presents
SOLUTIONS
to
IIT-JEE 2006
Disclaimer:
The following questions have been reconstructed with the help of memory recall of our students.
Brilliant Tutorials shall not take any responsibility for errors, omissions and modifications of any kind
in the questions given.
INSTRUCTIONS:
1.
This Question booklet contains 40 questions given in 5 sections to be attempted in two hours.
2.
Section−I contains 12 questions. Each question has only one correct answer. You will be awarded
3 marks for marking the correct answer and −1 for an incorrect answer. If you haven’t attempted the
question you will be awarded 0 marks.
3.
Section−II contains 8 questions. Each question has one or more than one correct answers. You will be
awarded 5 marks for marking all the correct answers and −1 for incorrect or incomplete answers. If
you haven’t attempted the question you will be awarded 0 marks.
4.
Section−III contains 4 passages containing 3 questions each. Each question has only one correct
answer. You will be awarded 5 marks for marking correct answer and −2 for an incorrect answer. If
you haven’t attempted the question you will be awarded 0 marks.
5.
Section−IV contains 4 numerical problems. You have to darken the correct numerical value in the
answer grid for each of these questions. The answer to each of the questions would lie between
0 to 9999. You will be awarded 6 marks for correct numerical value and 0 for an incorrect value.
6.
Section−V contains 4 questions. In each questions 4 statements namely A, B, C, D are given on one
side and p, q, r, s on the other side. You have to match each statement on one side with all the correct
matches on the other side. You will be awarded 6 marks if all the 4 statements have been correctly and
completely matched. For any other answer you will be awarded 0 marks.
7.
(*) marked questions correspond to syllabus of class XI (as per the syllabus for year 2005).
14
Chemistry
SECTION−I
1.
The reaction A to B is not feasible but on changing entropy through series of steps:
A ⎯→ C ⎯→ D ⎯→ B
∆S (A ⎯→ C) = 50 eu ; ∆S (C ⎯→ D) = 30 eu ; ∆S (B ⎯→ D) = 20 eu
The entropy change for A ⎯→ B would be
(a) 100 eu
(b) 60 eu
*2.
*3.
(c) −60 eu
(d) −100 eu
A colourless aqueous solution on adding water and on heating gave a white precipitate. This precipitate when reacted
with NH4Cl and NH4OH in excess resulted in dissolution of some of the precipitate and a gelatinous precipitate is obtained.
What is the hydroxide formed in aqueous solution?
(b) Mg(OH)2
(c) Al(OH)3
(d) Ca(OH)2
(a) Zn(OH)2
B(OH)3 + NaOH
Na[B(OH)4]
To keep the above reaction in forward direction, which reagent should be used?
(a) cis−1,2−diol
(b) trans−1,2−diol
(c) Borax
(d) Na2HPO4
4.
Benzenesulphonic acid and para−nitrophenol on reaction with sodium bicarbonate liberates which of the following gases?
(a) SO2, NO
(b) SO2, NO2
(c) SO2, CO2
(d) CO2, CO2
*5.
When CO2 is passed through water, which of the following species will be present in water?
(c) CO32−, CO2
(a) H2CO3, CO2, CO32−, HCO3− (b) HCO3−, CO32−
*6.
The IUPAC name of C6H5COCl is
(a) Benzoyl chloride
(b) Chloro phenyl ketone
(d) H2CO3, CO2
(c) Benzene carbonyl chloride (d) Phenyl chloro ketone
*7.
If for an ideal monoatomic gas, the ratio of pressure and volume is constant and is equal to one, then the molar heat capacity
of the gas is
3
4
5
(b) R
(c) R
(d) 0
(a) R
2
2
2
*8.
Haber−Bosch process for the formation of NH3 at 298 K is
2NH3(g) ; ∆H = −46.0 kJ ; Kp = 14
N2(g) + 3H2(g)
The correct statement is
(a) On adding N2, the equilibrium will spontaneously shift to the right and ∆S will be positive.
(b) Even though at 298 K, the yield is more, still industrial manufacture of NH3 is carried out at 400 K (Kp = 41). This is
because at 298 K, the catalyst increases the rate by factor of two for both forward and backward reaction. But at 400 K,
the iron catalyst increases the rate of forward reaction by a factor of two and that of backward reaction by a factor of 1.7.
(c) At equilibrium 2G NH 3 = G N 2 + 3G H 2 where G is the Gibbs free energy.
(d) Addition of catalyst does not alter Kp but changes ∆H.
9.
CH3NH2 + CHCl3 + KOH ⎯→ (?) + 3KCl + H2O.
The missing product is
⊕
(a) CH3−CN
10.
(b) CH3NHCl
(c) CH3−N≡C
⊕
(d) CH3−N≡C
Arrange the following compounds according to decreasing boiling points:
OH
OH
OH
OH
OH
(I)
(II)
(a) (IV) > (III) > (II) > (I)
*11.
(III)
OH
OH (IV)
(b) (III) > (IV) > (II) > (I)
(c) (I) > (II) > (III) > (IV)
(d) (II) > (II) > (I) > (IV)
NOCl
CH3−CH=CH2 ⎯⎯⎯→ (A). Which of the following is the structure of compound (A)?
(a) CH3−CHNO−CH2Cl
(c) CH3−CHCl−CH2NO
(b) ON−CH2−CH2−CH2Cl
(d) CH2Cl−CHNO−CH2CH3
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12.
Chemistry
Blue solution of CuSO4 on treatment with excess of KCN gives colourless solution due to the
(a) formation of CuCN
(b) formation of Cu(OH)2
(c) formation of [Cu(CN)4]2−
(d) Cu2+ is reduced by CN− to Cu+ which forms complex [Cu(CN)4]3−
SECTION−II
*13.
Ag+ + NH3
[Ag(NH3)]+
[Ag(NH3)] + NH3
14.
15.
Over all formation constant of the complex [Ag(NH3)2]+ would be
(b) 1.08 × 103
(c) 6.8 × 106
(a) 1.08 × 107
(d) 6.8 × 107
If CO bond length equals 1.128 Å, then the possible value of CO bond length in Fe(CO)5 is
(a) 1.128 Å
(b) 1.10 Å
(c) 1.13 Å
(d) 1.115 Å
Identify the major products [P] and [Q] in the following reaction
+ CH3CH2CH2Cl
16.
Kf = 6.8 × 103
;
[Ag(NH3)2]+ ; K ′f = 1.6 × 103
+
anhydrous AlCl3
[P]
(i) O2
(ii) H3O+, ∆
[Q] + PhOH
(a)
and CH3COCH3
(b)
and CH3CH2CHO
(c)
and CH3COCH3
(d)
and CH3CH2CHO
Which of the following compound/s on treatment with aqueous alkali and on further acidification gives the following lactone
O
O?
COOMe
(b)
(c)
CH3
17.
CHO
COOMe
(a)
CHO
(d)
CHO
COOH
COOH
There is a compound of lowest molecular mass of open chain ketone. This compound and next homologue in this series
reacts with NH2OH to forms oximes. Which of the following is/are correct?
(a) Two oximes are formed.
(b) Three oximes are formed.
(c) One oxime is optically active.
(d) Two oximes are optically active.
gas C
gas A
Z
ideal gas
gas B
1
*18.
P
Which of the following statement is wrong?
(a) For gas A, a = 0 and Z will linearly depend on pressure.
(b) For gas B, b = 0 and Z will linearly depend on pressure.
(c) Gas C is a real gas and we can find ‘a’ and ‘b’ if intersection data is given.
(d) All vander Waal gases will behave like gas C and gives positive slope at high pressure.
*19.
CH3
hν
CH−CH2−CH3 + Cl2 ⎯→ ‘N’ isomers of C5H11Cl
CH3
Fractional
distillation
‘n’ distilled products
The values of ‘N’ and ‘n’ are
(a) 6, 6
(b) 6, 4
(c) 4, 4
(d) 3, 3
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*20.
Chemistry
Ammonical solution of MgSO4 in presence of NH4Cl is heated with Na2HPO4 a white crystalline precipitate is formed of
(b) Mg3(PO3)2
(c) MgSO4.MgCl2
(d) MgSO4.MgPO4
(a) MgNH4PO4
SECTION−III
PASSAGE− I
Mechanism of Hoffmann bromamide reaction was given as follows:
O
O
C−NH2
Cl
22.
23.
O
..
C−N−Br
..
C−NHBr
Cl
(I)
21.
O
N=C=O
Cl
Cl
(II)
Cl
(IV)
(III)
NH−C−O M+
Cl
(V)
NH2
(VI)
Which of the following is the reagent used from step (I) to step (II)?
(a) Br2 + NaOH
(b) KBr + NaOH
(c) KBr + NaOCH3
(d) NBS
Which of the following is the rate determining step?
(a) Formation of (II)
(b) Formation of (III)
(d) Formation of (V)
(c) Formation of (IV)
Which are the products formed on Hoffmann bromamide reaction of the compounds (VII) and (VIII)?
15
CONH2
D
CONH2
(VIII)
(VII)
15
NH2
(a)
15
NH2
NH2
(b)
+
NH2
+
D
D
D
15
NH2
+
(c)
NH2
+
15
NH2
NH2
+
15
NHD
+
(d)
NH2
D
D
D
D
PASSAGE − II
Tollen’s test is given by aldehydes
o
;
E red = +0.800 V
Ag+ + e− ⎯→ Ag
C6H12O6 + H2O ⎯→ C6H12O7 + 2H+ + 2e− ; E ox = −0.05 V
o
[Ag(NH3)2]+ + e− ⎯→ Ag + 2NH3 ; E red = 0.373 V
o
24.
25.
⎛ F ⎞
Use ⎜
⎟ = 38.9 V−1
RT
⎝
⎠
Calculate (ln K) for C6H12O6 + 2Ag+ + H2O ⎯→ C6H12O7 + 2H+ + 2Ag
(a) 55.6
(b) 29.6
(c) 66
On adding NH3, pH of the solution increases to 11 then, identify the effect on potential of half−cell.
o
(a) Eox increased from E ox by 0.65 V
o
(c) Ered increased from E red by 0.65 V
26.
(d) 58.35
o
(b) Eox decreased from E ox by 0.65 V
o
(d) Ered decreased from E red by 0.65 V
NH3 is used in this reaction rather than any other base. What is the correct reason for this?
(a) [Ag(NH3)2]+ is a weaker oxidizing agent than Ag+.
(b) NH3 prevents the decomposition of gluconic acids.
(c) Ag precipitates gluconic acid as it silver salt.
(d) NH3 changes the standard reduction potential of [Ag(NH3)2]+.
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Chemistry
PASSAGE − III
KCN
Ni ⎯⎯⎯→ complex 1 ; Ni ⎯⎯ ⎯ ⎯→ complex 2.
Both the above complexes have coordination number 4.
2+
2+
excess KCl
27.
The IUPAC names for the complexes are, respectively
(a) Potassium tetracyanonickelate(II) and potassium tetrachloronickelate(II)
(b) Potassium tetracyanonickel(II) and potassium tetrachloronickel(II)
(c) Potassium tetracyanonickel and potassium tetrachloronickel
(d) Potassium tetracyanonickelate(II) and potassium tetrachloronickel(II)
28.
Which option is correct regarding the magnetic behaviour of the two complexes?
(a) Both are diamagnetic.
(b) Both are paramagnetic.
(c) The cyano complex is diamagnetic and the chloro complex is paramagnetic.
(d) The cyano complex is paramagnetic and the chloro complex is diamagnetic.
29.
Hybridisation of Ni in the complexes is/are
(a) dsp2 in both
(c) dsp2 in cyano and sp3 in chloro complex
(b) sp3 in both
(d) sp3 in cyano and dsp2 in chloro complex
PASSAGE − IV
Due to the neutron present in cosmic rays, the following reaction takes place by the bombardment of nitrogen with neutrons.
+ 0n1 ⎯→ 6C14 + 1H1 ; 6C14 ⎯→ 7N14 + −1e0
Cosmic rays form 6C14 which get circulated in the atmosphere as well as in living species. Whenever there is a nuclear
explosion, the concentration of C14 increases both in the atmosphere as well as in living species.
When a species dies the C−14 concentration decreases and hence this decrease can be measured and the time estimated as to
when the organism died. For this we require (1) half life of C−14 (5760 years) (2) activity of C−14 in living species
(3) activity of C−14 in fossil.
Beyond 30,000 years the age of fossil cannot be accurately determined as the activity then would be too low.
14
7N
30.
In radiocarbon dating for finding the age of fossils, the most appropriate statement is
(a) During the life time C14 assimilated by the human beings is in equilibrium with the C14 that decomposes by β− emission.
At any time in a human life C14/C12 is always constant at a particular instant.
(b) C14 dating method is inappropriate for finding the life of a given sample because C14 undergoes β− emission and hence,
the ratio of C14 in human beings is not constant.
(c) Radiocarbon dating is inappropriate to determine the age because C14 concentration varies from place to place.
(d) None of the above.
31.
A fossil whose age can be determined using carbon dating must be
(a) 6 years old
(b) 6000 years old
(c) 60, 000 years old
(d) can be accurately calculated for any age of the sample
32.
Two organisms died on the same day. One died at a place where nuclear explosion had taken place while the other died at a
place where no such explosion had occurred. The ratio of C14 during life to that present in the fossil at an instant is C1 for the
former and C2 for the latter. The age of the former was calculated as t1 and for the latter as t2. The correct answer is
C
1
(a) There is an increase in t1 w.r.t. t2. t1 − t2 = ln 1
λ C2
(b) There is decrease in t1 w.r.t. t2. t1 − t2 =
(c)
C
1
ln 1
λ C2
t1
C
= 1
t2
C2
(d) Cannot be determined.
SECTION−IV
33.
AgBr is sparingly soluble in water. Ksp of AgBr is 12 × 10−14. AgNO3 solution of 10−7 M is added to the saturated solution of
AgBr. Calculate the specific conductance (conductivity) of the resultant solution in terms of 10−7 S m−1. Molar conductance
at infinite dilution of NO 3− , Ag+ and Br− is 7 × 10−3, 6 × 10−3 and 8 × 10−3 S m2 mol−1 respectively.
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Chemistry
34.
Molar mass of constituting particle of a crystalline solid is 75 g/mole, edge−length of unit cell is 5 Å and density of the
crystal is 2 g/cc. Determine radius of atoms of lattice points in pm. (NAV = 6 × 1023)
35.
On adding 75.2 g of phenol in 1 kg of a solvent (Kf = 14), depression in freezing point was observed as 7 K. Determine
degree of dimerisation of phenol.
*36.
2CO(g) + O2(g) ⎯→ 2CO2(g), the reaction is carried out under constant volume (1 L). 2 moles of CO and 1 mole of O2
produces 1 mole of CO2. At constant temperature (500 K), ∆H = −560 kJ/mol. What will be the internal energy change (∆U),
given that if pressure initially is 70 atm and finally it becomes 40 atm. Assume that the gases largely deviate from the ideal
gas behaviour. Given: 1 L atm = 0.1 kJ.
SECTION−V
*37.
(A)
(B)
(C)
(D)
*38.
Vn represents potential energy of nth orbit, Kn represents kinetic energy of nth orbit and En represents total energy of
the nth orbit of hydrogen atom.
(A)
Self reduction
Carbon reduction
Reduction by heating its iodide
Complex formation followed by precipitation
(p)
(q)
(r)
(s)
Vn
=?
Kn
(p) −2
(B) rn is proportional to (En)x
(C)
Copper extraction
Lead extraction
Silver extraction
Boron extraction
(q) 0
1
is proportional to (Z)y
Vn
(r) −1
(D) Orbital angular momentum in lowest energy orbit. (s) 1
Bi2O3 ⎯→
AlO2− ⎯→
SiO44− ⎯→
B4O72− ⎯→
BiO+
Al(OH)3
Si2O72−
B(OH)3
39.
(A)
(B)
(C)
(D)
*40.
(A) C6H5CH2CD2Br on reaction with
C2H5O− gives C6H5−CH=CD2
(B) PhCHBrCH3 and PhCHBrCD3, both
react with the same rate.
(C) C6H5CHDCH2Br on treatment with C2H5O−
and C2H5OD gives C6H5CD=CH2
(D) C6H5CH2CH2Br reacts faster than
C6H5CD2CH2Br on reaction with C2H5O−
in ethanol.
(p)
(q)
(r)
(s)
Hydrolysis
Dilution with water
Acidification
Heating
(p) E1
(q) E2
(r) E1CB
(s) First order reaction
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Chemistry
SOLUTIONS
SECTION−I
1.
∆SA → B = ∆SA → C + ∆SC → D + ∆SD → B = 50 + 30 +(−20) = 60 eu. ; Correct choice: (b).
*2.
Al(OH)3 forms gelatinous mass. ; Correct choice: (c).
*3.
B(OH)3 + OH−
[B(OH)4]−
OH
HO
+2
B
HO
−
OH
HO−CH2
CH2−O
HO−CH2
CH2−O
O−CH2
−
+ 2H2O
B
O−CH2
Due to the formation of Lewis acid−base complex ion, the equilibrium shifts in the forward direction. Since cis−diol and
trans−diol are given, it means these are cyclic diols. Correct choice: (a).
4.
None of the options given are correct. Benzene sulphonic acid would liberate CO2 as it is more acidic than H2CO3 while para
nitrophenol will not liberate any gas, as it is less acidic than H2CO3. Of the given options, (d) is the closest choice.
Correct choice: (d).
*5.:
H2O + CO2
H2CO3
H+ + HCO 3− ; HCO 3−
H+ + CO 32−
Correct choice: (a).
*6.:
Benzoyl chloride. ; Correct choice: (a).
*7.:
Equation of process PV−1 = constant
Molar heat capacity for the process =
R
R
R
R
+
=
= 2R.
+
5
γ −1
1 − (−1)
2
−1
3
Correct choice: (b).
*8.
Correct choice: (c).
9.
This is carbylamine reaction and the product is alkyl isocyanide. ; Correct choice: (d).
10.
Boiling points of the given compounds are: Phenol = 181.7°C, Catechol = 245.5°C, Resorcinol = 281°C and Quinol = 285°C.
Increase in the number of hydroxyl groups increases the boiling points. Among isomeric dihydroxy benzenes, as the distance
between hydroxy groups increases, the repulsion between them decreases and hence boiling point increases.
Correct choice: (a).
*11.
NOCl
NO+ + Cl−
⊕ ..
:N=O :
Cl
CH3CH=CH2 + ON−Cl
CH3CH=CH2 + Cl
CH3CH−CH2
NO
NO+
Correct choice: (c).
12.
2CuSO4 + 4KCN ⎯→ 2CuCN + (CN)2 + 2K2SO4
CuCN + 4KCN
⎯→ [Cu(CN)4]3− + 4K+ + CN−
Correct choice: (d).
SECTION−II
*13.
K = Kf × K ′f = 6.8 × 103 × 1.6 × 103 = 1.08 × 107 ; Correct choice: (a).
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14.
15.
Chemistry
Bond length of CO in Fe(CO)5 is greater then C≡O bond, because bond order of CO in complex decreases due to dπ−pπ
metal to ligand back bonding ; Correct choice: (c).
⊕
..
+ CH3CH2CH2−Cl
.. : + AlCl3
⊕
CH3−CH2−CH2−Cl−AlCl3
CH3−CH2−CH2 + AlCl4
Hydride shift
⊕
CH3−CH−CH3
CH3
CH3
⊕
CH3
+ CH
CH
−H+
CH3
Cumene
[P]
CH3
O−O−H
CH3
OH
O
H3O+
O2
+ CH3−C−CH3
Acetone
[Q]
Cumene hydroperoxide
Correct choice: (c).
O−
O
C
16.
C
H
H
OH
intramolecular
Cannizaro rxn.
O
O
C−OH
H
C−H
C
O
O
C
H−O
O
C
−H2O
O
OH
⊕
OH
C
⊕
O−H
−H
+
CH2
CH2
O
OH
H
H
IMPE
C
O− Na
CH2−OH
H3O+
O
OH
..
O
..
H+
H
C
OH
CH2−OH
Correct choice: (c).
O
17.
Lowest molecular mass open chain ketone ⎯→ CH3−C−CH3
O
Next homologue in this series ⎯→ CH3−C−CH2−CH3
Formation of oxime with NH2OH
..
−H2O
C 2H5
C
CH3 C2H5
C
C2H5
CH3
anti (C2H5)
NH2OH
CH3−C=O
−H2O
CH3
CH3
CH3
OH
N
N
NH2OH
CH3−C=O
..
OH
syn.(C2H5)
OH
C=N
..
Correct choice: (b).
*18.
a
Pb
. Therefore, Z does not vary linearly with pressure. For gas A, a = 0, Z = 1 +
implies Z
VRT
RT
varies linearly with pressure. Given the intersection data for gas C, it is possible to find the values of ‘a’ and ‘b’. All vander
Waal gases like gas C give positive slope at high pressures ; Correct choice: (b).
For gas B, b = 0, Z = 1 −
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Chemistry
Cl
*19.
Cl2/ hν
CH3−CH−CH2−CH3
CH3−CH−CH2−CH2−Cl + CH3−CH−CH−CH3 + CH3−C−CH2−CH3
CH3
CH3
CH3 Cl
(d, l)
CH3
Cl
+
CH2−CH−CH2−CH3
CH3
(d, l)
Correct choice: (b).
*20.
White crystalline precipitate is MgNH4PO4.
Mg2+ + NH3 + HPO 24 − ⎯→ Mg(NH4)PO4↓
Correct choice: (a).
SECTION−III
PASSAGE− I
21.
Br2 + NaOH is the required reagent ; Correct choice: (a).
22.
Formation of compound (IV) will occur in the rate determining step. ; Correct choice: (c).
23.
Correct choice: (a).
PASSAGE − II
24.
o
o
E cell = E R − E L = 0.8 − (0.05) = 0.75 V
o
∆G cell = − 2FE cell = −RT ln K ; ∆G cell = −2F(0.75) = −RT ln K
o
o
o
2F
(0.75) = 2(38.9) (0.75) = 58.35.
RT
Correct choice: (d).
⇒
25.
ln K =
C6H12O6 + H2O ⎯→ C6H12O7 + 2H+ + 2e−
E
E° −
=
RT
2
2.303
ln[H + ] 2 = E° −
ln[H + ] = E° +
pH
2F
38.9 × 2
38.9
1.302
1.302
⎞
⎛ 2.303
× 11⎟ = E° +
; E − E° =
= 0.65 V.
E° + ⎜
38
.
9
2
2
⎠
⎝
Correct choice: (a).
E
26.
=
o
o
Since, E Ag + | Ag > E [ Ag ( NH
+
3 ]2 ] / Ag
.
Correct choice: (a).
PASSAGE − III
27.
⎯ ⎯KCl
⎯→ K2[NiCl4]
Ni2+ ⎯⎯⎯→ K2[Ni(CN)4] ; Ni2+ ⎯excess
Correct choice: (a).
28.
Ni2+
KCN
4s
3d
[Ni(CN)4]2−
4p
X
X
X
X
dsp2
Hence, [Ni(CN)4]2− is diamagnetic.
[NiCl4]2−
X
X
X
sp
X
3
2−
Hence, [NiCl4] is paramagnetic. ; Correct choice: (c).
29.:
Correct choice: (c).
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22
Chemistry
PASSAGE − IV
30.
(a) is incorrect because after nuclear explosion C14/C12 will not remain constant. (b) is incorrect because C14 will always
remain in equilibrium with the surroundings. ; Correct choice: (c).
31.
For effective carbon dating, the fossil must be minimum 1000 years old. ; Correct choice: (b).
1
1
ln C1 ; t2 =
ln C 2
λ
λ
1 C
t1 − t2 = ln 1
λ C2
t1 =
32.
Since due to nuclear explosion the initial activity AO would be greater ; it would take more time to reach the same value of
activity A. Correct choice: (a).
SECTION−IV
33.
In presence of 10−7 M AgNO3, Ag+ from AgBr would be calculated as
(x + 10−7) x = 12 × 10−14 ; x = [Ag+]AgBr = 3 × 10−7 M
∞
∞
Λ m (AgBr) = (6 × 10−3 + 8 × 10−3) = 14 × 10−3 S m2 mol−1. Λ m (AgNO 3 ) = (6 × 10−3 + 7 × 10−3) = 13 × 10−3 S m2 mol−1.
∞
κAgBr = Λ m (AgBr) × [AgBr] = 14 × 10−3 S m2 mol−1 × 3 × 10−7
∞
mol
dm
κ AgNO3 = Λ m (AgNO 3 ) × [AgNO3] = 13 × 10−3 S m2 mol−1 × 10−7
3
= 14 × 10−3 ×
mol
dm
3
=
3 ×10 −7
10
−3
13 ×10 −3 ×10 −7
10 −3
×
S m2
m3
= 42 × 10−7 S m−1
S m −1 = 13 × 10−7 S m−1
κsoln = κAgBr + KAgNO3 = (42 × 10−7 + 13 × 10−7) S m−1 = 55 × 10−7 S m−1.
34.
ρ=
n×M
N AV × a
3
; n=
For a BCC lattice,
35.
2 × 6 ×10 23 (5 ×10 −8 ) 3
=2 ;
75
3 a = 4r ;
r=
∴ Metal crystallizes in BCC structure.
3×5
= 2.165 Å = 216.5 pm.
4
Dimerization reaction of phenol would be represented as
2PhOH
(PhOH)2
α
(1 − α)
2
where ‘α’ is the degree of dimerization of phenol.
∆Tf = Kf × (molality)obs ; 7 = 14 ×
*36.
∆H = ∆U + ∆(PV) = ∆U + V∆P
75.2 ⎛ α ⎞
× ⎜1 − ⎟ ; α = 0.75.
94 ⎝ 2 ⎠
(Q ∆V = 0) ; ∆U = ∆H − V∆P = −560 − [1(40 − 70) × 0.1] = −560 + 3 = −557 kJ/mol.
SECTION−V
*37.
(A) :
(p), (q) ; (B) : (q) ; (C) : (s) ; (D) : (r), (p)
*38.
(A) :
(p) ; (B) : (r) ; (C) : (s) ; (D) : (q)
39.
(A) :
(p) ; (B) : (q) ; (C) : (s) ; (D) : (r)
*40.
(A) :
(q) ; (B) : (p), (s) ; (C) : (q) ; (D) : (q)
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East Delhi Centre : Ph.: 22792226-27, West Delhi Centre : Ph.: 25527517-18, North Delhi Centre : Ph.: 25221424-25, South Delhi Centre : Ph.: 26537392-95 Fax: 26537396