
[Ag(NH3)2]+ + e  Ag + 2NH3

= 0.373 V
Ereduction
Given that
21.
2.303 RT
F
= 0.0591 &
= 38.35
F
RT
Calculate equilibrium constant for the reaction in logarithimic form (n k)?
C6H12O6 + H2O + 2Ag+  Ag + C6H12O7 + 2 H+
Sol.
(A) 66.13
(C)
(B) 118.3
(C) 58.42
(D) 29.22
2Ag+ + 2e–  2Ag; G1° = – 2F × 0.8
C6H12O6 + H2O  C6 H12 O7 + 2H+ + 2e– ; G2° = – 1F × 0.05
–––––––––––––––––––––––––––––––––––––
2Ag+ + C6 H12 O6 + H2O  2Ag + C6 H12 O7 + 2H+ ; G° = ?
G° = – nFE°, n = 2, and E° = 0.75
G° = – 2 × 96500 × 0.75 = – 8.314 × 298 × n k
n k = 58.42
22.
Sol.
When concentrated ammonia is added to the solution, pH is raised to 11. Which of the electrode is affected by the
change in pH and by how much.
(A) Eoxidation increases over E° by 0.65 V
(B) Ereduction increases over E° by 0.65 V
(C) Eoxidation decreases over E° by 0.65 V
(D) Ereduction decreases over E° by 0.65 V
(A)
The anode is affected by change in pH
E  E 
[C H O ]
0.0591
log 6 12 6 [H  ] 2 `
2
[C 6H12 O 7 ]
E = E° – 0.0591 log [H+]
= E° – 0.0591 log [10– 11]
= E° + 0.0591 × 11
= E° + 0.65
23.
Sol.
Ammonia is always is added in this reaction. Which of the following must be incorrect?
(A) NH3 combines with Ag+ to form a complex.
(B) Ag(NH3)2+ is a stronger oxidising reagent than Ag+
(C) In absence of NH3 silver salt of gluconic acid is formed.
(D) NH3 has affected the standard reduction potential of glucose/gluconic acid electrode.
(D)
TM
RESONANCE
Page # 9
Passage - 2
KCN



HCN
NiCl2
complex A
HCl

excess
NiCl2
complex B
A & B complex has the co-ordination number 4
24.
Sol.
25.
Sol.
26.
Sol.
Then the IUPAC name of complex is
(A) Potassium tetracyanonicklate() and Potassium tetrachloronicklate()
(B) Potassium tetracyanonickel() and Potassium tetrachloronickel()
(C) Potassium cyanonicklate() and Potassium chloronicklate()
(D) Potassium cyanonickel() and Potassium chloronickel()
(A)
The product in both complex are
(A) both diamagentic
(B) diamagnetic (A) & (B) is paramagnetic with one inpaired e–
(C) A is diamagentic & B is paramagnetic with 2 unpaired e–
(D) both are paramagentic
(C)
The hybridisation of both complex are
(A) dsp2
(B) sp2 & dsp2
(C)
(C) dsp2 & sp3
(D) both sp3
Passage - 3
Carbon–14 is used to determine the age of organic material. The procedure is based on the formation of 14C by
neutron capture in the upper atmosphere.
14
1
7 N  0n

14
1
6 C  1n
14
C is absorbed by living organisms during photosynthesis. The 14C content is constant in living organism once the
plant or animal dies, the uptake of carbon dioxide by it ceases and the level of 14C in the dead being, falls due to the
decay which C14 undergoes
14
6C

14
7N 

The half life period of 14C is 5770 years. The decay constant () can be calculated by using the following formula 
=
0.693
t1/ 2
The comparison of the  – activity of the dead matter with that of the carbon still in circulation enables measurement
of the period of the isolation of the material from the living cycle. The method however, ceases to be accurate over
periods longer than 30,000 years. The proportion of 14C to 12C in living matter is 1 : 1012.
27.
Sol.
Which of the following option is correct?
(A)
Rate of exchange of carbon between atmosphere and living is slower than decay of C14.
(B)
Carbon dating can be used to find out the age of earth crust and rocks
(C)
Rate of exchange of C14 between atmosphere and leaving organism is so fast that an equilibrium is set up
between the intake of C14 by organism and its exponential decay.
(D)
Carbon dating can not be used to determine concentration of 14C in dead beings.
(C)
TM
RESONANCE
Page # 10
28.
Sol.
29.
What should be the age of fossil for meaningful determination of its age?
(A) 6 years
(B) 6000 years
(C) 60,000 years
(D) It can be used to calculate any age
(B)
A nuclear explosion has taken place leading to increase in concentration of C14 in nearby areas. C14 concentration
is C1 in nearby areas and C2 in areas far away. If the age of the fossil is determined to be T1 and T2 at the places
respectively then
(A) The age of the fossil will increase at the place where explosion has taken place and T1 – T2 =
(B) The age of the fossil will decrease at the place where explosion has taken place and T1 – T2 =
C1
1
ln C

2
C1
1
ln C

2
(C) The age of fossil will be determined to be same
T1 C1
(D) T  C
2
2
Sol.
(A)
Passage - 4
X


30.
Which reagent (X) is used to convert I to II
(A) KBr / NaOH
(C) NaHCO3
(B) Br2 / NaOH
(D) N-Bromo succinamide
Sol.(B)
31.
Which step is rate determining step
(A) Formation of II
(B) Formation of III
(C) Formation of V
(D) Formation of IV
Sol.(D) The migration of p-chlorophenyl group to N-atom is the slowest step. Therefore formation of isocyanate is rate
determining step.
TM
RESONANCE
Page # 11
32.
Sol.
+
under Hofmann conditions will give :
(A)
+
(B)
+
(C)
+
(D)
+
(A)
Since the overall reaction is intramolecular the product under Hofmann conditions will be
+
PART (D) :
Question no. 33 to 36 are subjective problems. Circle their correct answers. There is no negative marking
for it. Each question carries 6 marks.
33.
75.2 gm C6 H5 OH in 1kg of solvent depress the freezing point by 7k. If kf = 14 then find the % of phenol that
dimerises.
Sol.
2

7 = 14 × 0.8 
 2 
 = 0.75 = 75%
TM
RESONANCE
Page # 12
34.
The given reaction
2CO
Sol.
+
O2
 2CO2
H = – 560 kJ
2moles
1 mole
is carried and in one litre container, if the pressure in the container gets changes from 70 atm to 40 atm as reaction
gets completed. Calculate U of the reaction. [1L atm = 0.1 kJ]
H = U + (PV)
so,
U = H – (PV)
= – 560 – [40 – 70] (L atm)
= (– 560 + 30 × 0.1) kJ
= – 557 kJ
35.
For a unit cell edge length = 5Å, the element is of atomic mass 75, has density of 2gm/cc. Calculate atomic radius
of the element.
Sol.
d=
ZM
NA a3
Z = 20 × 10–1 = 2

so its is a bcc unit cell
hence
3
× 5Å = 216.5 pm
4
We have taken a saturated solution of AgBr. Ksp of AgBr is 12 × 10–14 . If 10–7 mole of AgNO3 are added to 1 litre of
this solution find conductivity (specific conductance) of this solution in terms of 10–7 S m–1 units.
3 a = 4R
36.
º
Given  ( Ag ) = 6 × 10–3
Sol.
so
R=
º
º
Sm2 mol–1  (Br  ) = 8 × 10–3 S m2 mol–1 ,  (NO3 ) = 7 × 10–3 Sm2 mol–1
The solubility of AgBr in presence of 10–7 molar AgNO3 is 3 × 10–7 M.
Therefore [Br–] = 3 × 10–4 m3, [Ag+] = 4 × 10–4 m3 and [NO3–] = 10–4 m3
Therefore total = Br  +  Ag  + NO3 = 55 Sm–1
PART (E) :
Question no. 37 to 40 carry 6 marks each. These may have more than one correct options. There is no
negative marking for these.
37.
Sol.
Match the extraction processes listed in Column I with metals listed in Column II:
Column I
Column II
(A) Self reduction
(P) Lead
(B) Carbon reduction
(Q) Silver
(C) Complex formation and displacement by metal
(R) Copper
(D) Decomposition of iodide
(S) Boron
A – P, R ; B – P, R ; C – Q ; D – S
TM
RESONANCE
Page # 13
38.
Match the following:
Column I
3+
(A) Bi
 (BiO)
Column II
+
(P) Heat
(B) [AlO2]–  Al(OH)3
(Q) Hydrolysis
(C) SiO44–  Si2O76–
(R) Acidification
(D) (B4O72–)  [B(OH)3]
(S) Dilution by water
Sol.
A – S; B – Q ; C – P ; D – Q, R
39.
According to Bohr’s theory,
En = Total energy
Kn = Kinetic energy
Vn = Potential energy
rn = Radius of nth orbit
Match the following:
Column I
(A) Vn / Kn = ?
(B) If radius of nth orbit  Enx , x = ?
(C) Angular momentum in lowest orbital
(D)
1
rn
 Zy , y  ?
Column II
(P) 0
(Q) – 1
(R) – 2
(S) 1
Sol.
A– R; B – Q; C – P; D – S
40.
Match the following (one term in colum-I may match with more than one terms in colum-II)
Sol.
A
B
C
D




Q
Q
R, S
P, S
TM
RESONANCE
Page # 14