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NEET UG 2013, Chemistry Question Paper
(Date of Examination: 05-05-2013)
Class
XI NEET
Unit No.
I
II
III
IV
V
VI
VII
VIII
IX
X
XI
XII
XIII
XIV
Unit name
Some Basic Concepts of Chemistry
Structure of Atom
Classification of Elements and Periodicity in Properties
Chemical Bonding and Molecular Structure
States of Matter: Gases and Liquids
Thermodynamics
Equilibrium
Redox Reactions
Hydrogen
s-Block Elements (Alkali and Alkaline earth metals)
Some p-Block elements
Organic Chemistry-Some Basic Principles and
Techniques
Hydrocarbons:
Alkanes
Alkenes
Alkynes
Aromatic Hydrocarbons
Environmental Chemistry
Question No.
46, 47, 52
58, 62, 77, 85
56
68, 71
73
63, 64, 75
74, 81
82, 83, 88, 90
Total No. of
Questions
0
03
0
04
01
0
02
0
01
0
03
02
0
0
0
04
0
Total = 20
I
II
III
IV
V
VI
XII NEET
VII
VIII
IX
X
XI
Solid State
Solutions
Electrochemistry
Chemical Kinetics
Surface Chemistry
General Principles and Processes of Isolation of
Elements
p-Block Elements
d and f -Block Elements
Co-ordination Compounds
Haloalkanes and Haloarenes
Alcohols Phenols and Ethers Alcohols
55, 57
54, 84
49, 51, 53
48, 50
02
02
03
02
0
0
60, 61, 66, 69
67, 72, 76
59, 70
04
03
02
0
0
0
01
Phenols
XII
Aldehydes, Ketones and
Carboxylic Acids
Ethers
Aldehydes Ketones
79
65, 89
Carboxylic Acids
XIII
XIV
XV
XVI
Organic Compounds Containing Nitrogen
Biomolecules
Polymers
Chemistry in Everyday life
02
0
87
80, 86
78
01
0
02
01
Total = 25
Total = 45
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The value of Planck’s constant is 6.63 × 10−34 Js. The speed of light is 3 × 1017 nm s−1. Which value is
closest to the wavelength in nanometer of a quantum of light with frequency of 6 × 1015 s−1?
(C) 50
(D) 75
(A) 10
(B) 25
Ans: (C)
Solution:
Given:
Speed of light, c = 3 × 1017 nm s−1
Frequency, ν = 6 × 1015 s−1
Planck’s constant = 6.63 × 10−34 Js.
To find:
Wavelength, λ = ? nm
c
Formula:
ν=
λ
Calculation: From formula,
46.
∴
λ=
3×1017 nm s −1
= 50 nm
6 ×1015 s −1
47.
What is the maximum numbers of electrons that can be associated with the following set of quantum
numbers?
n = 3, l = 1 and m = −1
(A) 1
(B) 6
(C) 4
(D) 2
Ans: (D)
Solution:
An orbital can accommodate a maximum of 2 electrons.
48.
What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to
35°C ? (R = 8.314 J mol–1 K–1)
(A) 342 kJ mol–1
(B) 269 kJ mol–1
(C) 34.7 kJ mol–1
(D) 15.1 kJ mol–1
Ans: (C)
Solution:
Given:
Initial temperature, T1 = 20° C = (20 + 273)K = 293 K
Final temperature, T2 = 35° C = (35 + 273)K = 308 K
Final rate of the reaction (K2) = 2 × Initial rate of the reaction (K1)
Gas constant R = 8.314 Jmol−1 K−1
To find:
Activation energy, Ea = ?
Formula:
log
E a ⎡ T2 − T1 ⎤
K2
=
⎢
⎥
K1 2.303R ⎣ T1 .T2 ⎦
Calculation: From formula,
log 2 =
Ea
2.303 × 8.314
0.3010 =
Ea =
⎡ 308 − 293 ⎤
⎢
⎥
⎣ 293× 308 ⎦
E a ×15
2.303× 8.314 ×10−3 × 293× 308
0.3010 × 2.303× 8.314 ×10−3 × 293× 308
15
= 34.67 kJ mol−1 ≈ 34.7 kJ mol−1
2
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49.
A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing
hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be
(A) 0.059 V
(B) 0.59 V
(C) 0.118 V
(D) 1.18 V
Ans: (B)
Solution:
Given:
pH of the HCl solution = 10
To find:
oxidation potential of electrode, Ecell = ?
0.059
1
log −10 = + 0.59 V
Calculation: Ecell =
1
10
50.
A reaction having equal energies of activation for forward and reverse reactions has
(A) ∆S = 0
(B) ∆G = 0
(C) ∆H = 0
(D) ∆H = ∆G = ∆S = 0
Ans: (C)
Solution:
When Energy of activation for forward reaction = Energy of activation for reverse reaction
∆G = 0, ∆H = 0 and
∴
∆S = 0
At 25°C molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 ohm–1 cm2 mol–1
and at infinite dilution its molar conductance is 238 ohm–1 cm2 mol–1. The degree of ionisation of
ammonium hydroxide at the same concentration and temperature is
(A) 2.080%
(B) 20.800%
(C) 4.008%
(D) 40.800%
Ans: (C)
Solution:
Molar conductance of 0.1 molar aqueous solution of ammonium hydroxide at 25°C,
Given:
λm = 9.54 ohm−1 cm2 mol−1
Molar conductance of 0.1 molar aqueous solution of ammonium hydroxide at infinite dilution,
λ ∞m = 238 ohm−1 cm2 mol−1
To find:
Degree of ionization of ammonium hydroxide at same concentration and temperature = ?
λ
Formula:
Degree of ionization = ∞m × 100
λm
Calculation: From formula,
λ
Degree of ionization = ∞m × 100
λm
9.54 ×100
=
238
= 4.008%
51.
⎛ Z2 ⎞
Based on equation E = –2.178 × 10–18 J ⎜ 2 ⎟ certain conclusions are written. Which of them is NOT
⎝n ⎠
CORRECT?
(A) The negative sign in equation simply means that the energy of electron bound to the nucleus is lower
than it would be if the electrons were at the infinite distance from the nucleus
(B) Larger the value of n, the larger is the orbit radius
(C) Equation can be used to calculate the change in energy when the electron changes orbit
(D) For n = 1, the electron has a more negative energy than it does for n = 6 which means that the electron
is more loosely bound in the smallest allowed orbit
Ans: (D)
Solution:
In the smallest allowed orbit, electron is strongly attracted by the nucleus.
52.
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NEET UG 2013 Chemistry Question Paper
A button cell used in watches functions as following 
2+
−
+ 2OH (aq)
Zn(s) + Ag2O(s) + H2O(l) ⎯→ 2Ag(s) + Zn (aq)
If half cell potentials are
2+
Zn (aq)
+ 2e– ⎯→ Zn(s) ; E° = –0.76 V
−
, E° = 0.34 V
Ag2O(s) + H2O(l) + 2e– ⎯→ 2Ag(s) + 2OH (aq)
The cell potential will be
(A) 1.10 V
(B)
(C) 0.84 V
(D)
Ans: (A)
Solution:
E 0cathode = 0.34 V
Given:
0.42 V
1.34 V
E 0anode = − 0.76 V
To find:
E 0cell = ?
Formula:
E 0cell = E 0cathode − E 0anode
Calculation: From formula,
E 0cell = E 0cathode − E 0anode
= 0.34 − (−0.76) = 1.1 V
54.
How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M HNO3?
The concentrated acid is 70% HNO3.
(A) 45.0 g conc. HNO3
(B) 90.0 g conc. HNO3
(C) 70.0 g conc. HNO3
(D) 54.0 g conc. HNO3
Ans: (A)
Solution:
Given:
M of conc. HNO3 = 2.0
V of conc. HNO3 = 250 mL.
250 × 2
M × V = No. of Moles of HNO3 =
= 0.5
1000
To find:
Grams of concentrated nitric acid = ?
100
Calculation: HNO3 required = 0.5 × 63 ×
= 45 g
70
55.
The number of carbon atoms per unit cell of diamond unit cell is
(A) 4
(B) 8
(C) 6
Ans: (B)
Solution:
Diamond has ZnS type structure i.e. 8 carbon atoms per unit cell.
(D)
1
56.
Maximum deviation from ideal gas is expected from
(B) N2(g)
(C) CH4(g)
(4) NH3(g)
(A) H2(g)
Ans: (D)
Solution:
Among the given gases, NH3 has the highest Tc value and thereby shows maximum deviation from ideal gas
behaviour.
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A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm–3.
The molar mass of the metal is (NA Avogadro's constant = 6.02 × 1023 mol–1)
(B) 30 g mol–1
(C) 27 g mol–1
(D) 20 g mol–1
(A) 40 g mol–1
Ans: (C)
Solution:
Given:
Metal has fcc lattice, ∴ Z = 4
The density of the metal, d = 2.72 g cm−3
The edge length of the unit cell, = 404 pm.
Volume, v = l3 = (4.04 × 10−8)3 m
Avogadro’s constant, NA = 6.02 × 1023 mol−1
To find:
The molar mass of the metal, M = ?
Z× M
Formula:
d=
V × NA
Calculation: From formula,
Z× M
d=
V × NA
4× M
2.72 =
3
( 4.04×10−8 ) × 6.02×1023
57.
M=
2.72 × (4.04)3 × 6.02 ×10−1
= 27 g/mol.
4
58.
Dipole-induced dipole interactions are present in which of the following pairs?
(A) H2O and alcohol
(B) Cl2 and CCl4
(C) HCl and He atoms
(D) SiF4 and He atoms
Ans: (C)
δ+
δ−
Solution:
H − Cl ...... ⊕
∴
He
HCl is polar and He is non-polar
They show dipole –induced dipole interaction.
59.
A magnetic moment of 1.73 BM will be shown by one among the following
(B) [Ni(CN)4]2–
(A) [Cu(NH3)4]2+
(D) [CoCl6]4–
(C) TiCl4
Ans: (A)
Solution:
Magnetic moment (µ) = n (n + 2)
1.73 = n (n + 2)
n=1
So, compound must contain one unpaired electron.
∴
The compound is [Cu(NH3)4]2+.
60.
Roasting of sulphides gives the gas X as a byproduct. This is a colorless gas with choking smell of burnt
sulphur and causes great damage to respiratory organs as a result of acid rain. Its aqueous solution is acidic
acts as a reducing agent and its acid has never been isolated. The gas X is
(B) SO2
(A) H2S
(D) SO3
(C) CO2
Ans: (B)
Solution:
The gas is SO2.
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NEET UG 2013 Chemistry Question Paper
Which is the strongest acid in the following?
(B) HClO3
(C) HClO4
(D) H2SO3
(A) H2SO4
Ans: (C)
Solution:
Among the oxo acids, higher the oxidation state of central atom, stronger is the acid.
In case of HClO4, Cl has the maximum oxidation state of +7. Thus HClO4 is the strongest acid.
62. Which of the following is paramagnetic?
(C) CN−
(A) CO
(B) O −2
Ans: (B)
Solution:
O −2 has one unpaired electron, in the antibonding molecular orbital.
∴
It is paramagnetic.
(D)
NO+
63.
Which of the following structure is similar to graphite?
(D) B2H6
(A) BN
(B) B
(C) B4C
Ans: (A)
Solution:
Boron nitride has structure same as that of graphite consisting of sheets made up of hexagonal rings of
alternate B and N atoms joined together.
64.
The basic structural unit of silicates is
(B) SiO 44 −
(A) SiO–
Ans: (B)
Solution:
Basic structural unit of silicates is SiO 44 −
65.
(C)
SiO32 −
(D)
SiO 24 −
Reaction by which Benzaldehyde CANNOT be prepared?
(A)
CH3
+ CrO2Cl2 in CS2 followed by H3O+
COCl
(B)
(C)
+ H2 in presence of Pd-BaSO4
+ CO + HCl in presence of anhydrous AlCl3
COOH
(D)
+ Zn/Hg and conc. HCl
Ans: (D)
Solution:
Zn/Hg and con. HCl cannot reduce −COOH group to −CHO
66.
Which of the following does NOT give oxygen on heating?
(B) Zn(ClO3)2
(A) KClO3
(D) (NH4)2Cr2O7
(C) K2Cr2O7
Ans: (D)
Solution:
On heating (NH4)2Cr2O7, the products formed are N2, Cr2O3 and H2O. Oxygen is not formed in this
reaction.
∆
(NH4)2Cr2O7 ⎯⎯
→ N2(↑) + Cr2O3 + 4H2O
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Which of the following lanthanoid ions is diamagnetic?
(At. nos. Ce = 58, Sm = 62, Eu = 63, Yb = 70)
(B) Sm2+
(C) Eu2+
(A) Ce2+
Ans: (D)
Solution:
Yb2+ has an electronic configuration of 4f14.
Since there is absence of unpaired electron, it is diamagnetic in nature.
W
67.
(D)
Yb2+
(D)
BaCl2
68.
Identify the CORRECT order of solubility in aqueous medium
(A) CuS > ZnS > Na2S
(B) ZnS > Na2S > CuS
(C) Na2S > CuS > ZnS
(D) Na2S > ZnS > CuS
Ans: (D)
Solution:
Solubility product values decreases in the order Na2S > ZnS > CuS.
69.
XeF2 is isostructural with
(B)
(A) TeF2
ICl−2
(C)
SbCl3
Ans: (B)
Solution:
XeF2 is isostrucutral with ICl−2 as both of them possess linear geometry.
70.
An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium(III) chloride.
The number of moles of AgCl precipitated would be
(A) 0.001
(B) 0.002
(C) 0.003
(D) 0.01
Ans: (A)
Solution:
[Cr(H2O)4Cl2]Cl has one ionizable Cl−.
∴
10−3 moles of the complex precipitates the same amount of AgCl.
71.
Which of these is least likely to act as a Lewis base?
(C)
(A) CO
(B) F–
Ans: (C)
Solution:
BF3 is a Lewis acid as it is electron deficient.
72.
BF3
(D)
PF3
(D)
SO2
(D)
PH3
KMnO4 can be prepared from K2MnO4 as per the reaction:
3MnO 24 − + 2H2O ⎯→ 2MnO −4 + MnO2 + 4OH−
The reaction can go to completion by removing OH– ions by adding
(A) HCl
(B) KOH
(C) CO2
Ans: (C)
Solution:
3K2MnO4 + 2CO2 ⎯→ 2KMnO4 + MnO2 ↓ 2K2CO3
73.
Which of the following is electron-deficient?
(B) (SiH3)2
(A) (CH3)2
Ans: (C)
Solution:
(BH3)2, Diborane is electron deficient.
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Structure of the compound whose IUPAC name is 3-Ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid is
OH
OH
COOH
(A)
(B)
COOH
(D)
COOH
OH
COOH
(C)
OH
Ans: (B)
OH
Solution:
3
6
5
2
1
COOH
4
3-Ethyl-2-hydroxy-4-methylhex-3-en-5ynoic acid
75.
Which of these is NOT a monomer for a high molecular mass silicone polymer?
(B) Me2SiCl2
(C) Me3SiCl
(D) PhSiCl3
(A) MeSiCl3
Ans: (C)
Solution:
(Me)3SiCl is not a monomer for a high molecular mass silicone polymer
76.
Which of the following statements about the interstitial compounds is INCORRECT?
(A) They retain metallic conductivity
(B) They are chemically reactive
(C) They are much harder than the pure metal
(D) They have higher melting points than the pure metal
Ans: (B)
Solution:
Interstitial compounds are chemically inert.
Which one of the following molecules contains no π bond?
(A) CO2
(B) H2O
(C) SO2
(D) NO2
Ans: (B)
Solution:
H2O molecule consists of two O − H σ bonds and two lone pair of electrons. There is no π bond present.
H
77.
O
H
78.
Antiseptics and disinfectants either kill or prevent growth of microorganisms. Identify which of the
following statements is NOT TRUE.
(A) A 0.2% solution of phenol is an antiseptic while 1% solution acts as a disinfectant
(B) Chlorine and Iodine are used as strong disinfectants
(C) Dilute solutions of Boric acid and Hydrogen Peroxide are strong antiseptics
(D) Disinfectants harm the living tissues
Ans: (C)
Solution:
Dilute solutions of boric acid and H2O2 are mild antiseptics.
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Among the following ethers, which one will produce methyl alcohol on treatment with hot concentrated HI?
(A) CH3 − CH2 − CH2 − CH2 − O − CH3
(B) CH3 − CH2 − CH − O − CH3
|
CH3
(C)
(D)
Ans: (C)
Solution:
CH3
|
CH3 − C − O − CH3
|
CH3
CH3 − CH − CH2 − O − CH3
|
CH3
CH3
CH3
|
|
CH3 − C − O − CH3 + hot conc. HI ⎯→ CH3OH + CH3 − C − I
|
|
CH3
Methanol
CH3
80.
Nylon is an example of
(A) Polyester
(C) Polyamide
Ans: (C)
Solution:
Nylons are polyamides
81.
(B)
(D)
Polysaccharide
Polythene
The structure of isobutyl group in an organic compound is
CH3
CH − CH2 −
(A)
CH3
(B)
CH3 − CH − CH2 − CH3
|
(C)
CH3 − CH2 − CH2 − CH2 −
(D)
CH3
|
CH3 − C −
|
CH3
Ans: (A)
Solution:
(CH3)2CH − CH2 − group is known as isobutyl group
82.
Nitrobenzene on reaction with conc. HNO3/H2SO4 at 80-100°C forms which one of the following products?
(A) 1, 2-Dinitrobenzene
(B) 1, 3-Dinitrobenzene
(C) 1, 4-Dinitrobenzene
(D) 1, 2, 4-Trinitrobenzene
Ans: (B)
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Solution:
−NO2 group is a meta directing group.
∴
Nitrobenzene on reaction with conc.HNO3/H2SO4 (i.e. Nitrating mixture) at 80-100°C forms
1,3-Dinitrobenzene.
NO2
NO2
cons.HNO3 /H 2SO4
⎯⎯⎯⎯⎯⎯
→
Nitrobenzene
NO2
1, 3-Dinitrobenzene
83.
Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating?
(A) − C ≡ N
(B) − SO3H
(C) − COOH
(D) − NO2
Ans: (D)
Solution:
−NO2 group is more deactivating than the other groups given.
6.02 × 1020 molecules of urea are present in 100 mL of its solution. The concentration of solution is
(A) 0.02 M
(B) 0.01 M
(C) 0.001 M
(D) 0.1 M
Ans: (B)
Solution:
Given:
6.02 × 1020 molecules of urea are present in 100 mL of its solution
To find:
The concentration of solution = ?
Calculation: 1 mole of urea contains 6.02 × 1023 molecules
⎛ 6.02 ×1020 ⎞
∴
6.02 × 1020 molecules of urea contains = ⎜
moles
23 ⎟
⎝ 6.02 ×10 ⎠
84.
∴
⎛ 6.02 ×1020 ⎞
moles of urea are present in 100 mL of solution.
⎜
23 ⎟
⎝ 6.02 ×10 ⎠
Concentration of the solution is measured in terms of 1dm3 i.e. 1000 mL
⎛ 6.02 ×1020 ⎞
⎜
⎟
6.02 ×1023 ⎠
= 0.01 M
Concentration = ⎝
⎛ 100 ⎞
⎜
⎟
⎝ 1000 ⎠
85.
Which of the following is a polar molecule?
(B) SF4
(C) SiF4
(D) XeF4
(A) BF3
Ans: (B)
Solution:
Due to the presence of a lone pair of electron on "S", SF4 has a distorted geometry. It is see-saw shaped and
has a resultant dipole moment. Therefore SF4 is a polar molecule.
86.
Which is the monomer of Neoprene in the following?
(A) CH2 = CH − C ≡ CH
(B) CH2 = C − CH = CH2
|
CH3
(C) CH2 = C − CH = CH2
|
Cl
(D) CH2 = CH − CH = CH2
Ans: (C)
Solution:
Chloroprene (CH2 = C − CH = CH2)
|
Cl
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NO2
87.
W
NO2
In the reaction
A
⎯⎯
→
Br
+
N2Cl
A is
(A)
(C)
Ans: (C)
Solution:
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HgSO4/H2SO4
H3PO2 and H2O
Br
−
(B)
(D)
NO2
Cu2Cl2
H+/H2O
NO2
H3PO2 / H 2 O
⎯⎯⎯⎯⎯
→
Br
+
N2Cl
Br
−
H3PO2 and H2O replaces N2Cl group by hydrogen.
88.
The radical
(A)
(B)
(C)
(D)
Ans: (A)
Solution:
89.
CH2 is aromatic because it has
6 p-orbitals and 6 unpaired electrons
7 p-orbitals and 6 unpaired electrons
7 p-orbitals and 7 unpaired electrons
6 p-orbitals and 7 unpaired electrons
6p orbitals and 6 unpaired electrons contributes to aromaticity.
The order of stability of the following tautomeric compounds is
OH
O
CH2 = C − CH2 − C − CH3
I
O
O
CH3 − C − CH2 − C − CH3
II
OH
O
CH3 − C = CH − C − CH3
III
(A)
(C)
Ans: (B)
Solution:
I > II > III
II > I > III
(B)
(D)
III > II > I
II > III > I
III is stabilised by intramolecular hydrogen bonding.
90.
Which of the following compounds will NOT undergo Friedal-Craft's reaction easily?
(A) Cumene
(B) Xylene
(C) Nitrobenzene
(D) Toluene
Ans: (C)
Solution:
Nitrobenzene does not undergo Friedel-Crafts reaction due to the strong deactivating effect of
−NO2 group.
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