AIEEE/2011/CHEMSITRY
S. No
Questions
1
Solutions
Q.1
the maximum covalent character is shown by the compound :
(a) SnCl2
(b) AlCl3
(c) MgCl2
(d) FeCl2
Sol: 1 (b)
Greater charge and small size of cation cause more polarization and more covalent
character
Q.2
The presence or absence of hydroxyl group on which carbon atom of sugar differentiates
RNA and DNA?
(a) 2nd
(b) 3rd
(c) 4th
(d) 1st
CCI3 CHO was subjected to Cannizzaro’s reaction by using NaOH. The mixture of the
products contains CCI3 COONa and another compound. The other compound is :
(a) Trichloromethanol
(b) 2, 2, 2-Trichloropropanol
(c) Chloroform
(d) 2, 2, 2-Trichloroethanol
Sol: 2 (a)
RNA → β – D – Ribose. DNA → β–D–2– deoxy Ribose
Q.4
C2 H5 ONa has reacted with CH3 COCl . The compound that is produced in the above
reaction is :
(a) 2-Butanone
(b) Ethyl chloride
(c) Ethyl ethanoate
(d) Diethyl ether
Sol: 4 (c)
Q.5
The reduction potential of hydrogen half cell will be negative if :
Sol: 5 (b)
Q.3
OH (−)
CCI3 COONa+CCI3 CH2 OH
Sol: 3 (d)
2CCI3 CHO
IUPAC Name is 2,2,2 – Trichloro ethanol
(a) p(H2) = 1 atm and [H+] = 1.0 M
(b) p(H2) = 2 atm and [H+] = 1.0 M
(c) p(H2) = 2 atm and [H+] = 2.0 M
(d) p(H2) = 1 atm and [H+] = 2.0 M
Q.6
The strongest acid amongst the following compounds is :
(a) HCOOH
(b) CH3CH2CH(CI)CO2H
(c)CICH2CH2CH2COOH
(d) CH3COOH
Sol: 6 (b)
Electron releasing groups (R - group) de stabilizes conjugate base.
The +I effect of C3 H7 is less than - I effect of Cl
Q.7
The degree of dissociation (α) of a weak electrolyte, AXBY is related to van’t Hoff factor
(i) by the expression :
Sol: 7 (d)
i−1
(a) a=x+y+1
x+y−1
(b) a=
i−1
x+y+1
(c) a=
i−1
i−1
(d) a=x+y−1
Ax By →xA+y +yB −X
n = x+y
i−1
So 𝛼 = x+y−1
Q.8
`a’ and `b’ are van der Waals’ constants for gases. Chlorine is more easily liquefied than
ethane because
(a) a and b for Cl2< a and b for C2H6
(b) a for Cl2< a for C2H6 but b for Cl2> b for C2H6
(c) a for Cl2> a for C2H6 but b for Cl2< b for C2H6
(d) a and b for Cl2 > a and b for C2H6
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Sol:8 (c)
a
ease of liquefaction ∝
b
∵ 𝐶2 𝐻5 (𝑎 = 5.49, 𝑏 = 0.0658)
C𝐼2 (𝑎 = 6.49, 𝑏 = 0.0562)
AIEEE/2011/CHEMSITRY
Q.9
Q.10
A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is
converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8
atm, the value of K is
(a) 3 atm
(b) 0.3 atm
(c) 0.18 atm
(d) 1.8 atm
Sol: 9 (d)
Boron cannot form which one of the following anions?
Sol: 10 (d)
B = 1𝑆 2 2𝑆 2 2𝑃1 (valence shell (2S, 2𝑃𝑥 , 2𝑃𝑦 ,2𝑃𝑍 )
−
(a) BH 4
Q.11
−
(b) B(OH) 4
−
(c) BO 2
3−
(d) BF
6
Which of the following facts about the complex [Cr(NH3)6]CI3 wrong ?
(a) The complex is paramagnetic
CO2 (g) + C
2
P
co
P co 2
Kp =
0.6×0.6
2CO(g)
= 1.8 atm
0.2
a maximum valency of four only. So (BF6 )3− not possible.
Sol: 11 (b)
Cr NH3
(b) The complex is an outer orbital complex
=
2
6
Cl3 involves d2 sp3 hybridization
(c) The complex gives white precipitate with silver nitrate solution
(d) The complex involves d2sp3 hybridization and is octahedral in shape.
Q.12
Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which
should be added to 4 kg of water to prevent it from freezing at −6oC will be :
[Kf for water = 1.86 K kg mol −1 , and molar mass of ethylene glycol = 62g mol −1 )
(a) 204.30g
Q.13
(b) 400.00 g
(c) 304.60 g
(d) 804.32g
Which one of the following order represents the correct sequence of the increasing basic
nature of the given oxides?
Sol: 12 (d)
W ×1000
∆Tf = K f × W2 ×m
1
2
𝑊1 & 𝑊2 = of solvent solute respecting
Sol: 13 (d)
Q.14
(a) Ng2O<K2O<MgO<AI2O3
(c) K2O<Na2O<AI2O3<MgO
(d) AI2O3<MgO<Na2O<K2O
The rate of a chemical reaction doubles for every 10oC rise of temperature. If the
temperature is raised by 50oC, the rate of the reaction increases by about :
(a) 24 times
(b) 32 times
(c) 64 times
(d) 10 times
Sol: 14 (b)
Q.15
The magnetic moment (spin only) of [NiCl4]2 – is
(a) 5.46 BM
Q.16
(b) 2.82 BM
(c) 1.41 BM
(d) 1.82 BM
−
+
+
The hybridization of orbitals of N atom in NO 3 , NO and NH are respectively :
2
4
(a) sp2, sp, sp3
(c) sp2, sp3, sp
(b) sp, sp3, sp2
(d) sp, sp2, sp3
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∆T
Temperature coefficient μ =2; μ 10 =
So 2
k
= 25 = 32 k 2
1
k2
k1
;
Therefore 32 k1 = k 2
Sol: 15 (b)
μ = n(n + 2) BM (∵ n = 2 for (Ni CI4 )2− )
= 2 2 + 2 = 2.82BM
Sol: 16 (a)
4000 ×62
Metallic strength
decreases
group
Increases (Metallic strength)
50
10
W 2 ×1000
period
In a periodic table
(a) MgO<K2O<AI2O3<Na2O
∆𝑇𝑓 = 00 − (-60 ) = 6 = 1.86 ×
𝑚2 = mw of solute
∴ W2 = 800g
AIEEE/2011/CHEMSITRY
Q.17
Q.18
In context of the lanthanoids, which of the following statements is not correct?
Sol: 17 (c)
(a) All the members exhibit +3 oxidation state
(b) Because of similar properties the separation of lanthanoids is
not easy.
(c) Availability of 4f electrons results in the formation of
compounds in +4 state for all the members of the series.
(d) There is a gradual decrease in the radii of the members with
increasing atomic number in the series.
Lanthanides general O.S = + 3, some of elements show only + 4 O.S.
A 5.2 molal aqueous solution of methyl alcohol, CH3 OH , is supplied. What is the mole
fraction of methyl alcohol in the solution?
Sol: 18 (b)
𝑋2 𝑎𝑞 =
(a) 0.190
Q.19
Q.20
Q.21
Q.22
Q.23
(b) 0.086
(c) 0.050
𝑚
𝑚+
1000
18
=
5.2
1000
18
5.2+
= 0.09
(d) 0.100
Which of the following statement is wrong?
Sol: 19 (d)
(a) Nitrogen cannot form dπ - pπ bond.
(b) Single N- N bond is weaker than the single P – P bond,
(c) N2O4 has two resonance structures
(d) The stability of hydrides increases fromNH3 to BiH3 in group
15 of the periodic table
The outer electron configuration of Gd (Atomic No : 64 is :
(a) 4f8 5d0 6s2
(b) 4f 4 5d4 6s2
7
1
2
(c) 4f 5d 6s
(d) 4f3 4d5 6s2
Which of the following statements regarding sulphur is incorrect?
(a) The vapour at 200oC consists mostly of S8 rings
(b) At 600oC the gas mainly consists of S2 molecules
(c) The oxidation state of sulphur is never less than +4 in its compounds
(d) S2 molecule is paramagnetic.
N
P
As
Sb
Bi
The structure of IF7 is :
(a) trigonal bipyramid
(b) octahedral
(c) pentagonal bipyramid
(d) square pyramid
Ozonolysis of an organic compound gives formaldehyde as one of the products. This
confirms the presence of :
(a) a vinyl group
Sol: 22 (c)
In 7 IF , I undergoes sp3d3 hybridisation
(b) an isopropyl group
(c) an acetylenic triple bond
(d) two ethylenic double bonds
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M – H bond energy decreases
Sol: 20 (c)
Sol: 21 (c)
S = 1𝑆 2 2𝑆 2 2𝑃6 3𝑆 2 3𝑃4 valence shell
So S( minimum O.S = 2)
Sol: 23 (a)
By ozonolysis of C𝐻2 = 𝐶𝐻 − 𝑔𝑖𝑣𝑒 𝑓𝑜𝑟𝑚𝑎𝑙𝑑𝑒𝑕𝑦𝑑𝑒.
3
AIEEE/2011/CHEMSITRY
Q.24
A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emissions
is at 680 nm, the other is at :
(a) 325 nm
(b) 743 nm
(c) 518 nm
(d) 1035 nm
4
Sol: 24 (b)
1
1
1
=λ +λ
λ
1
absorbed
1
2
1
1
⇒ 355 = 680 +λ
2
⇒ λ2 = 742.8 ≅ 743nm
Q.25
Silver Mirror test is given by which one of the following compounds?
(a)Acetone
(b) Formaldehyde
(c)Benzophenone
(d) Acetaldehyde
Sol: 25 (b, d)
The compound having – CHO group oxidized by tollen’s reagent and give silver
mirror.
Q.26
Which of the following reagents may be used to distinguish between phenol and benzoic
acid ?
(a) Tollen’s reagent
(b) Molisch reagent
(c) Neutral Fe Cl3
(d) Aqueous NaOH
Q.27
Phenol is heated with a solution of mixture of KBr and KBrO3. The major product
obtained in the above reaction is
(a) 3-Bromophenol
(b) 4-Bromophenol
Sol: 26 (c)
(neutral)
FeCI3
𝐶6 𝐻5 𝑂𝐻
violet coloured complex
(neutral)
FeCI3
𝐶6 𝐻5 𝐶𝐻2 𝑂𝐻
pale dull yellow ppt.
Sol: 27 (c)
In acidic medium, KBr + KBrO3 in turn produces Br 2 . Phenol reacts with Br2 (aq) to
give 2, 4, 6-trinitrophenol
(c) 2, 4, 6- Tribromophenol
Q.28
(d) 2-Bromophenol
In a FCC lattice, atom A occupies the corner positions and atom B occupies the face
centre positions. If one atom of B is missing from one of the face centred points, the
formula of the compound is:
(a) AB2
(b) A2 B3
(c) A2B5
(d) A2B
Sol: 28 (c)
1
for FCC ( corner group) effective no. (A) = 8 × 8 = 1
The entropy change involved in the isothermal reversible expansion of 2 moles of an
ideal gas from a volume of 10 dm3 to a volume of 100 dm3 at 27oC is :
Sol: 29 (d)
For an ideal gas, for isothermal reversible process,
(a) 35.8J mol −1K−1
(c) 42.3J mol−1K−1
∆S = 2.303 nR log
1
5
where as for B (center atom) = = 2 × 5 = = 2
So formula = A1 B5 or A2 B5
2
Q.29
(b) 32.3J mol−1K−1
(d) 38.3J mol−1K−1
V2
V1
= 2.303×2×8.314×log
−1
= 38.3 J mol
Q.30
.k
100
10
−1
Identify the compound that exhibits tautomerism.
Sol: 30 b, (b,c)
(a) Lactic acid
(c) Phenol
both 2-pentanone, phenol can exhibit tautomerism
(b) 2-Pentanone
(d) 2- Butene
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