1. Answer A(q), B(r), C(s), D(p)
total equivalent mass =
5
total equivalent mass = ⎟
⎠
total equivalent mass = ⎟
⎠
Factual type.
3. Answer A(s), B(p, s), C(q), D(r)
(A) [Co(NH3)4Cl2] exhibits geometrical isomerism
CO
cis-isomer trans isomer
AakashIIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 (B) Cis-[Co(en)2Cl2] exhibits optical isomerism
CO
(D) co-ordinationisomers ]
(See saw shaped) TeCl4
Cl
O–
Trigonal pyramidal with bond angle less than 109.5°
ClO3–
O O
I
Cl
Cl Angular with bond angle less than 109.5°
C H H
Trigonal planar O
ICl2+ H CO2
5. Answer A(s), B(r), C(q), D(p)
(A) Weak acid with strong base, solution is basic therefore pH > 7 pH = 21
[
pKw+pKa+logC]
(B) Strong acid and weak base, solution is acid therefore pH < 7 pH = 21
[
pKw−pKb−logC]
(C) Salt of weak acid with weak base pH =
[
pKw pKa pKb]
2
1 + −
(D) Salt of strong acid with strong base solution is neutral therefore pH = pKw
2 1
AakashIIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 6. Answer A(r), B(s), C(q), D(p)
(A) Na2S + (CH3COO)2Pb → PbS 2CH3COONa ppt
black ↓+
(B) Phenol gives colour with neutral FeCl3 solution (C) 3NaCNS + FeCl3→ Fe(CNS) 3NaCl
colour red blood
3 +
(D) NaI + AgNO3→
) solution OH NH in le lub inso
( 3
ppt
yellow 4
NaNO AgI↓ +
7. Answer A(r), B(q), C(s), D(p) (A) ΔG = – nFE
(B) ΔH° = RT2 T P
K log ⎥⎦⎤
⎢⎣⎡
∂
∂
(C) ΔS = – T P
G⎥⎦⎤
⎢⎣⎡
∂Δ
∂
(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
(B) + CH – Cl3 anhy AlCl3
CH3
(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) CH = CD–COOH + HBr2 CH – CD – COOH2
Br H
(optically active)
Reaction proceeds through carbocation intermediate.
(B) Addition of Br2 is trans addition and Cis-form gives racemic mixture.
(C)
(optically active)
C C H C2
H
C H2 5 H
+ Baeyer's Reagent Syn
addition C C H C2
H
C H2 5 OH OH H
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T
Cl 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
⇒ +1 + X – 4 = 0
+3 +5
–3
oxidation (reducing agent)
reduction (oxidising agent) x = +3
Both
(B) Cr
O O O
O O
x + 4(–1) –2 = 0
+6
0
reduction (oxidising agent)
x = +6 Cr (C) H2SO5
H – O – S – O – O – H O
O
x + 2(–2) + 2(–1) = 0
+6
–2
reduction (oxidising agent)
S x = +6
(D) N2O5
2x + 5(–2) = 0
+5
–3
reduction (oxidising agent) x = +5
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
AakashIIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 H – O – S – O – O – H
O
O
(peroxy linkage)
Marshall’s acid H2S2O8
HO – S – O – O – S – OH O
O
O
O
(peroxy linkage)
oxidation number of S = +6 (D) K2Cr2O7
– –
O – Cr – O – Cr – O O
O
O
O
(no peroxy linkage)
oxidation number of chromium = +6 K2CrO4, 2K+ + CrO42–
Cr – O– O
O O–
(no peroxy linkage)
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
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(diamagnetic)
(B) CO →
isoelectronic with N2 diamagnetic B.O. =
paramagnetic 17. Answer A(p, r), B(r), C(q, r), D(r, s)
AakashIIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 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 = 1.66 35 = For diatomic gas, r = 1.4
57 = For polyatomic gas, r = 1.33
34 = 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)
(A)
Fe C O2 4
2+ +3 +3 +4
–1e–
–2 × 1e–
2–
Fe + CO2
n-factor for FeC2O4 is 3.
AakashIIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 (B) Br2→ Br– + BrO3–
r2
B° → Br–, Equivalent weight = 2 M
+ −
° → 3
5
2 BrO
Br , Equivalent weight = 10
M
Total equivalent weight =
5 M 3 10
M 6 10
M 2
M+ = =
So, n-factor for Br2 is 3 5.
(C)
Fe S2+ –12 +3 +4
–1e– –2 × 5e–
Fe + SO2
n-factor for FeS2 is 11.
(D) Cu S2
+1 –2 +2 +4
–2 × 1e– –6e–
CuO + SO2
n-factor for Cu2S is 8.
25. Answer A(q), B(s), C(p), D(r)
(A) −
−→+ 4
7
O Cl Cl
Net change in oxidation number = 8
(B) 5
3 Cr6O Cr + → +
Net change in oxidation number = 3 (C)
0 2 1 –
2
2O O
H →
Net change in oxidation number = 2
(D) −
+ +
+ → 24
2 6 2 6
O Cr O
Cr
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) CH – CH == CHPh3
H+ CH – CH – CH – Ph3 2
Carbocation stabilize by resonance
Br CH – CH – CH – Ph3 2 Br
∴ Markownikov product
AakashIIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 (B) CH – C CH3 ≡
–H+ Hg+2
CH – C == CH3 2 H O2
–H+
CH – C – CH3 3
CH – C == CH3 2 O
Tautomerism
OH
Electrophilic addition
(C) Due to –I effect of R3N+ it gives anti-Markownikov addition product (D)
addition llic electrophi
2 3 HBr
2
2 CH CH CHBr
CH = ⎯⎯ →⎯ −
29. Answer A(p, q), B(r, s), C(r, s), D(p, q) (A) p → Oxidative ozonolysis
q → Oxidation with KMnO4
(B) and (C) C == C C — C
OH OH by OsO and cold dil KMnO4 4 (D) p → Oxidative ozonolysis
q → Oxidation with KMnO4/H+ 30. Answer A(p, s), B(p, r), C(p, r), D(q, r)
(A) – C – C – C – ONa – C – C – C – O + Na+ O
anode cathode
– C – C – C – O O
– C – C – C – O• O
– C – C + CO• 2
free radical
C == C + • – C – C – C – C – O
∴ p, s (B) Wurtz
Na
•
R – X R + NaX• free radical
R + R R – R
Alkane
•
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– C – C + Zn
– C – C – C – C – Br
– C – C•
free radical
Alkane
(D) Friedel craft
⎯→
⎯ +
−Cl AlCl3
R R AlCl+ 4
H
+ R+ AlCl4
R
+ AlCl + HCl3
Carbocation
31. Answer A(p, s), B(q, r, s), C(q, s), D(q, s)
(A) ≡
Benzyne
sp2 hybridization nucleophile (electron rich)
(B) Carbocation → – C – C sp2 hybridization electron deficient (electrophile) rearrangement is due to stability
(C) H – C – C H H
H H
free radical
electron deficient (electrophile), sp2 hybridization
(D) :CH2 → electron deficient (electrophile) sp2 hybridization 32. Answer A(q), B(p), C(s), D(r)
(A) CH – C – CH Cl3 2 H
H
alc
KOH CH – C == CH3 2 H
(B) CH3COONa⎯⎯electrolys⎯⎯⎯is→CH3−CH3+CO2+H2+NaOH
(C) C ==C O3
C – C O O
O
H O/Zn2
C == O + C == O + H O 2 2
(D) H2C=CH2+Cl2⎯⎯→Cl– C – C – Cl H H
H H
AakashIIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 33. Answer A(p, q), B(p, q, r),C(p, q, r, s), D(p)
(A)
CH OH CHOH CH OH
2
2
HIO4 2HCHO + HCOOH
(B) CHO (CHOH)
CH OH
4
2
HIO4 5 HCOOH + HCHO
(C)
CH OH CO (CHOH)
CH OH
2
4
2
HIO4 CHO
COOH
+ HCHO + HCOOH
(D)
CH OH CH OH
2
2
HIO4 2HCHO
34. Answer A(s), B(r, s),C(q, s), D(p) (A) 1° and 2° oxidised by CrO3/H+
(B) CH – C – R3 OH
H
gives iodoform test
(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.
AakashIIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 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+, d2sp3 hybridisation is present and it is paramagnetic.
(D) In [Fe(CN)6]4–,d2sp3 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 +
odourless) and(Colourless
CO2 ↑
(B) Na2SO3 + H2SO4→ Na2SO4 + H2O +
odour) pungent and(Colourless
SO2↑
(C) 2NaCl + H2SO4→ Na2SO4 +
odour) pungent and
s (Colourles2HCl↑
(D) 2NaNO3 + H2SO4→ Na2SO4 +
odour) pungent and
gas brown
(Light2NO2↑ + H2O + O2↑ 2 1
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