Unit IV Miscellaneous+Brainstorming
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Miscellaneous Questions
UNIT
4
Section A : Straight Objective Type 1.
Answer (1) No. of atoms in 1 gm of CO2 =
2.
3 N . 44 A
Answer (4) n factor of Cr2O72– is 6 equivalents of Cr2O72– = equivalents of N2H5+ = 0.136 moles of Cr2O72– =
3.
0.136 = 0.0227 6
Answer (3) Mass of ethanol needed = 1.2 × 46 = 55.2 Volume of ethanol measured out =
4.
mass of ethanol = 55.2 = 70 ml density of ethanol 0.7893
Answer (1) MnO2 + (NH4)2SO4 ⎯→ MnSO4 + (NH4)2S2O8 Let the moles of (NH4)2SO4 reacting with 1 mole of MnO2 are x equivalents of MnO2 = equivalents of (NH4)2 SO4 1×2=1×x x=2
5.
Answer (2) NaNO2 + NH4Cl ⎯→ NaCl + NH4NO2 Δ NH4NO2 ⎯⎯→ N2 + 2H2O
moles of NaNO2 = moles of NH4Cl =
10 limiting reagent 69 10 53.5
Volume of N2 at STP is = 6.
10 × 22400 ml = 3246.37 ml 69
Answer (3) Let the molar % of CH4 = x Maverage =
16 x + 32(100 − x ) = 20 100 16x + 3200 – 32x = 2000
16x = 1200
x = 1200 = 75% 16 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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7.
Answer (3) Br2 ⎯→ Br –,
equivalent mass =
M 2
Br2 ⎯→ Br O3–,
equivalent mass =
M 10
Net equilibrium mass =
6 × 160 = 96 gm 10
= 8.
M + M = 6M 2 10 10
Answer (1) KMnO4 + FeC2O4 ⎯→ Mn2+ + Fe3+ + CO2 n=5
n=3
equivalents of KMnO4 = equivalents of FeC2O4 moles × 5 = 1 × 3 moles of KMnO4 = 9.
3 5
Answer (3) KIO3 + KI ⎯→ I2 n=5 n=1 KIO3 + 5KI ⎯→ 3I2 moles of KIO3 =
2.14 = 0.01 214
Amount of KI consumed = 5 × 0.01 = 0.05 Moles of I2 formed = 0.01 × 3 = 0.03 10. Answer (3) SO3 + 2NaOH ⎯→ Na2SO4 + H2O equivalents of H2SO4 = equivalents of NaOH = 0.4 × 26.7 × 10–3 −3 mass of H2SO4 = 0.4 × 26.7 × 10 × 98 2
= 0.523 g Let the mass of H2SO4 is x g ∴ mass of SO3 is 0.5 – x SO3 + H2O
⎯→
H2SO4
80 g
98 g
(0.5 – x)
98 × (0.5 − x ) 80
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98 Total mass of H2SO4 = x + (0.5 – x) = 0.523 80 = x + 0.6125 – 1.225x = 0.523 0.225 x = 0.0895 x = 0.398 % of SO3 =
0.102 × 100 = 20.6% 0. 5
11. Answer (1) Since the given salt is isomorphous with K2SO4 ∴ Oxidation no. of element is +6 Hence atomic weight of element = 15 × 6 = 90 12. Answer (4)
n2 Z For first excited state n = 2. rn = 0.53 ×
For Li2+, Z = 3 2 r = 0.53 × 2 = 0.53 × 4 Å 3 3 13. Answer (4)
If we try to see it microwave or visible light that will be hopelessly inaccurate since wavelength of such lights are perhaps more than million times larger than a diameter of electron. So we might try to do better by using radiation of appropriately shorter wavelength in this case γ-ray. 14. Answer (1) −34 8 E = hc = 6.62 × 10 × −310× 10 λ 2450 × 10
= 8.10 × 10–19 J/atom = 488 kJ/mol = 116.2 kcal/mol Which is greater than bond energy of diatomic molecule hence bond will dissociate. 15. Answer (1) Magnetic moment =
n(n + 2) = 1.73
n=1 Therefore unpaired electrons is 1 Therefore configuration of Vx+ will be 4s 03d 1 i.e. V4+ 16. Answer (3) The capturing subatomic particle cause change in mass The reduced mass
mp .m s 1836m e × 207me m = m + m = 1836m + 207m p s e e m = 186 me
The ionisation energy E =
2π 2mk 2 e 4 h2
m 2π 2mk 2 e 4 · With reduced mass I.E = = 186 × 13.6 eV me h2
Increases by 186 times. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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17. Answer (3) The deflection of cathode rays by the electric and magnetic field clearly show that cathode rays consist of charged particle. 18. Answer (1) Cl(17) – 1s2, 2s2, 2p6, 3s2, 3p5 n = 3, l = 1, m = 1 Radial nodes
=n–l–1 =3– 1–1=1
Angular nodes = l = 1. 19. Answer (3) T∝
n3 Z2
TH = 2T nn3 ZH2
⇒
n3
=2· (1)3 (1) 2
Z2 = 2×
23 Z2
Z2 = 24 Z=4 20. Answer (4)
1
λ shortest
⎡ ⎤ = RZ 2 ⎢ 12 − 12 ⎥ ⎣⎢ n1 n2 ⎦⎥
⎡ ⎤ = 109678 × 1 ⎢ 12 − 12 ⎥ ∞ ⎦ ⎣2 = 3.647 × 10–5 cm = 3647 Å 21. Answer (2) Kinetic energy K = |E|
(E = total energy)
K = | – 3.4| = 3.4 eV de-Broglie wavelength λ=
h = P
h 2mK
~ 10 −10 m
22. Answer (4) −34 λ = h = 6.6 × 10 −24 mv 3.3 × 10
= 2 × 10–10 m =2Å Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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23. Answer (2) 2(2l + 1) where l is azimuthal quantum number. 24. Answer (3) Value of m cannot be greater than l. 25. Answer (2) No. of waves =
Circumfere nce = 2πr Wavelength λ
= 2πr = 2π (mvr ) h h / mv = 2π × nh h 2π n=3 26. Answer (1) Inert gases highest value of Ionisation energy. 27. Answer (4) Group 18 elements are called noble gas elements. 28. Answer (2) For isoelectronic species, the ionic radius increases as negative charge on ion increases. 29. Answer (2) Be(OH)2 and Al(OH)3 are amphoteric in nature. 30. Answer (3)
⎛ ch arg e ⎞ Polarising power ⎜ radius ⎟ is maximum for Al3+. ⎝ ⎠ 31. Answer (1) Li and Mg have diagonal relationship. This is due to similar polarising power of Li+ and Mg2+ ions. 32. Answer (3) Electronegativity increases in the period from left to right and decreases from top to bottom in group. 33. Answer (1) NH+4 , N = 5 + 4 − 1 = 4 attached atoms = 4 Tetrahedral bond angle 109º 28′. 2 2 NH3 , N = 5 + 3 = 4 attached atoms = 3 Trigonal pyramidal angle less than 109º 28′ due to presence of lone 2 2 pair
But in NF3 fluorine is more electronegative than Nitrogen therefore bond pair electron will be away from centre atom. ∴ Its bond angle is less than NH3. In case of N(CH3)3 due to steric hinderance its bond angle will be more than NH3 and NF3.
..
H | + N 9º 10
H
N H
′ 28
H
CH3
More than 107º less than 109º as methyl group is Bulkier than hydrogen
CH3
CH3
.. N F
.. less than 107º
F F
H
N 107º H H
34. Answer (3) In HNO2 and HNO3 nitrogen is sp2 hybridised state. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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35. Answer (1) As the molecule
Cl Cl
B–B
Cl Cl
is symmetrical and planar net dipole moment of Cl2 BBCl2 is zero.
36. Answer (4) Species
% s-character
BF4–
BF3
BF2+
sp3
sp2
sp
25
33
50
Order of increasing s-character is BF4– < BF3 < BF2+ 37. Answer (3) Fractional bond order means π-bond between the molecules has been delocalized e.g. benzene. In benzene bond order of C is 1.5 i.e. π-bond between the carbon atoms has been delocalised. 38. Answer (4) SCl4 has see saw shape whereas other has tetrahedral shape. 39. Answer (4) –
O | C
Bond order = 1 + –
O
O
1 = 1.33 3
In CO Bond order is 3. In CO2, O = C = O, Bond oder = 2. Higher bond order stronger bond and smaller bond length therefore the correct order of C – O bond length is CO < CO2 < CO32– 40. Answer (2) Due to polarity ICl has higher boiling point than Br2. 41. Answer (4) In PCl5, there is sp3d hybridisation orbitals used for hybridisation are s + p x + p y + pz + d z 2 . 42. Answer (1) Due to back bonding BF3 is weaker acid, among these back bonding is stronger in B – F. 43. Answer (2)
Cl | C Cl
Cl
Cl
net dipole moment μ = 0
Increasing order of dipole moment, CH3Cl > CH2Cl2 > CHCl3 > CCl4. 44. Answer (1) n1T1 = n2T2 n × 300 = n2 × 400 n2 =
3 n 4
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45. Answer (2) PV = nRT for unknown gas P × 0.44 =
0.2 × RT m
…(i)
P × 0.32 =
0.1 RT 44
…(ii)
dividing (i) by (ii) 44 2 = × 44 32 m
M = 64 ∴ gas is SO2. 46. Answer (4) Degree of freedom of mixture fmix =
n1f1 + n 2 f2 1× 3 + 1× 5 = n1 + n 2 2
f1, f2 are degree of freedom of gases. fmix = 4 γ mix = 1 +
2 2 = 1 + = 1.50 fmix 4
47. Answer (3) 3RT m
Vrms =
Vrms (H2 ) = Vrms (N2 )
7=
T=
TH2 2
×
TH2 × MN2 MH2 × TN2
28 TN2
TN2 = 2TH2 TH2 < TN2 48. Answer (2) Kinetic energy of an ideal gas depends only on temperature. 49. Answer (1)
PO2 =
w 32 w w + 16 32
⎛ 1⎞ × Ptotal = Ptotal ⎜ ⎟ ⎝3⎠
w = mass of methane = mass of oxygen ∴ fraction of total pressure is
1 3
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50. Answer (1) Preal < Pgas, due to force of attraction in gas molecules. 51. Answer (1) VH2
MCH4
=
VCH4
MH2
=
16 2
VH2 = 2 2 VCH4
VH2 > VCH4 ∴ Balloon will enlarge. 52. Answer (2) P=
dRT M
d=
PM RT
d ∝ P, d ∝
1 T
53. Answer (2) For any reaction be spontaneous, ΔG should be negative Now ΔG = ΔH – TΔS For given process ΔH is positive and ΔS is also positive therefore for ΔG to be negative ΔH < TΔS. 54. Answer (2) 3CO2(g) → 3C(g) + 2O3(g) ΔH = 3ΔHatomization C(s) → C(g) + 2ΔHf° O3(g) – 3ΔHf° CO2(g) = 3ΔHatomization C(s) → C(g) + 2ΔHf°O3(g) + 3 × 407 ∴ ΔH will be more than 1221 kJ. 55. Answer (2) Boiling point of liquid =
50000 = 454.5 K 110
56. Answer (1) w = P(V2 – V1) = nRT2 – nRT1 T2 – T1 =
w nR
Q = nCp (T2 – T1) 500 = nCp 500 = Cp =
w nR
Cp w R
500R 7 = 3.5R = R 142.5 2
Hence, gas is diatomic i.e. O2. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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57. Answer (4) During isothermal expansion work done W = nRT + loge
While during adiabatic expansion work done W =
Vf we also know for isothermal change Vi
⎛ Vf P ⎞ ⎜⎜ = i ⎟⎟ V P f ⎠ ⎝ i
Pf Vf − Pi Vi γ −1
We see in both these cases, workdone depends on pressure. 58. Answer (2) For triatomic gas γ =
4 3
TVγ – 1 = constant n=γ–1=
4 1 –1= = 0.33. 3 3
59. Answer (2) ΔH° = 2 × ΔHf° (HF) – 2 × ΔHf° (HCl) – 84.4 = 2 × (– 64.2) – 2 × ΔHf° (HCl) ΔHf° (HCl) = – 22 kcal/mol ΔHf° (HCl) = −
22 kcal/g 36.5
ΔHf° (HCl) = – 0.603 kcal/g 60. Answer (3) We know when reaction is carried out in a closed vessel of fixed volume ∴ ΔV = 0 W = PΔV = 0 Ca(s) + 2HCl(aq) → CaCl2(aq) + H2(g) From stoichiometry we know the number of moles of H2 produced =
Volume of hydrogen gas produced at 27°C =
4 moles = 0.1. 40
nRT 0.1× 0.082 × 300 = = 3 litre P 0.821
Change in volume during the reaction, ΔV = 3 – 0 = 3 litre Work done by the system = –PΔV = –0.821 × 3 = –2.463 atm × litre. 61. Answer (1) Number of equivalents of H2SO4 taken = Number of equivalents of KOH added =
0.2 × 2 × 400 = 0.16 1000
600 × 0.1 = 0.06 1000
Number of equivalents of acid and bases which neutralized each other = 0.06 Heat evolved = 0.06 × 57.1 = 3.426 kJ. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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62. Answer (2)
Initially At equilibrium
+
1
H2O (g)
H2
2
0
0
2–x
x
x/2
2 O 2 ( g)
1
⎛ x ⎞2 x⎜ ⎟ ⎝ 2 ⎠ = 6.4 × 10 −5 2−x
Dissociation of H2O is so small so that we can assume PH2O = 2 atm 2−x~ −2 x 3 / 2 = 2 2 × 6.4 × 10 −5
∴
x = 0.0032
0.0032 2 × 100 = 0.08% % of O2 exist in steam = 2 63. Answer (3) pH + pOH = 14 pOH = 4 [OH–] = 10–4 BOH
B+ + OH–
initially moles
0.2
0
0
at equilibrium
0.2 – x
x
x
x = 10–4 Kb =
[B + ][OH− ] (10 −4 )(10 −4 ) = [BOH] (0.2 − 10 − 4 )
~ 5 × 10–8.
64. Answer (3) Solution of NaCl is neutral pH = 7 Solution of NH4Cl is acidic pH < 7 Solution of NaCN is basic pH > 7 HCl is strong acid therefore its pH will be less among all of these. 65. Answer (2) Since ionic product of water at 100°C is 1 × 10–12 ∴ water will be neutral at pH = 6 since pH is 7 therefore solution will be alkaline. 66. Answer (4) No affect because buffer solution resist to change in its pH value. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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67. Answer (4) H2(g)
+
Br2(g)
2HBr
0.8
0.3
0.1
0.8 – x
0.3 – x
0.1 + 2x
Since Kc is very large x ~ 0.3 ∴ Amount of HBr at equilibrium = 0.7 moles. 68. Answer (3) OH–
NH4OH
NH4+
10–6
0
0
10–6 α
10–6 α
At t = 0
At equil. 10–6(1 – α)
(10 α )(10 α ) −6
2 × 10–5 =
+
10
−6
−6
(1 − α )
On calculation we get α = 0.954 [OH–] = 0.954 × 10–6 pOH = – log (0.954 × 10–6) = 6.02 pH = 14 – 6.02 = 7.98. 69. Answer (2) N2(g) Initially At equilibrium x = 1×
+
3H2(g)
2NH3(g)
2
6
0
(2 – x)
(6 – 3x)
2x
30 = 0 .3 100
Initial mol
=2+6+0=8
Mol at equilibrium = 1.7 + 5.1 + 0.6 = 7.4 Mol ∝ Volume Hence, Vf 7.4 = = 0.93. Vi 8
70. Answer (1) H3O+
H2O
+
H+
2H2O
H3O+
+
OH–
Kionisation K eq =
Kw (55.56 )2
--------------------------------------------------------------------------------------------------------------------------------------------
H+
H2O
+
OH–
Kw Kw = K ionisation × 55.56 (55.56) 2
Kionisation = 55.56 pKa for H3O+ = – log[H3O+] = – log 55.56 = –1.74. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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71. Answer (2) [Ca(OH)2]
=
5 × 10 −4 × 1000 = 5 × 10 −3 100
[OH–]
= 2 × 5 × 10–3 = 10–2 M
[H+]
=
pH
= – log 10–12 = 12.
10 −14 = 10 −12 M 10 − 2
72. Answer (2) Zn2+ + 2H2O Kh =
[ Zn(OH) 2 ][H+ ] [ Zn 2+ ]
Zn2+ + 2OH–
Zn(OH)2 Kb =
Zn(OH)2 + 2H+
[ Zn 2+ ][OH− ]2 [ Zn(OH)2 ]
Kw = [H+] [OH–] 2
Kh =
Kw . Kb
73. Answer (2) λeq (BaSO4)
=
1000 × K N
Normality
=
1000 × 8 × 10 −5 = 2 × 10 − 4 400
Molarity
=
2 × 10 −4 = 10 − 4 M i.e. [Ba2+] = [SO 24− ] = 10–4 M 2
Solubility product Ksp = [Ba2+] [SO24− ] = (10–4)2 = 10–8 M2 74. Answer (3) This is metal-insoluble salt-anion electrode. 75. Answer (2) The electroplating reaction would be Cr 3 + + 3e – ⎯⎯→ Cr
Let the current is passed for t hours ∴
30 × 75 t × 3600 1 × × × 52 = 39 100 96500 3
t = 2.68 hours Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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76. Answer (4) 2H2O(l ) → 2H2 + O2 Q 3 mole of gases are formed from 2 mole of water ∴ 1 mole of gases are formed from
2 mole of water 3
77. Answer (4) Fe3+ + 3e– → Fe; ΔG1o = –3F(–0.36)
…(i)
Fe2+ + 2e– → Fe; ΔGo2 = –2F(–0.439)
…(ii)
o Fe3+ + e– → Fe2+; ΔG3 = –1F E°
…(iii)
∴
ΔGo3 = ΔG1o − ΔGo2 –F E° = –3F(–0.36) + 2F(–0.439) E° = –3 × 0.36 + 2(0.439) E° = –0.202 volt
78. Answer (3) Equivalent mass of metal =
m(96500 ) C× t
79. Answer (4) Fe2+ + Sn → Fe + Sn2+ o o E ocell = ERP cathode − ERP anode = –0.44 – (–0.14) = –0.30 V.
80. Answer (2) Specific conductance of AgCl = 1.86 × 10–6 – 6 × 10–8 = 1.8 × 10–6
Λ =
N=
1000 × K N 1000 × 1.8 × 10 −6 1.8 × 10 −3 = = 0.013 × 10 −3 = 1.3 × 10 −5 mol / litre 137.25 137 .25
81. Answer (3) Pt H2(g)|HCl(sol)||AgCl(s)|Ag 82. Answer (1) Cu2+(aq) + 2OH–(aq)
Cu(OH)2(s)
Ksp = [Cu2+][OH–]2 [H+][OH–] = 10–14 ∴ [OH–] =
10 −14 10 −14
=1
[Cu2+] = 1 × 10–19 Cu2+ + 2e → Cu(s) Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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0.059 1 log 2 [Cu 2+ ]
E = E° –
= 0.34 –
0.059 1 ⎛ ⎞ log⎜ ⎟ −19 2 ⎝ 1× 10 ⎠
= 0.34 –
0.059 × 19 2
= –0.22 V. 83. Answer (2) H2 → 2H+ + 2e E=0–
0.059 [H+ ] 2 log 2 p H2
0.3 = +
0.059 pH 1
pH =
0.3 ~5 0.059
84. Answer (4) For the metal – insoluble metal salt anion electrode 0.059 log K sp 1
E = E° +
0.22 = 0.80 +
–0.58 =
0.059 log K sp 1
0.059 log K sp 1
logKsp = –9.83 ∴ Ksp = 1.5 × 10–10. 85. Answer (1) 2Na+ + SO42–
Na2SO4 2H2O
2H+ + 2OH–
At Anode
2OH– → 2OH + 2e 2OH → H2O +
At Cathode
1 O ↑ 2 2
2H+ + 2e– → 2H 2H → H2 ↑
So, products at cathode and anode respectively are H2 and O2. 86. Answer (1) On placing one body diagonal, 4 corners, 2 edges, 2 faces and 1 body atom is removed. Because (B) atoms are presents at opposite face centres then it may or may not be removed. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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87. Answer (2) ⎛ r + ⎞ 90 Radius ratio ⎜ − ⎟ = 195 = 0.46 ⎝r ⎠
This shows that co-ordination number is 6 and the position of an anion may be octahedral void. 88. Answer (2) For BCC structure,
r =
a 3 400 × 1.732 = 4 4
= 173.2 pm. 89. Answer (2) Q 63.5 g Cu contains
6.02 × 10 23 Unit cell 4
∴ 1 g Cu contains
6.02 × 10 23 × 1 4 × 63.5 = 2.37 × 1021 Unit cell.
90. Answer (1) 6 body diagonal planes and 4 body diagonal are present in a cube. 91. Answer (2) Factual type 92. Answer (4) 1 mol AlCl3 developed 2 mol of cation vacancies after doping in 1 mol NaCl. 93. Answer (3) 1 ×8 = 1 8 1 Number of Ag atom = × 6 = 3 2
Number of Cu atom =
Number of Au atom =
1 × 12 = 3 4
Hence, formula of alloy is CuAg3Au3. 94. Answer (2) Factual type 95. Answer (3) Frenkel defect shows increase in dielectric constant. 96. Answer (1) Factual type 97. Answer (3) x t For the completion of reaction, x = a then
For zero order reaction, K =
t=
a K
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98. Answer (3) N2 + 3H2
−
K1 K2
2NH3
1 d [NH3 ] = Rate 2 − Rate1 2 dt
= K2[NH3]2 – K1[N2][H2]3 −
d [NH3 ] dt
= 2K2[NH3]2 – 2K1[N2][H2]3
99. Answer (2) For nth fraction, x =
1 &a =1 n
K=
2.303 a log t (a − x )
K=
2.303 1 log 1⎞ t ⎛ ⎜1 − ⎟ n ⎝ ⎠
t=
2.303 n log K (n − 1)
100. Answer (1) K1 =
2.303 100 log 24 25
…(1)
K2 =
2.303 100 log 60 (100 − x )
…(2)
K1 = K2
2.303 2.303 100 log 4 = log 24 60 (100 − x ) log
100 = 1.505 (100 − x )
log100 – log(100 – x) = 1.505 So,
(100 – x) = Antilog of 0.505
101. Answer (1) After 30 minutes, The number of atoms of x will be 1/8th of initial amount and the number of atoms of y will be 1/4th of initial amount.
Number of atoms ' x' left 1/ 8 = = 0. 5 Number of atoms of ' y' left 1/ 4 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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102. Answer (3) Order of reaction is an experimental verification. 103. Answer (4) From equation (i),
[ A ]2 [A 2 ]
K eq =
[A] = Keq. [A2]½ Slow step is the rate determining step Rate = K [A] [B2] Rate = K · Keq. [A]½ [B2] Order of reaction =
1 + 1 = 1½ 2
104. Answer (4) Total time
= n × t½
96
= n × 24
n
=
96 =4 24
4
10 ⎛ 1⎞ N = 10 ⎜ ⎟ = 16 ⎝2⎠
N = 0.625 M 105. Answer (1) In experiment (1) & (3), rate becomes constant. So, the order w.r.t. B is zero. In experiment (2) & (3), the concentration of A decreases 3 times then rate also decreases 3 times. So, the order w.r.t. A is 1. Rate = K [A] 106. Answer (2) From Roult’s law, P − Ps 18 = m× Ps 1000
m=
(m = molality)
100 − 80 × 55.5 80
107. Answer (1) 1 M means that 1 mole NaNO3(85 g) is dissolved in 1000 ml solution. Mass of solution = 1000 × 1.25 = 1250 g Mass of solvent = 1250 – 85 = 1165 g Molality (m) =
1 1000 = = 0.86 m. 1165 1165 1000
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108. Answer (1) P = i × cRT 8.21 = i ×
1 × 0.0821 × 300 6
i = 2. For BaCl2 BaCl2
Ba2+ + 2Cl–
At t = 0,
1
0
0
At t = t,
(1 – x)
x
2x
i=
1 + 2x =2 1
x = 0.5 Hence, percentage degree of dissociation = 50%. 109. Answer (4) Higher is the freezing point lesser is the lowering in freezing point. So, in ‘B’ case lowering in freezing point is less in comparison to A. It shows that if ‘A’ is normal state then B will be in associated state. 110. Answer (3) 2CH3COOH
(CH3COOH)2
At t = 0,
1
0
At t = t
(1 – 2x)
x
i=
1+ x = 0.6 1
x = 0.4 Degree of dimerisation = 2x = 2 × 0.4 = 0.8 Percentage of dimerisation = 80%. 111. Answer (2) (i) Effective molarity = 0.01 × 2 = 0.02 (ii) Effective molarity = 0.05 × 1 = 0.05 (iii) Effective molarity = 0.01 × 3 = 0.03 (iv) Effective molarity = 0.02 × 2 = 0.04. Higher is the effective molarity, more is the boiling point of solution. Hence, order of boiling point becomes (ii) > (iv) > (iii) > (i). 112. Answer (3) Mole fraction of benzene and toluene in liquid phase are 0.5 and 0.5 because equal moles of benzene and toluene are mixed. According to Roult’s law, o o PT = Pbenzene × Xbenzene + Ptoluene × X toluene
PT = 700 × 0.5 + 600 × 0.5 TT = 350 + 300 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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PT = 650 mm of Hg. Mole fraction of benzene in vapour state =
Pbenzene 350 = = 0.54 PT 650
113. Answer (3) According to Freundlich adsorption isotherm, 1
x = K.P n m
114. Answer (2) A positively charged sol of hydrated ferric hydroxide is obtained due to adsorption of Fe3+ ions. 115. Answer (4) Moles of oxalic acid adsorbed = 50 × 0.5 × 10–3 = 2.5 × 10–2. Weight of oxalic acid = 2.5 × 10–2 × 126 = 3.15 g The amount of oxalic acid adsorbed per gm of carbon = 3.15 × 2 = 6.3 g. 116. Answer (2) NO2 group is electron withdrawing group. CH3 group is electron donating group. 117. Answer (2) Depends on stability of carbanion Allylic carbanion stabilize by resonance. 118. Answer (3) Depends on nature of leaving group, order of leaving ability of group I > Br > Cl > F 119. Answer (3) N2 is good leaving group. 120. Answer (4) III carbanion becomes aromatic & II carbanion becomes antiaromatic ∴ order is III > I > II 121. Answer (2) 2° carbocation is more stable than 1° carbocation. 122. Answer (4) (4) option antiaromatic obey condition of 4n rule. 123. Answer (2)
O
All ring are aromatic and max dipole moment due to C – O Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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124. Answer (3)
Br
Br Aromatic and stable
125. Answer (3) Planar structure with 2 ring 126. Answer (2) Obeys 4nπ rule with planar 127. Answer (1) Least acidic among all options
pk a ∝
1 acidic strength
128. Answer (4) Electrophile attack on CH2 == CH2 129. Answer (4) Phenoxide ion stabilize by resonance 130. Answer (1) Double bond attached with ring can’t form Cis-trans isomer. So, only two Cis-trans isomers are possible due to middle double bond. 131. Answer (4) Both show same configuration 132. Answer (4) Order of - I effect
NMe3 > NHMe2 > NMeH2 > NH3 133. Answer (2)
O–δ CH3 – C – H +δ
H +δ –δ + H – C ≡ CD
CH3 – C – C ≡ CD OH
134. Answer (1)
CH CH
Cu2Cl2 NH4Cl
CH2 == C – C ≡ C – H H
H2 / Pd / BaSO4
(B) CH + CH
Diels Alder Reaction
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135. Answer (1)
H2C == CH2 + Br2
H2C ==== CH2 + Br +δ Br Bromonium ion
CH3OH
136. Answer (3) Due to ring expansion and final product by Saytzeff’s rule 137. Answer (2)
H
H H2/Pd
CH2 == CH – C – COOH
C2H5 – C – CH3
C2H5
C2H5
Optically active (X)
Optically inactive
138. Answer (3) Anti Markownikov’s addition 139. Answer (3)
CH4 + 2O2 ⎯⎯→ CO2 + 1 V ( g) 2 V ( g) 1 V ( g) 20 CC 50 CC 0 0 10 CC 20 CC
2H2O (l )
Volume left = 10 + 20 = 30 140. Answer (3) 1.79 m substances evolve 1.34 ml of CH4 90 g will evolve
1.34 × 90 = 67.34 l 1.79
i.e. 3 times of 22.4 l
∴ 3 mole active H
141. Answer (3) CH3
CH3
CH3 – C == CH2
KMnO4 Heat
CH3 – C == O + CO2 + H2O
142. Answer (4) Peroxy acid give anti addition
CHCl3 + NaOH
CCl2 + NaCl + H2O dichloro carbonate
143. Answer (3) CH2 CH2
+ C
Cl Cl
CH2 C CH2
Cl Cl
Mg ether
H 2C C H 2C
144. Answer (4) Meso gives trans 2-butene Racemic mixture gives cis 2-butene Because debromination is anti elimination. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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145. Answer (2) I2 is stable than HI 146. Answer (4) Intermediate free radical is formed in all option 147. Answer (2) R – H + Cl2 → R – Cl + HCl (substitution). 148. Answer (3) Br is better leaving group than Cl and Allylic carbocation stabilize by resonance. 149. Answer (3)
NH2
N2Cl HNO2/DCl
H3PO2
CH3
CH3
+ N2 + HCl CH3
150. Answer (3)
COOH
COOH
COOH
COOH
(C)
(B)
(D)
151. Answer (4) Intermediate carbocation, rearranged product is obtained. 152. Answer (2) Bromination is more selective order of bromination for 3°H > 2° > 1°H 153. Answer (3) Option (3) decolourises Br2 water due to presence of –CH == CH2 group but does not give ppt. with Ammonical AgNO3 154. Answer (3) 2.0 g Br2 react with 0.7 g of hydrocarbon 160 g Br2 react with
0.7 × 160 = 56 g 2
155. Answer (3) +
–
I + Cl electrophilic substitution takes place
ICl
(electrophile)
156. Answer (3) Carbocation followed by rearrangement 157. Answer (1)
Cl H
Cl H +
Cl – C – C == CH2 Cl
H
Cl – C – C – CH2
H Cl
–
Cl3C – C – CH2Cl
Cl H
H
Stable carbocation
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158. Answer (1)
H2C == CH unstable carbocation 159. Answer (4) NO2 is ambident nucleophile and KNO2 is ionic +
K
O – N == O
160. Answer (2) Due to its symmetrical structure 161. Answer (4)
Cl H–C
OH
Cl Cl
H–C
+ KOH
HCOOH
OH OH
KOH
HCOOK
162. Answer (3) Depends on stability of carbocation by resonance and hyperconjugation 163. Answer (3)
CH
CH3
+ HBr
Zn dust
CHBr2
CH
CH3 CH == CH – CH3 But-2-ene
164. Answer (1)
Cl +
Cl
165. Answer (1)
OH +
H
H3C – C == CH2
CH3 – C – CH3
Ph
H2O +
–H
CH3 – C – CH3
Ph
Ph
166. Answer (4) Ketone with grignard reagent produces 3° alcohol 167. Answer (2)
Cl
Cl –
+ OH nucleophile
OH
OH –Cl
168. Answer (3) C2H5Os is less stable than CH3Os 169. Answer (2) Carbanion as intermediate Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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170. Answer (3) ClCH2 – O – CH == CH2 (chloromethyl vinyl ether) gives white ppt. with alc. AgNO3 171. Answer (4)
Ph HI
CH3 – O – CH – CH3
CH3I + CH – CH3 CH3 (Y)
CH3
NaOI
(X)
CH3I + CH3COONa 172. Answer (2)
O
O
(1)
CH2 — C O CH2 — C
(2) AlCl3
C – OH
+
(3) H2O/H
CH2 Zn(Hg)/HCl
C — CH2
O
(A)
O O O
O
H3PO4 –H2O
H CH2
(C)
CH2
C — OH CH2 (B)
173. Answer (1) R – X + NaOR +
–
R – O Na + X – R
R – O – R + NaX
174. Answer (1)
CHO
CH2OH
(CHOH)4 CH2OH
CO (CHOH)3
glucose
CH2OH fructose
both required 5HIO4 molecule for cleavage ∴ ratio 1 : 1 175. Answer (4) + Hg (OAC)2
CH3OH
OCH3
–
NaBH4/OH
OCH3
Hg – OAC Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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176. Answer (3)
O
H
O–H
H
O
O NH2OH
+
OH H2O
N
H
OH N—H
NH2OH
H
N – OH
H – N – OH
177. Answer (4) Cl is electron withdrawing group 178. Answer (2)
O
O
O
O
OH
O H N O H CH3
+ CH3NH2
O
N – CH3 O
H
179. Answer (2) Cl is good leaving group and CH3COCl is vigorous acylating group. 180. Answer (1) Intramolecular aldol condensation takes place. 181. Answer (1)
O Me – C – Cl + Nu
O
O
Me – C – Cl
Me – C + Cl Nu
Nu Cl– is a good leaving group 182. Answer (4)
–I effect producing group at α position increases acidic strength. 183. Answer (2) Friedel craft does not takes place in nitrobenzene due to –I effect & – R effect of NO2 group and acetophenone is M-directing Only 2 option is correct 184. Answer (4)
H C2H5 – C == O
H KCN H2SO4
C2H5 – C – OH
H LiAlH4
C2H5 – C – OH
CN
CH2NH2
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185. Answer (3)
O CH3 – C
(i)
C — OEt
C
O
H
(ii) Ph – C – C – CH3
Base
O
CH3 O
Base
CH2
H
H bonding
H
O
C
OEt
CH enal O
Ph – C – C – CH3
H
Stable carbanion NaOI
(iii) ICHr2COCH 2CH3 ⎯⎯⎯→ CHI3 + CH3CH2COONa Iodoform
186. Answer (4) PhCH2COOH + NaHCO3 ⎯⎯→ PhCH2COONa + H2O + CO2
187. Answer (2)
CH2COCl (CH2)2
CH2
CH2COOH
+
H3O
Δ –CO2 –H2O
(CH2)2
CH2COCl
CH2COOH
CH2
CH2
C == O CH2
188. Answer (4)
O O
H
C — NH2
NH2
C N
Br2
CH3
CH3COCl
KOH
189. Answer (3) It has 2 NH2 group and one acidic group 190. Answer (2) +
Δ H / H2O C6H5NH2 + CHCl3 + KOH ⎯⎯→ C6H5NC ⎯⎯ ⎯⎯→ C6H5NH2 + HCOOH
191. Answer (3)
sp2 hybridised N has high electronegativity so less basic 192. Answer (1)
O
O
O
–
R – C – NH2
OBr – OH
–
R – C – N – Br
R – C – N – Br
–
R – N == C == O
OH
–2
R – NH2 + CO3
H 193. Answer (1)
N
N
O
O lone pair of electron involve in resonance Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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194. Answer (3) Due to more stability of benzyl carbocation 195. Answer (4)
O
O
C
C N–K
R–X –KX
COOH N–R
C
C
O 196. Answer (3)
O
H2O
RNH2 + COOH
LiAlH4 reduces — CONH2 to — CH2NH2 excess of H2 reduces phenyl group and CONH2 197. Answer (1) +I effect of CH3 increases basic strength in option (1) 198. Answer (4) NaOBr C6H5CH2CONH2 ⎯⎯ ⎯ ⎯→ C6H5CH2NH2
199. Answer (1) C≡N
LiAlH4
CH2NH2
CH3
CH3l
N
CH3 I
Δ AgOH
CH2 + (CH3)3N + AgI
CH3
200. Answer (1)
O
O–H +
CH3 – C – NH2 + H
CH3 – C – NH2
OH CH3 – C – NH2
201. Answer (3) KCN + H2O ⎯⎯→ K + + CN−
CN – + H2O
HCN + OH–
KCN is a salt of strong base and weak acid it undergo salt hydrolysis 202. Answer (4) Amines are basic Amides are amphoteric Ketones are neutral (with α H acidic) Nitriles → acidic R – C ≡ N 203. Answer (2)
H N (a) N (b) H
HCl 1 mole
(b) is more basic ∴ HCl attack on (b) a lone pair of (a) involves in resonance with benzene ring ∴ less basic. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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204. Answer (3) CN
COOH
CH2Br
CH2OH
–CN group undergoes hydrolysis, by reaction with alkali 205. Answer (1) Acid - base reaction 206. Answer (4) Beckman’s rearrangement 207. Answer (1) α-keratin is a water insoluble fibrous protein present in nails, hair etc. 208. Answer (4) Poly ethylidine chloride does not have a chiral carbon so it does not exist in these forms 209. Answer (3) Natural rubber is a polymer of isoprene
CH2 == C – CH == CH2 i.e.
CH3
210. Answer (1) Monomers are condensed with elimination of small molecules eg H2O, NH3 etc. and it is polymerised by the formation of cation during chain growth polymerisation. 211. Answer (2)
HO galactose unit
CH2OH O H H H
H OH
OH
OH H
H
OH
O H
H
H
OH
O H CH2OH
glucose unit
Lactose 212. Answer (3) Fibroin has globular structure 213. Answer (3) Nucleic acids (DNA, RNA) has phosphate sugar - base - sequence 214. Answer (2) At isoelectric point there is no migration of ions towards their electrodes and solubility is minimum 215. Answer (3)
CH3 C == O HIO4
H – C – OH H3C – C == O
CH3COOH + HCOOH + CH3COOH
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216. Answer (4) The dielectric constants of pure H2O2 and 65% H2O2 solution are 93 and 120 respectively, which is greater than that of water (about 82). It is not used as a solvent because of its strongly oxidising nature and its readily decomposition even in the presence of traces of many heavy metal ions. MnO4– ion is a strong oxidising agent with which H2O2 acts as a reducing agent.
H2O2 + 2H+ + 2e − → 2H2O;
E° = 1.77 V
O 2 + 2H+ + 2e − → H2O 2 ;
E° = 0.68 V
−
HO 2 + H2O + 2e − → 3OH− ;
E° = 0.87 V
217. Answer (4) Q 1000 g aqueous solution contains 10 g of carbonate
∴ 106 g aqueous solution contains
10 × 10 6 of carbonate 1000
= 10,000 ppm. 218. Answer (1) Presence of MnO2, carbon or finely divided metals accelerate the decomposition of H2O2. Therefore, these substances are called positive catalysts. 219. Answer (3) The two O – H bonds in H2O2 are in different planes due to repulsion between different bonding and antibonding orbitals. 220. Answer (2) When anhydrous CuSO4 absorbs water then it becomes blue due to formation of CuSO4.5H2O. 221. Answer (3) H2 gas doesn’t reduce the oxides of highly electropositive metals. 222. Answer (2) Meq. of KI
= Meq. of H2O2 in 50 ml = Meq. of Na2S2O3
W × 1000 34 / 2
= 20 × 0.1
WH2O2 in 50 ml
=
20 × 0.1× 34 2000
= 0.034 g 0.034 × 1000 ∴ WH2O2 in 1000 ml = 50
= 0.68 g 223. Answer (4) CrO5 is blue. On standing CrO5 changes to Cr2(SO4)3 which is green. 224. Answer (2) Be does not form peroxide or superoxide. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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225. Answer (4) Among alkali metals, only Li forms stable nitride. 226. Answer (1) Δ Li2CO3 ⎯⎯→ Li2O + CO2 Δ Na 2CO 3 ⎯⎯→ no effect
CO2 is due to Li2CO3 only. Hence, 0.75 mol CO2 = 0.75 mol of Li2CO3. 75 mol% Li2CO3 25 mol% Na2CO3 227. Answer (2)
Na 2SO 4 + 2C → Na2S+ 2CO 2 (A)
Na [Fe( CN) NO]
Na 2S ⎯⎯2⎯ ⎯ ⎯5 ⎯⎯→ Na 4 [Fe(CN)5 NOS] (A)
(B ) CdCO
3 Na 2S ⎯⎯ ⎯⎯ → CdS
(A)
( C)
228. Answer (1) In the mixture, only NaHCO3 decomposed to give CO2, Δ 2NaHCO 3 ⎯⎯→ Na 2 CO3 + H2 O + CO 2 1 mole
2 mole
Thus, in the mixture, there is only one mole of Na2CO3. ∴ Mole percentage of Na2CO3 in the mixture =
1 × 100 3
= 33.33. 229. Answer (1) 2Mg + O 2 → 2MgO MgO + H2 O → Mg(OH)2 3Mg + N2 → Mg3N2 Mg3N2 + 6H2 O → 3Mg(OH)2 + 2NH3
230. Answer (1) MgCl2.6H2O is a colourless, crystalline, deliquescent substance soluble in water. On heating it undergoes partial hydrolysis with evolution of hydrogen chloride.
MgCl2.6H2O
Mg(OH)Cl + HCl Δ MgO
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231. Answer (2) In the vapour state at high temperature Be is sp-hybridised and hence, BeCl2 is linear. In the solid state, BeCl2 is polymeric. In this structure each Be atom has two covalent and two co-ordinate bonds. Therefore, in the solid state Be has sp3–hybridisation. 232. Answer (1) Alkali metal carbonates such as Na2CO3, K2CO3 don’t decompose to give CO2 on heating. CO2 is neither combustible nor supporter of combustion. Ca(OH)2 + CO2 - CaCO3 + H2O. 233. Answer (1) More the charge density of cation, more will be its hydration and less will be its ionic mobility and so less will be the ionic conductances. Among these Be2+ has highest charge density. 234. Answer (2) H2 S + K 2 Cr2 O 7 → Cr2 O 3 green
Cr2O 3 + Na2O 2 → Na 2CrO 4 ( X)
(Y)
Na 2 CrO 4 + BaCl 2 → BaCrO 4 ↓ 2NaCl yellow ppt.
Sodium gives yellow colour in flame test. 235. Answer (3) Factual Type. 236. Answer (3) Cassiterite (tin stone) is SnO2. It contains FeWO4, MnWO4 (wolframite) as a magnetic impurity. 237. Answer (3) Van Arkel process is generally applied for obtaining ultrapure metals. The impure metal is converted into a volatile compound while the impurities are not affected. The volatile compound is then decomposed electrically to get the pure metal. Ti, Zr, Hf, Si, etc. have been refined by this method. 238. Answer (4) Factual Type. 239. Answer (3) Originally Wilhelm Kroll produced Ti by reducing TiCl4 with Ca in an electric furnace. Later Mg was used. At this high temperature Ti is highly reactive and reacts readily with air or N2. It is, therefore, necessary to perform the reaction under an atmosphere of argan. 1000 −1150°C TiCl4 + 2Mg ⎯⎯ ⎯⎯⎯ ⎯→ Ti + 2MgCl2
240. Answer (1) –2
0
H2S + SO2
H2O + S Oxidation
Reducing agent
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241. Answer (4) +5
+2
HNO3 → NO dil.
Change in oxidation no. = 3 ∴ Eq. wt. of HNO3 =
63 3
= 21. 242. Answer (2) [Al(OH)4.(H2O)2]– 243. Answer (1) PbO2 is not a peroxide. 244. Answer (3)
O
OH P
HO O
O
O
P
P O
O OH
Cyclic metaphosphoric acid contains three P – O – P bonds. 245. Answer (1) The hybridisation of N in (SiH3)3N is sp2 and so triangular planar. 246. Answer (1) XeF2 + PF5 → [XeF]+ [PF6]– 247. Answer (3) XeF6 has sp3d3 hybridisation and has one lone pair of electrons. 248. Answer (2) XeF2 and CO2 both are linear. 249. Answer (1) The covalency of F is one only. It does not have vacant d-orbital in its outermost orbit. Further Cl-atom is larger in size than F. 250. Answer (2) This is brown ring test for nitrates. 251. Answer (4) In Cu, all the d-electron are paired (3d104s1). In Cr, all the d-electrons are unpaired (3d54s1). Hence, d–d electron repulsions in Cu are much greater than those in Cr. Hence, Cu-atom is larger in size than Cr. In Cu2+(3d9), d – d electron repulsions decrease due to presence of one unpaired d-electron. Moreover, the electrons are attracted by 29 protons of the nucleus whereas in Cr2+, three unpaired electrons are still present but they are attracted by only 24 protons of the nucleus. Thus, Cu2+ is smaller in size than Cr2+. 252. Answer (3) In presence of moist air, a thin film of green basic copper carbonate is formed on its surface and hence, copper corrodes. 2Cu + O2 + H2O + CO2 → CuCO 3 .Cu(OH)2 . ( green )
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253. Answer (3) 1E2 of Cr > 1E2 of Mn. This is because after the removal of 1st electron, Cr acquires a stable configuration (d5) and removal of 2nd electron is very difficult. 254. Answer (1) La(OH)3 is more basic than Lu(OH)3. As the size of the lanthanide ions decreases from La3+, the covalent character of the hydroxides increases. Hence, basic strength decreases from La(OH)3 to Lu(OH)3. 255. Answer (1)
[Co(H2O)4 Cl2 ]Cl.H2O + AgNO3 → AgCl ↓ 1mole
256. Answer (1) 3+
[Fe(H2O)6] + H2O 2+
2[Fe(H2O)5OH]
hydrolysis
2+
[Fe(H2O)5OH] + H3O
+
OH [(H2O)5Fe
Fe(H2O)5]
4+
OH dimer 257. Answer (4)
SnCl2 + 2HgCl2 → SnCl4 + Hg2Cl2 (white)
SnCl2 + Hg2Cl2 → SnCl4 + 2Hg (grey)
258. Answer (3) FeSO4.7H2O contains some ferric ions (Fe3+ ) due to aerial oxidation. 259. Answer (3) NH3 is a strong ligand and so pairing of electron takes place. In [Ni(NH3)4]2+, the hybridisation is dsp2. 260. Answer (3) Since, one mole of the compound gives 2 moles of AgCl, it means two Cl-atoms should remain outside the co-ordination sphere.
[Co(NH3 )5 (NO 2 )]Cl2 → [Co(NH3 )5 (NO 2 )]2+ + 2Cl− 144444244444 3 3 ions
261. Answer (3) II
II
III
II
FeCl 3 + 3K 4 [Fe(CN)6 ] → Fe 4 [Fe(CN6 ]3 + 12KCl Prussian blue III
II
FeCl3 + K 4 [Fe(CN)6 ] → KFe[Fe(CN6 ] + 3KCl Prussian blue
262. Answer (3) Square planar complexes having unsymmetrical bidentate ligands can also show geometrical isomerism. For example, platinum glycinato complex, [Pt(Gly)2], exhibits geometrical isomerism. 263. Answer (2) log K1.K2.K3 = 9.70
(Q K1.K2.K3 = K)
log K = 9.70 K = Antilog 9.70. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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264. Answer (1) Both CN– and CO are strong field ligands and form low spin complex. Both [Ni(CN)4]2– and [Ni(CO)4] have no unpaired electron and so their magnetic moment is zero. 265. Answer (3) 2NaOH + 2Na[Cr(OH) 4 ] + 3H2 O 2 → 2 NaCrO 4 + 8H2 O yellow solution
Na 2 CrO 4 + BaCl 2 → BaCrO 4 ↓+ 2NaCl yellow ppt.
266. Answer (4)
O CH3 – C = NOH NiCl2 + 2 CH3 – C = NOH (in NH4OH)
CH3 – C = N
H–O Ni
CH3 – C = N O–H
dimethyl glyoxime
N = C – CH3 N = C – CH3 O
cherry red ppt.
267. Answer (1) As2S3, SnS, Sb2S3 (II B – arsenic group) are soluble in yellow ammonium sulphide (YAS). 268. Answer (4) Chromyl chloride Test Δ KCl + H2SO 4 ⎯⎯→ KHSO 4 + HCl conc . Δ K 2Cr2 O 7 + 2H2SO 4 ⎯⎯→ 2KHSO 4 + 2CrO 3 + H2 O conc .
CrO3 + 2HCl → CrO2Cl2 + H2O [ X]
CrO2Cl2 + 4NaOH → Na 2CrO4 + 2NaCl + 2H2O [Y]
Na 2CrO 4 (CH3 COO ) 2 Pb → PbCrO 4 ↓ +2CH3 COONa yellow ppt.
269. Answer (2) Factual Type. 270. Answer (1) PbI2 yellow ppt.
,
AgI yellow ppt.
271. Answer (4) Cu2+ and Cd2+ are in group-II while Zn2+ is in group-IV. 272. Answer (2) Ag2S + NaCN → Na[Ag(CN)2] + Na2S Zn is more electropositive than Ag, So Ag is displaced by adding Zn to soluble [Ag(CN)2]–. 273. Answer (4) DMG test is for nickel salts. 274. Answer (3) Cu 2 + + 4NH3 → [Cu(NH 3 ) 4 ] 2 + aq.
275. Answer (2) CO gas burns with blue flame. CaC2O4 is reducing agent and reduces pink solution of acidified MnO4– to colourless Mn2+. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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Section - B : Multiple Choice Questions 1.
Answer (1, 2, 3)
+
+
–H
H
2.
Answer (1, 2, 3)
OH 3.
does not react with HI, test give iodobenzene
Answer (3, 4)
I
I + alc KOH
4.
+
I Answer (1, 2, 3) Electron withdrawing group increases the activity of diazonium salt towards coupling reaction.
5.
Answer (1, 3, 4) Both ethanoic acid and methanoic acid gives CO2 with NaHCO3 Only methanoic acid react with Tollen’s reagent Fehling solution and KMnO4
6.
Answer (1, 4) (i) Due to presence of –COOH (ii) Due to presence of phenolic group
7.
Answer (2, 4) (i) Hoffmann elimination (ii) Hindered base (CH3)3CO
8.
Answer (3, 4) Neutral FeCl3 gives purple colour with phenols
9.
Answer (1, 2, 4) NaHCO3 gives test with compounds which are more acidic than H2CO3
10. Answer (2, 3, 4)
CH3 S is mesitylene
H3C
CH3
11. Answer (1, 2, 3, 4) All give elimination reaction Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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12. Answer (1, 2, 3, 4) All form stable carbocation 13. Answer (1, 2) Benzaldehyde gives silver mirror test and acetophenone gives Bromoform reaction 14. Answer (1, 3)
O (i)
CH3 – C ≡ C – CH3 S
HgSO4/H2SO4 H2O
CH3 – C – CH2 – CH3 T O
CH3 – C ≡ C – CH3
BH3/THF
CH3 – C – CH2 – CH3
–
H2O2/OH
S
T O HgSO4 H2SO4
C≡ C
(ii)
C – CH2
S
T O BH3/THF
C≡ C S
C – CH2 T
–
H2O2/OH
15. Answer (1, 2, 3, 4)
+
Rearrangement
H
H
H
+
OH
16. Answer (2, 3, 4) Ag(NH3)2+ + 2H+ → Ag+ + 2NH4+ is not the example of disproportionation while others are the example of disproportionation reactions. 17. Answer (1, 2, 3) +5
0
−
BrO 3 + Br − → Br2 (nf = 5 )
(nf =1)
18. Answer (1, 2) Molarity and normality are temperature dependent while molality and number of equivalents are temperature independent. 19. Answer (1, 3) y y
x
dx2 – y2
x
dxy
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20. Answer (1, 2) If n = 4 then the value of ‘l’ may be 0, 1, 2 and 3. So, electron may be in a ‘f’ orbital and the electron may have an energy proportional to (4 + 2). 21. Answer (1, 2) Be and Al, Li and Mg show diagonal relationship due to similar polarising power. 22. Answer (1, 3, 4) Only Ca and Zn belong to same period i.e. 4th period. 23. Answer (1, 2, 4) IE1 of Cu is more than K due to poor shielding of d-orbital electrons. 24. Answer (1, 4) Co and Ni, Hf and Zr have similar atomic radii due to poor shielding of nuclear charge of d and f-orbital electrons. 25. Answer (3, 4) SnH4 > GeH4 > SiH4 > CH4 (enthalpy of vapourisation) NH3 > SbH3 > AsH3 > PH3 (melting point) 26. Answer (2, 3) Both are equivalent, but if one terminal nitrogen will be isotopic both canonical structures will be different. 27. Answer (1, 4) NF3, H3O+ and NH3 have sp3 hybridisation, tetrahedral geometry and pyramidal shape. While BF3 and NO3– have sp2 hybridisation and trigonal planar geometry. 28. Answer (3, 4) Factual type. 29. Answer (1, 3) According to Charles law, V∝T V = K.T Taking logarithm on both sides logV = logK + logT Compare with y = mx + c then we get,
log V
slope = 1
log T 30. Answer (1, 3) If z < 1 then it implies that gas is more compressible and ‘b’ is negligible but not ‘a’. 31. Answer (2, 3) Dalton’s law is applicable for non reacting gases only. 32. Answer (3, 4) Factual type. 33. Answer (2, 4) Factual type. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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34. Answer (1, 2) Factual type. 35. Answer (1, 2, 3) Equilibrium is not affected by catalyst. 36. Answer (1, 2) According to Le-chatelier’s principle. 37. Answer (1, 3) Option - I CH3COOH + NaOH ⎯⎯→ CH3COONa + H2O Initially
60 ml 0.1 M
20 ml 0.1 M
0
0
At t = ‘t’
40 ml 0.1 M
0
20 ml 0.1 M 20 ml 0.1 M
Resulting solution is acidic buffer solution. Option - II Only CH3COONH4 does not act as buffer. Aqueous solution of CH3COONH4 acts as buffer. Option - III NaCN
+
HCl
⎯⎯→
HCN
+
NaCl
Initially
40 ml 0.1 M
20 ml 0.1 M
0
0
At t = ‘t’
20 ml 0.1 M
0
20 ml 0.1 M
0.1 M
Resulting solution is acidic buffer solution. 38. Answer (1, 2) The discharge potential of F– is higher than OH–. 39. Answer (1, 2, 4)
Zn(s) Zn2+ Cu2+ Cu (M1 )
(M2 )
is not the example of concentration cell because cathode and anode are different. 40. Answer (1, 4) In CrO5 and H2O2, peroxide linkages are present while H2S2O7 and HNO2 have no peroxide linkage. 41. Answer (1, 4) H2SO5 acts as oxidising agent only while HCl acts as reducing agent only. 42. Answer (3, 4) For simple cubic – unoccupied space – 48% For bcc
– unoccupied space – 32%
For fcc and hcp – unoccupied space – 26% Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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43. Answer (2, 4) For body centre ‘B’ atom, co-ordination number is 8. For face centre B atom, co-ordination number is 6. 44. Answer (2, 3) In β-decay,
n n ratio decreases while in positron emission, ratio increases. p p
45. Answer (1, 2) Acetone and chloroform shows negative deviation while ethanol and water shows positive deviation from Roult’s law. 46. Answer (1, 2, 4) The reducing power of nascent hydrogen is much less than the atomic hydrogen. 47. Answer (1, 2, 3) Ionic hydrides are powerful reducing agents. Both MgH2 and H2O are covalent hydrides but the bond dissociation energy of H2O is much higher than that of MgH2. Therefore, the reducing character increases in the order : H2O < MgH2 < NaH. 48. Answer (1, 2, 3) O22– : KKσ(2s)2σ*(2s)2σ(2pz)2 π(2px) = π(2py)2 π*(2px)2 = π*(2py)2 Bond order = =
Nb − Na 2 8−6 2
= 1. 49. Answer (1, 2) The co-ordination number of Be2+ is not more than 4. 50. Answer (1, 2, 3, 4) All statements are correct. 51. Answer (1, 3) The solubility of sulphate of alkaline earth metals decrease on moving down the group carbonates of alkali metals (except Li2CO3) do not decompose to give CO2 on heating. 52. Answer (1, 2, 3, 4) BeCl2 and AlCl3 are Lewis acids. They readily undergo hydrolysis and give HCl. 53. Answer (2, 3, 4) Factual type. 54. Answer (1, 2, 3) Factual type. 55. Answer (1, 3) Fe and Pt do not form amalgam. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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56. Answer (1, 2, 3, 4) Crystal field splitting energy depend on the shape of complex. For example CFSE of tetra hydral complex 4 is times of the CFSE of octahydral complex. CFSE also depend type of ligand and size of metal and 9 electronic configuration. 57. Answer (1, 2, 3) Factual type. 58. Answer (1, 3) In two dimensional sheet silicates, three oxygen atoms of each tetrahedral are shared with adjacent SiO44– tetrahedral, such sharing forms two dimensional sheet structure with general formula (Si2O5)n2n–. 59. Answer (1, 2, 4) Brine is saturated NaCl solution, on its electrolysis, H2 is liberated at cathode and Cl2 gas is liberated at anode and thus NaOH is formed. 60. Answer (2, 3) 2ICl
I+ + ICl2–
61. Answer (1, 3, 4) In V(CH2 Si Me3)4, co-ordination number of the central atom is 4, while in others, the co-ordination number of the central atom is six. 62. Answer (1, 2) Eu, Yb exhibit +2 oxidation state. These states are accounted by the extra stability of half-filled and Completely filled f-orbitals. 63. Answer (1, 2, 3) Because at high temperature, Ti forms highly stable carbide and also metal is reactive towards oxygen nitrogen at elevated temperature. 64. Answer (2, 3, 4) NbF4 is paramagnetic while NbX4 (X = Cl, Br, I) are diamagnetic. 65. Answer (1, 4) In [Fe(CN)6]3–, there is one unpaired electron while in [Fe(CN)6]4–, there is no unpaired electron. H2O is a weak ligand. 66. Answer (2, 3, 4) CN– is a strong field ligand and so forms inner orbital complex. H2O and F– are weak field ligands and so form outer orbital complex. 67. Answer (1, 2, 3, 4) All statements are correct. 68. Answer (2, 3, 4) In octahedral crystal field, t2g orbitals (dxy, dxz and dyz) have lower energy. 69. Answer (1)
Fe 2+ → [ Ar ]3d6 3d
6
4 unpaired electron Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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On pairing by strong field ligand,
no unpaired electron 70. Answer (2, 3, 4)
d1, d2, d3, d8, d9 and d10 can not form both high-spin and low-spin in octahedral complexes. 71. Answer (2, 3) The square planar complexes of the type MA4, or MAB3 do not show geometrical isomerism. 72. Answer (1, 2, 3) In [Cu(NH3)4]2+, hybridisation of the central atom is dsp2, so it is square planar, while other have sp3 hybridisation. 73. Answer (1, 3) Factual type. 74. Answer (2, 3) Cl– is a weak field ligand and so form high spin complex. [CoCl6]4– x–6=–4 x = +2. 75. Answer (1, 2, 3, 4) All statements are correct. 76. Answer (1, 2) Conductivity of solution does not depend upon magnetic moment. 77. Answer (2, 3, 4) Dilute HCl does not react with MgSO4, Na2SO4 and NaNO3. 78. Answer (1, 2) NH4NO2 → N2 + 2H2O (NH4)2Cr2O7 → N2 + Cr2O3 + H2O 79. Answer (2, 3, 4) Cu2+, Hg2+ are in group II. Ni2+, Zn2+, Co2+, are group IV. 80. Answer (1, 2, 4) Na2CrO4 → Yellow Na2Cr2O7 → Orange CrO2Cl2 → Intense red vapour. 81. Answer (1, 2, 3) Na2S is water soluble. 82. Answer (3, 4) C2O42– and I– can be decomposed by concentrated H2SO4 but not by dil.HCl. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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Section - C : Linked Comprehension C1. 1. Answer (3) A(g) → B(g) + C(g) initial pressure
P1
at time t
P1 – P′ P′
P′
after long time
0
P1
After long time
P1 + P1 = P,
At time ‘t’
P1 – P′ + P′ + P′ = Pt,
K=
0
P1
0
So,
⎛P⎞ P1 = ⎜ ⎟ ⎝2⎠
P⎞ ⎛ P′ = ⎜ Pt − ⎟ 2⎠ ⎝
⎛ P1 ⎞ 2.303 ⎟⎟ log⎜⎜ t ⎝ P1 − P′ ⎠
P ⎛ ⎞ ⎟ 2.303 ⎜⎜ 2 ⎟ log = P⎟ t ⎜P ⎜ − Pt + ⎟ 2⎠ ⎝2
=
⎡ ⎤ 2.303 P log⎢ ⎥ t 2 ( P − P ) t ⎦ ⎣
2. Answer (2)
K=
2.303 ⎛ 10 ⎞ log⎜ ⎟ t ⎝ 2 ⎠
K=
2.303 log 5 t
Now in second case K=
2.303 ⎛ 5 ⎞ log⎜ ⎟ t ⎝5− x⎠
2.303 2.303 ⎛ 5 ⎞ log 5 = log⎜ ⎟ t t ⎝5−x⎠
5=
5 5−x
5–x=1 x=4 ∴ conc. remaining unreacted = 5 – 4 = 1 ∴ 80% of reactant changes to product. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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3. Answer (1) +
H CH3COOC2H5 + H2O ⎯⎯ ⎯→ CH3COOH + C2H5OH
Rate = K[CH3COOC2H5][H+] = K[H+] = K′ Since (H+) comes out from catalyst Rate = K′[CH3COOC2H5] →
HCl t=0
[H+]HCl
H+
+
Cl–
0.05 m
0
0
0
0.05 m 0.05 m
= 0.05 m
H2SO4 ⎯⎯→ H+
+
HSO4–
0.05 M
0
0
0
0.05 M
0.05 M
HSO4–
H+ + SO42–
0.05 – α
α
α
(H+)H2SO4 = (0.05 + α) M ∴
K H2SO4 > K HCl
C2. 1. Answer (4) w2 M2 P −P = x2 = 0 w w P 1 + 2 M1 M2 0
s
2 M2 74.66 − 74.01 = 100 2 74.66 + 78 M2
Solving M2 = 177.6 g mol–1 Now ratio of mass of C and H is wC : wH = 94.4 : 5.6 ratio of number of atoms of C and H will be n C : nH =
94.4 5.6 : =7:5 12 1
empirical formula C7H5 molecular formula = (C7H5)n = 89 n 89 n = 177.6 n∼2
∴ molecular formula = C14H10. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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2. Answer (3) number of moles of urea =
10 60
number of moles of glucose =
10 180
number of moles of sucrose =
10 342
osmotic pressure (π) ∝ number of moles ∴ π2 > π1 > π3 3. Answer (3) Van’t Hoff factor =
194 65.4
Ca(NO3)2
Ca2+ + 2NO3–
1
0
0
0
1–α
α
2α
i=
1 + 2α = 2.966 1
2α = 1.97 α = 0.98 ∴ % degree of dissociation = 98%. C3. 1. Answer (2) In experiment (i) and (iii) Rate ∝ [A]x [5]2 = 25 ∝ [5]x x=2 In experiment (iii) and (iv), concentration of B increases 5 times but rate is not changing. So, order w.r.t. B is zero. 2. Answer (3) Rate = K[A]2[B]0 4 × 10–4 = K[0.1]2[0.1]0 K = 4 × 10–2 litre mol–1s–1 3. Answer (3) Rate = K[A]2[B]0 Rate = 4 × 10–2[0.2]2[0.35]0 Rate = 16 × 10–4 mol litre–1s–1. C4. 1. Answer (2) Rate = (T.C.)Δt/10 Rate = (2)90/10 = (2)9 Rate = 512 times. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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2. Answer (4) At 25°C, the value of rate constant is maximum because activation is zero. So, at 50°C or any other temperature, the value of rate constant remains same. 3. Answer (4) If Ea = 0 or T = ∞ then K achieves the maximum value and 100% reactant will convert into the product. C5. 1. Answer (2) The temperature at which the vapour pressure of a solid becomes equal to the external pressure is called sublimation point ∴ T < 114°C 2. Answer (1) Rate of diffusion of loss of mass of solid. 3. Answer (3) Entropy change of sublimation (ΔSsub) is greater than that of entropy change of fusion (ΔSfus). C6. 1. Answer (4) ΔTb = mKb 0.52 = m × 0.52 m=1 moles of solute m = mass of solvent in kg
∴ m = 1 it means 1 moles of solute present in 1 kg of solvent ∴ mole fraction of urea = 2. Answer (1)
1 1 = = 0.018 1000 56.55 +1 18
ΔTb = imKb 2.08 = i × 1 × 0.52 i=4 K3[Fe(CN)6] → 3K+ + [Fe(CN)6]3– i = 4 for this salt because on ionisation it gives four ions. 3. Answer (3) For AB2 compound ΔTb = m·Kb ΔTb 6 m= K =M b 1
for the compound A2B ΔTb 9 = Kb M2
Let the atomic mass of A is x and B is Y 6 9 = x + 2y 2 x + y Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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12x + 6y = 9x + 18y 3x = 12y This is true only if x = 40 and y = 10. C7. 1. Answer (2) Ca 2B 6 O11 + 2Na 2CO 3 → Na 2B 4 O 7 + 2NaBO 2 + 2CaCO 3 ↓ ( C)
(A)
(D )
(B )
2. Answer (3) Compound (B) is precipitate ∴ CaCO3 3. Answer (3) Compound (A) is colemanite Ca2B6O11. C8. 1. Answer (1) 2. Answer (2) 3. Answer (1) 2FeCr2O4 + 8NaOH +
7 O → 4Na 2 CrO 4 + Fe 2 O 3 + 4H2 O 2 2 (A) ( inso luble ) ( so lub le )
2Na 2CrO 4 + H2SO 4 → Na 2Cr2 O 4 + Na 2SO 4 + H2O (A)
( sod. chromile )
(B )
Na 2 Cr2O 7 + 3C → Na 2 Cr2O 4 + 3CO ( X)
( sod. chromate )
Na2Cr2O4 + H2O → Cr2O 3 + 2NaOH ppt
Cr2O 3 + 2Al → Al2O3 + 2Cr [ Y]
Roasted mass is extracted with water when Na2CrO4 gas in solution leaving behind insoluble Fe2O3. The sodium chromate solution is treated with a calculate amount of sulphuric acid to convert chromate into dichromate. The solution is concentrated when less soluble Na2SO4 crystallises out leaving behind more soluble Na2Cr2O7 in solution. The solution is further concentrated to get solid Na2Cr2O7. C9. 1. Answer (2) M=
10 xd 10 × 9.8 × 1.05 = = 1.05 M Mt 98
2. Answer (1) M1V1 + M2V2 = M3V3 (nf for HCl and HBr = 1) 0.1 × 100 + 0.5 × 200 = M3 × 300 110 = M3 × 300 M3 =
11 M 30
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3. Answer (3) n-factor of Mohr salt = 4 N = W × 1000 = 39.2 × 1000 = 1.6 N E V 392 250 4
C10. 1. Answer (2) n-factor = 6 2. Answer (4) NaCl ––→ Na+ + Cl–
H 2O
H
+
+
OH
–
at cathode
2H+ + 2e → H2↑
at anode
2Cl– → Cl2↑ + 2e
3. Answer (3) W= E=
E Q F
27 =9 3
C11. 1. Answer (3) 3–C
– C ≡ C–
2σ and 2π 2. Answer (4) –2
O C
O,
O = C = O
O
C O
resonance 3. Answer (2) C
≡O
C
O
C12. 1. Answer (3) 2 × 10–7 M is concentration of I– in the solution when AgCl starts precipitation 2. Answer (2) Ksp = 4 × 10–12 S = 10–4 ∴ Ksp = 4S3 No. of ion = 3 M(OH)2
M+2 + 2OH–
3. Answer (2) Basic solution pH > 7 ⎡K sp = 4S3 = 5.5 × 10 −6 ⎤ ⎢ ⎥ ⎢⎣S = (OH− ) ≈ 10 − 2 pH = 2 pH ≈ 12⎥⎦ Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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C13. 1. Answer (1) Aromatic 2. Answer (2) Only (II) and (III) obeys Huckel rule 3. Answer (4) All obeys condition of aromaticity C14. 1. Answer (4)
O P O
O O O
P
P
O
O O P
O
O 2. Answer (3) No PP or OO bonds are present. 3. Answer (2) ∴ dibasic
H3PO3 n factor = 2
H3PO3 is reducing as well as oxidising in nature C15. 1. Answer (2) 2. Answer (1) 3. Answer (4) O2 FeS ⎯⎯ ⎯→ Fe2O3 + SO2 (A)
(B )
FeS+ H2SO4 ⎯⎯→ FeSO4 + H2S (D)
(C)
FeS O4 + K 3 (Fe(CN)6 ] ⎯⎯→ Fe3 [Fe(CN)6 ]2 (D)
(E )
C16. 1. Answer (2) Depends on ionic character 2. Answer (2) n factor of NaCl = 1 ∴ N = M or Λ e = Λ m 3. Answer (2) Λ cx = Λ CA + Λ BX − Λ BA = 120 + 198 − 140 = 178 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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C17. 1. Answer (2) k=
0.693 −1 hr 2.1
t = 2.303 log 100 = 2.303 × 2.1× 2 = 13.96 hrs. k 1 0.693
2. Answer (2) Volume of dry N2O =
22.4 × 6.4 × 99 64 100
= 2.22 litre. 3. Answer (4) t1/2 of first order does not depend on concentration C18. 1. Answer (4) K = Ae −E a / RT
All are correct 2. Answer (4) If Ea = 0 then K achieves the maximum value and temperature has no effect. 3. Answer (1)
ln k = 14.34 −
1.25 × 10 4 T
Ea = 1.25 × 10 4 RT
Ea = 1.25 × 104 × 2 cal = 25 kcal. C19. 1. Answer (3) K4[Fe(CN)6] α=
i −1 i −1 , 0.8 = n −1 5 −1
i = 4.2 2. Answer (2) For NaCl, ΔTb = Tb − Tbo = i.k b .m = 2 × 0.1 K b For Al2(SO4)3,
ΔTb = 5 × 0.03 Kb
For MgCl2, ΔTb = 3 × 0.02 Kb NaCl > Al2(SO4)3 > MgCl2 3. Answer (2) π = i CRT α=
i −1 , i = 2.8 n −1
π = 2.8 × 0.1 × 0.0821 × 300 = 6.89 atm. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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C20. 1. Answer (3) N1V1 + N2V2 = N3V3 0.1 × 10 + 0.2 × 20 = N3 × 1000 N3 = N 200
2. Answer (2) no. of equivalents =
=
N1V1 + N2 V2 1000 0.1× 10 + 0.2 × 20 = 0.005 . 1000
3. Answer (2) Mills equivalent of NaOH = Mills equivalent of Acid. W × 1000 = 5 40
W = 0.2 g. C21. 1. Answer (1) B.O. of O2+ = 2.5 B.O. of O2 = 2.0 2. Answer (1) B.O. of CO, NO+, CN–, N2 = 3.0 B.O. of NO– = 2 3. Answer (2) B.O. of He2, Be2 and H2–2 = 0 C22. 1. Answer (3) Z > 1 for positive deviation and gases are less compressible Q attraction force is less Z < 1 for negative deviation and gases are more compressible Since attraction force is dominant 2. Answer (3) At high temperature real gas show less deviation 3. Answer (2) Force of attraction is less at high pressure C23. 1. Answer (1) Most ionic ∴ has highest melting point 2. Answer (2) Maximum polarisability for Cu+2 3. Answer (4) MgCl2 is more ionic than AlCl3 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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C24. 1. Answer (2) A=
2 8 ×4 = 3 3
B=
1 1 ×8 + ×6 = 4 8 2
Unit cell formula = A8/3B4 Formula of compound = A8B12 i.e. A2B3 2. Answer (4) A +2 =
1 × 8 = 1 (Tetrahedra l) 8
B+3 =
1 × 4 = 2 (Octahedral ) 2
O −2 =
1 1 ×8 + ×6 = 4 8 2
formula AB2O4 3. Answer (2) Tetrahedral void is present on body diagonal surrounded by one corner sphere atom and 3 face centre sphere C25. 1. Answer (3) d=
MZ 74 × 4 = N V 6.023 × 10 23 × ( 4 × 10 −8 )3
In Frenkel defect atom occupy interstitial space ∴ density remains same 2. Answer (2)
4 3 πr 3 3 = π 3 8 4r ⎞ ⎟⎟ 3⎠
2× Packing fraction =
⎛ ⎜⎜ ⎝
3. Answer (1) d fcc =
dbcc =
M× 4 N × (3.5 × 10 −8 )3
M× 2 N × (3.0 × 10 −8 )3
dfcc 4 × (3 . 0 ) 3 = = 1.26 dbcc 2 × (3.5)3 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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C26. 1. Answer (3) n=
90 = 5 mole of H2O 18
2 mole of H2O produces 1 mole of O2 ∴ 5 mole of H2O produces 2.5 mole of O2 2. Answer (1) W =Zit =
E it F
W it × n factor = n moles = E F t=
n × F 2 × 2 × 96500 = = 1.93 × 10 5 s i 2
3. Answer (4) 22.4 L O2 molar mass 32 22.4 L O2 evolves from 32 g at STP 0.224 cc will evolve from 0.32 ×10–3 g of O2 W=
i=
E it F
0.32 × 10 −3 × 96500 = 3.86 A 8
C27. 1. Answer (1) Acetic acid undergo dissociation in aqueous solution 2. Answer (3) i for ammonium phosphate = 4 (NH4)3PO4 > CaCl2 > glucose = benzene 3. Answer (1) ΔTf =
ik f w × 1000 Mt × w
C28. 1. Answer (1) Hybridisation is sp3d, shape is T-shaped because the two lone pairs occupy the two equatorial position. 2. Answer (2) In XeO2F2, hybridisation of central atom is sp3d. It has one lone pair around central atom. Its shape is Seesaw or distorted trigonal pyramidal. 3. Answer (3) In XeF4, the hybridisation of the central atom is sp3d2 and it has two lone pairs, and so its shape is square planar. 2XeF4 + 3H2O → Xe + XeO3 + 3H2F2 + F2. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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C29. 1. Answer (2) Boron dissolves the fused alkalies, liberating H2 gas. 2B + 6NaOH → 2Na3BO3 + 3H2↑ 2. Answer (1) 3SiO 2 +
4B
↓ Reducing agent ( X )
→ 2B 2O 3 + 3Si (Y)
3. Answer (2) B + 3HNO3 → H3BO3 + 3NO2↑ 2NO2 + H2O → HNO2 + HNO3 C30. 1. Answer (3) H3PO3 is a dibasic acid, so its n-factor is 2 0.2 M H3PO3 ≡ 0.4 NH3PO3 N1V1 = N2V2 0.4 V1 = 1 × 300, V1 = 300 = 750 mL 0.4 2. Answer (3)
O P O H
O
O H
H 3. Answer (1) H3PO3 reduces HgCl2 to Hg2Cl2 C31. 1. Answer (3) Iron pyrite, FeS2, is mainly used for the manufacture of SO2. 2. Answer (4) Sulphide ores are concentrated by Froth-floatation process. 3. Answer (4) Carbon reduction process is generally called as smelting. The oxides of the less electropositive metals are reduced by strongly heating them with coal or coke. PbO + C → Pb + CO Fe2O3 + 3C → 2Fe + 3CO C32. 1. Answer (2) Ba gives apple green colour to the flame. This is chromyl chloride test. 2. Answer (3) When solid chloride is heated with concentration H2SO4 in presence of K2Cr2O7, deep red vapours of chromyl chloride, CrO2Cl2, are evolved. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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3. Answer (2) CrO2Cl2 + 4NaOH → Na2CrO4 + 2NaCl + 2H2O Yellow colour Na2CrO4 + Pb(CH3COO)2 → PbCrO4↓ + 2CH3COONa Yellow ppt.
C33. 1. Answer (1)
pH = pK a + log
[HCO3− ] [H2CO3 ]
or
7.4 = 6.1 + log
or
[HCO3− ] [H2CO3 ]
[HCO3− ] = 20 [H2CO3 ]
2. Answer (3) Blood is a buffer solution of mixture of HCO3– + H2CO3. 3. Answer (1) Serum protein acts as a buffer. C34. 1. Answer (1) Serpeck’s process – This process is used when silica is present in considerable amounts in bauxite ore. The ore is mixed with coke and heated at 1800º C in presence of nitrogen, where AlN is formed. Al2O3 + 3C + N2 → 2AlN + 3CO SiO2 + 2C → Si + 2CO 2. Answer (3) Al2O3 + 3C + N2 → 2AlN + 3CO↑ 3. Answer (2) Factual type. C35. 1. Answer (4) Factual type. 2. Answer (1) The process of conversion of Fe to FeO in smelting is a non spontaneous process while C to CO is a spontaneous process. 3. Answer (3) Factual type. C36. 1. Answer (1) Liquation process – The process is based on the difference in fusibility of the metal and impurities. When the impurities are less fusible than the metal itself, this process is used. Distillation – The process is used for those metals which are easily volatile. Zone Refining – When highly pure metals are required, this method is applied for purification. The method is based on the difference in solubility of impurities in molten and solid state of the metal. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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2. Answer (3) Zone refining method is used in purification of germanium, gallium, silicon etc. which are used as semiconductors. 3. Answer (3) Chromatography based on adsorption technique for removal of impurities. C37. 1. Answer (2) Cylinders of F2 are now commercially available. However, for many purposes F2 is converted into ClF3, which though very reactive is less unpleasant and easier to transport. 200 − 300 º C 3F2 + Cl2 ⎯⎯ ⎯ ⎯⎯→ 2ClF3
2. Answer (4) It was discovered in the 1940s that the isotopes of uranium could be separated by gaseous diffusion of UF6. This was important in preparing enriched uranium to make the first atomic bomb. 3. Answer (4) Tetrafluorethene, F2C = CF2 can be made as, 500 −1000 º C 2CHClF2 ⎯⎯ ⎯⎯⎯ ⎯→ CF2 = CF2 + 2HCl
Fluoroalkenes of this type can be polymerised either thermally, or using a free radical initiator. Depending on the degree of polymerisation, that is, on the molecular weight produced, the products may be oils, greases or a solid of high molecular weight called polytetrafluoroethylene or Teflon. C38. The compound (A) on strong heating gives two oxides of sulphur, it may be a sulphate. The solution (E) on treatment with thiocyanate ions gives blood red coloured compound (H) indicates that the solution (E) contains Fe3+ ions. Thus the compound (A) is FeSO4.7H2O. Heat Heat FeSO 4 .7H2 O ⎯⎯ ⎯→ FeSO 4 ⎯⎯ ⎯→ Fe 2 O 3 + SO 2 + SO 3 − 7H2 O
(A )
Fe 2O 3 + 6HCl → (B )
(C)
(B ) Blackish brown powder
(D)
+ 3H2O
2FeCl3 (E ) Yellow Solution
2FeCl3 + H2S →
+ 2HCl +
2FeCl2 ( G) Apple green solution
FeCl 3 + 3CNS − → (E )
Fe(CNS )3
S (F ) White turbidity
+ 3Cl −
Blood red coloured solution
1. Answer (2) SO2 is acidic oxide
SO 2 + H2O → H2SO3 (Y)
2. Answer (2) +3
+2
(E )
(G)
2FeCl3 + H2 S → 2FeCl 2 + 2HCl + S
Change in O.N. = 1 FFeCl3 =
M 1
3. Answer (1) +3
+2
4FeCl3 + 3K 4 [Fe(CN)6 ] → Fe 4 [Fe(CN)6 ]3 + 12KCl (E )
Pr ussian blue
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C39. 1. Answer (1) Fe3+ → [Ar]3d5 3d
5
Fe3+ has maximum magnetic moment because it has maximum number of unpaired electrons, i.e., 5. 2. Answer (2) The unpaired electron gives rise to a magnetic field because of its spin, and also because of the orbital angular momentum. The general equation for the magnetic moments of the first row of transition metal ions is, μ(S + L ) = 4S(S + 1) + L(L + 1) .μ B Where S is total of the spin quantum numbers and L is the resultant of the orbital angular momentum quantum numbers of all the electrons in the molecule. In many compounds of the first row transition elements, the orbital contribution is quenched by the electric fields of the surrounding atoms. As a slight approximation the orbital contributions can be ignored and the observed magnetic moment may be considered to arise only from unpaired spins. The spin only magnetic moment μs may be written. μ S = 4S(S + 1).μ B
The magnetic moment μ is related to the number of unpaired spins n by the equation. μ S = n(n + 2).μB
3. Answer (4) μ = n(n + 2)
or
5.92 = n(n + 2)
n=5 C40. 1. Answer (3)
AgCl + 2NH3 → [ Ag(NH3 )2 ]Cl aq
2. Answer (2) AgBr is a pale yellow solid. It is insoluble in water and concentrated acids. It is partially soluble in strong solution of NH4OH due to complex formation. AgBr + 2NH4OH → [Ag(NH3)2]Br + 2H2O AgBr dissolves in sodium thiosulphate.
AgBr + 2Na2S 2O3 → Na3 [ Ag(S 2O3 )2 ]+ NaBr ( So lub le )
AgI is insoluble in water as well as in NH4OH but soluble in Na2S2O3 due to complex formation. 3. Answer (2) AgCl, AgBr, AgI, Ag2SO4 etc. are insoluble. AgF, AgNO3, AgClO4 are soluble salts. C41. 1. Answer (3) H2O is weak field ligand and so give outer orbital complex. In [Fe(H2O)6]3+, the hybridisation of the central atom is sp3d 2. NH3 and CN– are strong field ligand and give inner orbital complex. 2. Answer (1) In [Ni(H2O)6]2+, there are two unpaired electrons. 3. Answer (2) [Fe(CN)6]4–, the hybridisation of the central atom is d2sp3. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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C42. 1. Answer (4) [Cr(NH3)5Br]SO4 can exhibit ionisation isomerism. 2. Answer (3) [Cr(H2O)6]Cl3 [Cr(H2O)5Cl]Cl2.H2O [Cr(H2O)4Cl2]Cl.H2O 3. Answer (4) Hydrate isomerism. C43. 1. Answer (2) en is a bidentate ligand and so is chelating ligand and chelation gives the stability of the complex. 2. Answer (1) Chelating effect gives stability. NH3, H2O and F– are monodentate ligand and monodentate ligand cannot be a chelating ligand. 3. Answer (3) The higher the overall stability constant value of the complex, the more stable it is. Alternatively, 1/K values called ‘instability constant’ explain the dissociation of the complex into metal ion and ligands in the solution. C44. 1. Answer (2) 2. Answer (3) 3. Answer (3) K [Fe( CN) ]
6 ( A ) ⎯⎯3 ⎯ ⎯ ⎯ ⎯ → blue ppt (C)
(A) has Fe2+, (C) is Turnbull’s blue, KFeII[FeIII(CN)6] BaCl 2 (A) ⎯⎯ ⎯ ⎯→ white ppt. (B) insoluble in concentration HNO3.
Thus (B) is BaSO4 and (A) has SO42– ion. K [HgI ]
4 ( A ) + NaOH ⎯⎯→ gas ⎯⎯2 ⎯⎯ → brown ppt.
(D)
Gas is NH3 and (A) contains NH4+ ion. Brown ppt (D) is Iodide of Millon’s base, HgO.HgI.NH2. Thus (A) has Fe2+, NH4+, SO42–, H2O. So (A) is FeSO4.(NH4)2SO4.6H2O (Mohr’s Salt) C45. (i)
CuSO 4 (aq) + BaCl 2 (aq) → BaSO 4 ↓+ CuCl 2 (aq) White ppt
(ii) 2CuSO 4 + 4KI → Cu 2I2 + 2K 2 SO 4 + I2 ( excess )
(iii) 2CuSO 4 + K 4 [Fe(CN)6 ] → Cu2 [Fe(CN)6 ] + 2K 2SO 4 Chocolate brown ppt.
1. Answer (1) Presence of SO42– ion. 2. Answer (2) Cu2[Fe(CN)6] 3. Answer (1) CuSO4 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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Section - D : Assertion - Reason Type 1.
Answer (1) The solubility product of ZnS is more.
2.
Answer (3) Ethylacetoacetate mainly exists in enolic form i.e., it gives violet colour with neutral FeCl3 solution.
3.
Answer (1) N-methyl acetamide exists in two geometrical forms due to restricted rotation about carbon nitrogen bond.
4.
Answer (4)
KMnO 4 + SO 3 (nf =5 )
2−
(nf = 2 )
+
H ⎯⎯ ⎯→ Mn2+ + SO42–
Hence, the molar ratio of KMnO4 and SO32– are 2 : 5 and number of moles of KMnO4 needed to react with 1 mole of sulphite ion in acidic medium are 5.
2 4 not . 5 5
Answer (2) In Frenkel defect, density remains same and doping of group 14 element with element of group 13 produces p-type semiconductors.
6.
Answer (3) Weak electrolytes are feebly ionised while strong electrolytes are more ionised at moderate concentration.
7.
Answer (2) For spontaneous reaction, ΔGsystem must be negative and ΔStotal must be positive.
8.
Answer (1) CH3COOH can’t give iodoform test because H atom of CH3 of CH3COOH can’t abstract by OH– base.
9.
Answer (1) Lone pair - bond pair repulsion makes the N – F bond longer. So, experimentally determined N – F bond length is greater than the sum of the single bond covalent radii of N of F.
10. Answer (1)
CH2 = CH – CH2 – CH3 butene-1
Br2
* – CH – CH CH2 – CH 2 3 Br Br (d or l containing one asymmetric C-atom)
11. Answer (3) Boric acid is a monobasic acid because in aqueous solution, it accepts one OH– and converts into [B(OH)4]– B(OH)3 + H2O
B(OH)4– + H+
12. Answer (4) HOCl ionises in aqueous medium to give H+ and OCl– ions and behaves as Arrhenius acid. 13. Answer (3) Ag2S is soluble in NaCN due to the formation of soluble complex, [Ag(CN)2]– not AgCN. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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14. Answer (2) For the interaction of molecules, collision of molecules in between closely present molecules take place. 15. Answer (4) When sodium acetate is added in acetic acid solution then pH of solution increases because ionisation of CH3COOH is suppressed due to common ion effect and [H+] becomes less. 16. Answer (4) The amount of solute that dissolve depends upon the nature, temperature and pressure of the substance. Some solute becomes more soluble and some are less soluble in water at higher temperature. 17. Answer (3) 1-butene on reaction with HBr in the presence of peroxide produces 1-bromo butane. It involves the formation of secondary free radical. 18. Answer (1) Due to small size, Li+ will be heavely hydrated and in aqueous solution, it moves slowly and shows least ionic conductance. 19. Answer (2) Rearrangement reactions produce isomers because molecular formula remains same and structure becomes different. 20. Answer (2) CH3 – CH = CH2 is more stable than CH2 = CH2 due to presence of more number of hyperconjugate bonds. 21. Answer (4) Dipole moment is a vector quantity compounds which have the polar bonds may or may not be polar. For example BF3 is non polar while it has polar B – F bond. 22. Answer (3) Radioactive decay follows the first order kinetics. It does not depend on the external factors like pressure, temperature etc. 23. Answer (2)
OH 2-propanol gives the positive iodoform test due to presence of – CH – CH3 group. 24. Answer (2) ⊕
Ph – CH2 – OH gives positive Luca’s test due to the formation of most stable C 6H5 − C H2 . 25. Answer (1) In ethanol, molecules are associated by intermolecular hydrogen bonding while in dimethyl ether, it is absent. So, boiling point of ethanol is higher than dimethyl ether. 26. Answer (2) In SN2 reaction, bond fission and bond formation takes place in a single step and transition state intermediate is formed. 27. Answer (3) 2-bromobutane on reaction with KOH in ethanol gives 2-butene as a major product because butene-2 is more stable than butene-1. 28. Answer (4) Lassaignes test can’t be used to detect nitrogen in hydrazine (H2N-NH2)) due to absence of carbon. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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29. Answer (2) →
FeCl3 + 3NaCNS
Fe(SCN)3
+
3NaCl
Diazonium salt does not give Lassaigne’s test because on heating it evolves N2 gas. 30. Answer (1) Due to ortho effect, o-methyl benzoic acid is more acidic than benzoic acid. 31. Answer (4) In H2SO5, the oxidation number of sulphur is +6 and it acts as an oxidising agent. 32. Answer (2) +2 –1
+3
Fe S2 + O2 – –1e
Fe2O3 + SO2
–2 × 5e
+4
–
Hence, n-factor for FeS2 is 11. 33. Answer (3) PCl3 + 3H2 O →
H3PO 3 + 3HCl phosphorou s acid
H3PO3 is diabasic acid. 34. Answer (4) In PCl5, two axial P – Cl bonds are longer than three equatorial P – Cl bonds. 35. Answer (1)
⊕CH 2 +
+ is more stable than
because aromaticity is maintained in the resonating form of
while
⊕CH 2 aromaticity is not maintained in the resonating form of
.
36. Answer (2) When a gas expands below inversion temperature, then it shows cooling effect Ti = 2 × Tb . 37. Answer (2) The bond order of CO+ is 3.5 while CO is 3. 38. Answer (3) K3[Fe(CN)6] is more stable than K4[Fe(CN)6] due to high charge on Fe in 1st complex. In both complexes, the hybridised state of Fe is d 2sp3. 39. Answer (2) CH3COOH and CH3COONa is the acidic buffer solution while NH4OH and NH4Cl is basic buffer solution. 40. Answer (1) o-hydroxyl benzoic acid is more stronger than benzoic acid due to intramolecular hydrogen bonding in o-hydroxy benzoate anion. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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41. Answer (1) Aniline undergoes nitration with concentrated HNO3 and H2SO4 to give 47% m-isomer because aniline is +
converted into anilinium ion and NH3 group deactivates the ring. 42. Answer (2) Alkynes are more reactive than alkenes towards catalytic addition of H2 because π electron cloud density is higher in alkynes than alkenes. 43. Answer (1) p-amino benzoic acid does not form Zwitter ion because –COOH is weaker acidic group than –SO3H. 44. Answer (3) 2° amines have higher boiling point than 3° amines because it forms hydrogen bonding. Gabriel pthalimide process is used for the preparation of 1° amine. 45. Answer (2) Benzaldehyde reacts with HCN to form two optical isomers and it also gives Cannizarro’s reaction in presence of concentrated base and Cannizarro’s reaction is an example of disproportionation reaction. 46. Answer (2) Aromatic diazonium salt is more stable than aliphatic diazonium salt due to resonance. 47. Answer (4) In pyrrole as well as in pyridine hybridised state of N is sp2. In pyrrole, lone pair of electron participate in resonance i.e. it is less basic than pyridine. 48. Answer (4) In benzophenone, α-hydrogen is absent and it gives only one oxime with NH2OH due to two same phenyl group in benzophenone. 49. Answer (4) The bond angle around O in diphenyl ether is fairly close to 109°28′ due to repulsion in between two phenyl group. 50. Answer (2) For phenol, keto form is less stable than enol form because in keto form, aromaticity disappeared. 51. Answer (3)
OH OH + EtMgBr → 2C2H6 (C6H6O2) Q 110 g C6H6O2 gives 2 × 22400 ml C2H6 at NTP ∴ 1.1 g C6H6O2 gives
2 × 22400 × 1.1 = 448 ml C2H6 110
So, compound C6H6O2 contains two active hydrogen atoms. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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52. Answer (1) N2 + 3H2 2 mole 3 mole
⎯⎯→ 2NH3
Here, H2 is the limiting reagent. So, product is formed according to H2. 53. Answer (2) The basicity of H3PO3 is 2 N = M × basicity = 0.3 × 2 = 0.6 In acidic medium, KMnO4 is converted into Mn2+. So, n-factor for KMnO4 is 5. 54. Answer (2) Principle quantum number determines the number of shells. Higher is the value of ‘n’, more is the energy. 55. Answer (4) Due to smaller size of fluorine there is great repulsion between F electrons than chlorine electrons. So, bond dissociation energy of F2 is less than Cl2. 56. Answer (4) The dipole moment of CH3F is less than CH3Cl because C – F bond length is much less than C – Cl bond length. However, F is more electronegative than Cl. 57. Answer (1) Rate of diffusion ∝
1 1 . ∝ V.D. M
58. Answer (2) van der Waal’s constant ‘a’ of NH3 is more than N2 due to presence of hydrogen bonding in NH3. 59. Answer (3) Work is not a state function. It depends on path. 60. Answer (2) Many endothermic reaction that are not spontaneous at room temperature becomes spontaneous at higher temperature because T. ΔS > ΔH. 61. Answer (4) The mixture of NH3 and HCl in 2 : 1 ratio acts as basic buffer not acidic buffer. 62. Answer (3) Catalyst has no effect on equilibrium. 63. Answer (4) In qualitative analysis, ZnS gets precipitated after PbS. 64. Answer (3) Fluorine is stronger oxidising agent than chlorine but electron affinity of fluorine is less than chlorine. 65. Answer (3) The mobility of sodium ion is lower than that of potassium ion because sodium ion is heavely hydrated than potassium ion. 66. Answer (2) Factual type. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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67. Answer (2) Y XA
−α ⎯⎯→ ⎯ YX−−24B
A and B are isodiaphers because (n – p) are same. 68. Answer (3) Threshold energy always greater than activation energy. 69. Answer (4) Catalyst does not change ΔH of reaction. 70. Answer (3) Ethanol and water shows positive deviation from Roult’s law and it forms minimum boiling azeotropes. 71. Answer (2) Gold sol is negatively charged colloidal solution. 72. Answer (2) ⊕
⊕
CH3 − O − C H2 is more stable ion than CH3 C H2 carbocation due to resonance. 73. Answer (3) Compound having D configuration not necessarily dextrorotatory. 74. Answer (2) If pH is greater than pKa then compound will exist primarily in basic form. R – COOH is more acidic than +
R − N H3 . 75. Answer (2) The addition of grignard reagent is nucleophilic addition. 76. Answer (4) Reaction is an example of electrophilic addition not the nucleophilic addition. 77. Answer (3) 1, 3-butadiene can show 1, 2 as well as 1, 4 addition but 1, 4 addition product is more stable at higher temperature. 78. Answer (1) R – H + I2
R – I + HI
To stop reversibility of reaction, iodic acid (HIO3) is used which oxidises HI into I2. 79. Answer (4) CHF3 is less acidic than CHCl3. 80. Answer (1) Weaker is the base, better is the leaving group. 81. Answer (2) Hoffmann elimination takes place. 82. Answer (1) Higher is the stability of intermediate, more is the rate of reaction. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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83. Answer (2) Ring expansion takes place. 84. Answer (3)
O
O
O
All compounds having – C – CH3 do not give iodoform test. For example HO – C– CH 3 , H2N – C– CH3 etc. do not give iodoform test. 85. Answer (3) In acidic medium, phenol phthalein has benzenoid structure (colourless) while in basic medium, phenolphthalein has quinonoid structure (pink). 86. Answer (3) Cl– is weaker base than N H2 . So CH3COCl is more reactive than CH3CONH2. 87. Answer (2) Methyl ketones can react with NaHSO3. 88. Answer (1) Due to participation of lone pair of electron in resonance, amides are less basic than amines. 89. Answer (1) Factual type 90. Answer (4) Factual type 91. Answer (2) H2O2 oxidises black PbS into white PbSO4. 92. Answer (4) CnH2n + 2 does not act as lewis acid due to unavailability of vacant orbital. 93. Answer (2) K2CO3 can’t be prepared by solvay ammonia soda process because KHCO3 is more soluble. 94. Answer (1) Sodium reacts violently with water to evolve H2 which catches fire. 95. Answer (1) PbO2 reacts with HCl and NaOH due to amphoteric nature. 96. Answer (1) Factual type 97. Answer (3) Ag and Au form soluble complex with NaCN, impurities remain insoluble. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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98. Answer (1) Factual type 99. Answer (2) All nitrates are soluble in water. 100. Answer (4) Na2HPO3 is not an acidic salt because it does not contain ionisable proton. 101. Answer (1) Bleaching powder exists as Ca2+(OCl–)Cl–. 102. Answer (1) SiCl4 can be hydrolysed due to presence of Vacant ‘d’ orbital in Si atom. 103. Answer (1) Factual type. 104. Answer (3)
O The structure of H3PO4 is H – O – P – O – H
O H 105. Answer (1) Factual type 106. Answer (4) Factual type 107. Answer (1) Factual type 108. Answer (2) The structure of [Ni(DMG)2] is
O H3C – C = N
H–O Ni
H3C – C = N O–H
N = C – CH3 N = C – CH3 O
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Section - E : Matrix-Match Type 1.
Answer A(q), B(r), C(s), D(p) −1
0
(A) I 2 ⎯⎯→ I− equivalent mass = +5
0
I 2 ⎯⎯→ IO 3
−
equivalent mass =
−3
(B) P 4 ⎯⎯→ PH3 equivalent mass = +1
0
P 4 ⎯⎯→ H2PO 2
−
M 10
M M 6M 3M + = = 2 10 10 5
total equivalent mass = 0
M 2
M 12
equivalent mass =
M 4
⎛ M M ⎞ 3M + M ⎛ M ⎞ =⎜ ⎟ total equivalent mass = ⎜ + ⎟= 12 ⎝ 4 12 ⎠ ⎝3⎠ +5
+7
+4
(C) KClO3 ⎯⎯→ HClO4 + ClO2 KClO3 ⎯⎯→ HClO4 equivalent mass = KClO3 ⎯⎯→ ClO2 equivalent mass =
total equivalent mass =
M 2
M 1
M M ⎛ 3M ⎞ + =⎜ ⎟ 2 1 ⎝ 2 ⎠
⎛M⎞ (D) Cu2S ⎯⎯→ Cu2+ + SO2 equivalent mass = ⎜ ⎟ ⎝8⎠ +1
+2
−2
+4
Cu 2 S ⎯⎯→ Cu2 + + SO 2
n = (1 × 2 + 6) = 8 2.
Answer A(r), B(s), C(q), D(p) Factual type.
3.
Answer A(s), B(p, s), C(q), D(r) (A) [Co(NH3)4Cl2] exhibits geometrical isomerism
NH3
NH3 Cl
NH3 CO NH3
CO Cl
NH3 cis-isomer
Cl
NH3
NH3
Cl NH3 trans isomer
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(B) Cis-[Co(en)2Cl2] exhibits optical isomerism
en
Cl
Cl
en
CO
CO en
Cl
Cl
en
d-form mirror (C) [Co(en)2(NO2)Cl]SCN
e-form
[Co(en)2 (NO 2 )SCN]Cl⎫ ⎬ionization isomers [Co(en)2 (SCN)Cl]NO 2 ⎭ [Co(NH3 )6 ] [Cr(CN)6 ]⎫ (D) [Co(CN) ] [Cr(NH ) ]⎬co-ordination isomers 6 3 6 ⎭
4.
Answer A(s), B(q), C(p), D(r)
TeCl4
–
Cl Cl
ClO3
–
O Te
Cl
Cl Cl (See saw shaped) ICl2
+
Cl
O
Trigonal pyramidal with bond angle less than 109.5° H2CO
I
O C
Cl Angular with bond angle less than 109.5°
5.
O
H H Trigonal planar
Answer A(s), B(r), C(q), D(p) (A) Weak acid with strong base, solution is basic therefore pH > 7 1 [pK w + pK a + log C] 2 (B) Strong acid and weak base, solution is acid therefore pH < 7
pH =
1 [pK w − pK b − log C] 2 (C) Salt of weak acid with weak base
pH =
1 [pK w + pK a − pK b ] 2 (D) Salt of strong acid with strong base solution is neutral
pH =
therefore pH =
1 pK w 2
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6.
Answer A(r), B(s), C(q), D(p) (A) Na2S + (CH3COO)2Pb → PbS ↓ + 2CH3 COONa black ppt
(B) Phenol gives colour with neutral FeCl3 solution (C) 3NaCNS + FeCl3 → Fe(CNS)3 + 3NaCl blood red colour
(D) NaI + AgNO3 → AgI ↓ + yellow ppt
7.
NaNO 3 ( insoluble in NH4 OH solution )
Answer A(r), B(q), C(s), D(p) (A) ΔG = – nFE ⎡ ∂ log K ⎤ (B) ΔH° = RT2 ⎢ ∂T ⎥ ⎣ ⎦P ⎡ ∂ΔG ⎤ (C) ΔS = – ⎢ ∂T ⎥ ⎣ ⎦P
(D) ΔG° = – RT log K. 8.
Answer A(s), B(r), C(p), D(q) (A) ArN2+ + X– + CuX → Ar° + N2 + CuX2 Ar° + CuX2 → ArX + CuX
CH3 + CH3 – Cl
(B)
+
anhy AlCl3
(electrophile is C H3 )
(C) In Claisen condensation the intermediate is carbanion (D) In Reimer Tiemann reaction the attacking species is dichloro carbene (iCCl2) 9.
Answer A(r) B(q), C(s), D(p) (A) Precipitation of colloidal solution is called coagulation. (B) Gold number is protective power of colloidal solution. (C) Peptisation – To convert the freshly prepared ppt into colloidal solution. (D) Tyndall effect is due scattering of light.
10. Answer A(p, r), B(q, r), C(r), D(r, s) (A) CH2 = CD–COOH + HBr
CH2 – CD – COOH Br
H
(optically active)
Reaction proceeds through carbocation intermediate. (B) Addition of Br2 is trans addition and Cis-form gives racemic mixture.
H2C
C2H5 C
(C)
H
+ Baeyer's Reagent
C H
Syn addition
H2C
C2H5 C
H
C
OH OH
H
(optically active)
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Cl + CH2
(D)
C – Cl
T
T
(optically active)
11. Answer A(s), B(p), C(q), D(r) Factual type. 12. Answer A(q, r), B(r, s), C(r, s), D(p, r) (A) HNO2
+5 oxidation (reducing agent)
⇒ +1 + X – 4 = 0 x = +3
+3 reduction (oxidising agent) –3
Both
O (B)
+6
O
O Cr O
reduction (oxidising agent)
O
x + 4(–1) –2 = 0
Cr 0
x = +6 (C) H2SO5
O
+6
H–O–S–O–O–H reduction (oxidising agent)
O x + 2(–2) + 2(–1) = 0
S –2
x = +6
+5
(D) N2O5 2x + 5(–2) = 0
reduction (oxidising agent)
x = +5
–3 13. Answer A(p), B(r), C(q, s), D(q) (A) CaOCl2 → Ca2+ + Cl– + OCl–, In which Cl has +1 as well as –1 oxidation number. (B) NH4NO3 →
NH4+
+
NO3–
x + 4 = +1
x – 6 = –1
x = –3
x = +5
(C) Caro’s acid → H2SO5 ⇒ x + 2(–2) + 2(–1) = 0 x = +6 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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O H – O – S – O – O – H (peroxy linkage) O Marshall’s acid H2S2O8
O
O
HO – S – O – O – S – OH (peroxy linkage) O
O
oxidation number of S = +6 (D) K2Cr2O7
O
O
–
–
O – Cr – O – Cr – O (no peroxy linkage) O
O
oxidation number of chromium = +6 K2CrO4, 2K+ + CrO42–
O – Cr – O (no peroxy linkage)
O
–
O
oxidation number of Cr = +6 14. Answer A(p, r), B(p, s), C(q, s), D(q, r) For 1s, n = 1, l = 0 radial nodes = n – l – 1 = 1 – 0 – 1 = 0 angular nodes = l = 0 for 2p, n = 2, l = 1 radial nodes = 2 – 1 – 1 = 0 angular nodes = 1 for 3p, n = 3, l = 1 radial nodes = 3 – 1 – 1 = 1 angular nodes = 1 for 2s, n = 2, l = 0 radial nodes = 2 – 0 – 1 = 1 angular nodes = 0 15. Answer A(q), B(s), C(p, q), D(r) Krypton is inert gas Strontium is s-block element Arsenic is p-block element Titanium is d-block element Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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16. Answer A(s), B(p, s), C(p, s), D(q, r)
(A)
O22–(18)
→
σ1s2σ*1s2σ2s2σ*2s2σ2px2
Bond order =
⎡π2p y 2 ⎤ ⎡π * 2p y 2 ⎤ ⎥ ⎥⎢ ⎢ 2 2 ⎢⎣ π2p z ⎥⎦ ⎢⎣ π * 2p z ⎥⎦
10 − 8 = 1 (diamagnetic) 2
(B) CO → Number of electron = 6 + 8 = 14 ⎫ ⎬ isoelectro nic Number of electron in N2 = 2 × 7 = 14 ⎭ ⎡π2p y 2 ⎤ 2 CO(14) → σ1s2σ*1s2σ2s2σ*2s2 ⎢ ⎥ σ2p x 2 ⎢⎣ π2p z ⎥⎦
1 (10 − 4) = 3 2
diamagnetic B.O. =
⎡π2p y 2 ⎤ NO+(14) – σ1s2σ*1s2σ2s2σ*2s2σ2px2 ⎢ ⎥ 2 ⎢⎣π2p z ⎥⎦ isoelectronic with N2 diamagnetic
1 (10 – 4) = 3 2
B.O. =
He2+ – σ1s2σ*1s1 B.O. =
2 −1 1 = paramagnetic 2 2
17. Answer A(p, r), B(r), C(q, r), D(r, s) Mg2C3 + 4H2O → 2Mg(OH)2 + CH3 – C ≡ CH 2Mg2+ + C34– → [2–C = C = C2–] contains 2π bonds CaC2 + 2H2O → Ca(OH)2 + CH ≡ CH Ca2+ + C22– [–C ≡ C–] contain 2π bonds CaCN2 + 3H2O → CaCO3 + 2NH3 CN2– ⇒ [–N = C = N–] contains 2π bonds. 18. Answer A(s), B(r), C(p), D(q)
⎛ N ⎞ 3 ⎜ 2 ⎟ (m ) PV m ⎠ (A) R = = ⎝ = JK −1 mol −1 Tn (mol) (kelvin)
(B) P =
n2a V2
unit of a =
PV 2 n2
=
atm × litre2
⎛ R (C) Boltzmann constant K = ⎜⎜ N ⎝ AV
(D) V = nb unit of b =
(mol)2 ⎞ ⎟⎟ ⎠
V ⎛ litre ⎞ =⎜ ⎟ n ⎝ mol ⎠
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19. Answer A(q), B(r), C(s), D(p) ΔH = ΔE + ΔnRT Δn = number of moles of gaseous products – number of moles of gaseous reactants (A) Δn = 0; ∴ ΔH = ΔE (B) Δn = 2 – (1 + 3) = –2;
ΔH = ΔE – 2RT
(C) Δn = (1 + 1) = 2;
ΔH = ΔE + 2RT
(D) Δn = 1 + 1 – 1 = 1;
ΔH = ΔE + RT
20. Answer A(q), B(p), C(r), D(s) Factual type. 21. Answer A(p, q), B(p, q), C(s) D(r) For monoatomic gas,
r=
5 = 1.66 3
For diatomic gas,
r=
7 = 1.4 5
For polyatomic gas,
r=
4 = 1.33 3
Use, CP – CV = R also. 22. Answer A(s), B(p), C(q), D(s) (A) Equimolar mixture of strong acid and a strong base can be acidic, alkaline or neutral. But Equi-equivalent mixture of strong acid and a strong base is neutral. (B) Due to strong acid, it is acidic. (C) Due to strong base, it is basic. (D) Depends upon the Ka and Kb values of weak acid and weak base. 23. Answer A(p), B(s), C(p), D(r) (A) Ozonolysis of alkenes is the oxidation process because aldehyde and ketone is formed as product. (B) Cannizzaro reaction is an example of disproportionation because oxidation as well as reduction takes place. (C) Cumene process is an oxidation process. (D) Nucleophilic substitution of chlorobenzene of aqueous NaOH is neither oxidation nor reduction. 24. Answer A(q), B(s), C(r), D(p) 2+ +3 2–
+3
Fe C2O4 Fe – –1e (A) – –2 × 1e
+
+4
CO2
n-factor for FeC2O4 is 3. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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(B) Br2 → Br– + BrO3– °
B r 2 → Br–, +5
°
Br2 → Br O3− ,
Equivalent weight =
M 2
Equivalent weight =
M 10
Total equivalent weight = 5 . 3
So, n-factor for Br2 is 2+ –1
+3
Fe S2
Fe –1e
(C)
M + M = 6M = 3M 2 10 10 5
+
+4
SO2
– –
–2 × 5e
n-factor for FeS2 is 11. +1 –2
+2
Cu2 S
+4
CuO + SO2 –
–2 × 1e
(D)
–
–6e
n-factor for Cu2S is 8. 25. Answer A(q), B(s), C(p), D(r) +7
(A) Cl− → Cl O −4 Net change in oxidation number = 8 +6
(B) Cr 3+ → Cr O5 Net change in oxidation number = 3 –1
0
(C) H2 O 2 → O 2 Net change in oxidation number = 2 +6
+6
(D) Cr O 22+ → Cr O 24− Net change in oxidation number = 0 26. Answer A(p, q, s), B(p, q, r), C(q, r), D(p, q, r, s) Factual type. 27. Answer A(p, s), B(p, r, s), C(p, r, s), D(p, s) All are aromatic obeys hackel rule and cyclic with sp2 hybridization. B and C are heterocyclic and A and D are homocyclic 28. Answer A(q, r), B(r), C(q, s), D(q) (A) CH3 – CH == CHPh
+
H
CH3 – CH2 – CH – Ph Carbocation stabilize by resonance
Br
CH3 – CH2 – CH – Ph Br
∴ Markownikov product Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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O (B) CH3 – C ≡ CH
CH3 – C – CH3
+
+2
–H
Hg
Tautomerism
OH
H2O
CH3 – C == CH2
+
CH3 – C == CH2
–H
Electrophilic addition +
(C) Due to –I effect of R 3 N it gives anti-Markownikov addition product HBr (D) CH2 = CH2 ⎯⎯ ⎯→ CH3 − CH2Br electrophi llic addition
29. Answer A(p, q), B(r, s), C(r, s), D(p, q) (A) p → Oxidative ozonolysis q → Oxidation with KMnO4
C == C
(B) and (C)
C—C OH
OH
by OsO4 and cold dil KMnO4 (D) p → Oxidative ozonolysis q → Oxidation with KMnO4/H+ 30. Answer A(p, s), B(p, r), C(p, r), D(q, r) anode
O
cathode
O
(A) – C – C – C – ONa
O +
– C – C – C – O + Na – C – C – C – O
O •
–C–C–C–O
•
– C – C + CO2 free radical
•
C == C + – C – C – C – C –
∴ p, s (B) Wurtz Na
R–X
R• + NaX free radical
•
•
R + R
R–R Alkane
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(C) Frankland •
– C – C + Zn
–C–C
Br
free radical
–C–C–C–C– Alkane
(D) Friedel craft +
R − Cl + AlCl3 ⎯⎯→ R AlCl4
H
R AlCl4
+
+ R
+ AlCl3 + HCl
Carbocation
31. Answer A(p, s), B(q, r, s), C(q, s), D(q, s)
≡
(A)
sp2 hybridization nucleophile (electron rich) Benzyne
(B) Carbocation →
sp2 hybridization electron deficient (electrophile)
–C–C
rearrangement is due to stability
H H (C) H – C – C free radical
H H electron deficient (electrophile), sp2 hybridization (D) :CH2 → electron deficient (electrophile) sp2 hybridization 32. Answer A(q), B(p), C(s), D(r)
H (A) CH3 – C – CH2Cl
alc KOH
CH3 – C == CH2
H
H
electrolys is (B) CH3COONa ⎯⎯ ⎯ ⎯⎯→ CH3 − CH3 + CO2 + H2 + NaOH
(C)
C == C
O3
C–C O
H2O/Zn
C == O + C == O + H2O2
O O H H
(D) H2 C = CH2 + Cl 2 ⎯⎯→ Cl – C – C – Cl H H Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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33. Answer A(p, q), B(p, q, r), C(p, q, r, s), D(p)
CH2OH (A) CHOH
HIO4
2HCHO + HCOOH
CH2OH CHO (B) (CHOH)4
HIO4
5 HCOOH + HCHO
CH2OH
CH2OH CO
HIO4
(C)
(CHOH)4
CHO
+ HCHO
COOH
+ HCOOH
CH2OH CH 2OH
(D)
HIO4
2HCHO
CH 2OH
34. Answer A(s), B(r, s), C(q, s), D(p) (A) 1° and 2° oxidised by CrO3/H+ OH
(B) CH3 – C – R gives iodoform test H
(C) Straight chain alcohols have highest B. Pt than isomeric branched chain compounds (D) 3° alc gives white turbidity with Luca’s reagent within few seconds 35. Answer A(r), B(s), C(p), D(q) (A) Reduction in acidic medium (B) NH4HS (Zinin reduction) (C) Reduction in alkaline medium (D) Reduction in neutral medium 36. Answer A(p, r), B(s), C(q, r), D(p, r) (A) Nylon-66 is synthetic fibre and formed by condensation polymerisation. (B) P.V.C. is the addition polymer of vinyl chloride. (C) Bakelite is thermosetting polymer and formed by condensation polymerisation. (D) Terylene is synthetic fibre and formed by condensation polymerisation. 37. Answer A(p, r, s), B(p, s), C(r), D(q, s) Factual type. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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38. Answer A(p, q), B(r, s), C(q, r), D(p, q, r) (A) Down’s process is used for the manufacture of Na by the electrolysis of molten NaCl using steel electrode as cathode and carbon electrode as anode. (B) Na2CO3 + Ca(OH)2 → CaCO3 + 2NaOH (Gossage process). (C) In Castner Kellner process, NaOH is formed by electrolysis of brine solution using Hg as cathode and carbon rod as anode. (D) Nelson cell is used for the manufacture of NaOH by electrolysis of brine solution using steel rod as cathode and carbon rod as anode. 39. Answer A(p), B(s), C(p, q, r), D(p, q, r) Factual type. 40. Answer A(r, s), B(p, q, r, s), C(p, r, s), d(q) (A) Cu2+ reacts with NH3 and KCN to form [Cu(NH3)4]2+, and [Cu(CN)4]3– respectively. (B) Zn2+ form amphoteric oxide (ZnO) and also reacts with NH3 and KCN to form complex. (C) Cr3+ forms amphoteric oxide (Cr2O3) and also reacts with NH3 and KCN to form complex. (D) Sc3+ is diamagnetic and forms colourless compound. 41. Answer A(q, s), B(p, r), C(p, r), D(q, s) [Co(NO2)3]3– and [Ni(CN)4]2– are diamagnetic and colourless while [Cr(CO)6] and [Cu(NH3)4]2+ are paramagnetic and coloured. 42. Answer A(p, q), B(q, s), C(r, s), D(p, r) (A) In [Ni(CN)4]2–, dsp2 hybridisation is present and it is diamagnetic. (B) In [Cu(NH3)4]2+,dsp2 hybridisation is present and it is paramagnetic. (C) In [Cr(NH3)6]3+, d 2sp3 hybridisation is present and it is paramagnetic. (D) In [Fe(CN)6]4–,d 2sp3 hybridisation is present and it is diamagnetic. 43. Answer A(p, q) B(p, s), C(p, s), D(r, s) (A) Na2CO3 + H2SO4 → Na2SO4 + H2O +
CO2 ↑ (Colourless and odourless)
(B) Na2SO3 + H2SO4 → Na2SO4 + H2O +
SO 2 ↑ (Colourless and pungent odour)
(C) 2NaCl + H2SO4 → Na2SO4 +
2HCl ↑
(Colourless and pungent odour)
(D) 2NaNO3 + H2SO4 → Na2SO4 +
2NO2 ↑ (Light brown gas and pungent odour)
+ H2O +
1O ↑ 2 2
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44. Answer A(p, q), B(s), C(r), D(r) (A) Pb 2+ + 2KI → PbI2 ↓ + 2K + Yellow ppt .
Pb
2+
+ K 2CrO 4 → PbCrO 4 ↓+ 2K + Yellow ppt .
3+ SnO 2 )3 ↓+ 6 Na + (B) 2Bi + 3Na 2SnO 2 → Bi 2 (Black ppt.
(C) Hg2 Cl2 + 2NH4 OH → [Hg(NH2 )]Cl + Hg ↓ + NH4 Cl + 2H2 O Black ppt.
(D) CuSO4 + 4NH4OH → [Cu(NH3 )4 ]SO 4 Deep blue soluble complex
Section - F : Subjective Type 1.
As KCN is 100% ionized therefore total no. of moles in the solution = 2 × 0.1892 = 0.3784
Kf =
0.704 = 1.86 K kg mol–1 .3784
Hg(CN)2 + 2CN− 0.095 − x
K=
0.1892 − 2 x
Hg(CN)24− x
x = 1.73 (0.1892 − 2x )2 (0.095 − x )
On solving we get x = 0.005 Total no. of moles = 0.09 + 0.1792 + 0.005 + 0.1892 = 0.4634 ΔT f = m · Kf = 1.86 × 0.4634 ΔT f = 0.8619 Freezing point of solution = –0.8619°C 2.
The solubility AB2(s) at 30°C be s mol/litre AB2(s)
A2+ (aq) + 2B– (aq)
P0 − P s n 31.82 − 31.78 3s = ⇒ = N 31.78 55.56 Ps
s = 0.0233 mol/litre Ksp (at 30°C) = 0.0233 × (2 × 0.0233)2 = 5.05 × 10–5 M3 Ksp (at 25°C) = 3.56 × 10–5 M3 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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Now log
K sp (35°C) K sp (25°C)
Success Magnet (Solutions)
=
ΔH ⎡ T2 − T1 ⎤ ⎢ ⎥ 2.303R ⎣ T1T2 ⎦
⎛ 5.05 × 10−5 ⎞ ΔH 5 ⎛ ⎞ ⎟ log⎜ ⎜ 3.56 × 10−5 ⎟ = 2.303 × 8.314 ⎜⎝ 303 × 298 ⎟⎠ ⎝ ⎠ ΔH = 52.5 kJ/mol
3.
⎛ 1⎞ We know that Nt = No ⎜ ⎟ ⎝2⎠
n
Therefore, amount of undecomposed drug left in the body of patient immediately after having sixth dose = Left over of first dose + left over of second dose +.....+ left over of 6th dose 5
4
1
⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ = 200 × ⎜ ⎟ + 200 × ⎜ ⎟ + ...... 200 × ⎜ ⎟ + 200 2 2 ⎝ ⎠ ⎝ ⎠ ⎝2⎠ ⎡⎛ 1 ⎞ 5 ⎛ 1 ⎞ 4 ⎤ ⎢⎜ ⎟ + ⎜ ⎟ + ...... + 1⎥ (Summation in 9 geometric progression) 200 = ⎢⎣⎝ 2 ⎠ ⎝2⎠ ⎥⎦
⎤ ⎡ ⎧ 5 ⎢ 1 ⎪1 − ⎛ 1 ⎞ ⎫⎪ ⎥ ⎢ 2 ⎨ ⎜⎝ 2 ⎟⎠ ⎬ ⎥ ⎪ ⎪⎭ ⎥ 63 = 393.75 mg + 1 = 200 × = 200 ⎢ ⎩ ⎥ ⎢ ⎛ 32 1⎞ ⎥ ⎢ ⎜1 − ⎟ ⎥⎦ ⎢⎣ ⎝ 2 ⎠
4.
Eclipsed ethane
Staggered ethane ΔG = –1.7 kcal mol–1
ΔG = –2.303 RT log Kc –1.7 × 103 = –2.303 × 2 × 300 log Kc Kc =
[Staggered form] = 17 [Eclipsed form]
∴ mole percentage of staggered form of ethane =
and mole percentage of eclipsed form of ethane =
5.
[H+] =
17 × 100 = 94.4% 18
1 × 100 = 5.6% 18
–7 1× 10 −14 = 10 M
and [D+] =
–8 3 × 10 −15 = 5.48 × 10 M
Thus [H+] > [D+]. So H2 is liberated much faster than D2 at cathode Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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6.
(A) NH4NO3, (B) N2O, (C) H2O, (D) N2, (E) O2, (F) NH3 Reaction involved are Δ NH 4NO 3 ⎯⎯→ N2 O+ 2H2 O (B )
(A)
(C)
Δ 2N2O ⎯⎯→ 2N2 + O 2 (D )
(E )
NH4NO3 + NaOH ⎯⎯→ NaNO3 + NH3 + H2O (F )
Al + NaOH + H2O ⎯⎯→ NaAlO 2 + 3(H) NaNO 3 + 8(H) ⎯⎯→ NaOH + NH3 + 2H2 O
7.
The co-ordination no. of cobalt is six since this compound does not give any precipitate with Ca(NO3)2. So oxalate should remain confined in co-ordination sphere. Considering these facts we can conclude that the compound having complex as cation should be [Co(C2O4)(NH3)4]Br and complex containing the co-ordination complex as anion should be NH4[Co(C2O4)(NH2)(NH3)2Br] Both are ionization isomer of each other.
8.
(i) + 1 (ii) Triaqua hydroxo peroxo titanium (IV) ion.
9.
Where x is amount of gas adsorbed on mass m at constant pressure.
x m
Absolute temp This graph shows that chemical adsorption first increases then decreases to attain constancy with rise in temperature 10. (i) Oxygen has no d-orbital and forms simple diatomic molecules and exists as a gas which sulphur has unoccupied d -orbital which allows some paired electron to unpair themselves so that it can extend its valency to +6. Where by it forms a complex molecule i.e. a ring of eight atoms (S8) and exists as a solid. (ii) Nitrites oxidizes iodide ion to iodine and thus liberated iodine gets dissolved in KI. Solution to intensity its colour forming KI3. On the other hand sulphites are themselves oxidized by I2 of solution and thus reducing I2 to discharge the colour of solution. 11. (A) NH4NO2, (B) NO2, (C) NH3, (D) N2 Reaction involved
NH4NO2 + HCl ⎯⎯→ NH4 Cl + HNO2 (A)
3HNO2 ⎯⎯→ HNO3 + H2O + 2NO
2NO + O 2 ⎯⎯→ 2NO2 ↑ (B )
NH4NO2 + NaOH ⎯⎯→ NaNO2 + NH3 ↑ + H2O ( C) Δ NH 4NO 2 ⎯⎯→ N2 + 2H2 O (D )
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12.
82 Pb
212
λ1 ⎯⎯→
83Pb
Success Magnet (Solutions) 212
λ2 ⎯⎯→ ⎯
λ1 =
0.693 −1 hr = 0.0654 hr −1 10.6
λ2 =
0.693 × 60 −1 hr = 0.6873 hr −1 60.5
The time required for maximum activity for
212 83Bi
is
λ2 2.303 2.303 0.6873 log t = λ − λ log λ = 0.6873 − 0.0654 0.0654 2 1 1
= 3.783 hours. 13. (i)
2KO 2 + 2H2 O ⎯⎯→ 2KOH + H2 O 2 + O 2
(ii) MgCl 2 · 6H2O ⎯⎯→ Mg(OH)Cl + HCl + 5H2 O Mg(OH)Cl ⎯⎯→ MgO ↓ + HCl ↑
(iii) SnCl 2 + Na 2 CO 3 ⎯⎯→ SnO ↓ + 2NaCl + CO 2 ↑ (iv) 3Ca(OCl)2 + 4NH3 ⎯⎯→ 3CaCl 2 + 6H2 O + 2N2 ↑ 14. Let reaction are taking place at a temperature T −E / RT K =A e a
B
−E / 2RT 2K = A e a
K
Comparing equation (ii) by (i) we get
A
A = 4K Overall velocity constant of compound (D) = K + 2K = 3K
2K
C
Let overall activation energy of compound D be Ea′ −E ′ / RT 3K = A e a
taking ln both sides we get −E a ′ 3K = ln RT A
Ea′ = Rt ln
A 3K
Ea′ = Rt ln
4 (as A = 4K) 3
=
⎛ Ea ⎞ ⎛4⎞ E ⎜ ⎟ × ln⎜ ⎟ from (i) and (ii) we have RT = a ln 4 3 ⎝ ⎠ ln 4 ⎝ ⎠
Ea′ = 0.21 Ea. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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15. Let the molar mass of first compound be M. That of second compound is therefore equal to M minus twice the formula mass of ClO4– plus twice the formula mass of SCN– M–(2 × 99.5) + 2 × 58 = M – 83 The % of carbon in the first compound is 12 x × 100 = 30.15 M
In the second compound which has 2 mole of carbon in the anion per mole of compound the % of carbon in this compound is 12(12 + x ) × 100 = 40.46 M − 83
Solving by simultaneous equation (i) and (ii) x = 14 and M = 557 The % of hydrogen in first compound is 1.0 y × 100 = 5.06 557
y = 28 The total formula mass of all the elements other than nitrogen is 56 units therefore 56 units must represents the nitrogen in one formula unit. Thus there are four nitrogen atoms per formula unit and the complete formula's are [Pd C14H28N4](ClO4)2 and [Pd C14H28N4](SCN)2 16. Let the rate of accumulation of SO3 in the environment be
dx at any instant (t) this given by dt
dx 1× 10 −6 − Kx = dt 60
Where K is the decay constant and x is the no. of moles of SO3 present in a litre of air dx .693 −8 x = 1.25 × 10 − dt 60
at the state of equilibrium the rate of accumulation will become zero as the rate of enrichment is same as that of decay dx =0 dt
x =
1.25 × 10 −8 × 60 mol / L 0.693
The final concentration of SO3 in air =
1.25 × 10 −8 × 60 × 80 0.693
= 8.65 × 10–5 g/litre Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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N2 (g) + 3H2 (g)
17. Initial moles
a
Final moles
2NH3 (g)
3a
0
(a – x) (3a – 3x)
2x
Given x = 0.6 a moles of N2 left = a – 0.6a = 0.4a moles of H2 left = 3a – 1.8a = 1.2a Moles of NH3 formed = 2 × 0.6a = 1.2a Total moles after the reaction = 0.4a + 1.2a + 1.2a Before the reaction 1.1 atm T = 298 K Total moles = 4a Since reaction took place in vessel of constant volume so applying gas law P1 P2 1 .1 P = = n1T1 n2 T2 ⇒ 4a × 298 2.8a × 573
P = 1.48 atm Therefore the final pressure of the reaction mixture will be 1.48 atm 18. The total number of moles = 6 Thus we have PV = n R T 3×V=6×RT
.......(i)
It now Cl2 is added at same pressure and temperature to double the volume we have 3 × 2V = n R T
.......(ii)
From (i) and (ii) n = 12 Previously the partial pressure PCl5, PCl3 and Cl2 were each 2 × 3 = 1 atm 3
∴ Kp =
PPCl3 × PCl2 PPCl5
=
1× 1 = 1 atm 1
Suppose x moles of Cl2 were added. Part of this would have combined with PCl3. Taking this to be y moles. No. of moles of Cl2 present = (2 + x – y) No. of moles of PCl3 present = (2 – y) No. of moles of PCl5 present =
(2 + y)
⎤ ⎡2 + x − y × 3⎥ atm Partial pressure of Cl2 = ⎢ ⎦ ⎣ 12 ⎤ ⎡2 − y × 3 ⎥ atm Partial pressure of PCl3 = ⎢ ⎦ ⎣ 12 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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⎡2 + y ⎤ × 3⎥ atm Partial pressure of PCl5 = ⎢ 12 ⎣ ⎦
Total no. of moles = 2 + x – y + 2 – y + 2 + y = 6 + x – y = 12 x–y=6 KP =
PPCl3 × PCl2 PPCl5
⎛2−y ⎞⎛ 8 ⎞ × 3 ⎟⎜ × 3 ⎟ ⎜ ⎝ 12 ⎠⎝ 12 ⎠ =1 = 2+y ×3 12 ⎡2 − y ⎤ ⎢ ⎥×2 = 1 ⎣2 + y ⎦
y=
2 3
x–y=6 x = 6+
2 20 = moles of Cl2 were added. 3 3
19. Total number of faradays passed =
2 × 10−3 × 16 × 60 96500
= 1.9896 × 10–5
1.9896 × 10−5 2 Since absorbance was reduced to 50% of its original value Moles of Cu2+ ions deposited
=
The initial moles of Cu2+ would be two times of moles of Cu2+ reduced
1.9896 × 10 −5 × 2 = 1.9896 × 10 −5 2 The conc. of CuSO4 in the solution = 1.9896 × 10–5 × 4 Initial moles of Cu2+ =
= 7.958 × 10–5 mol/litre 20.
0 0 E = EFe 3 + 2+ − E − − − Fe I /I 3
[Fe 2+ ][I3− ] 0.059 log − 3 2 [I ] [Fe3 + ] 2
At equilibrium E = 0 0 = 0.77 – 0.54 –
0.059 log K 2
On solving K = 6.26 × 107 M–2 Roasted 21. 2CuFeS2 + O2 ⎯⎯ ⎯⎯→ Cu2S + 2FeS + SO2↑
2Cu2S + 3O2 ⎯⎯→ 2Cu2O + 2SO2 ↑ 2Cu2S + 5O2 ⎯⎯→ 2CuSO4 + 2CuO 2Cu2O + Cu2S ⎯⎯→ 6Cu + SO2 ↑ Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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22. Moles of SrCO3 =
Success Magnet (Solutions)
121 = 0.817 148
Moles of CO2 required to make the first reaction at equilibrium =
PV 4×5 = = 0.817 RT 0.0821 × 298
Therefore SrCO3 would completely dissociate Since carbon (s) is added it will consume CO2 to produce CO and the reaction would try to reach equilibrium C(s) + CO2 (g)
2CO (g)
Initially
4
0
at equilibrium
4–x
2x
Kp = x
(2x )2 =3 4−x
= 1.39
PCO = 2.78 atm PCO = 4 – 1.39 = 2.61 atm 2
23. In buffer solution of pH = 8 the [OH¯] is equal to 10–6 M Pb2+ + 2OH¯
Pb(OH)2 (s)
s Ksp of Pb(OH)2 =
4s 3
2s
= 4 × (6.7 × 10–6)3 = 1.2×10 –15 M3
Solubility of Pb(OH)2 in a buffer solution of pH = 8 would be Ksp = [Pb2+] [OH¯]2 [Pb2+] =
K sp [OH− ] 2
=
1.2 × 10 −15 (10 − 6 )2
[Pb2+] = 1.2 × 10 –3 M 24. The density of the silicon lattice is given by Z×M d = N × a3 AV
Let x% of tetrahedral sites are occupied by silicon atoms in its fcc lattice. The No. of atoms in tetrahedral voids are double the no. of effective atoms in a lattice n=4+
8x = 4 + 0.08 x 100
4 + 0.08x =
2.23 × 6.023 × 1023 × (0.55 × 10 −7 )3 28
x = 49.75 ≈ 50 Thus 50% of tetrahedral voids are occupied by silicon atoms in this lattice The Si-Si bond length is equal to the sum of radii of 2 silicon atoms
2 Si =
3a = 4
3 × 0.55 = 0.238 nm 4
25. (A) CHO – (CHOH)4 – CH2OH (B) CN – (CHOH)5 – CH2OH (C) COOH – (CHOH)5 – CH2OH Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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OH 26. (1)
(2)
(3)
(4)
(5)
Br
O
CH3 27. P =
CH3
Br
CH3
N – C – C2H5 ,
Q=
CH3
Br
O
CH3 N – H,
R = C2H5COONa,
S=
CH3
N – C – CH3
CH3 T=
N – N == O
CH3
28. S → C2H5 – ND2 T → C2H5NH2
MgI 29. (A)
CH2CH2OH (B)
CH2CH2COOH (E)
CH2CH2Br (C)
CH2CH2CN (D)
CH2CH2 – COCl (F)
(G)
O OH CHO
OH CHO 30. (A) C6H5OH
(B)
(C)
(D)
CHO COOH (E)
31. (i) Tl+ will not disproportionate. (ii) A : Ca2B6O11 . 5H2O E : B2O3 32. (i)
B : CaCO3
C : Na2B4O7
D : NaBO2
F : Co(BO2)2
2Mg + CO2 ⎯⎯→ 2MgO + C
Mg reduces CO2 and continues to burn in CO2. (ii) In the silicon, the 3d-orbitals form coordinate bonds with water extending the coordination number to six. Therefore SiCl4 gets hydrolysed but CCl4 does not have vacant orbitals. (iii) Pb4+ is unstable due to inert pair effect. The small size of Pb4+ does not stabilize the four I– or Br – ions around it. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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33. (i)
Success Magnet (Solutions)
Δ Fe(CO)5 + 2NO ⎯⎯→ [Fe(CO)2 (NO)2 ] + 3CO
Δ (ii) 3KNO 3 + KNO 2 + Cr2O 3 ⎯⎯→ 2K 2 CrO 4 + 4NO
(iii) 3Cr 2 + + NO + 3H+ ⎯⎯→ NH2OH + 3Cr 3 + (iv) SO 2 + 2NO + H2O ⎯⎯→ N2O + H2SO 4 (v) 3MnO −4 + 5NO + 4H+ ⎯⎯→ 5NO3− + 3Mn2+ + 2H2O (vi) SO32− + 2NO2− + 2H+ ⎯⎯→ SO 24− + 2NO + H2O 34. (i)
K 2 Cr2 O 7 + 4H 2 SO 4 ⎯⎯→ K 2 SO 4 + Cr2 (SO 4 ) 3 + 4H 2 O + 3[O] [H 2 O + SO 2 + [O] ⎯⎯→ H 2 SO 4 ] × 3 K 2 Cr2 O 7 + H 2 SO 4 + 3SO 2 ⎯⎯→ K 2 SO 4 + Cr2 (SO 4 ) 3 + 4H 2 O green
(ii) In SO2, oxidation state of S is +4 hence it can be oxidized to +6 and reduced to any lower state by a suitable oxidizing agent. 35. (i) H2SO4 oxidizes HI to I2. (ii) CaOCl 2 + CO 2 (air ) ⎯⎯→ CaCO 3 + Cl2 6CaOCl 2 ⎯⎯→ 5CaCl 2 + Ca(ClO 3 ) 2
Hence, the bleaching activity is lost. 36. (i)
2 XeF4 + 3H2O ⎯⎯→ Xe + XeO3 + 3H2F2 + F2
(ii) 2 XeF6 + SiO 2 ⎯⎯→ 2 XeOF4 + SiF4
2 XeOF4 + SiO 2 ⎯⎯→ 2 XeO2F2 + SiF4 2 XeO2F2 + SiO2 ⎯⎯→ 2 XeO3 + SiF4 exp losive
(iii) XeF6 + 8NH3 ⎯⎯→ Xe + 6NH4F + N2 (iv) XeF6 + 3H2O ⎯⎯→ XeO3 + 6HF (v)
XeF6 + SbF5 ⎯⎯→[ XeF5 ]+ [SbF6 ] −
37. (i) Compound (D) is formed by distillation of Ca-acetate as well as by oxidation of (A) by KMnO4 and therefore (D) and (A) may be ketone say acetone and sec alcohol say 2-propanol respectively. Δ 2(CH3COO)2 Ca ⎯⎯→ 2CaCO3 + 2CH3 COCH3
(D ) Acetone
[O ] ⎯→ CH3COCH3 CH3 CH(OH)CH3 ⎯⎯
( A ) propanol − 2
(D )
(ii) (A) on treatment with conc. H2SO4 gives (B), an alkene conc . H2SO 4 ⎯ ⎯→ CH3 − CH = CH2 CH3 − CHOH CH3 ⎯⎯ ⎯ ⎯ −H O
(A)
2
(B ) Propene
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(iii) (B) reacts with Br2 water to give dibromide which on dehydrobromination by NaNH2 gives (C) NaNH
2 ⎯⎯ → CH3 − C ≡ CH CH3 – CH = CH2 + Br2 → CH3 − C H − C H2 ⎯⎯ −2HBr
⎜
Propyne (C)
⎜
Br
Br
(iv) Action of H2SO4 and HgSO4 on (C) gives acetone. H SO4 → CH3 COCH3 CH3—C ≡ CH + H2O ⎯⎯2 ⎯⎯ HgSO 4
(D ) Pr opanone
Thus A, B, C and D are
CH3 CHOHCH3 , CH3 − CH = CH2 , CH3 − C ≡ CH, CH3CO CH3 (A)
(B )
(C)
(D )
38. Since dibasic acid (A) reacts with two moles of acetyl chloride and four moles of HI. Hence it has two alcoholic – OH. 2CH3COCl –2HCl
C4H6O6 (A)
4HI Δ
C4H4O6(OCCH3)2
CH2COOH CH2COOH Succinic acid
Acid (C) reacts with one mol of CH3COCl and two moles of HI, hence it contains one alcoholic – OH. 2HI Δ
C3H6O3
CH3COCl –HCl
CH3CH2COOH
C3H5O3COCH3
Thus (A) is tartaric acid and (C) is lactic acid, since both (B) and (C) forms iodoform, hence (B) has keto group at C2 while (C) has alcoholic – OH at C2. Tartaric acid on heatings with KHSO4 eliminates CO2 and H2O gives pyruvic acid.
C HOH COOH ⎜
CHOH COOH (A)
KHSO 4 ⎯⎯ ⎯⎯→ CH3 COCOOH + CO + H O 2 2 Δ (B ) Pyruvic acid
2 [H ] CH3 CO − COOH ⎯⎯⎯→ CH3 − C H − COOH
⎜
(B )
OH (C) Lactic acid
CHOH COOH CHOH COOH (A) 4HI
+ 2CH3COCl →
CH2COOH CH2COOH
CH(OCOCH3)COOH CH(OCOCH3)COOH
+ 2H2O + I2
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CH3COCl
CH3CHOH COOH (C)
CH3CH(OCOCH3)COOH
– HCl
2HI
CH3CH2COOH + H2O + I2
Δ
Δ CH3COCOOH + 3I2 + 5NaOH ⎯⎯→ CHI3 + 3NaI + 3H2O +
(B )
C OONa ⎜
COONa
Δ CH3CHOH COOH + 4I2 + 6NaOH ⎯⎯→ CHI3 + 4NaI + 4H2O +
(C)
C OONa ⎜
COONa
Hence A, B & C are tartaric acid, pyruvic acid and lactic acid respectively. 39. The general formula of saturated hydrocarbon is Cn H2n+2 Molecular wt. of hydrocarbon = 12n + 2n + 2 = 58 or 12n + 2n = 58 – 2 = 56 or 14n = 56 or n = 4 Hence molecular formula of alkane C4H10 There are two isomers of this formula CH 3 CH3 CH2 CH2CH3
CH3 – CH – CH3
n − butane
2-methyl propane
But-2-methyl propane having tertiary carbon atom could explain the observed facts.
CH3
CH3
|
CH2Cl
|
CH3 – C – H
Cl
2 ⎯⎯ ⎯ →
|
hv
|
CH3 – C – Cl + CH3 – C– H |
CH3
|
CH3
CH3
(A) (A) Isomer : Since alcohol gives positive Lucas test immediately, hence it is a tertiary-alcohol and thus monochloro derivative is tertiary-alkyl chloride.
CH3 CH3 – C – Cl
CH3 NaOH aq.
CH3 (B)
CH3
CH3 – C – OH
conc. ZnCl2 + HCl (Lucas Test)
CH3 (C)
CH3 – C – Cl CH3 (cloudiness)
(C) Dehydrated easily to alkene (D).
CH3 CH3 – C – OH CH3
CH3 conc. H2SO4 Δ
CH3 – C – CH2 + H2O (D)
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CH3
CH3
CH3 – C = CH2
(i) O3
H
CH3 – C = O + H – C = O
(ii) Zn/H2O
(D)
(E)
(E) On distillation with H2SO4 gives mesitylene.
CH3 Conc. H SO
4 3 CH3 CO CH3 ⎯⎯ ⎯⎯2 ⎯⎯ →
Δ
CH3
CH3
Mesitylene (F) CH3
CH3
Hence (A)
CH3
|
|
CH3 – C– H
(B)
|
CH3 – C– Cl
(C) CH3 – C– OH
|
|
|
CH3
CH3
CH3
CH3 O
CH3
||
|
(D) CH3 – C = CH2
(E) CH3 – C– CH3
(F)
CH3
40.
CH – CH3 C ≡ C – CH3
CH3
CH – CH3
dil. H2SO4 HgSO4
O = C – CH2 – CH3
(A)
(Monoketone)
(A) Decolourises Br 2 /Cl 4 & alk. KMnO 4 but no reaction with ammonical AgNO 3 , since it has no terminal ≡ CH.
CH – CH3 C ≡ C – CH3
CH3COOH + Ethanoic acid
* CH – CH3 COOH
(i) O3 (ii) H2O2 Oxidative hydrolysis
Optically active acid Δ –CO2
CH2 – CH3 Ethyl benzene
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41. Addition of one mole of Br2 indicates the presence of one ethylenic double bond which on hydrolysis gives vic-diol (C)
C6H5 CH3 C6H5 – C = C – H
C6H5 CH3 Br2
C6H5 CH3 Hydrolysis
C6H5 – C — C – H Br (B)
(A)
C6H5 – C — C – H
Br
OH (C)
OH
The oxidation of (A) gives a ketone (D) and acetic acid, thus (A) must have the following structure which explains all the reactions. C 6H5 CH3 ⏐
C6H5
⏐
C = O + CH3COOH C 6H5 – C == C− H ⎯⎯ ⎯→ C H (A) 6 5 [O ]
C 6H5 CH3 |
|
(i ) O
3 ⎯ → C 6H5 − C == C− H ⎯⎯ ⎯
(ii) H2O
(A)
C6H5 C = O + CH3COOH C6H5
C 6H5 CH3
C 6H5 |
⏐
⏐
⏐
⏐
OH
OH
C 6H5 – C — C− H
⎯⎯ ⎯ 30 ⎯%⎯ ⎯ ⎯→ H2SO 4 Pincol−pinacolone rearrangement
C 6H5 − C H – C – CH3 ||
O Ketone (D)
42. As (A) gives blue colour in Victor meyer test, hence it is 2º-nitroalkane. It also gives effervescence with NaHCO3, hence it also contains – COOH group. Hence (A) is 2-nitropropanoic acid which explains all the reactions.
NO2 CH3 – CH COOH (A) The different reactions are as follows : Sn + HCl
HNO
2 CH3 – C H – NO 2 ⎯⎯ ⎯ ⎯→ CH3 − C H − NH2 ⎯⎯ ⎯ ⎯ → CH3 – CH – OH
|
|
COOH (A)
COOH (B)
COOH (C)
B gives carbyl amine reaction with CHCl3 & KOH due to presence of –NH2 group.
CH3 – CH – OH
I2 –2HI
CH3 – COCOOH
COOH (C)
NaOH
CH3 – CO – COONa
COONa CHI3 +
I2 + NaOH
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43. The compound (A) is basic and reacts with benzene sulphonyl chloride and diethyl oxalate which is an example of Hinsberg reaction & Hoffmann reaction of 1°-amine.
NH2 + ClSO2
NHSO2
(A) 1°-amine
KOH
Soluble
(B) sulphonamide of 1°-amine
NH2
COOC2H5
CONH – C6H5
COOC2H5
CONH – C6H5
Diethyl oxalate
(C) Oxamide (solid)
+ NH2 2-moles
(B) is soluble in KOH, because it contains acidic H on nitrogen. 44. From % composition its molecular formula is C3H4. Since decolourises Br2 - water and gives red precipitate with ammonical Cu2Cl2. Thus (A) is a terminal alkyne. The only structure of terminal alkyne containing three carbon is possible. Hence A is CH3 − C ≡ CH Pr opyne
By this all the reactions and B,C & D can be formulated.
CH3 Red Hot tube
CH3 – C ≡ CH
CH3
CH3
(A)
(B)
(i) O3 (ii) Zn/NaOH
(i) O3 (ii) Zn/NaOH
CH3 – C – CHO O (C) Compound C
CH3 – C – CHO O Gives dioxime, reduces Tollens reagent and gives Iodoform, since O both have – C – CH3 group
45. Since (B) on heating alone gives ethanoic acid, thus (B) is malonic acid. The formation of B certainly comes by the hydrolysis of propane 1, 3 dichloride followed by oxidation. On the basis of this other reactions can be explained.
CH2Cl CH2
CH2Cl
NaOH (aq.)
CH2OH CH2
CH2OH
[O] K2Cr2O7/H2SO4
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Miscellaneous Questions
COOH CH2
Success Magnet (Solutions)
COOC2H5
C2H5OH/H2SO4
CH2
Esterification
COOH
Knovengal reaction
(C)
(D)
Pyridine
COOC2H5
(B)
CH3 – CH = C(COOC2H5)2
CH3CHO
CH3
(i) KOH/H2O (ii) H/H2O (iii) Decarboxylation
COOH C=C (Cis)
H
+
CH3 H
H
H C=C (Trans)
(E)
COOH
(F)
O
O
CHC6H5
46. A.
B.
OH
CHC 6H 5
CHC6H5 C.
D.
CH2CHO E.
CH2
F. C6H5CHO
CH2CO – CHO 47.
(i ) O
3 C 8H10 ⎯⎯ ⎯⎯ → C 4H 6 O 2
( X)
(ii ) Zn / H2O
(Y )
Since compound (x) adds one mole of O3 hence it should be ether a >C=C< or a – C ≡ C – bond. If it was alkene its formula should be C8H16 (CnH2n) and if it was alkyne it should have the formula C8H14; it means it is neither a simple alkene nor simple alkyne. Since compound is unsaturated, thus it should be cyclosubstituted alkyne, like
CH2
(i ) O
3 ⎯⎯ ⎯ ⎯ →
CH – C ≡ C – CH CH2
CH2
CH2
CH – C – C – CH
(ii) H2O warm
CH2
CH2
(X)
O O
CH2
CH2
→
CH2
CH2
CH – COOH (Y)
Compound (Y) can be prepared from (Y) cyclopropyl bromide as follows :
CH2
CH2 Mg / ether
CH2
CH2 CO2
⎯→ CH – Br ⎯⎯ ⎯ ⎯
CH – MgBr ⎯⎯⎯→ CH2
(Z)
H2O CH – COOMgBr ⎯⎯ ⎯→
Δ
CH2
CH2 CH – COOH CH2 Thus X, Y & Z are
C≡ C (X)
,
COOH and (Y)
Br (Z)
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48. Reaction (1) indicates that (A) contains Cl– ions because, it gives white ppt soluble in NH4OH. It is again confirmed because it gives chromyl chloride test. The colour of oxidising and reducing flames indicate that (A) also contains Ni+2 ions. Hence, (A) is NiCl2 The different reactions are (I)
NiCl2 + 2AgNO3 → 2AgCl + Ni(NO3 )2 (A)
AgCl + 2NH3 → [Ag(NH3 )2 ] Cl Soluble
[Ag(NH3)2]Cl + 2HNO3 → AgCl ↓ + 2NH4NO3 White ppt (B)
The equations of chromyl chloride tests are, NiCl2 + Na2CO3 → 2NaCl + NiCO3 4NaCl + K2Cr2O7 + 6H2SO4 → 4NaHSO4 + 2KHSO4 + 3H2O + 2CrO 2Cl 2 (Re d )
CrO2Cl2 + 4NaOH → Na 2CrO 4 + 2NaCl + 2H2O Yellow solution (C)
Na2CrO4 + (CH3COO)2Pb → PbCrO 4 + 2CH3COONa Yellow ppt
Δ (II) Na2B4O710H2O ⎯⎯→ Na2B4O7 + 10H2O Δ Na2B4O7 ⎯⎯→
2NaBO
+B O
23 3 1442244 Transparen t bead
Δ NiO + B2O3 ⎯⎯→
Ni(BO 2 )2 Nickel metaborate (Brown) in oxidising flame
Δ Ni(BO2)2 + C ⎯⎯→
Ni
Grey in reducing flame
(III) NiCl2 + H2S → NiS ↓
+ B2O3 + CO
+ 2HCl
(Black ppt)
NiS + 2HCl + [O] → NiCl2 + H2S (A )
(IV) NiCl2 + 2NaHCO3 → NiCO3 + 2NaCl + CO2 + H2O (A)
Δ 2NiCO3 + 4NaOH + [O] ⎯⎯→ Ni2O 3 ↓ + 2Na2CO3 + H2O
Black ppt (D)
(V) NiCl2 + 2KCN → Ni(CN) 2 + 2KCl (A)
Green ppt (E )
Ni(CN)2 + 2KCN → K 2 [Ni(CN)4 ] (F )
NaOH + Br2 → NaOBr + HBr Δ 2K2[Ni(CN)4] + 4NaOH + 9NaOBr ⎯⎯→ Ni2O 3 ↓ + 4KCNO + 9NaBr + 4NaCNO
(D)
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Miscellaneous Questions Δ ⎯⎯→
(A)
49. (I)
Success Magnet (Solutions)
White powder
+
(B) Colourless gas turns lime water milky
( C) A residue yellow in hot and white in cold
K [Fe( CN)6 ] dil. HCl (II) (C) ⎯⎯ ⎯ ⎯→ solution ⎯⎯4 ⎯ ⎯ ⎯ ⎯→ a white ppt. dil HCl (III) (A) ⎯⎯ ⎯ ⎯→ solution + (B)
Excess NH4OH +H2S
(D) white ppt
The solution of (A) in dil. HCl NaOH
NaOH
dissolves
H2S
(E)
(E) white ppt
From reaction (I) the colourless gas is CO2, because it turns lime water milky, due to formation of insoluble CaCO3 CO 2 + Ca(OH)2 → CaCO3 + H2O (B )
lime water
The compound (C) is zinc oxide because it is yellow in hot and white in cold and hence (A) is zinc carbonate (ZnCO3). From reaction (II) (C) is a salt of Zn+2 which dissolves in dil. HCl and white ppt obtained after addition of K4[Fe(CN)6] is due to zinc ferrocyanide, a test of Zn+2 cation. From (III), it is proved that (A) is ZnCO3 because on treatment with dil HCl it gives gas (B) i.e. CO2, while Zn+2 goes in solution i.e. ZnCl2 on passing H2S gas in presence of NH4OH, it gives white ppt of ZnS (D). ZnS on heating with dil. H2SO4 evolves H2S gas, which is used for precipitation of sulphides of group (II) in acidic medium and of group (IV) in alkaline medium. ZnCl2 react with NaOH to give a ppt of Zn(OH)2 which dissolves in NaOH, as Zn(OH)2 is amphoteric in nature. The solution Na2ZnO2 again gives ZnS on passing H2S gas into it. Chemical reactions involved are : Δ ZnCO3 ⎯⎯→ ZnO + CO 2 (C)
(A)
(B )
ZnO + 2HCl → ZnCl 2 + H2O (C)
Solution
2ZnCl2 + K4[Fe(CN)6] → Zn 2 [Fe(CN)6 ] + 4KCl white ppt .
Δ ZnCO3 + 2HCl ⎯⎯→ ZnCl2 + CO 2 + H2O (A)
dil
(B )
ZnCl2 + H2S → ZnS
White ppt (D )
+ 2HCl
Δ ZnS + H2SO4 ⎯⎯→ ZnSO4 + H2S
ZnCl2 + 2NaOH → Zn(OH)2 ↓ + 2NaOH (E )
Zn(OH)2 ↓ + 2NaOH → Na 2 ZnO 2 + 2H2O so lub le
(E )
Na2ZnO2 + H2S → ZnS ↓ + 2NaOH white ppt
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50. On heating (A) with aluminium powder and NaOH, a gas is liberated, as it gives fumes with HCl and brown ppt. with Nessler’s reagent, hence it should be NH3. (I) NH3 + HCl → NH4 Cl White fumes
(II) 2HgI42– + NH3 + H2O → H 2N
I + 7I– + 3H+
O Hg
Hg
(Brown ppt.)
(III) NH3 on passing over heated CuO gives free Cu Δ 2NH3 + 3CuO ⎯⎯→ N2 + 3H2O + 3Cu
While the liberated N2 on reaction with Mg gives a solid compound i.e. magnesium nitride. N2 + 3Mg → Mg3N2
White solid
Above results indicates that original compound (A) should contain nitrate as it is reduced to NH3 by Al and NaOH. 8Al + 3NO3– + 21OH– → 8AlO33– + 6H2O + 3NH3 It is given that, Δ ( A ) ⎯⎯→ (B)+ (C) gas
solid
Gas (B) must be O2, because it is essential for living beings. (C) must have Pb+2 ions as on dissolving in HNO3 followed by the addition of HCl, it gives a white ppt which is soluble in hot water, but reappears on cooling. It is characteristics of PbCl2. Thus compound (A) is lead nitrate Δ 2Pb(NO3 )2 ⎯⎯→ 2PbO+ 4NO 2 + O 2 (C)
(A)
(B )
O 2 + 6PbO → 2Pb 3 O 4 (C)
Re d lead
PbO + 2HNO3 → Pb(NO3)2 + H2O Pb(NO3)2 + 2HCl → PbCl2 + 2HNO3. 51. According to the question
(A) Scarlet Compound
conc. HNO3
(B) Chocolate Brown
Filterate
(i) NaOH (ii) KI
(C) Yellow ppt.
Mn(NO3)2 HNO3
Pink coloured solution The compound (B) should be powerful oxidising agent which converts Mn2+ to a pink coloured MnO4– ions. Normally PbO2 is selected for this purpose. The compound (C) may be PbI2 which is yellow in colour Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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All the given reactions can be explained as :Δ Pb 3 O 4 + 4HNO3 ⎯⎯→ PbO2 + 2Pb(NO3)2 + 2H2O
(B ) Chocolate brown ppt.
(A ) Scarlet
Pb(NO3)2 + 2KI →
PbI2 (C) Yellow ppt .
+ 2KNO3
Δ ⎯→ Pb(MnO 4 )2 + 4Pb(NO3)2 + 2H2O 5PbO2 + 2Mn(NO3)2 + 4HNO3 ⎯
(D ) Pink coloured
(B )
Thus,
A = Pb3O4 B = PbO2 C = PbI2 D = Pb(MnO4)2
52.
222 4 Ra 266 88 → Rn 86 + 2 He
N 0 = 1gm – atom, t½ = 1600 year, t = 800 year.
Q t=
N 2.303 log 0 λ N 2.303 × t ½ N log 0 0.693 N
or
t=
or
800 =
2.303 × 1600 1 log 0.693 N
N = 0.707gm – atom. ∴ Amount of Ra decayed = 1 – 0.707 = 0.293 gm - atom ∴ Mole of Rn formed = 0.293 Mole of He formed = 0.293 ∴ Total moles of gases = 0.293 + 0.293 = 0.586
Q PV = nRT ∴ Total pressure of He and Rn is given as n ⋅ RT V
P =
=
0.586 × 0.0821 × 300 atm 5
P = 2.887 atm Partial pressure of He (pHe) = mole – fraction of He × total pressure =
1 × 2.887 atm = 1.443 atm. 2
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53. Let m' be the molality of the solution after the ice separates out at –3.534°C. Now we have, ΔTf = kF.m' ∴
m′ =
ΔTf 3.534 = = 1.9 Kf 1.86
Initially the molality is 1 m and wt. of solution is 1000 g. 1 mole of sucrose is dissolved in 1000 gm of H2O or 342 g of sucrose is dissolved in 1000 g of H2O. ∴ 1342 g of solution contains 342 g of sucrose ∴ 1000 g of solution contains
342 × 1000 g = 254.84 g of sucrose 1342
∴ Amount of H2O = (1000 – 254.84) g = 745.16 g Now, when ice separates out, the molality is 1.9 and the weight of sucrose remains the same as before.
Q (1.9 × 342)g of sucrose is present in 1000 g of H2O. ∴ 254.84 g of sucrose should be in
1000 × 254.84 = 392.18 g of H2O 1.5 × 342
Thus, amount of ice separated =
(745.16 – 392.18) g
=
352.98 g
54. (i) The effective radius of any nucleus is the distance from the centre of atom at which K.E. of α -particle is exactly equalled by the P.E. of interaction between positively charged nucleus and positively charged α -particle resulting into reversal of the direction of α -particle in Rutherford’s experiment. The P.E. of α -particle with charge 2e unit at a distance r from the nucleus of charge Ze is as given below.
P.E. =
2kze 2 r
The K.E. of α -particle (mass = m, and velocity = v) is as given below K.E. =
1 mv 2 2
At the point away from the nucleus by effective radius (r) we have
1 2kze 2 mv 2 = 2 r
or
r=
4kze 2 mv 2
e = 1.6 × 10–19 C, m = Mass of 2p + Mass of 2n ≈ 4 amu = 4 × 1.66 × 10–27 Kg Putting these data and those given in the question, we get
r=
4 × 9 × 10 9 × 79(1.6 × 10 −19 )2 4 × 1.66 × 10 −27 (2 × 107 )2
= 2.72 × 10 −14 m
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(ii) For Balmer series we have ν=
1 ⎛1 1 ⎞ = RH Z 2 ⎜ − 2 ⎟ λ ⎝4 n ⎠
For H – atom: Z = 1 and for longest wavelength transition in Balmer series n must be 3.
or
⎛ 1 1⎞ = RH ⎜ − ⎟ ⎝4 9⎠
1
Thus
6564 .7 × 10 −8
RH =
36 6564.7 × 10
−8
= 109677.5 Cm −1
When RH = Rydberg constant for hydrogen Addition of deuterium nucleus into hydrogen nucleus will produces He+ ion for which Z = 2 So
1 1 ⎛ 1 1⎞ = 2 2 RH ⎜ − ⎟ = 4 × λ 6564 .7 × 10 −8 ⎝4 9⎠
or
λ=
6564.7 × 10 −8 = 1641.1× 10 −8 Cm = 1641.1 Å 4
55. Let the solubility of AgCN be x mole/litre. On dissolution silver remains as Ag+ but CN– ions are converted mostly to HCN due to fixed acidity of the buffer. Ka =
or
[H+ ] [CN− ] [HCN]
[HCN] [CN− ]
=
[H+ ] 1× 10 −3 = = 1.6 × 10 6 − 10 Ka 6.2 × 10
...(i)
As each CN– ion hydrolyses to yield one HCN. x = [Ag+] = [CN–] + [HCN] [HCN]
from the ratio
from equation (i), it is clear that [CN –] < < [HCN]
[CN − ]
∴ x = [Ag +] = [HCN] Thus, from equation (i)
[CN− ] =
[HCN] 1.6 × 10 6
=
x 1.6 × 10 6
Since, KSP = [Ag+] [CN–] or 2.2 × 10–16 = x ×
x 1.6 × 10 6
∴ 2.2 × 10 –16 × 1.6 × 106 = x 2 ∴
x = 1.9 × 10 −5 M
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56. When the cell acts as electrolytic cell, Cu being anode and Zn being cathode, the concentration of Cu2+ will increase due to dissolution (Cu → Cu2+ + 2e) and Zn2+ concentration will decrease due to deposition (Zn2+ + 2e → Zn). Eq. of Zn deposited at cathode = Number of Faraday of electricity passed. =
number of coulomb 96500
=
0.48 × 10 × 60 × 60 96500
= 0.18 Eq. of Cu dissolved at anode = 0.18 Hence, mole of Zn2+ removed from the cathodic compartment =
0.18 = 0.09 and mole of Cu2+ gone to 2
anodic compartment = 0.09. ∴ Mole of Zn2+ initially present = M × V = 1 × 0.1 = 0.1 and mole of Cu2+ initially present = 1 × 0.1 = 0.1 ∴ Mole of Zn2+ present after electrolysis = 0.1 – 0.09 = 0.01 and mole of Cu2+ present after electrolysis = 0.1 + 0.09 = 0.19 ∴ [Zn2+] = 0.1M and [Cu 2+] = 1.9 M As the electrolytic cell is now allowed to act as a galvanic cell as represented below Zn | Zn2+ (0.1)| |Cu2+ (1.9)| Cu, Ecell = ECu2+,
– EZn2+,
Cu
Zn
0.0591 0.0591 ⎫ ⎧ ° ⎫ ⎧ log[Cu 2 + ]⎬ − ⎨E °Zn2 + , Zn + log( Zn 2 + )⎬ = ⎨E Cu2 + , Cu + 2 2 ⎭ ⎩ ⎭ ⎩ 0.0591 0.0591 ⎫ ⎧ ⎫ ⎧ log1.9⎬ − ⎨− 0.76 + log 0.1⎬ = ⎨+ 0.34 + 2 2 ⎭ ⎩ ⎭ ⎩
0.0591 × 0.27875 + 0.76 + 0.02955 = 1.137 Volt. 2
= +0.34 + 57. At STP
Before diffusion
D 2 = 1.12 lit. = 0.2 g⎫ ⎬ in I bulb H2 = 2.24 lit. = 0.2 g⎭
After diffusion
H2 = 0.1 g
(I)
(II)
or H2 diffuses from I = 0.1 g Now for diffusion of D2 and H2 rD2 rH2
Q
⎛ MH = ⎜ 2 ⎜ MD ⎝ 2
WD2 t D2
×
t H2 WH2
⎞ ⎟ ⎟ ⎠ ⎛ MD = ⎜ 2 ⎜ MH ⎝ 2
⎞ ⎟ ⎟ ⎠
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WD2
∴
t
×
Success Magnet (Solutions)
t ⎛4⎞ = ⎜ ⎟ 0 .1 ⎝2⎠
∴
WD2 = 0.14 g
∴
Wt. of gases II bulb = wt. of H2 + wt. of D2 = 0.10 g + 0.14 g = 0.24 g
Q
% D2 by wt. =
0.14 × 100 = 58.33% 0.24
% H2 in bulb II = 41.67% 58. We know, mole fraction in vapour phase YA =
x A PA0 x A PA0 + x BPB0
Where xA = mole fraction in liquid phase of x xB = mole fraction in liquid phase of y PA0PB0
= vap. pressure in pure state of liquid x and y
Given, YA = 0.4, PA0 = 0.4 atm, PB0 = 1.2 atm ∴
0.4 =
x A × 0.4 x A × 0.4 + (1 − x A ) 1.2
∴ xA = 0.667 xB = 0.333 Now P = PA + PB = xAPA0 + xBPB0 = 0.667 × 0.4 + 0.333 × 1.2 = 0.667 atms 59. Used meq. of HCl = Number of meq. of MgO and Mg3N2 = 60 – 12 = 48 2Mg + O2 → 2MgO (from air) 3Mg + N2 → Mg3N2 (from air) MgO + 2HCl → MgCl2 + H2O But Mg3N2 + 8HCl → 3MgCl2 + 2NH4Cl NH4Cl + NaOH → NaCl + H2O + NH3 Number of meq. of NH4Cl ≡ No. of meq. of NH3 = (10 – 6) = 4 No. of meq. of Mg3N2 =
1 4 [No. of millimole of NH4Cl] = =2 2 2
No. of millimole (or meq.) of HCl used by Mg3N2 = 2 × 8 = 16 Thus, number of meq. of HCl used by MgO = 48 – 16 = 32 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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∴ Number of meq. of MgO = 16 × 2 = 32 Number of meq. Mg burnt to MgO = 32 32 × 12
∴ Weight of Mg burnt to MgO = 1000 = 0.384 g Mg3N2
From equation, 3Mg + N2
2 millimoles of Mg3N2 = 6 millimoles of Mg Weight of Mg burnt to Mg3N2 =
6 × 24 = 0.144 g 1000
Total weight of Mg = 0.384 + 0.144 = 0.528 g % of Mg burnt to Mg3N2 =
0.144 × 100 = 27.27% 0.528
60. Sample of hard water contains 96 ppm SO42– and 40 ppm Ca2+ (CaSO4). Also it contains 183 ppm HCO3– and 60 ppm. Ca+2 [Ca(HCO3)2] To remove Ca(HCO3)2 from 103 kg or 106 g sample of hard water which contains 243 g Ca(HCO3)2 or 3/2 mole of Ca(HCO3)2, CaO required is 3/2 mole Ca(HCO3)2 + CaO → 2CaCO3 + H2O Thus, mole of CaO required = 3/2 or 1.5 Also Ca+2 ions left in solution are of CaSO4 i.e., 40 ppm Now 1 litre water contains Ca2+ after removal of Ca(HCO3)2 =
40 × 10 3 10 6
[Ca2+] =
or
= 40 × 10 −3 g
40 × 10 −3 = 10 −3 40
If there Ca2+ are exchanged with H+, then [H+] in solution = 2 × 10–3 pH = –log 2 × 10–3 = 2.6989 = 2.69 61. Using pOH = pKb + log [OH–] = 1.8 × 10–5
[Salt] [Base]
0.05 0.25
= 3.6 × 10–6 [Mg+2]
[Al+3]
=
=
6 ×10 −10 (3.6 × 10 − 6 )2 6 ×10 −32 (3.6 × 10 − 6 )3
= 46.29 M
= 1.28 × 10–15 M
62. (I) A → Products (II) B → Products
Q t1/2 for (I) at 310 K = 30 minute ∴ K(I) at 310 =
0.693 = 0.0231 min −1 30
....(i)
Q rate = K [ ] and both reactions are of Ist order Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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Also given, K I at 310 =2 K I at 310
....(ii)
Also given, K II at 310 =2 K I at 310
....(iii)
Also we have, E aII E aI
=
1 2
....(iv)
For I K I at 310 E aI ⎡ 310 − 300 ⎤ 2.303 log10 K at 300 = R ⎢ 310 × 300 ⎥ ⎣ ⎦ I
....(v)
For II K II at 310 E aII ⎡ 310 − 300 ⎤ 2.303 log10 K at 300 = R ⎢ 310 × 300 ⎥ ⎣ ⎦ II
....(vi)
Dividing Equation (v) by (vi)
∴
K I at 310 E aI K I at 300 = = 2 By eq. (iv ) K II at 310 E a II log10 K II at 300
or
log10
or
K I at 310 ⎡ K II at 310 ⎤ =⎢ ⎥ K I at 300 ⎣ K II at 300 ⎦
log10
....(vii)
⎡ K at 310 ⎤ K I at 310 = 2 log10 ⎢ II ⎥ K I at 300 ⎣ K II at 300 ⎦ 2
....(viii)
By equation (ii) and (viii) 2
or
⎡ K II at 310 ⎤ ⎢ ⎥ =2 ⎣ K II at 300 ⎦
or KII at 310 =
2 K II at 300
....(ix)
By equation (iii) and (ix) 2 × KI at 310 = or KII at 300 =
2 (K II at 300 )
....(x)
2 × K 1 at 310 2
By Equation (i) and (x) K =
2 × 0.0231
K = 3.27 × 10-2 min–1 K = 0.0327 min–1 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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63. (i) For A2CrO4 [CrO42–]
1.1 × 10 −12
=
(0.01)2
= 1.1 × 10–8 For BCrO4 2.2 × 10 −10 (0.01)
[CrO42–] =
= 2.2 × 10–8 (ii) Because, [CrO42–] in A2CrO4 is less in comparison to BCrO4 in saturated solution. So, A+ ion precipitates first. (iii) [A+] remaining
K sp A 2CrO4
=
[CrO24− ] 1.1 × 10 −12
=
2.2 × 10 −18
= 7.07 × 10–3 M (iv) The addition of CrO4–2 is not a practical method for separation of A+ and B+2 Na2S2O3 KI+HCl 64. Bleaching powder ⎯⎯ ⎯ ⎯→ I2 ⎯⎯ ⎯ ⎯→ I− + Na2S4O6
m.eq. of available Cl2 in 25 ml of bleaching powder solution = meq. of Na2S2O3 used 1 = 24.35 × 10 equivalents of available Cl2 in 500 ml of bleaching powder
solution
= 24.35 ×
1 × 500 × 1 = 48.7 10 25 1000 1000
ω = 48.7 35.8 1000 Cl2 = 1.729 gm.
% of available Cl2 in bleaching powder = 30%. H+ + H2PO4–
65. H3PO4
Let [H+] = [H2 PO4– ] = x, [H3PO4] = 0.01 – x 7.1 × 10–3 = H2PO4– [HPO4–2] = HPO42– [PO43–] =
x2 , x = 0.0056 = [H+] = [H2PO4–] 0.01– x H+ + HPO4–2 [6.3 × 10 −8 ] [0.0056 ] = 0.0056
6.3 × 10–8
H+ + PO43– [ 4.5 ×10 −13 ] [6.3 ×10 −8 ] = 5.1 × 10–18 [5.6 × 10 −3 ]
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Brain Storming Comprehensions
UNIT
5
Linked Comprehension Type C1. 1. Answer (3) Concentration of H2CO3 in rain water = 3.53 ×
20 10
6
= 7.06 × 10 −5 M
When NaOH is added the following reaction takes place H2CO3 + NaOH
NaHCO3 + H2O
initial conc.
7.06 × 10–5 M
after reaction conc.
(7.06 × 10–5 – 3 × 10–5) M 3 × 10–5 M = 4.06 × 10–5
Resulting solution will contain NaHCO3 and H2CO3 hence it will act as an acidic buffer pH = pKa + log
[salt] [acid]
= 6.35 + log
(3 × 10 −5 ) ( 4.06 × 10 −5 )
= 6.35 – 0.13 = 6.22 2. Answer (3) The equilibria that co-exist are Cd2+ + 2OH −
Cd(OH2)(s)
(x−y)
Cd2+ + S 2O 3 2− x−y
a−y−z
…(ii)
Cd(S 2O 3 ) y−z
[Cd(S 2O3 )2 ]2−
Cd(S 2O3 ) + S 2O 3 2− y−z
…(i)
2x
a−y−z
…(iii)
z
Let x be the moles of Cd(OH)2 that dissolves Ksp = (x – y)(2x)2 4.5 × 10–15 = (5 × 10–4 – y)(10–3)2 y = 4.99 × 10–4 ~ 5 × 10–4 [Cd2+] left to maintain the equilibrium is K sp [OH− ] 2
=
4.5 × 10 −15 (10 −3 ) 2
= 4.5 × 10 −9 M
Using (ii) and (iii) equation K1K2 =
[Cd(S 2O3 )2 ] 2− [Cd2+ ][S 2O 3 2− ] 2
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Let the conc. of thiosulphate at equilibrium be b mole/litre 8.3 × 103 × 2.5 × 102 =
z ( 4.5 × 10 −9 )b 2
As the equilibrium constant for reaction of complexation of Cd2+ with thiosulphate is high therefore it can be assumed that z ~ 5 × 10–4 b2 =
5 × 10 −4 2.075 × 10 6 × 4.5 × 10 −9
= 0.0535
b = 0.231 M Since value of y and z are very small (i.e., 5 × 10–4) therefore amount of Na2S2O3 needed to dissolve 5 × 10–4 mole of Cd(OH)2 is 0.231 moles. 3. Answer (1) pH = pKa + log
[salt ] 0.25 = 4.74 + log = 4.438 [acid] 0. 5
pOH = 9.562 [OH–] = 2.74 × 10–10 Ionic product of Mg(OH)2 = [Mg2+][OH–]2 = (0.1)(2.74 × 10–10)2 = 7.5 × 10–21 Ionic product of Al(OH)3 = (0.1)(2.74 × 10–10)3 = 2.05 × 10–30 Ionic product of Fe(OH)3 = 0.1(2.74 × 10–10)3 = 2.05 × 10–30 In case of only Mg(OH)2 ionic product < Ksp C2. 1. Answer (3) Electronic configuration of Be+ is 1s2, 2s1 Energy of last electron in Be+ is given by
Z 2eff . E = − 13.6 2 n –17.98 = − 13.6
Z 2eff . n2
∴ Zeff. = 2.3 Extent of shielding by inner electrons = 4 – 2.3 = 1.7 unit for Li, Zeff. = 3 – 1.7 = 1.3 1.3 × 1.3 eV = –5.746 eV 2× 2 Ionisation energy of lithium = 5.746 eV.
E = − 13.6 ×
2. Answer (3) |F| =
ke2 r4
du ke 2 = dr r4 =
mv 2 r
and mvr =
nh 2π
…(i)
…(ii) …(iii)
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Equating (ii) and (iii)
r=
ke 2 4π 2m 2 2
n h
= k1
m n2
1 − ke 2 − ke 2 − ke 2n 6 = = (potential energy) = 3 2 6k13m 3 6r 3 ⎛k m⎞ 6⎜⎜ 1 ⎟⎟ ⎝ n2 ⎠
Total energy =
Total energy ∝ n6 Total energy ∝ m–3. 3. Answer (3) Because of 6 different lines are observed in H-atom spectrum highest energy level is n = 4. Also the radiations responsible for photoelectric effect are due to transition form n = 4 to n = 1 and n = 3 to n = 1 1⎤ ⎡ ΔE1 = 13.6 ⎢1 − ⎥ = 12.75 eV 16 ⎦ ⎣ ⎡ 1⎤ ΔE2 = 13.6 ⎢1 − ⎥ = 12.1 eV ⎣ 9⎦
K1 = ΔE1 – φ, K2 = ΔE2 – φ 5=
12.75 − φ 12.1 − φ
φ = 11.94 eV C3. 1. Answer (2) We know that yA =
p oA x A p oA x A + pBo x B
yA − xA =
yA − xA =
yA − xA =
yA − xA =
p oA x A p oA x A + pBo x B
− xA
p oA x A − p oA x 2A − pBo x A x B p oA x A + pBo x B p oA x A − p oA x 2A − pBo x A (1 – x A ) p oA x A + pBo (1 – x A ) x A (p oA − pBo ) − x 2A (p oA − pBo ) x A (p oA − pBo ) + pBo
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For yA – xA to be maximum d(y A − x A ) =0 dx A
Solving we will get xA =
p oA pBo − pBo p oA − pBo
2. Answer (4)
p S = p oA x A + pBo x B
p S = p oA x A + pBo (1 − x A ) p S = pBo + x A (poA − pBo ) p oA pBo − pBo
= pBo +
(p oA − pBo )
(p oA − pB )
= pBo + p oA pBo − pBo = p oA pBo
3. Answer (1) We know that
p oA x A = y A p Total
...(1)
pBo x B = y Bp Total
...(2)
Dividing (2) by (1) ⇒
p oA x A pBo x B
=
yA yB
p oA ⎛ x A ⎜ pBo ⎜⎝ 1 − x A
⎞ yA ⎟⎟ = ⎠ 1− y A
1 − x A pBo 1 − y A = x A p oA yA ⎛ 1 ⎞ po 1 ⎜⎜ − 1⎟⎟ oB = −1 ⎝ xA ⎠ pA yA po 1 = 1 + oB yA pA
⎛ 1 ⎞ ⎜⎜ − 1⎟⎟ x ⎝ A ⎠
p o 1 ⎛⎜ pBo 1 = oB ⋅ + 1− y A p A x A ⎜⎝ p oA
⎞ ⎟ ⎟ ⎠
comparing with y = mx + c m=
pBo p oA
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C4. 1. Answer (4) There are 7 tens Hence, 27 = 128 2. Answer (3) Decrease in Ea increases the rate of reaction. 3. Answer (3) In Arrhenius equation, the factor e –Ea /RT represents the fraction of molecules having energy greater than activation energy. C5. 1. Answer (2) since θ = 120° −E a + log A 2.303RT y = mx + C log K =
Compare both equation tanθ = −
log K
Ea 2.303 R
tan 120° = −
θ 1/T
Ef 2.303 × 8.314
−E f 19 .147 Ef = 33.16 J − 3 =
∴ ΔH = Ef – Eb (for Endothermic reaction) 13.16 = 33.16 – Eb ∴ Eb = 20 J 2. Answer (3) For half life period we need rate constant K logK =
∴
–E a + log A 2.303 RT
1.732 × 1 + log 10 519.6 = 0.003 + 1 ≅ 1
log K =
Q log10 K = 1 ∴ K = 10 sec–1 ∴
t 1/2 =
0.693 0.693 ⇒ K 10
t1/2 = 6.93 sec 3. Answer (1) We know that for nth order r = K(a – x)n
log r
∴ logr = logK + nlog(a – x)
θ
y = mx + C ∴ m(slop) = n
log (a – x)
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Q reaction is first order ∴ n=1 ∴ slop(tanθ) = 1 and C = log K (Intercept) C=1 Q tanθ = 1 ∴ θ = 45° C6. 1. Answer (2) In propane, number of 1ºH atoms are 6 and number of 2º H atoms are 2. Product (x) is formed by ⎛x⎞ 3 replacement of 1ºH and product (y) is formed by replacement of 2ºH. So, the product ratio ⎜⎜ y ⎟⎟ is . ⎝ ⎠ 1
2. Answer (3) In case of Br, the abstraction of 3ºH is 1600 times in comparison to 1ºH. So, the % of x product is about 99%. 3. Answer (1)
CH3
CH3
CH3 – CH – CH3 + Cl2
hν
CH3
CH3 – CH – CH2 – Cl + CH3 – C – CH3
(I) = No. of 1ºH × reactivity of 1ºH
Possibility of (I)
Cl (II)
=9×1=9 Possibility of (II)
= No. of 3ºH × reactivity of 3ºH
=1×5=5 % of (I) =
9 × 100 = 64.3% 14
% of (II) =
5 × 100 = 35.7% 14
CH2
Hence, major product is CH 3 – CH – CH 2 – Cl C7. 1. Answer (2) Removal of H2O and formation of carbocation is the R.D.S. of pinacol-pinacolone rearrangement. 2. Answer (1)
CH3 CH3
CH3 CH3 CH3
H3C OH NH2
NaNO2 HCl
CH3
H3C OH
OH
O Now, the product undergoes pinacol-pinacolone rearrangement and finally gives H3C – C
CH3 C – CH3 . CH3
3. Answer (2) In both reactants, products are obtained by the shifting of phenyl group during pinacol-pinacolone rearrangement. So, only two products are obtained. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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C8. 1. Answer (1)
C6H5
O
O
C
C
–
H + C6H5
H
O
O
C
C OH
OH 2. Answer (4)
O
OH O
–
COO
3. Answer (3)
O
O
OH
C
C
C
–
COO
C9. 1. Answer (3) 2. Answer (1) 3. Answer (2)
O CH2 CH2
CH
CH2OH
CrO3/Δ
CH2
CH
CH
C
CH
CH2
CHO
(A) N HC
CH2 CN
CH2
CHO
AgNO3/NH4OH (B)
CH2
CH
COOH
(C) (D)
C10. 1. Answer (3) Due to presence of unsymmetrical e– repulsion by 3d32 and 3dx2y2. 2. Answer (4) Electronic configuration is symmetrical, so repulsion by e– is symmetrical. 3. Answer (1) In case of Fe+3, with strong or weak ligand, E.C. is symmetrical. C11. 1. Answer (2) This is according to M.O. diagram of CO which shows that HOMO (Highest Occupied Molecular Orbital) which acts as the electron pair donor to a metal is slightly antibonding between C and O. Thus donation of e– density from this orbital should increase the C – O strength. 2. Answer (3) Higher will be back bonding is N3 – CO 3. Answer (1) In CO, donor group reside closer to the C and acceptor antibonding lies closer to O and hence. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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C12. 1. Answer (4) In nitrogen family on moving down the group both thermal and electrical conductivity increases due to increase in delocalisation of electron from nitrogen to bismuth. 2. Answer (4) Greater electronegativity and higher oxidation state in responsible for greater acidic character. 3. Answer (4) Due to presence of empty d-orbital in Sb. It can expand covalency upto 6.
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