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IIT JEE 2010 Paper 1 and solution by Pawan Babel (head of chemistry dept.,)
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PART I : CHEMISTRY PAPER – I
SECTION – I Single Correct Choice 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.
1. In the reaction OCH3HBr⎯⎯⎯→ the products are
(A ) OCH3Br and H2
(B) Br and CH3Br
(C ) Br and CH3OH
(D) OH and CH3Br
Key: ( D) Sol.: HBr H Br+ −⎯⎯→ +
O CH3H⊕
⎯⎯→ O CH3
H2
BrSN
−
⎯⎯⎯→
OH
CH3Br
2. Plots sh owing the variation of the ra te constant (k) with t emperature (T) ar e gi ven be low. The p lot tha t following Arrhenius equation is
(A ) k
T
(B) k
T
(C ) k
T
(D) k
T Key: ( A)
Sol.: aE
RTK Ae−
= Rate constant K increases exponentially with the rise in temperature. Since rate const. K also depends upon
orientation factor A hence its maximum value is not at all infinity rather limited to an optimal value.
3. The species which by definition has ZERO standard molar enthalpy of formation at 298 K is (A ) Br2 (g) (B) Cl2 (g) (C ) H2O (g) (D) CH4(g)
Topper's Choice Pawan Sir
4
Key: (B ) Sol.: Bromine and water exist in liquid state at 298 K. Methane is not an elemental species.
4. The ionization isomer of ( ) ( )2 24Cr H O Cl NO Cl⎡ ⎤⎣ ⎦ is
(A ) ( ) ( )2 2 24Cr H O O N Cl⎡ ⎤⎣ ⎦ (B ) ( )2 2 24[Cr H O Cl ](NO )
(C ) ( ) ( )2 4Cr H O Cl ONO Cl⎡ ⎤⎣ ⎦ (D) ( )2 2 2 24
Cr H O Cl (NO ) H O⎡ ⎤ ⋅⎣ ⎦
Key: (B ) Sol.: ( ) ( ) ( ) ( )ionization
2 2 2 24 4Cr H O Cl NO Cl Cr H O Cl NO Cl
+ −⎡ ⎤ ⎡ ⎤ +⎣ ⎦ ⎣ ⎦
( ) ( ) ( )Ionization2 2 2 2 2 24 4
Cr H O Cl NO Cr H O Cl NO+
⎡ ⎤ ⎡ ⎤ +⎣ ⎦ ⎣ ⎦ .
5. The correct structure of ethylenediaminetetraacetic acid (EDTA) is
(A)
CH2COOH
N CH CH
CH2COOH
NCH2
CH2
COOH
COOH
(B) HOOC
N CH2 CH2
HOOC
NCOOH
COOH
(C )
CH2COOH
N CH2 CH2
CH2COOH
NCH2
CH2
COOH
COOH
(D) CH2COOH
N CH CH
H
NH
CH2 COOH
H2CCOOH
CH2COOH
Key: ( C) Sol.: Based on facts
6. The bond energy (in kcal mol–1) of a C—C single bond is approximately (A) 1 (B) 10 (C ) 100 (D) 1000. Key: ( C) Sol.: C – C single bond dissociation energy ranges between 88 to 150 K cal mol–1.
7. The synthesis of 3 –octyne is achieved by adding a bromolkane i nto a mixture o f so dium amide a nd an alkyne. The bromoalkane and alkyne respectively are
(A) 2 2 2 2 3BrCH CH CH CH CH and 3 2CH CH C CH≡ (B ) 2 2 3BrCH CH CH and 3 2 2CH CH CH C CH≡ (C ) 2 2 2 2 3BrCH CH CH CH CH and 3CH C CH≡ (D) 2 2 2 3BrCH CH CH CH and 3 2CH CH C CH≡ . Key: ( D) Sol.: CH 3 – CH2 – C ≡ C – C – CH2 – CH2 – CH3 3 – octyne
3 2 2 22
CH CH CH CH BrNaNH
3 2 3 2CH CH C CH CH CH C C− − − −
− − ≡ ⎯⎯⎯→ − − ≡ ⎯⎯⎯⎯⎯⎯⎯⎯→
3 2 2 2 2 3CH CH CH CH C C CH CH− − − − ≡ − −
3 octyne
8. The correct statement about the following disaccharide is
O
OH
H
OCH2
H
OH
H
H
CH2OH
OH
H
O
OHH2CO
HOH2C
CH2OH
H
H
OH
H
(a) (b)
Topper's Choice Pawan Sir
5
(A) Ring (a) is pyranose with α –glycosidic link (B) Ring (a) is furanose with α –glycosidic link (C) Ring (b) is furanose with α –glycosidic link (D) Ring (b) is pyranose with β –glycosidic link. Key: ( A) Sol.: Ring (a) is pyranose whereas ring(b) is furanose. α–anomeric form of ring (a) is attached through glycosidic
bond.
SECTION – II Multiple Correct Choice Type
This section contains 5 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct.
9. In the Newman projection for 2, 2–dimethylbutane
Y
CH3 CH3
X
H H
X and Y can respectively be (A) H and H (B) H and C2H5 (C ) C2H5 and H (D) CH3 and CH3. Key: ( B, D)
Sol.: CH3 C
CH3
CH3
CH2 CH32
2, 2-dimethyl butane
1 3 4
C1 – C2 rotation
H
H H
HH H
CH3 CH3
C2H5
CH3CH3
C2H5
180ºrotation
X and Y become H and C2H5
CH3 C
CH3
CH3
CH2 CH3
H H
CH3
CH3CH3
CH3
180rotation⎯⎯⎯→
CH3
H H
H3C CH3
CH3 X and Y become CH3 and CH3.
10. The regent(s) used for softening the temporary hardness of water is (are) (A) 3 4 2Ca (PO ) (B ) ( )2Ca OH (C ) 2 3Na CO (D) NaOCl .
Topper's Choice Pawan Sir
Key: (B,C,D)Sol.: Temporary hardness is due t o t he prese nce of b icarbonates of Ca an d Mg. T emporary h ardness ca n b e
removed by clarke’s process which involves the addition of slaked lime ( ) ( )3 3 22 2
Ca HCO Ca OH 2CaCO 2H O+ ⎯⎯→ +
Washing soda removes both the temporary and permanent hardness. ( )3 2 3 3 32
Ca HCO Na CO CaCO 2NaHCO+ ⎯⎯→ + .
11. Among the following, the intensive property is (properties are) (A) molar conductivity (B) electromotive force (C) resistance (D) heat capacity.Key: (A) Sol: Electromotive force, resistance and heat capacity depends upon the amout of the matter present hence they are extensive properties.
12. Aqueous solutions of HNO3, KOH, CH3COOH, and CH3COONa of identical concentrations are provided. The pair(s) of solutions which form a buffer upon mixing is(are)
(A ) 3HNO and 3CH COOH (B ) KOH and 3CH COONa (C ) 3HNO and 3CH COONa (D ) 3CH COOH and 3CH COONa . Key: (C, D) Sol.: Mixture of weak acid and its salt are known as acidic buffer. 3 3 2 3 2 3
strong acid weak acidH NO CH CO Na CH CO H Na NO+ − − + + −+ ⎯⎯→ +
In an acid–based reaction. The equilibrium shifts to the direction which results in the formation of weaker acid.
13. I n the reaction
OH
( ) 2NaOH aq /Br⎯⎯⎯⎯⎯→ the intermediate(s) is (are)
(A)
O
Br
Br
(B)
O
Br Br
(C )
O
Br
(D)
O
Br
Key: ( A, C)
Sol.:
OH
is strongly activating towards EAS reaction and it is ortho–para directing
Topper's Choice Pawan Sir
7
O
Br
Br
O
H Br
O
Br
SECTION – III Linked Comprehension Type
This section contains 2 paragraphs. Based upon the first paragraph, 3 multiple choice questions and based upon the second paragraph 2 Multiple choice questions have to be answered. Each of these questions have four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
Paragraph for Questions 14 to 16 Copper is the most noble of the first row tr ansition metals and occurs in small deposits in several countries. Ores of copper include ch alcanthite (Cu SO4. 5H 2O), at acamite (Cu 2Cl(OH)3), c uprite (Cu2O), c opper glance (Cu 2S) an d malachite (Cu2(OH)2CO3). However 80% of the world copper production comes from the ore chalcopyrite (CuFeS2). The extraction of copper from chalcopyrite involves partial roasting, removal of iron and self–reduction.
14. Partial roasting of chalcopyrite produces (A) Cu2S and FeO (B) Cu2O and FeO (C ) CuS and Fe2O3 (D) Cu2O and Fe2O3 Key: (B ) Sol: Cu FeS2 + O2 → Cu2S + 2FeS + SO2 The sulphites of copper and iron are partially oxidized 2 22FeS 3O 2FeO 2SO+ ⎯⎯→ +
2 2 2 22Cu S 3O 2Cu O 2SO+ ⎯⎯→ + .
15. Iron is removed from chalcopyrite as (A ) FeO (B) FeS (C ) Fe2O3 (D) FeSiO3 Key: ( D) Sol: Fe is removed in the form of FeSiO3. 2 3FeO SiO FeSiO+ ⎯⎯→
16. In self–reduction, the reducing species is (A ) S (B) 2O − (C ) 2S − (D ) SO2 Key: ( C) Sol: 2 2 2Cu S 2Cu O 6Cu SO+ ⎯⎯→ +
2S− oxidized into 4S+ hence it is reducing species .
Paragraph for Questions 17 to 18 The concentration of po tassium ions ins ide a bio logical cell is a t lea st twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is :
( ) ( ) ( ) ( )M s | M aq; 0.05 molar || M aq; 1molar | M s+ +
For the above electrolytic cell the magnitude of the cell potential cellE 70= mV.
Topper's Choice Pawan Sir
8
17. For the above cell (A) cellE 0; G 0< ∆ > (B ) cellE 0; G 0> ∆ < (C) cellE 0; G 0< ∆ ° > (D ) cellE 0; G 0> ∆ ° < Key: (B )
Sol: Ecell = 2.303RT 0.05logF 1
− = a positive value
= 70 mV (given) He nce ∆G < 0. 18. If the 0.05 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell
potential would be (A) 35 mV (B) 70 mV (C) 140 mV (D) 700 mV Key: ( C)
Sol: cell2.303RT 0.0025E log
F 1−
=
= ( )22.303RT log 0.05F
−
= 2 × 70 mV = 140 mV.
SECTION – IV Integer Answer Type
This Section contains TEN questions. The answer to each question is a Single Digit Integer ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled.
19. The value of n in the molecular formula BenAl2Si6O18 is Key: ( 3) Sol.: [ ]12
6 18Si O − 3 2 6 18Be Al Bi O
[ ]126 18Si O −
-
-
-
-
-
-
20. The total number of basic groups in the following form of lysine is
H3N CH2 CH2 CH2 CH2
CH
NH2
C
O
O Key: ( 2) Sol.:
H3N CH2 CH2 CH2 CH2
CH
NH2 CO
O
* Group are basic.
Topper's Choice Pawan Sir
9
21. Based on VSEPR theory, the number of 90 degree F – Br – F angles in BrF5 is Key: (0)
Sol.: F
F
F
F
Br
F
The structure of BrF5 is square pyramidal. The number of FBrF angles having the value of 90º is eight (8). Due to trivial distortion, however, the bond angles (F—Br—F) are slightly less than 90º(85º).
22. Amongst the following, the total number of compounds whose aqueous solution turns red litmus paper blue is
KC N K 2SO4 (NH4)2C2O4 Na Cl Zn(NO3)2 Fe Cl3 K2CO3 NH4NO3 Li CN Key: ( 3) Sol.: KC N, K2CO3, LiCN are basic salt can convert red litmus to blue.
23. Amongst the following, the total number of compounds soluble in aqueous NaOH is
NCH3CH3 COOH OCH2CH3
CH2OH
OH
NO2 OH
NCH3 CH3
CH2CH3
CH2CH3
COOH
Key: ( 4)
Sol.:
COOH OH OH
NCH3 CH3
COOH
, , ,
are soluble in aq. NaOH.
24. A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL, and 25.0 mL. The number of significant figures in the average titre value is
Key: (3)
Sol.: A verage = 25.2 25.25 25.03
+ +=
= 75.45 / 3 = 25.15 ≈ 25.1. No. of significant figure = 3.
25. The number of neutrons emitted when 23592 U undergoes controlled nuclear fission to 142
54 Xe and 9038 Sr is
Key (3)
Topper's Choice Pawan Sir
10
Sol.: 235 142 90 192 54 38 0U Xe Sr 3 n⎯⎯→ + + .
26. In the scheme given below, the total number of in tramolecular aldol condensation products formed from ‘Y’ is
( )3
2
1. NaOH aq1. O2. Zn, H O 2. heatY⎯⎯⎯⎯→ ⎯⎯⎯⎯⎯→
Key: ( 1)
Sol.: 3
2
1. O2. Zn /H O⎯⎯⎯⎯→
O
O
( )NaOH aq⎯⎯⎯⎯→
OH
O
heat⎯⎯⎯→
O
27. The concentration of R in the reaction R → P was measured as a function of time and the following data is obtained :
[R] (moloar) 1.0 0.75 0.40 0.10 t (min.) 0.0 0.05 0.12 0.18
The order of the reaction is Key: (0) Sol.: R ⎯→ P
dcdt
− after 0.05 min = 0.25 50.05
= M min–1
dcdt
− after 0.12 min = 060 50.12
= M min–1
dcdt
− after 0.18 min = 90 50.18
= M min–1
The av erage rate r emains sa me throu ghout. Thi s implies tha t ra te is in dependent of concentration (zero order).
28. The total number of cyclic isomers possible for a hydrocarbon with the molecular formula C4H6 is
Key (5)
Sol.: Cyclic isomers C4H6
CH3
CH3
CH2
Total isomers = 5
Topper's Choice Pawan Sir
