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JEE MAIN EXAMINATION - 2014
QUESTIONS WITH SOLUTIONS
PAPER CODE - F
JEE MAIN Examination(201[4C) (HCoEdeM - FI)STRY] CODE-F (Page # 1)
1. Consider separate solutions of 0.500 M 4. The metal that connnot be obtained by elec-
C H OH(aq), 0.100 M Mg (PO ) (aq), 0.250 trolysis of an aqueous solution of its salts is
2 5 3 4 2
M KBr(aq) and 0.125 M Na PO (aq) at 25ºC. (1) Ca
3 4
Which statement is true about these solu- (2) Cr
tions, assuming all salts to be strong elec- (3) Ag
trolytes? (4) Cu
(1) 0.125 M Na PO (aq) has the highest os- Sol. 1
3 4
motic pressure. Eº Eº
(2) 0.500 M C H OH(aq) has the highest os- H2o/H2 Ca2/Ca
2 5 Ca is more electro +ve so aqueous solu-
motic pressure.
tions cannot be use.
(3) 0.100 M Mg (PO ) (aq) has the highest
3 4 2
osmotic pressure.
5. On heating an aliphatic primary amine with
(4) they all the have the same osmotic pres-
chloroform and ethanolic potassium hydrox-
sure
ide, the organic compound formed is :
Sol. 4
(1) An alkyl cyanide
i ic
(2) An alkanol
0.5 M C H OH (aq.) 1 0.5
2 5 (3) An alkyl isocyanide
0.10 M Mg (PO ) 5 0.5
3 4 2 (4) An alkanediol
0.25 M KBr 2 0.5
Sol. (3)
0.125 M Na PO 4 0.5
3 4
All have same colligative prop. R—NH + CHCl +KOH R–NC
2 3
1º-amine chloroform Alkyl isocyanide
2. Which one of the following bases is not it is carbyl amine reaction or isocyanide test.
present in DNA?
(1) Cytosine (2) Thymine
1
(3) Adenine (4) Quinoline
6. For the reaction SO + O (g) SO ,
Sol. (4) 2(s) 2 2 3(g)
if K = K (RT)x where the symbols have usual
p C
meaning then the value of x is: (assuming
(Quinoline) ideality) ?
(1) – 1/2
(2) 1
Quinoline is not present into DNA.
(3) –1
Cytosin, Thymine and Adenine are present in
(4) 1/2
DNA.
Sol. (1)
1
3. Considering the basic strength of amines in SO2(g) 2O2(g) SO3(g)
aqueous solution, which one has the small-
Kp = K (RT)ng
est pK value? c
b
(1) (CH )N 1 1
3 3 n = 1 – (1 + ) = –
(2) CHNH g 2 2
6 5 2
(3) CH NH Kp = K (RT)–1/2
3 2 c
(4) (CH )NH
3 2 7. Given below are the half - cell reactions
Sol. (4)
Mn2+ + 2e–  Mn : Eº = – 1.18 V
2(Mn3 + e–  Mn2+) : Eº = + 1.51 V
Delocalised lone pair electrons. The Eº for 3Mn2+  Mn + 2Mn3+ will be
(1) – 0.33 V ; the reaction will not occur
(least Basic) (2) – 0.33 V ; the reaction will occur
Order of basic strength in aqueous solution (3) – 2.69 V ; the reaction will occur
 2º > 1º >3º amine > aniline. (4) – 2.69 V ; the reaction will not occur
1
Basic strength  K 
b pkb
(CH ) NH is most basic so it has smallest pK .
3 2 b
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Sol. (4) 10. Among the following oxoacids, the correct
Mn2+ + 2e–  Mn ; Eº = – 1.18 V decreasing order of acid strength is :
2Mn3+ + 2e–  2Mn2+ ; Eº = +1.51 V (1) HClO > HClO > HClO > HOCl
2 4 3
(1) – (2)
(2) HClO > HClO > HClO > HOCl
Mn2+ + 2e– + 2Mn2+  2Mn3+ + 2e–1 + Mn 4 3 2
(3) HOCl > HClO > HClO > HClO
Eº = – 1.18 – 1.51 ( n is same) 2 3 4
(4) HClO > HOCl > HClO > HClO
 Eº = – 2.69 V 4 2 3
Sol. (2)
(3) The reaction will nor occur
11. The ratio of masses of oxygen and nitrogen
8. If Z is a compressibility factor, van der waals
in a particular gaseous mixture is 1 : 4. The
equation at low pressure can be written as ?
ratio of number of their molecule is :
a
(1) Z = 1 – (1) 7 : 32 (2) 3 : 16
VRT
(3) 1 : 4 (4) 1 : 8
Pb
(2) Z = 1 + Sol. (1)
RT
O : N
RT mass 12 : 4 2
(3) Z = 1 +
Pb
1 4
Moles :
Pb 32 28
(4) Z = 1 –
RT 7 : 32
Sol. (1)
According to vander waal’s equation for 12. Which one of the following properties is not
one mole gas shown by NO?
(1) It combines with oxygen to form nitrogen
 a 
P (Vb)RT dioxide
 V2 (2) It's bond order is 2.5
at low pressure volume will be high (3) It is a neutral oxide
 b can be neglected, (4) It is diamagnetic in gaseous state
so, Sol. (4)
(1) NO + 1/2 O  NO
P a (V)RT (2) bond Order 2of NO i2s 2.5
 V2 (3) It is paramagnetic in gaseous state
(4) It is neutral oxide
a
 PV RT
V 13. Which series of reactions correctly represents
chemical relations related to iron and its com-
PV a
 Z = = 1 – pound ?
RT VRT
(1) Fe Cl,heat FeCl heat, air FeCl
2 3 2
9. In the reaction, Zn Fe
CH3COOH LiAlH4 A PCl5 B Alc. KOH C, (2) Fe dil H2SO4 FeSO4 H2SO4,O2
the product C is ?
Fe(SO ) heat Fe
(1) Acetyl chloride 4 3
(2) Ethylene (3)Fe O2, heat Fe3O4 CO, 600ºC FeO
(3) Acetaldehyde CO, 700ºC Fe
(4) Acetylene
(4) Fe O heat FeO dil HSO FeSO
Sol. (2) 2, 2 4 4
heat Fe
CH–CH–OH Sol. (3)
3 2
This reaction is used in blast furnace.
CH–CH–Cl
3 2
CH =CH (Ethylene)
2 2
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14. For which of the following molecule signifi- 15. CsCl crystallises in body centred cubic lat-
cant  0 ? tice. If 'a' is its edge length then which of
the following expressions is correct ?
Cl CN
(1) r r  3a
Cs Cl
(a) (b) 3
(2) r r  a
Cs Cl 2
Cl CN
(3) r r 3a
Cs Cl
OH SH
3a
(4) r r 
Cs Cl 2
Sol. 2
(c) (d)
Cl– forms SC
OH SH Cs+ goes to centre of the cube
(1) Only (c) 3a
 r r 
(2) Only (a)  – 2
(3) (c) and (d)
(4) (a) and (b)
16. Which one is classified as a condensation
Sol. (3)
polymer ?
(1) Teflon (2) Acrylonitrile
(3) Neoprene (4) Dacron
Sol. (4)
Terylene or Dacron
(a) (b)
(Dacron)
Other than dacron are addition polymers.
(c) (d) 17. For the estimation of nitrogen, 1.4 g of an
organic compound was digested by kjeldahl
method and the evolved ammonia was ab-
M
sorbed in 60 mL of sulphuric acid. The
10
M
unreacted acid required 20 mL of sodium
10
hydroxide for complete neutralization. The
percentage of nitrogen in the compound is ?
(1) 10 % (2) 5 %
(3) 6 % (4) 3 %
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Sol. (1)
1.4 gm  NH 20. The most suitable reagent for the conversion
3
of R – CH – OH  R – CHO is :
M M 2
 60mL HSO 20mL NaOH (1) CrO
10 2 4 10 3
(2) KMnO
Let’s Assume x % N 4
(3) PCC (Pyridininum Chlorochromate)
x (4) K Cr O
1.4 2 2 7
100 1 1  20  1  60 2 Sol. (3)
14 10 1000 10 1000
14100 R—CH—OH
x (0.012–0.002) 2
1.4 PCC (pyridinium chloro chromate) is mild
= 10 % oxidising agent.
18. For complete combustion of ethanol, 21. In which of the following reactions H O acts
2 2
C H OH(l) + 3O (g)  2CO (g) + 3H O(l), as a reducing agent?
2 5 2 2 2
the amount of heat produced as measured in (a) H O + 2H+ + 2e–  2H O
2 2 2
bomb calorimeter, is 1364.47 kJ mol–1 at 25ºC. (b) H O – 2e–  O + 2H+
2 2 2
Assuming ideality the Enthalpy of combustion, (c) H O + 2e–  2OH–
2 2
 H, for the reaction will be : (d) H O + 2OH– –2e–  O + 2H O
C (1) (a2),2 (c) 2 2
(R = 8.314 kJ mol–1)
(2) (a), (b)
(1) –1460.50 kJ mol–1
(3) (b), (d)
(2) –1350.50 kJ mol–1
(4) (c), (d)
(3) –1361.95 kJ mol–1
Sol. 3
(4) –1366.95 kJ mol–1
As a reducing agent
Sol. (4)
0
H = U + n RT HO 1 O 2e–
g 2 2 2
= – 1364.47 + (–1) × 8.314 × 10–3 × 298
= –1364.47 – 2.47
(b) & (d)
= – 1366.95 kJ mol–1
22. The correct statement for the molecule, CsI ,
19. The octahedral complex of a metal ion M3+ 3
is ?
with four monodentate ligands L , L , L and
1 2 3 (1) It contains Cs+ and I – ions.
L absorb wavelengths in the region of red, 3
4 (2) It contains Cs+, I– and lattice I mol-
green, yellow and blue, respectively. The in- 2
ecule.
creasing order of ligand strength of the four
(3) It is a covalent molecule.
ligands is ?
(4) It contains Cs3+ and I– ions.
(1) L < L < L < L
1 3 2 4 Sol. (1)
(2) L < L < L < L
1 2 4 3
(3) L < L < L < L
4 3 2 1
(4) L < L < L < L
3 2 4 1 23. The major organic compound formed by the
Sol. (1)
reaction of 1,1,1-trichloroethane with silver
(a) According to spectro chemical series
powder is ?
more absorption frequency stronger the
(1) 2 - Butene
ligand
(2) 2-Butyne
V I B G Y O R (3) Acetylene
(4) Ethene
E
L < L < L < L
1 3 2 4
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Sol. (2) = 0.25 × 10–2  –1 - cm–1
1000
 
m c
 10000.2510–2
= 0.5
CH—CC—CH + 6AgCl = 5 –1-cm2 mole–1
3 3
2-butyne = 5 × 10–4 S-m2 mol–1
24. The equivalent conductance of NaCl at con- 26. Sodium phenoxide when heated with CO un-
centration C and at infinite dilution are  der pressure at 125º C yields a product w2hich
C
and respectively. The correct relationship on acetylation produces C ?


between  and  is given as ? ONa
C  125º H+
(where the constant B is positive) + CO B C
2 5 Atm AcO
2
(1) C =  + (B) C
(2) C =  – (B) C The major product C would be;
(3)  =  + (B) C OH
C 
COOCH
(4)  =  – (B) C 3
C  (1)
Sol. (1)
 =  — B C
c 
OCOCH
25. Resistance of 0.2 M solution of an electro- 3
lyte is 50 . The specific conductance of
the solution is 1.4 S m–1. The resistance of (2)
0.5 M solution of the same electrolyte is 280 COOH
. The molar conductivity of 0.5 M solution
of the electrolyte in S m2 mol–1 is ? OH
(1) 5 × 103
(2) 5 × 102 COCH
3
(3) 5 × 10–3
(3)
(4) 5 × 10–4
Sol. 4 COCH
3

R =  OCOCH
a 3
COOH
1  (4)
50 = 
1.4 a
 Sol. (4)
70m1
a

R = 
a
1
280 = 70

 = 0.25 Sm–1
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29. For the non-stoichiometre reaction 2A + B 
C + D, the following kinetic data were ob-
tained in three separate expreiments, all at
298 K
Initial Initial Initialrateof
Concentration Concentration formationofC
(A) (B) (molLS)
0.1M 0.1M 1.2103
0.1M 0.2M 1.2103
0.2M 0.1M 2.4103
the rate law for the formation of C is
dc
(1) = k[A] [B]2
dt
dc
(2) = k[A] [B]
dt
dc
(3) = k[A] z
dt
dc
(4) = k[A]2 [B]
dt
27. The equation which is balanced and repre- Sol. (3)
sents the correct product(s) is? 2A + B  C + D
(1)[CoCl(NH )]+ + 5H+ CO2+ + 5NH+ + Cl– (1) & (3)  [A]1
(2) CuSO + 345KCN  K [Cu(CN) ] + K4SO (1) & (2)  [B]º
(3) Li O +4 2KCl  2LiCl 2+ K O 4 2 4  r = k[A]1
2 2
(4) [Mg(H O) ]2++(EDTA)4– excess NaOH
2 6 30. In S 2 reactions, the correct order of reac-
[Mg(EDTA)]2+ + 6H O N
2 tivity for the following compounds ?
Sol. (1)
CH Cl, CHCH Cl, (CH )CHCl and (CH )CCl is
This reaction is used to estimate Mg2+ in hard 3 3 2 3 2 3 3
(1) CHCl > (CH)CHCl > CHCHCl > (CH)CCl
water. 3 3 2 3 2 3 3
(2) (CH)CHCl > CHCHCl > CHCl > (CH)CCl
3 2 3 2 3 3 3
(3) CHCHCl > CHCl > (CH)CHCl > (CH)CCl
28. The correct set of four quantum numbers for 3 2 3 3 2 3 3
(4) CHCl> CHCHCl > (CH)CHCl > (CH)CCl
the valence electrons of rubidium atom (Z = 3 3 2 3 2 3 3
Sol. (1)
37) is ?
S 2
(1) 5, 1, 1 + 1/2 N
(2) 5, 0, 1, + 1/2 1
Reactivity of SN2 
(3) 5, 1, 0, + 1/2 Steric hindrance
(4) 5, 0, 0, + 1/2
Order of reactivity towards S 2.
Sol. (4) N
CHCl > CH–CH–Cl > (CH)CH–Cl > (CH)C–Cl
3 3 2 32 33
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SECTION – A Sol. [2]
Single Correct
p q ~q p ~q ~p ~q p  q
T T F F T T
sin(cos2x) T F T T F F
31. Lim is equal to :
x0 x2 F T F T F F
F F T F T T

(1)  (2)
2
(3) 1 (4) –
34. Let  and  be the roots of equation
Sol. [1]
px2 + qx + r = 0, p  0. If p, q, r are in A.P.
sin(c'x)
lim 1 1
x0 x2 and + = 4, then the value of | – | is:
 
sin(c2x) (1cx)
lxim0 (c2x)  x2 (1cx) (1) 2 13 (2) 61
9 9
1 2 17 34
= 1 .  . . 2 (3) (4)
2
9 9
Sol. [1]
= 
q r
Given  = –  =
32. Let the population of rabbits surviving at a p p
time t be governed by the differential equation
p, q, r  A. P. , then
dp(t) 1 q – p = r – q  2qrp ...(1)
= p(t) – 200.
dt 2
1 1
If p(0) = 100, then p(t) equals : again  4
 
(1) 400 – 300 e–t/2 (2) 400 – 300 et/2
(3) 300 – 200 e–t/2 (4) 600 – 500 et/2 
= 4
Sol. [2] 
2dp(t)
  dt
p(t)400 q4r ...(2)
2 log|p(t) – 400| = t + c ...(1)
(1) & (2) gives
t = 0, p = 100
– 8r = r + p
2 log (300) = C
 p 9r ...(3)
From (1)
2log|p(t) – 400| = t + 2 log (300) Now, | – | = ()2 4
|p(t) – 400| = et/2 . elog(300)
q2 4pr
p(t)400et/2(300) =
|p|
16r2 36r2
33. The statement ~ (P  ~q) is :
=
(1) a fallacy |9r|
(2) equivalent to p  q
52 2 13
(3) equivalent to ~p  q = 
9 9
(4) a tautology
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35. Let PS be the median of the triangle with Sol. [3]
vertices P(2, 2), Q(6, –1) and R(7, 3). The AA' = A'A
equation of the line passing through (1, –1) B = A–1A'
and parallel to PS is : B' = A(A–1)'
(1) 2x – 9y – 11 = 0 (2) 4x – 7y – 11 = 0 B . B' = A–1 (A' A) (A–1)'
(3) 2x + 9y + 7 = 0 (4) 4x + 7y + 3 = 0 = A–1 (A A') (A–1)1
Sol. [3] = (A–1 A) (A') (A')–1
Slope of PS = I
21 1
m = 
13 9 x1 y3 z4
2  38. The image of the line = =
2 2 3 1 5
in the plane 2x – y + z + 3 = 0 is the line :
2
m =  x3 y5 z2
9 (1) = =
3 1 5
Eqn of line
x3 y5 z2
2 (2) = =
y + 1 =  (x – 1) 3 1 5
9
9y + 9 = 2x + 2 x3 y5 z2
(3) = =
2x + 9y + 7 = 0 3 1 5
x3 y5 z2
(4) = =
3 1 5
36. The area of the region described by
A = {(x, y) : x2 + y2  1 and y2  1 – x} is : Sol. [2]
 2  4 x1 y3 z4
(1) + (2) +  
2 3 2 3 3 1 5
 4  2
(3) – (4) –
2 3 2 3
Sol. [2]
A = 2 1 1x dx
1 0
A = 2 02t2dt
1 1
1 x1 y3 z4 2343
A = 4 .    2
1 3 2 1 1 411
 4 x1 y3 z4
Area     2
2 3 2 1 1
x= 1 – 4, y = 3 + 2, z = 4 – 2
P' (–3, 5 ,2)
37. If A is an 3 × 3 non - singular matrix such
Line is || to the plane
that AA' = A'A and B = A–1 A', then BB' equals:
Equation of image line
(1) (B–1)' (2) I + B
(3) I (4) B–1 x3 y5 z2
 
3 1 5
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39. If x = – 1 and x = 2 are extreme points of Sol. [1]
f(x) =  log |x| + x2 + x then :
1 1
(1)  = 2,  = (2)  = – 6,  =
2 2
1 1
(3)  = – 6,  = – (4)  = 2,  = –
2 2
Sol. [4]
 C C = 1 + y
f'(x) = 2x1 1
x (1 – 0)2 + (1 – y)2 = (1 + y)2
at x = –1, 2  f'(x) = 0 1 + 1 – 2y + y = 1 + y2 + 2y
4y = 1
 – – 2+ 1 = 0 ....(1)
1
 y 
+ 4 + 1 = 0 .... (2) 4
2
1
 = 2,  =  42. If the coefficients of x3 and x4 in the
2 expansion of (1 + ax + bx2) (1 – 2x)18 in
powers of x are both zero, then (a, b) is
40. If a  R and the equation equal to :
– 3(x – [x])2 + 2(x – [x]) + a2 = 0
 272  251
(where [x] denotes the greatest integer  x) (1) 16,  (2) 16, 
has no integral solution, then all possible  3   3 
values of a lie in the interval :
 251  272
(1) (–, –2)  (2, ) (2) (–1, 0)  (0, 1) (3) 14,  (4) 14, 
(3) (1, 2) (4) (–2, –1)  3   3 
Sol. [2]
Sol. [1]
x – [x] = {x} = t  [0, 1) (1 + ax + bx2) (1 – 2x)18
–3t2+ 2t + a2 = 0 coeff. of x3 = 18C (–2)3 + a. 18C (–2)2
 a2= 3t2 – 2t  [0, 1) 3 + b . 18C (2–2)1
Since eqn cannot have integral 1
root : t  0
 a2  (0, 1) = 6528a61236b0
 a  (–1, 0)  (0, 1) coeff. of x4 = 18C (–2)4 + a . 18C (–2)3 + b 18C (–2)2
4 3 2
= 48960a.(6528)b6120
41. Let C be the circle with centre at (1, 1) and
272
radius = 1. If T is the circle centred at (0, We get (16, )
y), passing through origin and touching the 3
circle C externally, then the radius of T is
equal to : 43. If z is a complex number such that |z|  2,
(1) 1 (2) 3 then the minimum value of z 1 :
4 2 2
3 5
3 1 (1) is strictly greater than but less than
(3) (4) 2 2
2 2
5
(2) is equal to
2
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