IITJEE-2011-Paper 2-CPM-11
PAPER-2 [Code - 8]
IITJEE 2012
PART II: CHEMISTRY
SECTION I: Single Correct Answer Type
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

 The major product H of the given reaction sequence is  CH₃-CH₂-CO-CH₃ ──(⊝CN)──→ G ──(95% H₂SO₄, Heat)──→ H  (A)
CH₃-CH=C-COOH
|
CH₃  (B)
CH₃-CH=C-CN
|
CH₃  (C)
OH
|
CH₃-CH₂-C-COOH
|
CH₃  (D)
CH₃-CH=C-CO-NH₂
|
CH₃  Sol. (A)                OH                             OH
              |                              |
H₃C-CH₂-C-CH₃ ──→ H₃C-CH₂-C-CH₃ ──(95% H₂SO₄)──→ H₃C-CH₂-C-CH₃ ──(Heat, -H₂O)──→ H₃C-CH=C-COOH
||                |                              |                          |
O                 CN                             COOH                       CH₃
(G)                                                        (H)  NiCl₂{P(C₂H₅)₂(C₆H₅)}₂ exhibits temperature dependent magnetic behaviour (paramagnetic/diamagnetic). The coordination geometries of Ni²⁺ in the paramagnetic and diamagnetic states are respectively
(A) tetrahedral and tetrahedral
(B) square planar and square planar
(C) tetrahedral and square planar
(D) square planar and tetrahedral  Sol. (C)
In both complexes Ni exists as Ni²⁺.
In sp³ (tetrahedral)
Ni²⁺:
3d: [⥮][⥮][⥮][↿ ][↿ ]   4s: [×]   4p: [×][×][×]
× represents electron pair donated by the ligands
sp³ & paramagnetic  In dsp² (square planar geometry)
Ni²⁺:
3d: [⥮][⥮][⥮][⥮][×]   4s: [×]   4p: [×][×][  ]
dsp² & hence diamagnetic  In the cyanide extraction process of silver from argentite ore, the oxidising and reducing agents used are
(A) O₂ and CO respectively
(B) O₂ and Zn dust respectively
(C) HNO₃ and Zn dust respectively
(D) HNO₃ and CO respectively  Sol. (B)
The reactions involved in cyanide extraction process are:
Ag₂S + 4NaCN ⇌ 2Na[Ag(CN)₂] + Na₂S
(Argentite ore)  IITJEE-2012-Paper 2-PCM-12  4Na₂S + 5O₂ + 2H₂O ──→ 2Na₂SO₄ + 4NaOH + 2S
(oxidising agent)  2Na[Ag(CN)₂] + Zn (reducing agent) ──→ Na₂[Zn(CN)₄] + 2Ag↓  The reaction of white phosphorous with aqueous NaOH gives phosphine along with another phosphorous containing compound. The reaction type; the oxidation states of phosphorus in phosphine and the other product are respectively
(A) redox reaction; -3 and -5
(B) redox reaction; +3 and +5
(C) disproportionation reaction; -3 and +5
(D) disproportionation reaction; -3 and +3  Sol. (C)
The balanced disproportionation reaction involving white phosphorus with aq. NaOH is
Oxidation (0 to +1)
┌────────────────────────┐
▼                        │
P₄⁰ + 3NaOH + 3H₂O ──→ PH₃ + 3NaH₂PO₂
▲                       │
└───────────────────────┘
Reduction (0 to -3)  However, as the option involving +1 oxidation state is completely missing, one might consider that NaH₂PO₂ formed has undergone thermal decomposition as shown below:
2NaH₂PO₂ ──→ Na₂HPO₄ + PH₃
Although heating is nowhere mentioned in the question, the "other product" as per available options seems to be Na₂HPO₄ (oxidation state = +5).  The shape of XeO₂F₂ molecule is
(A) trigonal bipyramidal
(B) square planar
(C) tetrahedral
(D) see-saw  Sol. (D)
Hybridization = sp³d
F
|
: ── Xe = O
/    O    F
Shape = see-saw  For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0.76 K kg mol⁻¹)
(A) 724
(B) 740
(C) 736
(D) 718  Sol. (A)
B → Solute; A → Solvent
WB = 2.5 g, WA = 100 g
ΔTb = 2°
(p° - ps) / p° = XB = nB / (nB + nA)
(p° - ps) / p° = nB / nA  ∵ nB ≪ nA  IITJEE-2011-Paper 2-CPM-13  (p° - ps) / p° = nB / nA
(760 - Psoln) / 760 = (2.5 / M) / ((100 / 18) × (1000 / 1000)) = (m × 18) / 1000   ...(i)
and from boiling point elevation,
2 = 0.76 × m
m = 2 / 0.76   ...(ii)
on equating (i) and (ii)
Psoln = 724 mm  The compound that undergoes decarboxylation most readily under mild condition is  (A)COOH//|  |── CH₂COOH
/  (B)COOH//|  |═ O
/  (C)COOH//|  |── COOH
/  (D)CH₂COOH//|  |═ O
/  Sol. (B)
β-keto acids undergoes decarboxylation easily.
COOH
/
α/  |  |═ O (β)
/
(β-keto acid)  *28. Using the data provided, calculate the multiple bond energy (kJ mol⁻¹) of a C≡C bond in C₂H₂. That energy is (take the bond energy of a C-H bond as 350 kJ mol⁻¹)
2C(s) + H₂(g) ──→ C₂H₂(g)       ΔH = 225 kJ mol⁻¹
2C(s) ──→ 2C(g)                 ΔH = 1410 kJ mol⁻¹
H₂(g) ──→ 2H(g)                 ΔH = 330 kJ mol⁻¹
(A) 1165
(B) 837
(C) 865
(D) 815  Sol. (D)
(i) 2C(s) + H₂(g) ──→ H-C≡C-H(g)   ΔH = 225 kJ mol⁻¹
(ii) 2C(s) ──→ 2C(g)               ΔH = 1410 kJ mol⁻¹
(iii) H₂(g) ──→ 2H(g)              ΔH = 330 kJ mol⁻¹
From equation (i):
225 = [2 × ΔHC(s)→C(g) + 1 × BE_H-H] - [2 × BE_C-H + 1 × BE_C≡C]
225 = [1410 + 1 × 330] - [2 × 350 + 1 × BE_C≡C]
225 = [1410 + 330] - [700 + BE_C≡C]
225 = 1740 - 700 - BE_C≡C  IITJEE-2012-Paper 2-PCM-14  225 = 1040 - BE_C≡C
BE_C≡C = 1040 - 225 = 815 kJ mol⁻¹  SECTION II: Paragraph Type
This section contains 6 multiple choice questions relating to three paragraphs with two questions on each paragraph. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.  Paragraph for Questions 29 and 30
In the following reaction sequence, the compound J is an intermediate.
I ──((CH₃CO)₂O / CH₃COONa)──→ J ──((i) H₂, Pd/C; (ii) SOCl₂; (iii) anhyd. AlCl₃)──→ K
J (C₉H₈O₂) gives effervescence on treatment with NaHCO₃ and a positive Baeyer's test.  The compound I is  (A)CHO|/|   |
\ /
|
OCH₃  (B)CHO|/|   |
\ /
|
OH  (C)CHO|/|   |
\ /
|
H  (D)CHO|/|   |── CH₃
\ /
|
H  Ans. (C)  Sol. 29-30
(CH₃CO)₂O
Ph-CHO ────────────────────────→ Ph-CH=CH-COOH
(I)           CH₃COONa                (J)
(Perkin condensation)   (Cinnamic Acid)
│
├─(cold, alk. KMnO₄ / Baeyer's reagent)─→ Positive Test
│
└─(NaHCO₃)─→ Effervescence  The compound K is  (A)/|   |───┐
\ /    |
|     |═ O
└───--┘  (B)/|   |───┐
\ /    |═ O
|     |
└───--┘  (C)/|   |───┐
\ /    |
|     |
└───--┘  (D)/|   |───┐
\ /    |═ O
|     |═ O
└───--┘  Ans. (A)  IITJEE-2011-Paper 2-CPM-15   CH=CH-COOH              CH₂-CH₂-COOH            CH₂-CH₂-COCl
  /                       /                       /
/ \         H₂/Pd-C     / \         SOCl₂       / \
|   |     ────────────→ |   |     ────────────→ |   |\ /                     \ /                     \ /(J)││ anhyd. AlCl₃│ (Friedel-Crafts Acylation)▼/|   |───┐
\ /    │
│     │═ O
└─────┘
(K)  Paragraph for Questions 31 and 32
The electrochemical cell shown below is a concentration cell.
M | M²⁺ (saturated solution of a sparingly soluble salt, MX₂) || M²⁺ (0.001 mol dm⁻³) | M
The emf of the cell depends on the difference in concentrations of M²⁺ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.  The value of ΔG (kJ mol⁻¹) for the given cell is (take 1F = 96500 C mol⁻¹)
(A) -5.7
(B) 5.7
(C) 11.4
(D) -11.4  Sol. (D)
At anode: M(s) + 2X⁻(aq) ──→ MX₂(aq) + 2e⁻
At cathode: M²⁺(aq) + 2e⁻ ──→ M(s)
n-factor of the cell reaction is 2.
ΔG = -nFE = -2 × 96500 × 0.059 = -11387 J/mole = -11.387 kJ/mole ≈ -11.4 kJ/mole  The solubility product (Ksp; mol³ dm⁻⁹) of MX₂ at 298 K based on the information available for the given concentration cell is (take 2.303 × R × 298 / F = 0.059 V)
(A) 1 × 10⁻¹⁵
(B) 4 × 10⁻¹⁵
(C) 1 × 10⁻¹²
(D) 4 × 10⁻¹²  Sol. (B)
M | M²⁺(sat.) || M²⁺(0.001 M) | M     (Ksp = ?)
emf of concentration cell,
Ecell = (-0.059 / n) log([M²⁺]a / [M²⁺]c)
0.059 = (0.059 / 2) log(0.001 / [M²⁺]a)
[M²⁺]a = 10⁻⁵ = S (solubility of salt in saturated solution)
MX₂ ⇌ M²⁺ + 2X⁻(aq)
(S)   (S)   (2S)
Ksp = 4S³ = 4 × (10⁻⁵)³ = 4 × 10⁻¹⁵  Paragraph for Questions 33 and 34
Bleaching powder and bleach solution are produced on a large scale and used in several household products. The effectiveness of bleach solution is often measured by iodometry.  IITJEE-2012-Paper 2-PCM-16  *33. Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of that oxoacid is
(A) Cl₂O
(B) Cl₂O₇
(C) ClO₂
(D) Cl₂O₆  Sol. (A)
Ca(OCl)Cl ──→ Ca²⁺ + OCl⁻ + Cl⁻
(Bleaching Powder)
HOCl ──→ H⁺ + OCl⁻
(oxo acid)
2HOCl ──(-H₂O)──→ H₂O + Cl₂O
Anhydride of oxoacid (HOCl) is Cl₂O.  *34. 25 mL of household solution was mixed with 30 mL of 0.50 M KI and 10 mL of 4 N acetic acid. In the titration of the liberated iodine, 48 mL of 0.25 N Na₂S₂O₃ was used to reach the end point. The molarity of the household bleach solution is
(A) 0.48 M
(B) 0.96 M
(C) 0.24 M
(D) 0.024 M  Sol. (C)
CaOCl₂(aq) + 2KI ──→ I₂ + Ca(OH)₂ + KCl
25 mL        30 mL
(M)molar     0.5(M)  I₂ + 2Na₂S₂O₃ ──→ Na₂S₄O₆ + 2NaI
48 mL
0.25(N) = 0.25 M  So, number of millimoles of I₂ produced = 48 × (0.25 / 2) = 24 × 0.25 = 6
In reaction;
Number of millimoles of bleaching powder (n_CaOCl₂) = n_I₂ produced = (1/2) × n_Na₂S₂O₃ used = 6
So, (M) = n_CaOCl₂(millimoles) / V(in mL) = 6 millimoles / 25 mL = 0.24  SECTION III: Multiple Correct Answer(s) Type
The section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.  *35. The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statement(s) is (are) correct?  P ▲│ (P₁, V₁, T₁)│    ●││      \   isothermal│       \───────────● (P₂, V₂, T₂)││         \ adiabatic
│          \────────● (P₃, V₂, T₃)
│
└────────────────────────► V  (A) T₁ = T₂
(B) T₃ > T₁
(C) wisothermal > wadiabatic
(D) ΔUisothermal > ΔUadiabatic  IITJEE-2011-Paper 2-CPM-17  Sol. (A, C, D)
T₁ = T₂ because process is isothermal.
Work done in adiabatic process is less than in isothermal process because area covered by isothermal curve is more than the area covered by the adiabatic curve.
In adiabatic process expansion occurs by using internal energy hence it decreases while in isothermal process temperature remains constant that's why no change in internal energy.  For the given aqueous reactions, which of the statement(s) is (are) true?  excess KI + K₃[Fe(CN)₆] ──(dilute H₂SO₄)──→ brownish-yellow solution
│
│ ZnSO₄
▼
white precipitate + brownish-yellow filtrate
│
│ Na₂S₂O₃
▼
colourless solution  (A) The first reaction is a redox reaction.
(B) White precipitate is Zn₃[Fe(CN)₆]₂.
(C) Addition of filtrate to starch solution gives blue colour.
(D) White precipitate is soluble in NaOH solution.  Sol. (A, C, D)
K₃[Fe⁺³(CN)₆] + KI(excess) ──→ K₄[Fe⁺²(CN)₆] + KI₃ (redox reaction, Brownish yellow solution)
I₃⁻ (Brownish yellow filtrate) + 2Na₂S₂O₃ ──→ Na₂S₄O₆ + 2NaI + I⁻ (Clear solution)
K₄[Fe(CN)₆] + ZnSO₄ ──→ K₂Zn₃[Fe(CN)₆]₂ (White ppt.) ──(NaOH)──→ Na₂[Zn(OH)₄] (Soluble)  With reference to the scheme given, which of the given statement(s) about T, U, V and W is (are) correct?    O
  ║
/   \
|     O|     |H₃C─/(T)
│
│ LiAlH₄
▼
(U)
┌────┴────────────────┐
│ CrO₃/H⊕             │ excess (CH₃CO)₂O
▼                     ▼
(V)                   (W)  (A) T is soluble in hot aqueous NaOH
(B) U is optically active
(C) Molecular formula of W is C₁₀H₁₈O₄
(D) V gives effervescence on treatment with aqueous NaHCO₃  Sol. (A, C, D)  IITJEE-2012-Paper 2-PCM-18        O
      ║
    /   \
   |     O
   |     |
H₃C─/(T)││ LiAlH₄▼/── OH|    OH|   /H₃C─/───┘(U) (no chiral centre)││ CrO₃/H⊕              \ excess (CH₃CO)₂O│▼                        ▼
/── OH                      /── OCOCH₃
|    ║                      |    OCOCH₃
O    O                      |   /
|   /                  H₃C─/───┘
H₃C─/───┘                          (W) (C₁₀H₁₈O₄)
(V) (Effervescence with NaHCO₃)  Which of the given statement(s) about N, O, P and Q with respect to M is (are) correct?  HO                       HO                      Cl\                       \                     /C ── H                  C ── H              C ── OH/                       /                   /Cl                       CH₃                HO\                        \                    C ── H                   C ── OH            C ── H/                        /                  /HO                       Cl                 CH₃\                        \  CH₃                      H                  H
(M)                     (N)                (O)  CH₃                     CH₃│                       │H──┼──OH                HO─┼──H/│\                     /│  HO─┼─┼─┼─H              HO─┼─┼─┼─H
\ │ /                   \ │ /
\│/                     \│/
Cl                      Cl
(P)                     (Q)  (A) M and N are non-mirror image stereoisomers
(B) M and O are identical
(C) M and P are enantiomers
(D) M and Q are identical  Sol. (A, B, C)
Converting all the structure in the Fischer projection   Cl                      Cl                      Cl                      Cl                      Cl
  |                       |                       |                       |                       |
HO ──┼── H (R)         HO ──┼── H (R)         HO ──┼── H (R)          H ──┼── OH (S)       HO ──┼── H (R)
|                       |                       |                       |                       |
HO ──┼── H (S)          H ──┼── OH (R)        HO ──┼── H (S)          H ──┼── OH (R)        H ──┼── OH (R)
|                       |                       |                       |                       |
CH₃                     CH₃                     CH₃                     CH₃                     CH₃
(M)                     (N)                     (O)                     (P)                     (Q)  M and N are diastereoisomers
M and O are identical
M and P are enantiomers
M and Q are diastereoisomers
Hence, the correct options are A, B, C.  With respect to graphite and diamond, which of the statement(s) given below is (are) correct?
(A) Graphite is harder than diamond.
(B) Graphite has higher electrical conductivity than diamond.
(C) Graphite has higher thermal conductivity than diamond.
(D) Graphite has higher C-C bond order than diamond.  IITJEE-2011-Paper 2-CPM-19  Sol. (B, D)
Diamond is harder than graphite.
Graphite is good conductor of electricity as each carbon is attached to three C-atoms leaving one valency free, which is responsible for electrical conduction, while in diamond, all the four valencies of carbon are satisfied, hence insulator.
Diamond is better thermal conductor than graphite. Whereas electrical conduction is due to availability of free electrons; thermal conduction is due to transfer of thermal vibrations from atom to atom. A compact and precisely aligned crystal like diamond thus facilitates fast movement of heat.
In graphite, C-C bond acquires double bond character, hence higher bond order than in diamond.  The given graphs / data I, II, III and IV represent general trends observed for different physisorption and chemisorption processes under mild conditions of temperature and pressure. Which of the following choice(s) about I, II, III and IV is (are) correct?  ┌────────────────────────────┐  ┌────────────────────────────┐│ (I)             P constant │  │ (II)            P constant ││ ▲                          │  │ ▲                          ││ │\                         │  │ │                  ╭───    ││ │ \                        │  │ │                ╭─╯       ││ │  \                       │  │ │              ╭─╯         ││ │   ╰───                   │  │ │           ╭──╯           ││ │       ╰────────          │  │ │      ╭────╯              ││ └────────────────────────► │  │ └──────┴─────────────────► ││ Amount of gas adsorbed  T  │  │ Amount of gas adsorbed  T  │└────────────────────────────┘  └────────────────────────────┘┌────────────────────────────┐  ┌────────────────────────────┐
│ (III)                      │  │ (IV)                       │
│ ▲                200 K     │  │ ▲                          │
│ │            ╭───────      │  │ │             Eact         │
│ │         ╭──╯   250 K     │  │ │          ┌──┴──┐         │
│ │      ╭──╯  ╭───────      │  │ │       ╭──┴─────┴──╮      │
│ │   ╭──╯  ╭──╯             │  │0├───┬───┴───────────┴────► │
│ │╭──╯  ╭──╯                │  │ │   │ Distance of molecule │
│ └┴─────┴─────────────────► │  │ │  ╭╯ from the surface     │
│ Amount of gas adsorbed  P  │  │ │  │ ΔHads = 150 kJ mol⁻¹  │
│                            │  │ │  ╰─────────              │
│                            │  │ Potential Energy           │
└────────────────────────────┘  └────────────────────────────┘  (A) I is physisorption and II is chemisorption
(B) I is physisorption and III is chemisorption
(C) IV is chemisorption and II is chemisorption
(D) IV is chemisorption and III is chemisorption  Sol. (A, C)
Graph (I) and (III) represent physisorption because, in physisorption, the amount of adsorption decreases with the increase of temperature and increases with the increase of pressure.
Graph (II) represent chemisorption, because in chemisorption amount of adsorption increase with the increase of temperature. Graph (IV) is showing the formation of a chemical bond, hence chemisorption.  

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