03-iit Screening-2006 (chemistry)

IIT-JEE -2006 MEMORY BASED QUESTIONS

CHEMISTRY

SECTION- A Single Choice: Questions no. 1 to 12 has only one correct option. You will be awarded 3 marks for right answer and –1 mark for wrong answer 1.

B(OH)3 + NaOH → NaBO2 + Na[B(OH)4] + H2O addition of which would cause equilibrium to shift towards right (a) Cis 1, 2, diol

(b) trans 1, 2 diol

(c) Borax

(d) Na 2HPO 4

Sol.: (a) Cis 1, 2 diol on reaction with weak Boric acid converts it in compartivly strong monobasic acid thus favours forward reaction (a) COCl

2.

IUPAC Name of the

(a) Chlorophenyl ketone

(b) Chloroaceto benzene

(c) Benzenecarbonyl chloride

(d) Benzoyl chloride

Sol.: (c) 3.

Which gelatinous ppt. is soluble in NH4OH + NH4Cl (a) Zn(OH)2

(b) Al(OH)3

(c) Ca(OH)2

(d) Mg(OH)2

Sol.: (a) Zn2+ forms [Zn(NH3)4]++ complex thus precipitate dissolves.

S O 3H 4.

+ NaHCO3 →

OH + NaHCO3 →

NO2

The gases produced are respectively (a) CO2 and CO2

S O 3H Sol.: (a)

(b) CO2 and SO2

+ NaHCO3 →

(c) SO2 and NO2

S O 3N a + H2O + CO2-

OH O 2N

(d) CO1 and NO1

ONa + NaHCO3 →

NO2

+ H2O + CO2-

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Chemistry

5.

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Monoatomic ideal gas, P/V is constant find molar heat capacity (a) 3/2R

(b) 4/2R

(c) 5/2R

(d) 0

(c) 60 e.u.

(d) –100 e.u.

 PdV 

Sol.: (b) C = CV +  ndT    P = aV

∴

6.

From

PV = nRT

and

P = aV

thus

PdV R = ndT 2

C = CV +

C

D

A

B

R 3 R 4 = R+ = R 2 2 2 2

∆ S (A → C) = 50 e.u. ∆ S (C → D) = 30 e.u. ∆ S (B → D) = 20 e.u.

Calculate ∆ S (A → B) (e.u. is unit of entropy) (a) 100 e.u. Sol.: (c) 7.

(b) –60 e.u.

∆SA → B = [ ∆ S A → C + ∆ SC → D − ∆ SB → D ] = 60 e.u.

Ag + NH3

[Ag(NH3)]+

[Ag(NH3)]+ + NH3

[Ag(NH3)2]+

K1 = 6.8 × 103 K2 = 1.6 × 103

Calculate equilibrium constant for final complex formation (a) 1.08 × 107

(b) 1.08 × 109

(c) 1.08 × 105

(d) 1.08 × 104

Sol. (a) K = K1 × K2 = 1.6 × 6.8 × 106 = 1.08 × 107 8.

In blue solution of copper sulpahte excess of KCN is added then solution becomes colourless due to the formation of (a) [Cu(CN)4]2–

(b) Cu2+ get reduced to form [Cu(CN)4]3–

(c) Cu(CN)2

(d) CuCN

Sol. (b) Cu + 2CN → Cu(CN)2 2Cu(CN)2 → 2CuCN + (CN)2 +2

–

CuCN + 3 KCN → [Cu(CN)4]3– + 3K+

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IIT-JEE -2006

9.

Chemistry

NOCl CH2 = CH – CH3  → X, X is

(a)

CH 2 – CH – CH 3 | | Cl NO

(c) ON – CH2 –CH2 – CH2 – Cl

(b)

CH 2 – CH – CH 3 | | NO Cl

(d)

ON– CH – CH 2 – CH 3 | Cl

Sol. (b) NO+ is positive and Cl– is negative part of addendum which attache as per markownikoff’s rule. 10. In the Haber Basch process, for the preparation of ammonia, in presence of activated ferrous and molybdenum, which of the following is correct (a) The condition for equilibrium is 2 ÄG NH 3 = 3 ÄG H 2 + ÄG N 2 where G is Gibbs free energy per mole of the gaseous species measured at that partial pressure. The condition of equilibrium is unaffected by the use of catalyst, which increases the rate of both the forward and backward reactions to the same extent. (b) The equilibrium will shift to forward direction because according to IInd law of thermodynamics the entropy must increases in the direction of spontaneous reaction. (c) Catalyst will not alter the rate of either of the reaction. (d) Catalyst at 298 K doesn’t affect reaction rate as it increase, the forward and backward reaction by 2 and catalyst at 443 K increase reaction rate in the forward direction by 2 and backward direction by 1.7 Sol.: (a) 11.

When CO2 is passed through water, which of the following are present (a) CO2, H2CO3, HCO3–, CO32–

(b) CO2, H2CO3

(c) HCO3–, CO32–

(d) H2CO3, HCO3–

Sol.: (a) Dissolved CO2 combines with water to give H2CO3, which on dissociation give HCO3–and CO32. 12. The increasing order of boiling points of the following OH

OH I.

OH

OH II.

OH

III.

IV.

OH

(a) I < IV < III < II Sol.: (c)

(b) IV < III < II < I

(c) I < II < III < IV

OH

(d) I < II < IV < III

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Chemistry

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SECTION- B Question no. 13 to 20 has one or more than one correct option(s). You will be awarded 5 marks if you answer all correct options and only correct option(s) and –1 mark will be awarded for wrong answer. CH3 Fractional Cl 2   → [F] [N]  13. C H CH–CH2–CH3    → distillati on 3 [N], [F] are no. of isomers formed of C5H5Cl (a) 6, 6 (b) 6, 4 (c) 4, 4

(d) 3, 3

Cl

Sol.: (b)

Cl

Cl

Cl (+ , – )

(+ , – )

Thus total possible isomers are 6 and possible fractions obtained are 4 (+, – forms can not be separated by distillation) 14. When MgSO4 in presence of NH4Cl is treated with Na2HPO4. The white ppt. formed is of (a) MgSO4 (b) Mg(NH4)PO 4 (c) Mg 3(PO 4)2 (d) MgCl2 Sol.: Ans. (b) 15. The given graph represents the variation of Z(compressibility factor = PV/nRT) versus P, for three real gases A, B and C. Identify the only incorrect statement. A C Id e al g as A B

C

1 Z

B 0

P (atm ) (a) For the gas A, a = 0 and its dependence on P is linear at all pressure. (b) For the gas B, b = 0 and its dependence on P is linear at all pressure. (c) For the gas C, which is typical real gas for which neither a nor b = 0. By knowing the minima and the point of intersection, with Z = 1, a and b can be calculated. (d) At high pressure, the slope is positive for all real gases. 

Sol.: (b)  P+ 

a   (V – b) = RT V2 

b=0 Z=1–

a RTV

thus graph is hyperbolic. 16. The product ‘P’ containing nitrogen is in the following reaction CH3 – NH2 + KOH + CHCl3 → P + KCl + H2O .. ⊕ (a) C H 3 – N ≡ C (b) CH 3 − NHCl (c) ..

⊕

C H3 –N ≡ C :

(d) CH3 – C º N:

Sol.: Ans. (c) Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42 -4-

IIT-JEE -2006

Chemistry

17. The compound of lowest molecular mass of ketones and its next homologue is treated with hydroxyl amine (NH2–OH) then (a) the mixture will have two oximes

(b) the mixture will have three oximes

(c) one oxime is optically active

(d) none oxime is optically active

Sol.: (b)

CH3 CH3

N H 2O H

C =O

CH3 CH3

.. C =N

OH

(L o w e st k e to n e)

CH3

C =O

N H 2O H

C 2H 5

CH3

CH3

.. C =N

and

OH

C 2H 5

C 2H 5

C = N. .

OH

(N ex t h om o log u e o f lo w e st k e to n e)

Oxime formation is not stereo specific thus both syn. and anti oximes are obtained. 18. Bond length of C – O in Fe(CO)5 given that the length of CO in carbon monoxide 1.128 Å (a) 1.128

(b) 1.12

(c) 1.52

(d) 1.72

Sol.: (b) Due to bonding between CO and metal, CO bond length contracts thus (b)

O O conc . aq . alkali Acidification 19. A     →    →

, Which of the following compound is/are A?

L ac to n e This lactone is the only product of the reaction?

CHO

COOH

(a)

(b)

CHO

CH O Sol.: (a)

CHO

(c)

COOMe

OH

−

 → Cannizaro's

COOMe

COOH

esterification

COO

CH3

O

C H 2O H –

(d)

CHO

 →

O

(i ) O , heat anhyd .  → II + Phenol. products I and II are  → I  2 + CH3CH2CH2Cl AlCl 3 (ii ) H O + heat

20.

3

(a)

(c)

and CH3COCH3

and CH3COCH3

(b)

(d)

and CH3CH2CHO

and CH3CH2CHO

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Chemistry

IIT-JEE -2006 –

:H ⊕ ⊕ AlCl3  → Sol.: (a,c) CH3CH2CH2Cl  → CH 3CH 2 CH 2  CH 3 CHCH 3 shift

→ + CH 3CH 2 CH ⊕ 2

(P) ⊕

+ CH 3 CHCH 3 → (P) Cumene

give phenol and acetone on oxidation followed by Hydrolysis.

SECTION- C Question no. 21 to 32 are based on small write up first go through it then answer these questions. You will be awarded 5 marks for right answer and –2 marks for wrong answer. Comprehension-1 H N

C

H

Step−1

→

Cl

C

O

N

Br

Step −2

  →

Cl

O

O

Step−3

Step −4 N=C=O   → C l

 → C l

C

NH2

Due to formation of RCONHBr this reaction is named as Hoffmann Bromamide reaction 21. Reagents used in Step-1 are (a) KBr + NaOH Sol.: Ans. (c)

(b) KBr + NaOCH3

22. Rate determining step of the reaction (a) Step-2 (c) formation of isocyanate (step-3) Sol.: Ans. (c)

(b) Step-1 (d) formation of amine (step-4)

CON H2 ,

D

Products 15

NH2

15

N H2

(a)

(d) KBrO3 + KOH

15

CONH2 23.

(c) Br2 + NaOH

,

,

15

NH2

N H2 (b)

N H2 ,

D D

D NH2

15

N H2

(c)

,

NH2

NH2

(d)

,

D Sol.: (b) Because reaction is intramolecular. Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42 -6-

: :

Cl

H

N

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Chemistry

Comprehension-2 Carbon–14 is used to determine the age of organic material. The procedure is based on the formation of 14C by neutron capture in the upper atmosphere. N + 0n1 → 146C + 1H1 14 C is absorbed by living organisms during photosynthesis. The 14C content is constant in living organ-

14 7

ism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of 14C in the dead being, falls due to the decay which C14 undergoes 14 6

C →

14

N + β−

7

The half life period of 14C is 5770 years. The decay constant ( λ ) can be calculated by using the 0.693 following formula λ = t1 / 2 The comparison of the β − activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method however, ceases to be accurate over periods longer than 30,000 years. The proportion of 14C to 12C in living matter is 1 : 1012. 24. Which of the following option is correct? (a) In living organisms, circulation of 14C from atmosphere is high so the carbon content is constant in organism (b) Carbon dating can be used to find out the age of earth crust and rocks (c) Radioactive absorption due to cosmic radiation is equal to the rate of radioactive decay, hence the carbon content remains constant in living organism (d) Carbon dating can not be used to determine concentration of 14C in dead beings. Sol. (c) 25. What should be the age of fossil for meaningful determination of its age? (a) 6 years

(b) 6000 years

(c) 60,000 years

(d) It can be used to calculate any age

Sol. (b) 26. A nuclear explosion has taken place leading to increase in concentration of C14 in nearby areas. C14 concentration is C1 in nearby areas and C2 in areas far away. If the age of the fossil is determined to be T1 and T2 at the places respectively then (a) The age of the fossil will increase at the place where explosion has taken place and 1 C1 T1 – T2 = ln λ C2 (b) The age of the fossil will decrease at the place where explosion has taken place and 1 C1 T1 – T2 = ln λ C2 (c) The age of fossil will be determined to be same (d)

T1 C = 1 T2 C2

Sol. (a) Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42 -7-

Chemistry

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Comprehension-3 KCN → P NiCl2  HCN

HCl excess NiCl2    → Q

P and Q are complex having co-orrdination number 4 27. Then the IUPAC name of complex is (a) Potassium tetracyanonicklate (II) and Potassium tetrachloronicklate(II) (b) Potassium tetracyanonickel(II) and Potassium tetrachloronickel(II) (c) Potassium cyanonicklate(II) and Potassium chloronicklate(II) (d) Potassium cyanonickel(II) and Potassium chloronickel(II) Sol. (a) 28. Magnetic properties of complexes respectively are (a) both diamagentic (b) diamagnetic (P) & (Q) is paramagnetic with one unpaired e– (c) P is diamagentic & Q is paramagnetic with 2 unpaired e– (d) both are paramagentic Sol.: Ans. (c) 29. Hybridisation of the central atom in the two complexes (a) Chloro is dsp2 and cyano is sp3

(b) Chloro is sp3, cyano is dsp2

(c) Chloro is dsp2, cyano is dsp2

(b) Chloro is sp3, cyano is sp3

Sol.: Ans. (b) Comprehension-4 Ag+ + e– → Ag(s)

E0 red. = 0.8 V

Glucose → Gluconic acid + 2H+ + 2e–

E0 ox. = – 0.05 V

Ag(NH3)2 + e– → Ag (s) + 2NH3

E0 red. = 0.337

2.303 RT/F = .0591 F/RT = 38.35 30. Calculate lnK for equilibrium cell reaction by redox 1 and 2 2Ag+ + C6H12O6 + H2O → 2Ag(s) + C6H12O7 + 2H+ (a) 66.13 Sol.: (b) lnK = n ×

(b) 57.52

(c) 28.30

(d) 46.29

F × E0cell RT

= 2 × 38.35 × 0.75 = 57.52 Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42 -8-

IIT-JEE -2006

Chemistry

31. When ammonia is added to the solution, pH is raised to 11. Which half-cell reaction is affected by pH and by how much? (a) Eoxd will increase by a factor of 0.65 from Eoxd (b) Eoxd will decrease by a factor of 0.65 from Eoxd (c) Ered will increase by a factor of 0.65 from Ered (d) Ered will decrease by a factor of 0.65 from Ered Sol.: (a) On increasing concentration of NH3, the concentration of H+ ion decreases. Therefore, Eoxd of glucose/gluconic acid increases. 32. Ammonia is always added in this reaction. Which of the following must be incorrect? (a) NH3 combines with Ag+ to form a complex (b) Ag(NH3)2+ is a stronger oxidising reagent than Ag+ (c) In absence of NH3 silver salt of gluconic acid is formed. (d) NH3 has affected the standard reduction potential of glucose/gluconic acid electrode. Sol.: (d) Standard reduction potential is not influenced by change in concentration. SECTION- D Question no. 33 to 36 are subjective problems. Circle their correct answers. There is no negative marking for it. Each question carries 6 marks. 33. 75.2 gram of phenol is dissolved in 1 kilogram of benzaldehyde. Depression in freezing point is 7 K. Kf for solvent is 14 K/kg mol. Calculate percentage of dimerization. Sol.: 75% ∆ Tf = i K f × m 75.2 1 × × 1000 94 1000

7 = i × 14 × i = 0.625 For dimerization i=1– α+

α 2

α = 0.75

34.

ρ = 2 gm/cm3

edge length = 5 Å Atomic mass = 75, N = 6 × 1023 The atom is involved in cubic system. Find the atomic radius in picometres.

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Chemistry

Sol.: ñ =

IIT-JEE -2006

ZM a3N0

Z

ρ a3 N0 M 2 × (5 × 10 −8 ) 3 × (6 × 10 23 ) = =2 75

=

3a ~ 217 4 −

thus unit cell is BCC r = 35. The given reaction 2CO

+

2 moles

O2 → 2CO2

ÄH = –560 kJ

1 mole

is carried and in one litre container, if the pressure in the container gets changes from 70 atm to 40 atm as reaction gets completed. Calculate ÄU of the reaction. [1L atm = 0.1 kJ] Sol. ÄH so

= Ä U + ∆(PV) = Ä U = ∆ H – ∆(PV) = – 560 kJ – [40 – 70] (L atm) = (– 560 + 30 × 0.1) kJ = – 557 kJ

36. We have taken a saturated solution of AgBr. Ksp of AgBr is 12 × 10–14 . If 10–7 mole of AgNO3 are added to 1 litre of this solution find conductivity (specific conductance) of this solution in terms of 10–7 S m–1 units. 0 0 Given ë + = 6 × 10–3 Sm2 mol–1 ë

(Br − )

(Ag )

0 = 8 × 10–3 Sm2 mol–1, ë (NO3− ) = 7 × 10–3 Sm2 mol–1

Sol. The solubility of AgBr in presence of 10–7 molar AgNO3 is 3 × 10–7 M. Therefore [Br–] = 3 × 10–4 m3, [Ag+] = 4 × 10–4 m3 and [NO3–] = 10–4 m3 ˆ −+ K ˆ ++ K ˆ NO = 24 + 24 + 7 = 55 Sm–1. ˆ total = K Therefore K 3 Br Ag SECTION- E Question no. 37 to 40 carry 6 marks each. These may have more than one correct options. There is no negative marking for these. Match the following questions 37. (i)

CH3CH(Br)CD3 on treatment with alc. KOH

A. E1 reaction

gives CH2 = CHCD3 as a major product. (ii) PhCH(Br)CH3 reacts faster than PhCH(Br)CD3

B. E2 reaction

(iii) PhCH2CH2Br on treatment with C2H5OD/C2H5O–

C. E1 CB reaction

gives PhCD = CH2 as the major product (iv) PhCH2CH2Br and PhCD2CH2Br react with same

D. The reaction is unimolecular

rate. Sol.: (i) B

(ii) B

(iii) C, D

(iv) A, D

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38. (i)

Chemistry

SiO44– → Si2O76–

A. Acidification

(ii) AlO2– → Al(OH)3

B. Hydrolysis

(iii) B2O72– → B(OH)3

C. Dilution in water

(iv) Bi3+ → (BiO)+

D. Heating

Sol.: (i) D 39. (i)

(ii) A & C Ag

(iii) A & B

(iv) B

A. C reduction

(ii) B

B. Self reduction

(iii) Pb

C. Complex formation followed by ppt. of metal

(iv) Cu

D. tetra iodide

Sol.: (i) C

(ii) A

(iii) B

(iv) B

40. Vn = PE; Kn = KE; En = TE; rn = radius (i)

Vn =a Kn

A. 1

(ii) b = angular momentum in ground state in 1s orbital (iii) rn = (iv)

1 rn

B. 0

(En)c

C. –1

= (Z)d

D. –2

a, b, c, d are related as Sol.: (i) D

(ii) B

(iii) C

(iv) A

>=?@

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