Test Venue:
Lajpat Bhawan, Madhya Marg,
Sector 15-B, Chandigarh
Dr. Sangeeta Khanna Ph.D
1
CHEMISTRY COACHING CIRCLE
G:\Mega Scholarship Test-NON-MEDICAL.doc
Dr. Sangeeta Khanna Ph.D
READ THE INSTRUCTIONS CAREFULLY
1.
2.
3.
4.
5.
6.
7.
8.
The test is of 2 hours duration.
The maximum marks are 181.
This test consists of 60 questions.
For each question in Section – A you will be awarded 3 marks and Section – B will be awarded 4
marks for each question if you have darkened only the bubble corresponding to the correct answer &
zero mark if no bubbles are darkened. Minus one (-1) mark will be awarded for wrong answer
For each question in Section D (More than One Answer), you will be awarded 4 marks for each
correct answer. There is no negative mark awarded for incorrect answer(s) in this Section.
For each question in Section E (Matrix Match), you will be awarded 2 marks for each row in which
you have darkened the bubble(s) corresponding to the correct answer. Thus, each question in this
section carries a maximum of 8 marks. There is no negative marks awarded for incorrect answer(s)
in this Section.
For each question in Section F (Integer Type Answer), you will be awarded 4 marks for each correct
answer. There is no negative mark awarded for incorrect answer(s) in this Section.
Keep your mobiles switched off during Test in the Halls.
Section – A (Single Correct Choice Type)
This Section contains 31 multiple choice questions. Each question has four choices A), B), C) and D)
out of which ONLY ONE is correct. (Mark only One choice)
(31 × 3 = 93 Marks)
1.
Match the reaction (in List I) with the equivalent weight of underlined (in List II).
List I (Reaction)
2.
List II (Equivalent weight)
A
N2 + 3H2  2NH3
1
M
B
2H2 O 2  2H2O + O2
2
M/3
C
H3PO 2 H2PO 2 +H+ 3
M/2
D
MnO 4  Mn2+
M/5
4
A B C D
A B C D
A B C D
A B C D
a. 1 2 3 4
b. 2 3 1 4
c. 3 1 2 4
d. 4 1 2 3
B
H3BO3 on heating decomposes in two ways:
(I) H3BO3  HBO2 + H2O
(II) 2H3BO3  B2O3 + 3H2O
If 9 moles of H3BO3 is taken some part decomposed like (I) and remaining like l(II). If total 11 moles of
water are formed, the moles of B2O3 formed is:
a. 6
b. 5
c. 3
d. 2
D
Sol. Let x moles of H3BO3 decomposed in 1 way and (a – x) moles decomposed in II way so
H3BO 3  HBO 2  H2 O
x
x
2 H3BO 3  B 2 O 3  3H2 O
( a x )
3
(a x )
2
Total moles of water =
3
(a  x )  x  11
2
=x=5
 Moles of H3PO3 decompose in II reaction = 4
Dr. Sangeeta Khanna Ph.D
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CHEMISTRY COACHING CIRCLE
G:\Mega Scholarship Test-NON-MEDICAL.doc
Dr. Sangeeta Khanna Ph.D
3.
2 moles of H3BO3
1 mole of B2O3
4 moles of H3BO3
2 moles of B2O3
1.525 g of an organic compound was Kjeldahlised and the ammonia so produced was passed into
N
30 ml of 1 N HCl solution. The remaining HCl was further neutralized by 120 ml of
NaOH solution.
10
The percentage of nitrogen in the compound is:
a. 16.52%
A
b. 5.50 %
c. 0.5%
d. 20.4%
1

 30  1  120  
10  14  10
Sol. Percentage N = 

1000
1.525
= 16.52%
4. Match the column I with column II and mark the appropriate choice.
(A)
(B)
(C)
(D)
Column – I
State function
H = q
U = q
Intensive property
(i)
(ii)
(iii)
(iv)
Column – II
At constant pressure
Specific heat
Entropy
At constant volume
a. (A)  (iii), (B)  (i), (C)  (iv), (D)  (ii)
b. (A)  (ii), (B)  (iv), (C)  (i), (D)  (iii)
c. (A)  (ii), (B)  (iv), (C)  (iii), (D)  (i)
d. (A)  (iii), (B)  (ii), (C)  (i), (D)  (iv)
A
Sol. (A) Entropy is a state function.
(B) H = q at constant pressure.
(C) U = q at constant volume.
(D) Specific heat is an intensive property.
5. In Lassaigne’s test, which of the following organic compounds would produce a blood-red colour when
its sodium extract is treated with FeCl3 solution?
a. CH 3  C H  SO 3H
b. H2SO4
|
Cl
CH3
|
c. H2N
d. CH2  C H  SO 3H
CO2H
|
NH 2
6.
D
Which of the following order of stability of Intermediate is not correct
CH2
CH3
CH3
|
|

|
a. CH 3  C H  CH 3  C 

(Free radical)
CH3

b. C H2




C Cl 2
C F2 (Singlet Carbene)



c. CH3 – CH = CH - C H2 < C6H5 – CH = CH - C H2 < C6H5 – CH = CH - C H - CH3 (free radical)
CH3
CH3
|
|
|


d. CH 3  C   CH 3  C H  CH 3  CH 2  CH 2 (Carbocation)
CH3
D
Dr. Sangeeta Khanna Ph.D
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CHEMISTRY COACHING CIRCLE
G:\Mega Scholarship Test-NON-MEDICAL.doc
Dr. Sangeeta Khanna Ph.D
7.
Which of the following is correct order of stability of alkene?
a. CH3 – CH = CH2 <
<
b. CH3 – CH = CH2 <
<
c.
<
<
<
< CH3 – CH = CH2
<
< CH3 – CH = CH2 <
<
d.
B
Sol. As the number of  - H increases stability of alkene increases.
8. Temperature at which the CO2 has the same RMS velocity as that of H2 at STP?
a. 6600 K
b. 6060 K
c. 6006 K
d. 6.006 K
C
T
T
273 T2
Sol. 1  2
273 × 22 = T2 ; T2 = 6006 K

2
44
M1 M2
9. It takes 1.92  103 equivalents of KOH to neutralize 0.094 g of H2XO4. The atomic mass of X will be
(assuming H2XO4 as dibasic)
a. 8
b. 14
D
Sol. ngmeq KOH = ngmeq H2XO4
0.094
1.92 × 10–3 =
2
M.Wt
0.094  2
M.Wt 
1.92  10  3
= 98
At. Mass of X = 98 – 2 – 4 × 16
= 32
c. 18
C = CH – CH3
10. The IUPAC name of
d. 32
is
H5C2
a. 3-cyclopropyl-3-ethyl-2-propene
b. 1-cyclopropyl-1-ethypropene
c. 3-cyclopropyl-3-pentene
d. (1-ethyl-1-propenyl) cyclopropane
C
11. Which of the following substituted benzene derivatives would produce three isomeric products when
one more substituent is introduced?
Cl
Cl
Cl
Cl
Cl
Cl
I
II
III
a. I, II and III
b. I
D
12. The correct structure of trans-2-hexanal is
a.
c.
B
IV
c. II and IV
CHO
CHO
d.
4
d. I and III
CHO
b.
CHO
Dr. Sangeeta Khanna Ph.D
Cl
CHEMISTRY COACHING CIRCLE
G:\Mega Scholarship Test-NON-MEDICAL.doc
Dr. Sangeeta Khanna Ph.D
13. In which the following molecules -electron density in ring is maximum?
NO2
a.
O
NH2
b.
c.
OCH3
d.
B
Θ
Sol. O is better donor than – NH2 and – OCH3
14. What is the order of stability of N2 and its ions?
a. N2  N2  N2  N22
b. N2  N2  N2  N22 
c. N2  N2  N2  N22 
A
d. N22   N2  N2  N2
Sol. Bond order of N2 = 3, N2 = 2.5, N2 = 2.5 and N22  is 2. Higher the bond order, more is the stability.
H
H5C2
alc.KOH
CH3
15.
Br2/h
H
(P)
(A)
H
(CH3)3COK
H
(Q)
Which is correct statement for above reaction?
a. P and Q are constitutional isomers
c. P & Q both are not formed by same mechanism
A
CH3
CH3
alc.KOH
E2
Sol. C2H5 – C – CH3
Br
E2
(CH3)3COK
b. P and Q are geometrical isomers
d. All are correct
CH3 – CH = C – CH3
(P)
CH3
CH3 – C = CH2
(Q)
16. What volume of 5 M Na2SO4 must be added to 25 mL of 1 M BaCl2 to produce 10 g BaSO4? [M.Wt. of
BaSO4 = 233]
a. 8.58 mL
b. 7.2 mL
c. 10 mL
d. 12 mL
A
Sol. Na2SO4 + BaCl2  BaSO4 + 2NaCl
m
10

 0.0429
No. of moles of BaSO4 =
M 233
No. of moles of Na2SO4 needed = 0.0429
MV
5 V
0.0429 
or 0.0429 
1000
1000
V = 8.58 mL
17. In a reaction container, 100 g hydrogen and 100 g Cl2 are mixed for the formation of HCl gas. What is
the limiting reagent and how much HCl is formed in the reaction? [M.Wt. of Cl2 = 71]
a. H2 is limiting reagent and 36.5 g of HCl are formed.
b. Cl2 is limiting reagent and 102.8 g of HCl are formed.
c. H2 is limiting reagent and 142 g of HCl are formed.
d. Cl2 is limiting reagent and 73 g of HCl are formed.
B
Dr. Sangeeta Khanna Ph.D
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CHEMISTRY COACHING CIRCLE
G:\Mega Scholarship Test-NON-MEDICAL.doc
Dr. Sangeeta Khanna Ph.D
Sol. H2  Cl 2  2HCl
2g
71 g
73 g
2g H2 reacts with 71 g Cl1
71
100 g H2 will react with
×100 = 3550 g Cl2
2
Hence, Cl2 is the limiting reagent
71 g Cl2 produces 73 g HCl
73
100 g Cl2 will produce
 100  102.8 g HCl
71
18. Match the column I with column II and mark the appropriate choice.
Column – I
Mass of H2 produced when 0.5 mole of zinc reacts with (i)
excess of HCl.
(B) Mass of all atoms of a compound with formula C70H22
(ii)
(C) Number of molecules in 35.5 g of Cl2
(iii)
(D) Number of molecules in 64 g of SO2
(iv)
(A)
Column - II
3.01 × 1023 molecules
6.023 × 1023 molecules
1.43 × 10-21 g
1g
a. (A)  (ii), (B)  (i), (C)  (iv), (D)  (iii)
b. (A)  (i), (B)  (ii), (C)  (iii), (D)  (iv)
c. (A)  (iv), (B)  (iii), (C)  (i), (D)  (ii)
d. (A)  (iv), (B)  (iii), (C)  (ii), (D)  (i)
C
Sol. (A) : Zn + 2HCl  ZnCl2 + H2
1 mole of Zn produces 2 g of H2
0.5 mole of Zn will produce 1 g of H2
(B) : C70H22
Molar mass = 862
Mass of all atom is a molecule = 862/6.023 × 1023 = 1.43 × 10-21 g
(C) : 70 g of Cl2 = 6.023 × 1023 molecules
35.5 g of Cl2 = 3.01 × 1023 molecules
(D) : Molar mass of SO2 = 64 = 1 mole
64 g of SO2 = 6.023 × 1023 molecules
19. Strength of 10 volume hydrogen peroxide solution means
a. 30.35 g L-1
A
Sol. 2H2O2  2H2O 
68 g
b. 17 g L-1
c. 34 g L-1
d. 68 g L-1
O2
22.4 L at STP
22.4 L O2 is given by 68 g of H2O2
68
10 L O2 is given by
 10  30.35 g L-1
22.4
20. The mobilities of the alkali metal ions in aqueous solution are Li+ < Na+ < K+ < Rb+ < Cs+ because
a. greater is the degree of hydration larger the size of cation, lesser is the mobility in aqueous medium
b. larger the size of cation, greater is the mobility in aqueous medium
c. larger the size of cation, lesser is the mobility of ions in aqueous medium
d. lesser the degree of hydration, lesser is the mobility of ions in aqueous medium
A
Sol. Smaller the size of the ion, greater is the degree of hydration and hence lesser is the mobility in
aqueous medium.
X
21.
Residue + Colourless gas
heating
Z
excess of
CO2
water
Y
Dr. Sangeeta Khanna Ph.D
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CHEMISTRY COACHING CIRCLE
G:\Mega Scholarship Test-NON-MEDICAL.doc
Dr. Sangeeta Khanna Ph.D
Identify X, Y and Z
X
a. Ca(HCO3)2
c. CaCO3
B
CaCO3
(X)
Sol.
Y
CaCO3
CaO
heat
X
b. CaCO3
d. CaCO3
Y
Ca(OH)2
CaO
Z
Ca(HCO3)2
Ca(HCO3)2
CaO + CO2
H2 O
heating
Ca(HCO3)2
(Z)
Z
Ca(OH)2
Ca(OH)2
excess of
CO2
Ca(OH)2
(Y)
22. It is observed that H2 and He gases always show positive deviation from ideal behaviour i.e., Z > 1.
This is because
a. the value of a is very large due to high surface forces
b. the weak intermolecular forces of attraction due to which a is very small and a/V2 is negligible
c. the value of b is very large due to large size of the molecules
d. both a and b are very small and negligible
B
Sol. Due to weak intermolecular forces of attraction, H2 and He gases show the value of Z > 1
23. What are the oxidation states of phosphorus in the following compounds?
H3PO2, H3PO4, Mg2P2O7, PH3, HPO3
a. +1, +3, +3, +3, + 5
b. +3, +3, +5, +5, +5
c. +1, +2, +3, +5, +5
d. +1, +5, +5, -3, +5
D
Sol. H3PO2 : +3 + x – 4 = 0  x = +1
H3PO4 : + 3 + x – 8 = 0  x = +5
Mg2P2O7 : +4 + 2x – 14 = 0  x = +5
PH3 : x + 3 = 0  x = - 3
HPO3 : + 1 + X – 6 = 0  x = +5
24. The values of coefficients to balance the following reaction are
Cr(OH)3 + ClO– + OH–  CrO 24  + Cl– + H2O
a.
b.
c.
d.
D
Cr(OH)3
2
2
2
2
ClO–
3
4
4
3
CrO 24 
3
3
4
2
Cl–
3
2
2
3
Cr(OH)3 + 5OH–  CrO 24  + 4H2O + 3e- ] × 2
ClO– + H2O + 2e–  Cl– + 2OH–] × 3
2Cr(OH)3 + 4OH– + 3ClO–  2CrO 24  + 3Cl– + 5H2O
25. What in the pH of a solution obtained by mixing 10 mL of 0.1 M HCl and 40 mL of 0.2 M H 2SO4 1 ml
water.
Sol.
a. 0.74
b. 7.4
D
Sol. Millimoles of H+ from HCl = 0.1 × 10 = 1
Millimoles of H+ from H2SO4 = 0.2 × 40 × 2 = 16
Mil lim oles
16  1
17 1



Conc. of H+ =
Volume
40  10  1 51 3
pH = -log[H+] = -log 3 = 0.4771
c. 4.68
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CHEMISTRY COACHING CIRCLE
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d. 0.47
Dr. Sangeeta Khanna Ph.D
26. Solution of a monobasic acid has a pH = 5. If one mL of it is diluted to 1 litre, what will be the pH of the
resulting solution?
a. 3.45
B
Sol. pH = 5, [H+] = 10–5 M
b. 6.96
c. 8.58
d. 10.25
10 5
 10  8 M
1000
Total [H+] = 10-8 + 10–7 = 1.1 × 10–7
pH = -log [H+] = -log (1.1 × 10-7) = 6.96
27. For a reaction, 2SO2(g)+O2(g)
2SO3(g), 0.75 moles of SO2 and 1 mole of O2 are taken in 1 lit. vessel.
At equilibrium the concentration of SO3 was found to be 0.35 mol L–1. The Kc for the reaction would be
After dilution =
a. 0.93 L mol–1
A
Sol.
b. 1.4 L mol–1
2SO2(g) +
0.75 M
0.75 – 0.35
= 0.4
Initial conc.
At. Equi.
c. 0.6 L mol–1
O2(g)
1M
1 – 0.175
= 0.82
d. 2.95 L mol–1
2SO3(g)
0
0.35 mol/lit
= 0.35
(0.35) 2
0.1225

 0.9336
0.4  0.4  0.82 0.16  0.82
28. Graphs between pressure and volume are plotted at different temperatures. Which of the following
isotherms represents Boyle’s law as PV = constant?
KC 
T1
T2
T3
P
P
T1
T2
T3
1/V
logP
P
V
(i)
T3
T2
T1
PV
logV
(iii)
(ii)
(iv)
a. Only (ii) is correct representation of Boyle’s law.
b. Only (iv) is correct representation of Boyle’s law.
c. All are correct representations of Boyle’s law.
d. None of these representation is correct for Boyle’s law
C
29. Match the column I with column II and mark the appropriate choice.
Column – I
(A) Fe(OH)3
(B) Ag2CrO4
(C) CH3COOAg
(D) Ca3(PO4)2
Column – II
Ksp = s2
Ksp = 27s4
Ksp = 108s5
Ksp = 4s3
(i)
(ii)
(iii)
(iv)
a. (A)  (iii), (B)  (ii), (C)  (iv), (D)  (i)
c. (A)  (i), (B)  (iii), (C)  (ii), (D)  (iv)
B
Fe 3   3OH - ; K sp  27s 4
Sol. (A) Fe(OH)3
(B) Ag2CrO4
(3s)3
s
(C) CH3COOAg
b. (A)  (ii), (B)  (iv), (C)  (i), (D)  (iii)
d. (A)  (iv), (B)  (i), (C)  (iii), (D)  (ii)

( 2s)2

CH 3 COO  Ag ; K sp  4s
s
Dr. Sangeeta Khanna Ph.D
2Ag  CrO 24- ; K sp  4s3
3
(D) Ca3(PO4)2
3
( 3s )
s
8
3Ca
s
2
2PO 34  ; K sp  108s5
( 2s)2
CHEMISTRY COACHING CIRCLE
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Dr. Sangeeta Khanna Ph.D
30. Among the following in which pair, the second ion is more stable than first?
a.
b.
and
c.
CH2
and
and
d.
and
D
31. In which of the following, the order is not in accordance with the property mentioned.
a. Li < Na < K < Rb – Atomic radius
c. Si < P < S < Cl – Electronegativity
D
Sol. Electronegativity decreases down the group
b. F > N > O > C – Ionisation enthalpy
d. I < Br < F < Cl – Electronegativity
Section – C (Comprehension)
This Section contains 2 Comprehension. Each of these questions has four choices A), B), C) and D) out of
which ONLY ONE is correct.
6 × 4 = 24 Marks
Passage – 1
Orthoboric acid, H3BO3 reacts partially with water to form H3O+ and B(OH) 4 .H3BO3 cannot be titrated
satisfactorily with NaOH, as a sharp end point is not obtained. If any Cis-diol or glycerol is added to the
titration mixture, then H3BO3 behave as a strong acid. It can now be titrated with NaOH, and the end
point is detected using phenolphthalein as indicator.
1.
Orthoboric acid
2.
a. behaves as a monoprotic acid
b. behaves as a strong monoprotic acid
c. is a weak triprotic acid
d. accepts OH– ion in water
D
The product formed when H3BO3 reacts with NaOH is
3.
a. NaH2BO3
b. Na3BO3
c. Na[B(OH)4]
C
Which of the following statement is incorrect about orthoboric acid
a.
b.
c.
d.
C
d. Na2B3O7
Orthoboric acid contains triangular BO 33  units
In the solid the B(OH)3 units are hydrogen bonded together into two-dimensional sheets.
When heated to 100°C, orthoboric acid forms its anhydride Boron sequioxide B2O3.
It is formed when borax is neutralized by HCl(aq)
100
Sol. H3BO3 

HBO 2
Metaboric acid
Re d
 B 2O3
hot
Paragraph -2
Ozonolysis is one of the important method of determination of olefinic bonds in a compound. It is a
cleavage reaction in which double bond is completely broken and alkene molecule is converted into
two or more smaller molecules depending on the number of olefinic bonds with a carbonyl group.
Ozonolysis is carried out in two stages. First is addition of ozone to give ozonide second step involve
hydrolysis of ozonide in presence of reducing agent (usually zinc) to give product of reductive
ozonolysis. In presence of zinc it yields aldehydes and ketones while presence of Ag 2O, H2O or
peracids form acids and/ or ketones. Further ozonides of alkynes yield carboxylic acid by cleavage
through diketone which is oxidized. Acetylene give however mixture of glyoxal as well as formic acid.
Dr. Sangeeta Khanna Ph.D
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Dr. Sangeeta Khanna Ph.D
4.
An organic compound C6H8 (M.F.) on reductive ozonolysis given 2 moles of CHOCH2CHO. The
possible structure will be
CH3
CH2
a.
b.
c. H3C
d.
H3C
5.
A
An alkene on ozonolysis produces glyoxal (CHO – CHO) and methanal. The probable structure of
alkene is
a. CH3 – CH = CH – CH = CH2
c. CH3 – C  C – CH3
B
b. CH2 = CH – CH = CH2
d. CH2 = CH – C  CH
6.
CH3
On oxidative ozonolysis (assume benzene ring not affected) the above compound produces
CHO
|
a. Benzoic acid and 2 moles of
CO 2H
CHO
|
b. Benzoic acid,
and CH3CO2H
CO 2H
CO 2H
|
c. Benzoic acid and 2 moles of
CO 2H
CO 2H
|
d. Benzoic acid,
and CH3CO2H
CO 2H
D
SECTION – D (More than One Answer Type)
This Section contains 5 multiple choice questions. Each question has four choices A), B), C) and D)
out of which ONE OR MORE may be correct.
5 × 4 = 20 Marks
1.
In which of the following molecules are all the carbon atoms in sp2 hybridization state?
2.
a. Graphite
c. C60-Fullerene
A,B,C
Entropy increases in case of the reaction:
b. 1, 3, 5-Hexatriene
d. Diamond
a. N2(g) + 3H2(g)  2NH3 (g)
b. 2HI (g)  H2(g) + I2(g)
c. AgNO3 (aq) + NaCl (aq)  AgCl (s) + H2O () d. CaCO3(s)  CaO (s) + CO2 (g)
B,D
Sol. (a) No. of moles decreases. Hence, entropy decreases.
(b) No of substances increases. Hence, entropy increases
(c) Reactants give large number of ions in the solution. As the product AgCl is solid and H2O is almost
unionized, the no. of ions decreases. Hence, entropy decreases.
(d) As one of the products is gaseous, entropy increases
3. Among the following, which are tautomers?
O
||
OH
|
b. CH3 – CH2 – C  N and CH2 = CH – CN
d. CH3CH2COOH & HCOOCH2CH3
a. NH 2  C NH 2 and NH 2  C  NH
c. CH3 – CHO and CH2 = CH – OH
A,C
Dr. Sangeeta Khanna Ph.D
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Dr. Sangeeta Khanna Ph.D
4.
Which of the following compounds have electron deficiency at ortho and para position?
O
O
NH – C – CH3
a.
OH
O
O
N
b.
NH2
O=S=O
c.
d.
B,C,D
O
||
Sol.  NO 2 , - C- NH 2 , -SO3H are electron withdrawing group
5. Which of the following process have value by equilibrium constant. Keq. > 1.
CH3
R
c.
R
R
C=C
H
H
C=C
H
H
H
H
R
H3C
CH3
b.
H
CH3
H
O
O
O
c. CH3 – C – H
d.
CH2 = CH – OH
A,B,D
Sol. Product is more stable than reactant, equilibrium will be more shifted to formed & K will be greater than
one Kequ. > 1.
SECTION – E (Matrix Type)
This Section contains 3 questions. Each question has four choices (A, B, C and D) given in Column I and
five statements (p, q, r, & s) in Column II. (More than one Match may be Answer). No negative Marking
8 × 3 = 24 Marks
1.
Match the following: (Single Match)
Column I
Column II
(A)
Cyclic process
(p) T = 0
(B)
Isothermal
(q) P = 0
(c)
Isobaric
(r) S = 0
(d)
Adibatic
(s) Q = 0
Sol. A – p, q, r; B – p, C – q, D – s ,r
In isothermal process, T  0
In isobaric process, P  0
In adiabatic process, q = 0
In cyclic process; T = 0, S = 0, E = 0, P = 0
2. Match the following: (More than One Match)
Column I (Reaction)
Column II (Characteristic)
OH
|
A
(p) Intermediate is carbocation
H O
CH 3  C H  CH  CH 2 2 CH 3  C  CH 2  CH 3
|
CH3
B
H
|
CH3
HBr
CH 3  C H  CH  CH 2  CH 3  C H  CH 2  CH2
|
CH3
Dr. Sangeeta Khanna Ph.D
Peroxide
|
(q)
Rearrangement
|
CH3
11
Br
CHEMISTRY COACHING CIRCLE
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Dr. Sangeeta Khanna Ph.D
C
2RCOONa (aq) 
R  R  H2  2CO 2  NaOH
electroly sis
(r)
Intermediate is free radical
(s)
Markovnikoff;s Rule followed
CH3
CH – CH3
D
And.
+ CH3 – CH2 – CH2 – Cl 
+ HCl
AlCl3
Sol. A - (p, q, s), B – (r); (c) – (r), D – (p, q)
3. Match Column – I with Column – II. (More than One Match)
Column (I)
Column (II)

(i)
CH 3  C H  CH 3
(P) Pyramidal structure
Θ
(b) C H3
(Q) Planar geometry

(c) CH 3  C H  CH 3
(R) Electrophile
(d) Singlet carbene
(S) Nucleophile
Sol. A  Q, R; B  P, S; C  P, R; D  Q, R
Section – F (Integer Type)
This Section contains 5 questions. The answer to each question is a single digit integer ranging
from 0 to 10. The correct digit below the question number in the OMR is to be bubbled. No negative
Marking.
5 × 4 = 20 Marks
1. The number of B – O – B bonds in borax is
Sol. 5
2. How many structural isomers (acylic) of compound with molecular formula C 6H12 shows geometrical
isomerisms?
Sol. 4
2-Hexene, 3-Hexene; 3-Methyl-2-pentene; H3 C  HC  CH  C H  CH 3 ; 4-methyl-2-pentene
|
CH3
C  C  C C  C
|
C
3. How many of following compound have sp3 hybridised central atom.
NH3, H2O, H2S, SF4, PCl3, SO3; NCl3; XeO3; XeF2 ; IF7
Sol. 6
NH3; H2O, H2S, PCl3, NCl3, XeO3
4. For a reaction NH4COONH2(s)
2NH3(g) + CO2(g), the equilibrium pressure is 3 atm. Kp for the
reaction will be
Sol. 4
NH4COONH2(s)  2NH 3( g)  CO 2(g)
3P = 3;
P=1
2p
p
Kp = (2 × 1)2 (1) = 4
5. An orbital is described with the help of a wave function. Since many wave functions are possible for an
electron, there are many atomic orbitals. When atom is placed in a magnetic field the possible number
of orientations for an orbital of azimuthal quantum number 3 is
Sol. 7
When  = 3, magnetic quantum number has 7 values m = (2 + 1). These values are represented as
- 3, -2, -1, 0, +1, +2, +3
Dr. Sangeeta Khanna Ph.D
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