CHEMISTRY
JEE(Advanced) 2014 Final Exam/Paper-1/Code-8
PART - II : CHEMISTRY
SECTION–1 : (One or More Than One Options Correct Type)
This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and
(D) out of which ONE or MORE THAN ONE are correct.
21.
For the reaction
I¯ + ClO¯
+ H2SO4 ® Cl¯ + HSO¯
+ I2
3
4
The correct statement(s) in the balanced equation is / are :
(A) Stoichiometric coefficient of HSO¯
is 6
4
(B) Iodide is oxidized
(C) Sulphur is reduced
22.
A
LL
EN
(D) H2O is one of the products
Ans. (A, B, D)
Sol. Oxidation half reaction :
2I– ® I2 + 2e–
...... (1)
Reduction half reaction
6H+ + ClO3– + 6e– ® Cl– + 3H2O ........(2)
Multiplying equation (1) by 3 and add in (2)
6I¯ + ClO¯
+ 6H+ ® Cl¯ + 3I2 + 3H2O
3
6I¯ + ClO¯
+ 6H2SO4 ® Cl¯ + 3I2 + 3H2O + 6HSO¯
3
4
The pair(s) of reagents that yield paramagnetic species is / are :
(A) Na and excess of NH3
(B) K and excess of O2
(C) Cu and dilute HNO3
(D) O2 and 2-ethylanthraquinol
Ans. (A,B,C) / (B,C)
Sol. If ammonia considered as a gas then reaction will be :
1
2
(A) Na + NH3 ¾¾
® NaNH 2 + H 2
(excess)
1
2
( NaNH2 + H 2 are diamagnetic)
If ammonia considered as a liquid then reaction will be
M + (x+y)NH3 ® [M(NH3)x]+ + [e(NH3)y]–
ammoniated e–
blue colour
paramagnetic
S.R.A.
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JEE(Advanced) 2014 Final Exam/Paper-1/Code-8
+
(B) K + O2
(excess)
Paramagnetic
(C) 3Cu + 8HNO3 ¾¾
® 3Cu(NO3)2
+
Para magne tic
2NO
+
4H2O
Paramagnetic
O
OH
Et
(D)
23.
–
¾¾
® KO2 (K , O2 )
Et
O2
+ H2 O2
O
OH
In the reaction shown below, the major product(s) formed is / are :
NH2
acetic anhydride
NH2 ¾¾¾¾¾
¾
® product(s)
CH 2 Cl2
A
LL
EN
O
H
N
NH2
CH3
O
NH2 + CH3COOH
(A)
H
N
(B)
O
O
H
N
Å
CH3 + CH3COOH
O
1
NH3CH3COO
CH3
O
(C)
H
N
O
Ans. (A)
CH3 + H2O
H
N
(D)
O
O
CH3
O
O
CH2 NH C CH3
CH2 NH2
O
O
+ CH3 C O C
CH3
CH 2Cl2
C NH2
C NH2
O
O
–CH2–NH2 is more nucleophilic then
C–NH2
O
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24.
CHEMISTRY
In a galvanic cell , the salt bridge (A) Does not participate chemically in the cell reaction
(B) Stops the diffusion of ions from one electrode to another
(C) Is necessary for the occurence of the cell reaction
(D) Ensures mixing of the two electrolytic solutions
Ans. (A,B)
Note : We feel option (C) is incorrect because in some type of concentration cells, salt bridge is not required.
Which can be confirmed from NCERT (XII-Chemistry, Part-1) in Sub section 3.2 Galvanic Cell.
"The electrolytes of the two half-cells are connected internally through a salt bridge as shown in Fig.
3.1. Sometimes, both the electrodes dip in the same electrolyte solution and in such cases we do not
require a salt bridge. "
25.
Upon heating with Cu2S, the reagent(s) that give copper metal is/are
(A) CuFeS2
(B) CuO
(C) Cu2O
(D) CuSO4
Ans. (C)
26.
A
LL
EN
D
Sol. Cu2S + 2Cu2O ¾¾
® 6 Cu + SO2
Hydrogen bonding plays a central role in the following phenomena
(A) Ice floats in water
(B) Higher Lewis basicity of primary amines than tertiary amines in aqueous solutions
(C) Formic acid is more acidic than acetic acid
(D) Dimerisation of acetic acid in benzene
Ans. (A,B,D)
Sol. Hint
Þ Ice floats in water due to the low density of ice as compare to water which is due to open cage
like structure (formed by intermolecular H-bonding)
Þ Dimerisation of acetic acid in benzene is due to intermolecular hydrogen bonding
O
H3C—C
O
27.
H
O
C—CH3
H
O
Þ Basic strength of RNH2 > R3N it also explained by hydrogen bonding.
The reactivity of compound Z with different halogens under appropriate conditions is given belowOH
Mono halo substituted derivative when X2=l2
X2
C(CH3)3
di halo substituted derivative when X2=Br2
tri halo substituted derivative when X2=Cl2
The observed pattern of electrophilic substitution can be explained by (A) The steric effect of the halogen
(B) The steric effect of the tert-butyl group
(C) The elctronic effect of the phenolic group
(D) The electronic effect of the turt-butyl group
Ans. (A, B, C)
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Sol.
OH
I
X2 = I2
C(CH 3)3
OH
OH
Sol.
X2 = Br2
Br
C(CH3)3
Br
OH
Cl
C(CH 3)3
Cl
X2 = Cl2
C(CH 3)3
Cl
28.
A
LL
EN
Orientation in electrophilic substition reaction is decided by
(A) The steric effect of the halogen
(B) The steric effect of the tert-butyl group
(C) The electronic effect of the phenolic group
The correct combination of names for isomeric alcohols with molecular formula C4H10O is/are(A) tert-butanol and 2-methylpropan-2-ol
(B) tert-butanol and 1, 1-dimethylethan-1-ol
(C) n-butanol and butan-1-ol
(D) isobutyl alcohol and 2-methylpropan-1-ol
Ans. (A,B,C,D)
Sol. The combination of names for isomeric alcohols with molecular formula C4H10O is/are
Formula
Names
CH3CH2CH2CH2OH
n-butyl alcohol / n-butanol / butan-1-ol
CH3–CH–CH2–OH
isobutyl alcohol / 2-methyl propan-1-ol
CH3
CH3–CH2–CH–OH
Secondary butyl alcohol / butan-2-ol
CH3
CH3
CH3–C–OH
CH3
Tertiary butyl alcohol / tert butanol / 2-methyl propan-2-ol
/ 1,1-dimethyl ethan-1-ol
Reference : National Institute of standards and technology (NIST)
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29.
An ideal gas in thermally insulated vessel at internal pressure = P1, volume = V1 and absolute temperature
= T1 expands irrversibly against zero external pressure, as shown in the diagram. The final internal
pressure, volume and absolute temperature of the gas are P2, V2 and T2, respectively. For this expansion,
Pext=0
Pext=0
Irreversible
P1,V1,T1
P2,V2,T2
Thermal insulation
30.
A
LL
EN
(A) q = 0
(B) T2 = T1
(C) P2V2 = P1V1
(D) P2V2g = P1V1g
Ans. (A,B,C)
Sol. Process is adiabatic
q=0
Pext = 0
w=0
DU = q + w = 0
The change in internal energy of an ideal gas depends only on temperature therefore DT = 0
hence process is isothermal therefore
T2 = T1 & P2V2 = P1V1
D is incorrect, it is valid for adiabatic reversible process
The correct statements(s) for orthoboric acid is/are-
(A) It behaves as a weak acid in water due to self ionization
(B) Acidity of its aqueous solution increses upon addition of ethylene glycol
(C) It has a three dimensional structure due to hydrogen bonding.
(D) It is a weak electrolyte in water
Ans. (D)
Sol. (A) It does not self ionized in water and ionized in water as follows
–
+
ˆ ˆ†
H3BO3 + H2O ‡
ˆˆ B(OH) 4 + H
(B) Acidity of the aq.solution of boric acid not affected by ethylene glycol
(C) In boric acid due to hydrogen bonding two dimensional sheet structure is formed.
(D) In water the pKa value of H3BO3 is 9.25
–
+
ˆ ˆ†
H3BO3 + H2O ‡
ˆˆ B(OH) 4 + H
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CHEMISTRY
SECTION–2 : (One Integer Value Correct Type)
This section contains 10 questions. Each question, when worked out will result in one integer from
0 to 9 (both inclusive).
1
2
31.
In an atom, the total number of electrons having quantum numbers n = 4, |ml| = 1 and ms = – is
Ans. (6)
Sol. For n = 4, orbitals are
–1 0 +1
0
–2 –1 0 +1 +2
–3 –2 –1 0 +1 +2 +3
Total number of orbitals having {|ml| = 1} = 6
1
2
Total number of electrons having {|ml| = 1 and ms = - } = 6
32.
Ans. (1)
Sol.
A
LL
EN
The total number of distinct naturally occurring amino acids obtained by complete acidic hydrolysis
of the peptide shown below is
O
H
H
N
H
O
O
O
O
N
N
N
N
N
N
O
O
N
CH2 O
H CH2
H
H
O
O
O
H
O
N
N
O
NH
CH2
C
H
O
N
NH
O
O
N
H
N
H
N
CH2 O
O
H3 O
+
CMe3
|
+
+
H3N–CH2–CO¯2 + O2¯C—CH—NH3
(B)
(A)
+
CO¯2
H3 N
HO–C–CH–NH–CH2–C–OH
CH2
(C)
+
O CH2
O
(D)
A Þ is glycine which is only naturally occuring amino acid.
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33.
If the value of Avogadro number is 6.023 × 1023 mol–1 and the value of Boltzmann constant is
1.380 × 10–23 JK–1, then the number of significant digits in the calculated value of the universal gas
constant is
Ans. (4)
Sol. Universal gas constant R = kNA
where k= Boltzman constant and NA = Avogadro number
\ R = 1.380 × 10–23 × 6.023 × 1023 J/K-mole
= 8.31174
@ 8.312
So significant figures = 4
34.
A compound H 2X with molar weight of 80 g is dissolved in a solvent having density of
0.4 g mol–1, Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is
A
LL
EN
Ans. (8)
Sol. Molarity = 3.2 M
Let volume of solution = 1000 ml = volume of solvent
Mass of solvent = 1000 × 0.4 = 400 gm
nsolute = 3.2 mole
3.2
=8
Molality (m) = 400
1000
35.
MX2 dissociates into M2+ and X– ions in an aqueous solution, with a degree of dissociation (a) of
0.5. The ratio of the observed depression of freezing point of the aqueous solution to the value of
the depression of freezing point in the absence of ionic dissociation is
Ans. (2)
Sol. MX2 ¾® M2+ + 2X–
( DTf )observed
( DTf )Theoretical = i = 1 + a (n–1)
\
i = 1 + 0.5 (3–1)
i=2
36.
Consider the following list of reagents :
Acidifeid K2Cr2O7, alkaline KMnO4, CuSO4, H2O2 , Cl2, O3, FeCl3, HNO3 and Na2S2O3.
The total number of reagents that can oxidise aqueous iodide to iodine is
Ans. (7)
Sol. Acidified K2Cr2O7, CuSO4, H2O2, Cl2, O3 FeCl3, HNO3 oxidise aq. iodide to iodine.
Alkaline KMnO4 oxidise aq. iodide to IO3–
No reaction between iodide & Na2S2O3
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37.
The total number(s) of stable conformers with non-zero dipole moment for the following compound
is (are)
Cl
Br
Br
CH3
Cl
CH3
Ans. (3)
Cl
Sol. Br
Br
CH3
Cl
Br
= Br
Cl
CH 3
CH3
Cl
CH3
Cl
Br
Cl
Br
=
H3C CH
3
A
LL
EN
Stable conformer (with m ¹ 0)
Cl
Br
CH3
CH3
Cl
Br
Cl
Br
Cl
Br
Br
Cl
Cl
CH3
CH3
(Me-Me) gauche
38.
Br
CH3
(Br-Me) gauche
CH3
(Cl-Me) gauche
Among PbS, CuS, HgS, MnS, Ag2S, NiS, CoS, Bi2S3 , and SnS2 the total number of BLACK coloured
sulphides is
Ans. (6) / (7)
Sol. PbS, CuS, HgS, Ag2S, NiS, CoS are black
MnS – dirty pink/Buff
SnS2 – yellow
Bi2S3 – brown / black (brownish black)
39.
Consider all possible isomeric ketones including stereoisomers of MW = 100, All these isomers are
independently reacted with NaBH4 (NOTE : stereoisomers are also reacted separately). The total number
of ketones that give a racemic product(s) is/are
Ans. (5)
Sol. M. wt 100 of ketone
So m. formula = C6H12O
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(1) n-Butyl –C–CH3
O
(2) Isobutyl –C–CH3
OH
Isobutyl –CH–CH3
NaBH4
O
(3) 3º butyl –C–CH3
all are
± racemic mixture
OH
3ºbutyl–CH–CH3
O
CHEMISTRY
n-butyl CH–CH3
OH
CH3
CH 3
(4) CH3–CH2–CH–C–CH3
CH 3–CH2–CH–CH–CH 3
O
R(2ºbutyl)
CH3
NaBH4
(5) CH3–CH2–CH–C–CH3
O
OH
2-alcohols (R,R) & (R,S)
diastereomeric pair
CH3–CH2–CH–CH–CH3
S(2ºbutyl)
A
LL
EN
CH 3 OH
2-alcohols (S,S) & (S,R)
diastereomeric pair
(6) n-propyl –C–Et
O
(7) Isopropyl –C–Et
O
NaBH4
n-propyl –CH–Et
– racemic
OH (+)
mixture
Isopropyl CH–Et
(+) racemic
OH –
mixture
(1 ; 2 ; 3 ; 6 ; 7)
40.
A list of species having the formula XZ4 is given below :
XeF4, SF4, SiF4, BF4–, BrF4–, [Cu(NH3)4]2+, [FeCl4]2–, [CoCl4]2– and [PtCl4]2–.
Defining shape on the basis of the location of X and Z atoms, the total number of species having
a square planar shape is
Ans. (4)
Sol. XeF4, BrF4– , [Cu(NH3)4]2+ , [PtCl4]2– are square planar
SF4 – Sea saw
SiF4 , BF4–, [FeCl4]2–, [CoCl4]2– are tetrahedral
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