1. Answer (2, 4)
The substances having same composition of atoms and similar crystal structure are isomorphous.
2. Answer (2, 3)
%C in C2H5OH = 100 52% 1
16 5 24
24 × =
+ + +
%C in C6H12O6 = 100 40% 96
12 72
72 × =
+ +
%C in CH3COOH = 100 40%
1 32 12 3 12
24 × =
+ + + +
%C in C2H5NH2 = 100 53% 2
14 5 24
24 × =
+ + + 3. Answer (1, 4)
O H NaF NaOH
HF+ ⎯⎯→ + 2
moles of HF = 0.01 1000 1 100 .
0 × =
moles of NaOH = 0.01 1000 1 100 .
0 × =
number of moles of NaF formed = 0.01 M
05 . 0 1000
200 01 . ] 0 NaF
[ = =
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-KCrO3Cl = 1 + x + 3 × –2 + (–1) = 0 x = +6
Cr O O O
O O CrO5
Oxidation number of Cr = +6 5. Answer (2, 3)
Those substance can be oxidised and reduced, in which central element is neither in lowest nor in highest oxidation state.
For Cl, range of oxidation number is from –1 to 7.
In HCl, Cl is present in lowest oxidation state In HClO4, Cl is present in highest oxidation state.
6. Answer (1, 2, 3)
MnO2 + (NH4)2SO4→ MnSO4 + (NH4)2S2O8 n = 2 n = 1
equivalents of MnO2 = equivalents of (NH4)2SO4 1 × 2 = 1x
x = 2
7. Answer (1, 2, 3)
Resultant normality of solution N1V1 + N2V2 + N3V3 = N4V4 5 × 1 + 20 ×
2
1 + 30 × 3
1= N4(1000)
N4
100025 = 40N N4 = 1
Resultant [H+] = 40
1 = 0.025
Normality = mass/equivalentvol(litre) mass
40 x
401 × molecular mass of NaOH = 40 x = 1 gm
8. Answer (1, 2, 3)
SO2 + H2O2⎯→ H2SO4
m. equivalents of SO2 = m. equivalents of H2SO4 = m. equivalents of NaOH
= 20 × 0.1 = 2 n factor of SO2 = 1
22 =
Volume of SO2 at STP = 22400 × 10–3 = 22.4 ml.
Concentration of SO2 in air is 22.4 pm
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In the above balance equation is it clear that only two moles of NO3– undergo change in oxidation state while six moles remain in same oxidation state 2HNO3 + 6H+ + 6e → 2NO + 2H2O
Total 8 moles of HNO3 exchange 6 mole of electrons 1 mole of HNO3 exchanges
8 6 or
4
3 mole of electrons
∴ n factors of HNO3 =
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Since in the Y electron is in higher state therefore energy required to change (X) to (Y) Since in X element there is 19 electron which represent the K-atom.
15. Answer (1, 2, 3, 4)
Number of scattered α-article in Rutherford experiment is inversely proportional to square of kinetic energy of incident α-particles.
17. Answer (1, 2, 3, 4)
An acceptable solution of schrodinger wave equation must satisfy the following condition (i) It should be single valued
(ii) It should be continuous
(iii) The function should be normalised i.e.
∫
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-19. Answer (2, 3, 4)
There are three possible values of spin quantum number it means an orbital can accommodate 3 electrons.
So 1s – 1s3, first period would have 3 vertical columns.
20. Answer (1, 3, 4)
Kinetic energy of the ejected electron depends on the frequency of incident radiation not on the intensity.
Intense and weak beam are having more or less number of photons.
21. Answer (1, 2)
Since only six different wavelengths are emitted therefore highest excited state is n = 4 therefore (1) is correct.
In the emitted radiation two wavelength are shorter than λ0 it means that initially atoms were in excited state therefore (2) is also correct.
Transition corresponding 4 → 1, 3 → 1, 2 → 1 belongs to lyman series.
23. Answer (2, 3)
Energy of orbital of hydrogen depend upon n and not on l.
24. Answer (2, 3, 4)
Cu and Al, Si and Ge donot show inert pair effect.
26. Answer (1, 2, 3, 4)
(1) KF combines with HF and forms KHF2 with exists as K+ + [F --- H —— F]–.
(2) Due to smaller size of Li+ ions it has high polarising power therefore predominantly covalent in nature.
(3) CsBr3 exists as Cs+ + Br3–.
(4) Sodium sulphate is soluble in water but BaSO4 is sparingly soluble because hydration energy of BaSO4 is less than its lattice energy.
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-Rb+ – ionic radius 1.48 Å O2– – ionic radius 1.40 Å and
Li+ – ionic radius – 0.68 Å Mg2+ – ionic radius – 0.60 Å 28. Answer (2, 3)
Electron affinity of O(g) and S(g) are negative therefore involve emission of energy.
29. Answer (1, 2, 3)
Alkali metal have lowest I.E. energy because after releasing one electron these acquires noble gas configuration.
Electron affinity of nitrogen is less than oxygen because nitrogen has half filled p-orbital therefore it is more stable.
F– is weakest reducing agent among halide due to maximum stability due to highest hydration energy.
30. Answer (1, 2, 4)
Tl+ is more stable than Tl3+
Ga3+ is more stable than Ga+ Pb4+ is less stable than Pb2+
Bi3+ is more stable than Bi5+
These are due to inert pair effect.
31. Answer (2, 3, 4)
Electronegativity, ionisation energy and oxidizing power increases from iodine to fluorine.
32. Answer (1, 2, 3)
Low ionisation energy, high electron affinity and high lattice energy favours the ionic bond formation.
33. Answer (1, 3, 4)
Increasing metallic character increase the electron donating tendency of metal.
34. Answer (1, 4)
PCl5 in solid form exists as [PCl4+][PCl6–] and PBr5 exists as [PBr4+][Br–] in solid state.
35. Answer (1, 4) 2 5
2 8 2
XeOF2N + =
= attached atoms = 3; ∴ T-shaped
2 4 1 2 5 2
NH2−N= + + = attached atoms = 2; ∴ Angular 36. Answer (2, 3)
B – C C – B≡ F F F
F sp2 sp sp sp2 ∴ planar
N – SiH3 SiH3
SiH3
Lone pair of nitrogen is involve in pπ-dπ bonding there fore delocalised therefore nitrogen is sp2 hybridsed and planar.
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-37. Answer (2, 4)
–N = C = O – linear S = C = S – linear 38. Answer (1, 2, 4)
(a) On decreasing the electronegativity of central atom bond angle decreases (b) Bond angle of NH3 – 107°
Bond angle of H2O – 104.5°
Bond angle of F2O – 103°
39. Answer (1, 2, 3, 4)
Electron affinity of anion is positive and non spontaneous due to electron –2 repulsion.
40. Answer (1, 2, 3)
O CH3
O CH3 (μ ≠ 0)
N
N O O
O O (μ 0)=
Br
Cl (μ 0)~–
C
C CH3 H
H C H2 5 (μ ≠ 0) 41. Answer (2, 3, 4)
Statement 2 and 3 are facts
Unit of P = Unit of 2
2
V a n
∴ Unit of a = atm L2mol–2 42. Answer (3, 4)
Mutual attraction of molecules known as van der Waal intermolecular force.
43. Answer (3, 4)
RT Z=PV
Average RT
2 KE= 3
44. Answer (2, 3, 4)
Above critical temperature gas cannot be liquified.
45. Answer (1, 3, 4)
) V (
) V ( V Z V
id m ideal molar
=
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-m id
Vm = Vid Z = 1 Vm < Vid Z < 1
If force of attraction dominates then Z < 1 46. Answer (2, 3)
Mg2C3⎯⎯→ 2 Mg2+ + C34–
2 Mg2+ + [2–C C C2–] Two sigma and two pi bonds CaCN2 ⇒ Ca2+ + CN22–
[–N C N–] Two sigma and two pi bonds 47. Answer (1, 2, 3)
In solid state N2O5 exist as [NO2+][NO3−] therefore hybridisation of each nitrogen is sp and sp2 respectively.
In gaseous state N2O5 exists as N O N O
O
O
O
therefore hybridisation of each nitrogen is sp2.
N2O5 is called anhydride of nitric acid because in reaction with H2O, N2O5 forms nitric acid H2O + N2O5⎯⎯→ 2 HNO3
48. Answer (1, 2, 3, 4) KK (σ2s2) (σ*2s2) (π 2px
2 = π 2py 2)
Four electrons are present in 2π molecular orbitals that’s why double bond contains both π bonds.
49. Answer (1, 3, 4)
(1)
Cl P
Cl Br
Br Cl
dipole moment 0μ ≠
(2)
Cl P
Br Br
Br Cl
dipole moment = 0μ
(3)
O O
CH3 CH3
dipole moment is not zero due to following structure
(4) N
H
H O
H dipole moment is not zero
50. Answer (1, 3, 4)
Order of acidic strength H3PO2 > H3PO3 > H3PO4, hybridisation of phosphorus in all acids are not sp3. In H3PO3, H3PO2 there is P—H bond present, therefore these are reducing in nature.
51. Answer (2, 4)
Under the critical condition gases does not follow ideal behaviour for Z is not equal to 1 at absolute zero temperature the kinetic energy of gas molecules will be zero.
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-52. Answer (1, 2, 3, 4)
van der Waal’s constant a measure the intermolecular force of attraction b is called excluded volume Vc = 3b.
53. Answer (1, 3)
Kinetic energy for 1 mole gas = RT 2 3
1 mole of gas has molecules = Nav. 54. Answer (1, 2)
On the expansion the volume increases therefore pressure decreases as the temperature is constant therefore kinetic energy of gas molecule remain same.
55. Answer (1, 2)
Pc = 2 b 27
a
Vc = 3b.
56. Answer (1, 3, 4)
224 . 1 : 128 . 1 : 1
M RT : 3 M RT : 8 M RT 2
u : u
:
ump average rms
π
57. Answer (2, 4)
For spontaneous process ΔG < 0, ΔH < 0 and ΔE < 0 58. Answer (1, 4)
Absolute values of entropy and internal energy cannot be calculated.
59. Answer (2, 4)
Statement of IInd law of thermodynamics.
60. Answer (1, 2, 3, 4)
All are well known relations.
61. Answer (2, 3, 4)
If P, Q are arbitrarily chosen intensive variables then P/Q, PQ are intensive variables and dQ
dP is intensive property.
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-ΔE is a state function
∴ It is zero is cyclic process internal energy of ideal gas depends only on temperature
∴ In isothermal process ΔE = 0.
63. Answer (1, 2, 4)
In isothermal process T = constant P1V1 = P2V2 (Boyle’s law at constant T) ΔU = 0
ΔH1 = ΔH2 64. Answer (2, 4)
Standard heat of formation of all elements in their standard states is zero ΔHf(O) ≠ 0 and
ΔHf (diamond) ≠ 0 because these are not standard state.
65. Answer (1, 3, 4)
State function depends only on initial and final position therefore Enthalpy, Entropy, Gibbs free energy are state function.
66. Answer (2, 3)
During the streching of rubber band the long flexible macromolecules get uncoiled the uncoiled arrangement has more specific geometry and more order thus entropy decreases.
67. Answer (1, 3, 4)
Work done in reversible process is more than work done in irreversible process at equilibrium ΔG is zero.
68. Answer (3, 4)
BOH + HCl → BCl + H2O
x x
4
3 0 0
4x
x−3 0 ⎟
⎠
⎜ ⎞
⎝
⎛ 4
x
3 ⎟
⎠
⎜ ⎞
⎝
⎛ 4
x 3
pOH =
x 4
4 x log3 ] 5
BoH [
] Salt log[ pKb
× + ×
= +
pH = 14 – 5 – log 3 = 8.523.
68.(a). Answer (3) (IIT-JEE 2008)
At equivalence point
(acid)2 2 (base)1V1 N V
N =
2.5 × 5 2 =
15 2 × V
V = 7.5 ml
∴ Milli equivalents of salt = 1
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-pH = 7 – 2 1 pkb –
2 1 log C
= 7 – 6 – 2 1 log
10 1
= 7 – 6 + 0.5 = 1.5 (H+) = 10–1.5 = 3.2 × 10–2 M 69. Answer (1, 3)
At equilibrium ΔG = 0 ΔG° = – 2.303 RT Log K – nF E°cell = – 2.303 RT log K At 25°C
K n log 0591 .
Ecell° = 0 .
70. Answer (1, 2, 3, 4)
Pressure favours the forward reaction. The temperature at which atmospheric pressure is equal to vapour pressure is called boiling point if pressure boiling point will be increased .
71. Answer (1, 3)
On increasing the ammonia the partial pressure of NH3 increases where as increasing the temperature favours the dissociation of NH4HS therefore more NH3 will be formed.
72. Answer (1, 2, 4) NaCN Na+ + CN–
CN– + H2O HCN + OH– (basic) CH3COONa CH3COO– + Na+
CH3COO– + H2O CH3COOH + OH– (basic) Na2CO3 2Na+ + CO32–
CO32– + 2H2O H2CO3 + 2OH– (basic) 73. Answer (1, 2)
N2(g) + 3H2(g) 2NH3(g) ΔH = negative
Since the no. of moles of gases decreases in product sides therefore on increasing the pressure forward reaction favours. Catalyst increases the rate of reaction, therefore NH3 formation will be fast.
74. Answer (1, 2, 3)
Electron deficient species are called lewis acid therefore BF3, Ag+ are electron deficient
SnCl4 can expand its octet due to vacant d-orbital therefore behaves as a lewis acid.
75. Answer (2, 3)
The aqueous solution of NH4Cl and CuSO4 are acidic.
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-CH3COOH is weak acid its concentration of H+ ions is less than 10–6 M therefore pH > 6 CH3COOH + NaOH → CH3COONa + H2O
initial m. mol 2 6 0 0
after reaction 0 4 2 2
Since solution is basic therefore pH > 7.
The aqueous solution of CH3COONH4 is generally neutral
∴ pH > 6.
77. Answer (3, 4)
NO3– is a conjugate base of HNO3 which is strong acid HSO4– is a conjugate base of H2SO4 which is strong acid 78. Answer (1, 2, 4)
Due to common ion effect solubility of AgCl will be less than water in NaCl, AgNO3 and CaCl2 solution.
79. Answer (1, 2) Hln H+ + In–
Ka = ] HIn [
] In ][
H [ + −
[H+] =
] base [
] acid [ Ka
for 75% red [H+] = 5 3 10 5 25
10− 75⎥⎦⎤= × −
⎢⎣⎡ pH = 4.52
for 75% blue [H+] = 5 10 5 3 1 75
10− 25⎟= × −
⎠
⎜ ⎞
⎝
⎛ = 5.47
80. Answer (2, 3)
On increasing the temperature ionic product of water increases so pH and pOH decreases but water will remain neutral.
81. Answer (1, 2, 4)
⎥⎦
⎢ ⎤
⎣
⎡ −
= Δ
2 1 1
2
T 1 T
1 R 303 . 2
H K
logK
Since in the option (3) ΔH = 0 because (Eaf = Eab)
∴ Therefore only this reaction is independent of temperature and (K2 = K1) on the other hand there is not ΔH = 0 therefore equilibrium constant depends on temperature.
82. Answer (2, 3, 4)
N2(g) + 3H2(g) 2NH3(g)
initial moles 3 9 0
at equilibrium 3 – x 9 – 3x 2x
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-t = -t1 the reaction attains the equilibrium therefore the amount of NH3 remins constant after t1 time 2x = 2
because initial molar ratio is 1 : 3.
⎟⎠
Upto phenolphthalein NaOH is fully neutralised and Na2CO3 will be converted to NaHCO3. In next step NaHCO3 coming from Na2CO3 neutralised by HCl using methyl orange indicator. So y ml should be less than x ml that required for phenolphthalein end point.
84. Answer (2, 3)
Due to smaller size of Li+ is more solvated than Na+ ion therefore conductivity is less than Na+ ion.
85. Answer (1, 3)
on increasing the concentration of [Cu2+] and decreasing the concentration of [Cd2+] Ecell will be more positive.
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Molar conductance of an electrolyte depends upon its degree of dissociation with increase in dilution the molar conductance increases due to increase in dissociation specific conductance decreases upon dilution because number of current carrying ions per unit volume of solution decreases.
89. Answer (1, 2)
The value of the constant A for a given solvent and temperature depends on the type of electrolyte i.e., charge on cation and anion produced on the dissociation of the electrolyte in the solution.
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-92. Answer (1, 2, 3)
] A nF log[
RT 303 . E 2
EA/An+ = °A/An+ − n+ from this equation it is clear that
+
An
/
EA decreases with increasing [An+]
] A [ log 1 nF
RT 303 . E 2
EAn+/A = °An+/A− n+
93. Answer (1, 2, 3)
If a given metal ion has negative reduction potential H+ will be reduced by metal. Similarly if reduction potential is positive metal will be reduced by H2. If metal ion with negative potential is coupled H-electrode the hydrogen half cell should function as cathode thus metal electrode will be negative half cell (anode). In aqueous solution containing Zn2+, Na+ and Mg2+ the H+ will get preferentially reduced while in aqueous solution of Cu2+, Ag+, Au3+ these ions will be discharged ahead of H+.
94. Answer (1, 4)
As cell proceeds Ecell tend to zero to attain equilibrium state reaction quotient also increases to reach the state of equilibrium.
95. Answer (1, 2, 3)
Mole of Fe3+ = 0.1 × 1 = 0.1
Mole of electron = 0.149mole 96500
4 3600× =
Fe3+ + e → Fe2+
0.100 mol electron required to reduce all the Fe3+ to Fe2+ 0.049 mol electron to reduce the Fe2+ to Fe Fe2+ + 2e → Fe
Mole of iron formed = 0.049 0.025moleFe 2
1× =
96. Answer (1, 2)
H – O – S – O – O – H O
O
(peroxy linkage)
H – O – S – O – O – S – OH O
O
(peroxy linkage) O
O
97. Answer (2, 3)
2 4 0 4
n 2 2
SO Cu S
Cu
+
=
−
+ ⎯⎯→ + equivalent mass =
4 M
2 4 2
n 2 0
O H O
−
= ⎯⎯→ equivalent mass =
4 M
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-Factual type 99. Answer (3, 4)
Original Formula
Number of A atoms =
2 4 1 8 1× =
Number of B atoms = 2 1 2 1× =
Number of C atoms = 12 1 4
1× + = 4
Hence, the formula of compound is A1/2 BC4 or AB2 C8. On placing body diagonal plane, 2 corner atoms, 2 edge atoms, 1 body atoms are removed but 2 face atoms may or may not be removed.