JEE Main - 2014Paper - I

CHEMISTRYCODE- E

CHEMISTRY Analysis

Sl.No. UNIT NAME Q. Nos. Correct Wrong

1. SOME BASIC CONCEPTS IN CHEMISTRY 47

2. STATES OF MATTER 32

3. ATOMIC STRUCTURE 31

4. CHEMICAL BONDING & MOLECULAR STRUCTURE 44,57

5. CHEMICAL THERMODYNAMICS 36

6. CARBAXILIC COMPOUNDS

7. EQUILIBRIUM 39

8. REDOX REACTIONS & ELECTROCHEMISTRY 35,37,42,48

9. ARROMATIC HYDROCARBON

10. SURFACE,NUCLEAR CHEMISTRY

11. CLASSIFICATION OF ELEMENTS & PERIODICITY IN PROPERTIES

12. CHEMICAL KINETICS 40

13. HYDROGEN, GENERAL ORGANIC CHEMISTRY

14. BLOCK ELEMENTS (ALKALI & ALKALINE EARTH METALS)

15. p - BLOCK ELEMENTS, III A,IVA ELEMENTS

16. d and f - BLOCK ELEMENTS

17. CO-ORDINATION CHEMISTRY 43,50

18. ENVIRONMENTAL CHEMISTRY

19. PURIFICATION & CHARACTERISATION OF ORGANIC COMPOUNDS 34

20. SOME BASIC PRINCIPLES OF ORGANIC CHEMISTRY

21. HYDEROCARBONS 54

22. ORGANIC COMPOUNDS CONTAINING HALOGENS 51

23. ORGANIC COMPOUNDS CONTAINING OXYGEN 53,55,60

24. ORGANIC COMPOUNDS CONTAINING NITROGEN 52,56

25. POLYMERS 58

26. BIO MOLECULES , METALLURGY 59,49

27. SOLUTIONS 38

28. SOLID STATE 33

29. VII A GROUP ELEMENTS 41,46

30. HYDROGEN & ITS COMPOUNDS 45

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31. The correct set of four quantum numbers for the valence electrons of rubidium atom ( )37Z = is :

1)1

5,0,0,2

+ 2)1

5,1,0,2

+ 3)1

5,1,1,2

+ 4)1

5,0,1,2

+

Sol : 1

15R b s→

5n =

0l =

0m =

1

2s = +

32. If Z is a compressibility factor, van der Waals equation at low pressure can be written as :

1) 1RT

ZPb

= + 2) 1a

ZVRT

= − 3) 1Pb

ZRT

= − 4) 1Pb

ZRT

= +

Sol : 2

At low pressure V more b is neglegible

( )2

aP v b RT

v + − = vanderwaal’s equation for 1 mole of gas b is negligeble

2

aP V RT

v + =

aPV RT

v+ =

Deviding either side with RT

PV a RT PVZ

RT VRT RT RT+ = =

1a

ZVRT

+ =

1a

ZVRT

= −

[CODE-E]

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33. CsCl crystallises in body centred cubic lattice. If ‘a’ is its edge length then which of the followingexpressions is correct ?

1) 3Cs Clr r a−+ = 2)

3

2Cs Cl

ar r −+ = 3)

3

2Cs Clr r a−+ = 4) 3Cs Cl

r r a−+ =

Sol : 3

In BCC edge length “a” is equal to ( ) 3

2

ar r+ −+ =

34. For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and

the evolved ammonia was absorbed in 60 mL of10

Msulphuric acid. The unreacted acid required

20 mL of10

Msodium hydroxide for complete neutralization. The percentage of nitrogen in the

compound is :

1) 6% 2) 10% 3) 3% 4) 5%

Sol : 2

W = wt of organic compound = 1.4 gm, for sulphric acid2

10N N=

60V ml=

milliequivalents of sulphuric acid2

60 1210

= × =

for sodium hydroxide1

10N N=

20V ml=

milliequivalents of sodium hydroxide1

20 210

= × =

unreacted acid milliequivalents = milliequivalents of NaOH reacted

reacted acid milliEquivalents 12 2 10= − =

MilliEquivalents of 3 10NH =

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1000 10Wt

Equivalents weight× =

wt of 3 0.17NH gm=

% Nitrogen0.17 14 100

1017 1.4

×= × =

35. Resistance of 0.2 M solution of an electrolyte if 50 .Ω The specific conductance of the solution is

11.4 .S m− The resistance of 0.5 M solution of the same electrolyte is 280 .Ω The molar

conductivity of 0.5 M solution of the electrolyte in 2 1S m mol− is :

1) 45 10−× 2) 35 10−× 3) 35 10× 4) 25 10×Sol : 1

50R = Ω11.4 .K s m−=

1 lR

k a= ×

150

1.4

l

a= ×

1 170 0.7l

m cma

− −= →

280 ; 0.5R M= Ω =

10.7l

cma

−=

1 lR

k a= × ;

1280 0.7

K= ×

1 10.7

280K ohm cm− −=

1000k

M ×= 0.7 1000 7000

5280 0.5 280 5

×= ⇒ =× ×

1 2 15 . .ohm cm mol − −=

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4 1 2 15 10 . .ohm m mol − − −= ×

or 4 2 15 10 . .s m mol − −= ×

36. For complete combustion of ethanol, ( ) ( ) ( ) ( )2 5 2 2 23 2 3 ,C H OH l O g CO g H O l+ → + the amount

of heat produced as measured in bomb calorimeter, is 11364.47 kJ mol− at 25 .C Assuming

ideality the Enthalpy of combustion, ,c H∆ for the reaction will be :

( )18.314R kJ mol −=

1) 11366.95kJ mol−− 2) 11361.95kJ mol−−

3) 11460.50kJ mol−− 4) 11350.50kJ mol−−

Sol : 1

( ) ( ) ( ) ( )2 5 22 9 2 92 2 3l lC H OH CO CO H O→ → +

1364.47 /E KJ Mole∆ = −

?H∆ =

2 3 1H E nRT n∆ = ∆ + ∆ ∆ = − = −

( )31364.47 1 8.314 10 298H −∆ = − + − × × ×

1364.47 2.477H∆ =− −11366.95H kJmol−∆ =−

it is combustion recation which is exothermic

1361.93 /H kJ mole∆ = −

37. The equivalent conductance of NaCl at concentration C and at infinite dilution are C and ∞

respectively. The correct relationship between C and ∞ is given as :

( Where the constant B is positive )

1) ( )C B C ∞= + 2) ( )C B C ∞= − 3) ( )C B C ∞= − 4) ( )C B C ∞= +

Sol : 3

Acc to debye Huckel onsagar equation

c b c ∞= −

( )c B c ∞∴ = −

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38. Consider separate solutions of 0.500 M ( )2 5 ,C H OH aq ( ) ( )3 4 20.100 ,M Mg PO aq

( )0.250M KBr aq and ( )3 40.125M Na PO aq at 25 .C Which statement true about these

solutions, assuming all salts to be strong electrolytes?

1) They all have the same osmotic pressure.

2) ( ) ( )3 4 20.100M Mg PO aq has the highest osmotic pressure.

3) ( )3 40.125M Na PO aq has the highest osmotic pressure.

4) ( )2 50.500M C H OH aq has the highest osmotic pressure.

Sol : 1

2 50.5 0.5 0.5 1 0.5M C H OH i⇒ × ⇒ × ⇒

( )3 4 20.1 0.1 0.1 5 0.5M g PO i⇒ × ⇒ × ⇒

0.25 0.25 0.25 2 0.5KBr i⇒ × ⇒ × ⇒

3 40.125 0.125 0.125 4 0.5M Na PO i⇒ × ⇒ × ⇒

∴ all process same osmatic pressure.

CST i = ×

39. For the reaction ( ) ( ) ( )'2 2 3

1

2g g gSO O SO+ if ( )x

P CK K RT= where the symbols have usual

meaning then the value of x is :

( assuming ideality)

1) 1− 2)1

2− 3)

1

24) 1

Sol : 2

( ) ( ) ( )2 2 3

1

2g g gSO O SO+

( )x

p cK K RT x n∴ = = ∆

n∆ = No. of moles of gaseous proudcts - No.of moles of gaseous reactants.

11 1.5 0.5

2n∆ = − ⇒ − ⇒ −

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40. For the non - stoichiometry reaction 2 ,A B C D+ → + the following kinetic data were obtained in

three separate experiments, all at 298 K.

Initial Concentration ( )A Initial Concentration ( )B Initial rate of formation of C

( )mol L S− −

0.1M 0.1M 31.2 10−×

0.1M 0.2 M 31.2 10−×

0.2 M 0.1M 32.4 10−×The rate law for the formation of C is :

1) [ ][ ]dck A B

dt= 2) [ ] [ ]2dc

k A Bdt

= 3) [ ][ ]2dck A B

dt= 4) [ ]dc

k Adt

=

Sol : 4

2A B C D+ → + by increasing con’c of B rate of reaction does not changes i.e, Rate of reaction independent of

[ ]B

When con’c of A is doubled rate of reaction also doubled i.e rate of reaction [ ]A

[ ]dcK A

dt∴ =

41. Among the following oxoacids, the correct decreasing order of acid strength is :

1) 2 3 4HOCl HClO HClO HClO> > > 2) 4 2 3HClO HOCl HClO HClO> > >

3) 4 3 2HClO HClO HClO HOCl> > > 4) 2 4 3HClO HClO HClO HOCl> > >

Sol : 3

In crlorine oxoacids as the oxidation no.of chlorine increates acidic strength increases.

∴ acidic strength decreasing order

4 3 2HClO HClO HClO HOCl> > >

42. The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is :

1) Ag 2) Ca 3) Cu 4) Cr

Sol : 2

The active metals like Na, K, Li, Mg & Ca -have high oxidation potential

Electrolysis of their molten salts produce the metal aqueous salts evolve 2H gas.

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43. The octahedral complex of a metal ion 3M + with four monodentate ligands 1 2 3, ,L L L and 4L

absorb wavelengths in the region of red, green , yellow and blue, respectively. The increasing orderof ligand strength of the four ligands is :

1) 4 3 2 1L L L L< < < 2) 1 3 2 4L L L L< < <

3) 3 2 4 1L L L L< < < 4) 1 2 4 3L L L L< < <

Sol : 2

Wavelengths orders

Red Yellow Green Blue< < <Energy order

( ) ( ) ( ) ( )1 3 2 4

ReL L L L

d Yellow Green Blue> > >

Increasing order of ligand strength

1 3 2 4L L L L< < <

44. Which one of the following properties is not shown by NO ?

1) It is diamagnetic in gaseous state

2) It is a neutral oxide

3) It combines with oxygen to form nitrogen dioxide

4) It’s bond order is 2.5Sol : 1

NO - Paramagnetic ,neutral oxide, bond order is - 2.5

2 22 2NO O NO+ →

45. In which of the following reactions 2 2H O acts as a reducing agent ?

a) 2 2 22 2 2H O H e H O+ −+ + → b) 2 2 22 2H O e O H− +− → +

c) 2 2 2 2H O e OH− −− → d) 2 2 2 22 2 2H O OH e O H O− −+ − → +

1) ( ) ( ),a b 2) ( ) ( ),c d 3) ( ) ( ),a c 4) ( ) ( ),b d

Sol : 4

2 2 22 2H O e O H− ⊕− → +

2 2 2 22 2 2H O OH e O H O−+ − → +

In these reactions electrons are lost ∴ 2 2H O acts as reducing agent.

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46. The correct statement for the molecule, 3,CsI is :

1) it is a covalent molecule. 2) it contains Cs+ and 3I − ions,

3) it contains 3Cs + and I − ions. 4) it contains ,Cs I+ − and lattice 2I molecule.

Sol : 2

3 3io n is a t io nC s I C s I+ − → +

47. The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1: 4 . The ratio ofnumber of their molecule is :

1) 1: 4 2) 7 :32 3) 1:8 4) 3:16

Sol : 2

No.of moles of oxygen2

2

oW

M wt of o=

No.of moles of nitrogen2

2

NW

Mwt of N=

2

2

1

4O

N

W

W=

Mwt of 2 28N =

Mwt of 2 32O =

2

2

1 28 7

4 32 32

Moles of O

Moles of N= × =

no.of moles ratio 7 :32=no.of molecules ratio 7 :32=

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48. Given below are the half - cell reactions :

2 02 ; 1.18Mn e Mn E V+ −+ → = −

( )3 2 02 ; 1.51Mn e Mn E V+ − ++ → = +

The 0E for 2 33 2Mn Mn Mn+ +→ + will be :

1) 2.69 ;V− the reaction will not occur 2) 2.69 ;V− the reaction will occur

3) 0.33 ;V− the reaction will not occur 4) 0.33 ;V− the reaction will occur

Sol : 1

2 2Mn e Mn+ −+ →

01

0 01 1

1.18

2

E V

G FE

= −

∆ = −

3 22 2 2Mn e Mn+ − ++ →

02

0 02 2

1.51

2

E V

G FE

= +

∆ = −

Given Reaction

2 3 03 2 oMn Mn Mn G nFE+ +→ + ∆ = −

from above three equations

0 01 2G G G∆ = ∆ − ∆

( ) ( )0 01 22 2G FE FE∆ = − − −

( )0 01 22G F E E∆ = − −

( )2 1.18 1.51G F∆ = − − −

( )2 2.69G F∆ = − −

( )2 2.69onFE F− = − −

( )2 2 2.69oE− = − −

2.69oE V= −

0,oE is ve G Ve− ∆ = + hence reaction will not occur

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49. Which series of reactions correctly represents chemical relations related to iron and its compound ?

1) ( )2 4 2 4 2,4 2 4 3

dil H SO H SO O heatFe FeSO Fe SO Fe→ → →

2) 2 2 4,4

O heat dil H SO heatFe FeO FeSO Fe→ → →

3) 2 , ,3 2

Cl heat heat air ZnFe FeCl FeCl Fe→ → →

4) 2 , ,600 ,7003 4

O heat CO C CO CFe Fe O FeO Fe→ → →

Sol : 4

2 3 46 4 2Fe O Fe O+ →

6003 4 23CFe O CO Feo CO+ → +

7002

CFeO CO Fe CO+ → +

50. The equation which is balanced and represents the correct product (s) is :

1) 2 22 2Li O KCl LiCl K O+ → +

2) ( ) 23 45

5 5CoCl NH H Co NH Cl+ + + + − + → + +

3) ( ) ( ) ( )2 24

2 266excess NaOHMg H O EDTA Mg EDTA H O

+ +− + → +

4) ( )4 2 2 444CuSO KCN K Cu CN K SO + → +

Sol : 2

( ) 1

3 5CoCl NH

+ is a complex ion by additon of acid it does not dissociate into its ions

51. In 2NS reactions, the correct order of reactivity for the following compounds :

( )3 3 2 3 2, ,CH Cl CH CH Cl CH CHCl and ( )3 3

CH CCl is :

1) ( ) ( )3 3 3 2 32 3CH Cl CH CHCl CH CH Cl CH CCl> > >

2) ( ) ( )3 3 2 3 32 3CH Cl CH CH Cl CH CHCl CH CCl> > >

3) ( ) ( )3 2 3 3 32 3CH CH Cl CH Cl CH CHCl CH CCl> > >

4) ( ) ( )3 3 2 3 32 3CH CHCl CH CH Cl CH Cl CH CCl> > >

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Sol : 2

2SN Reactivity order

Methylhalide0 0 0

1 2 3halide halide halide> − > − > −

( ) ( )3 3 2 3 32 3H C Cl H C CH Cl H C CH Cl H C C Cl− > − − > − > −

52. On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, theorganic compound formed is :

1) an alkanol 2) an alkanediol 3) an alkyl cyanide 4) an alkyl isocyanide

Sol : 4

3

02

1 min

CHCl

KOH alkyl Isocyanidealiphatic a e

R NH R NC−− −

− → −

53. The most suitable reagent for the conversion of 2R CH OH R CHO− − → − is :

1) 4KMnO 2) 2 2 7K Cr O

3) 3CrO 4) PCC ( Pyridinium Chlorochromate)

Sol :4

02

1

PCC

aldehydealcohol

R CH OH R CHO−

− − → −

54. The major organic compound formed by the reaction of 1,1,1 -trichloroethane with silver powder is :

1) Acetylene 2) Ethene 3) 2 - Butyne 4) 2 - Butene

Sol :3

1,1,1 Trichloroethane− - 3 3H C HCCl−

3 3 3 3 3 32

6 6butyne

H C CCl Ag Cl C CH H C C C CH AgC−

− + + − → − ≡ − +

55. Sodium phenoxide when heated with 2CO under pressure at 125 C yields a product which on

acetylation produces C.

ONa 125° H+

5 Atm+ CO2 Ac O2

B C

The major product C would be :

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1)C O O H

O C O C H 3

2)

3) 4)

Sol :1

56. Considering the basic strength of amines in aqueous solution, which one has the smallest bpK

value ?

1) ( )3 2CH NH 2) 3 2CH NH 3) ( )3 3

CH N 4) 6 5 2C H NH

Sol :1

Basic nature order

( ) ( )3 3 2 3 6 5 22 3H C NH H C NH H C N C C NH> − > > −

strong base has the lowest bPk

57. For which of the following molecule significant 0? ≠ .

a)

Cl

Cl

b)

CN

CN

c)

O H

O H

d)

S H

S H

1) Only ( )a 2) ( )a and ( )b 3) Only ( )c 4) ( )c and ( )d

Sol :4

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O H

O H

S H

S H

0

0.83D

≠

∴ = 0 ≠

58. Which one is classified as a condensation polymer ?

1) Dacron 2) Neoprene 3) Teflon 4) Acrylonitrile

Sol :1

Dacron is a condeusation polymer formed by the following monomers-

2 2

| |

OH OH

Ethylene glycol H C CH−

tereptthalic acid HOOC COOH

59. Which one of the following bases in not present in DNA ?

1) Quinoline 2) Adenine 3) Cytosine 4) Thymine

Sol :1

In DNA - The bases are Thymine , cytosine , Adenine & Guanine

60. In the reaction,

54 .3 ,PClLiAlH Alc KOHCH COOH A B C→ → → the product C is :

1) Acetaldehyde 2) Acetylene 3) Ethylene 4) Acetyl chloride

Sol :3

43 3 2

5

2 2 3 2

| |LiAlH

alc KoHHCl

O

H C C OH H C H C OH

PCl

H C CH H C CH Cl−−

− − → − −

↓

= ← − −