jee main - 2014 paper - i chemistry · 32. if z is a compressibility factor, van der waals equation...
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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−−
− − → − −
↓
= ← − −