Test Venue:
Lajpat Bhawan, Madhya Marg,
Sector 15-B, Chandigarh
Dr. Sangeeta Khanna Ph.D
1
CHEMISTRY COACHING CIRCLE
G:\Mega Scholarship Test - MEDICAL.doc
Dr. Sangeeta Khanna Ph.D
READ INSTRUCTIONS CAREFULLY
1. Please mention your Name, Roll No. on the OMR Answer Sheet.
2. Mention/Fill bubbles carefully on your OMR Answer Sheet – Booklet – A, Booklet – B,
Booklet – C or Booklet – DB as mentioned on your Question Paper. Otherwise your answer
sheet will not get properly evaluated.
3. The test is of 2 hours duration.
4. This test consists of 60 questions.
5. The maximum marks are 180
6. For each question in Section A, B & C you will be awarded 3 marks if you have darkened only
the bubble corresponding to the correct answer and zero mark if no bubbles are darkened.
Minus one (-1) mark will be awarded for wrong answer
7. 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.
Which of the following statements is incorrect about internal energy?
a.
b.
c.
The absolute value of internal energy cannot be determined
The internal energy of one mole of a substance is same at any temperature or pressure
The measurement of heat change during a reaction by bomb calorimeter is equal to the internal
energy change
Internal energy is an extensive property
d.
B
Sol. Internal energy of a substance depends upon temperature and pressure.
2. For combustion of 1 mole of benzene at 25°C, the heat of reaction at constant pressure is – 780.9 kcal.
What will be the heat of reaction at constant volume?
1
C6H6(( )  7 O2(g)  6CO 2(g)  3H2O( )
2
a. – 781.8 kcal
b. – 780.0 kcal
B
Sol. qp = qV + ngRT, qv=qp-ngRT (H=E + ngRT)
qv = -780.9 – (- 1.5 × 2 × 298 × 10-3) = - 780 kcal
3. G in the process of melting of ice at -10 °C is
c. 781.8 kcal
d. 780.0 kcal
a. > 0
b. < 0
c. = 0
d. None
A
Sol. Melting of ice is non-spontaneous at 10°C, so G = +ve
4. In an experiment, NO2 gas is prepared and taken into 3 test tubes X, Y and Z. NO 2 gas which is brown
in colour dimerises into N2O4 which is colourless. Test tube X is kept at room temperature, Y is kept in
ice and Z is kept in hot water. What colour changes will you observe in the test tubes and why?
2NO 2(g)
N2O 4(g) ; H  -57.2 kJ mol -1
Brown
Colourless
Ice
NO2
Hot water
NO2
X
Dr. Sangeeta Khanna Ph.D
Y
Z
2
CHEMISTRY COACHING CIRCLE
G:\Mega Scholarship Test - MEDICAL.doc
Dr. Sangeeta Khanna Ph.D
a.
b.
c.
d.
In test tube X, brown colour intensifies since backward reaction is favoured at low temperature
In test tube Y, brown colour intensifies since backward reaction takes place at room temperature
In test tube Z, brown colour intensifies since high temperature favours the backward reaction.
Brown colour of test tubes X, Y and Z remains same since there is no effect of change in
temperature on the reaction
C
Sol. It is an exothermic reaction, hence the forward reaction is favoured at low temperature which means
colourless N2O4 will be formed resulting in decrease in intensity of brown colour. High temperature
favours backward reaction resulting in formation of NO2, thus intensifying brown colour.
5. 0.05 mole of NaOH is added to 5 litres of water. What will be the pH of the solution?
a. 12
A
b. 7
d. 10
0.05
= 0.01 or 10–2 M
5
Sol. Conc of NaOH =
6.
c. 2
[OH–] = 10–2 M
Kw
10 14
[H ] 

 10 12

2
[OH ] 10
pH = - log [H+] = - log 10–12 = 12
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
7.
D
Which of the following compounds has wrong IUPAC name?
a. CH3 – CH2 – CH2 – COO – CH2CH3  Ethyl butanoate
b. CH 3  C H  CH 2  CHO  3-Methyl-butanal
|
CH 3
c. CH 3  CH  C H  CH 3  2-Methyl-3-butanol
|
|
OH
CH3
O
||
d. CH 3  C H  C  CH 2  CH 3  2-Methyl-3-pentanone
|
CH3
C
Sol. The correct name is 3-methyl-2-butanol
8. A mixture of C2H6, C2H4 and C2H2 is bubbled through alkaline solution of copper (I) chloride, contained
in Woulf’s bottle. The gas coming out is
9.
a. original mixture
b. C2H6
c. C2H6 and C2H4 mixture
d. C2H4 and C2H2
C
Which of the following property of hydrogen does not resemble to alkali metals?
a. Unipositive ion
c. Ionization energy
C
Dr. Sangeeta Khanna Ph.D
b. Formation of oxides
d. Formation of sulphides
3
CHEMISTRY COACHING CIRCLE
G:\Mega Scholarship Test - MEDICAL.doc
Dr. Sangeeta Khanna Ph.D
10. If the solubility of salt A2B3 is x mol L – 1, then its solubility product is equal to:
a. 12 x3
b. 108x3
c. 36 x5
d. 108x5
D
Sol. A2B3
2A+3
+ 3B–2
x
2x
3x
Ksp = (2x)2 (3x)3 = 108 x5
11. The ratio of area of orbit of first excited state of electron to the area of orbit of ground level, for
hydrogen atom, will be:
a. 16 : 1
b. 4 : 1
A
Sol. For first excited state, n = 2
A 2 r22 n 24 16



A 1 r12 n14
1
c. 8 : 1
d. 2 : 1
n2


 0.529 Å 
r 
Z


12. Diborane has
a. 2 two - centre – two – electron bonds and 4 three – centre – three – electron bonds.
b. 4 two – centre – two – electron bonds and 2 three – centre – two – electron bonds
c. 3 two – centre – two – electron bonds and 3 three – centre – three – electron bonds
d. all six identical bonds.
B
Hb
Ht
Sol.
Ht
Ht
B
or
B
Ht
B
Ht
Ht
Hb
97°
1.33Å
1.19Å
Hb
Ht
B
121.5°
Ht
Hb
1.77Å
Thus the diborane molecule has four two-centre-two-electron bonds (2c – 2e bonds) also called usual
bonds and two three-centre-two-electron bonds (3c – 2e) also called banana bonds. Hydrogen
attached to usual and banana bonds are called Ht (terminal H) and Hb (bridged H) respectively.
13. Which of the following is correct order of stability of alkene?
a. CH3 – CH = CH2 <
<
b. CH3 – CH = CH2 <
c.
<
<
<
<
< CH – CH = CH
3
2
<
< CH3 – CH = CH2 <
<
d.
B
Sol. As the number of  - H increases stability of alkene increases.
14. Temperature at which the CO2 has the same RMS velocity as that of H2 at STP?
a. 6600 K
C
T
T
Sol. 1  2
M1 M2
b. 6060 K
c. 6006 K
d. 6.006 K
273 T2

2
44
273 × 22 = T2 ;
Dr. Sangeeta Khanna Ph.D
T2 = 6006 K
4
CHEMISTRY COACHING CIRCLE
G:\Mega Scholarship Test - MEDICAL.doc
Dr. Sangeeta Khanna Ph.D
15. A liquid (v.p. = 100 m Hg) is transferred from a vessel of 10 ml capacity to 1000 ml capacity, hence its
vapour pressure
1
th times
10
d. Remain as such
a. increases 10 times
b. Decreases
c. Increases 100 times
D
16. Which of the following substituted benzene derivatives would produce three isomeric products when
one more substituent is introduced?
Cl
Cl
Cl
Cl
Cl
Cl
Cl
I
II
III
IV
a. I, II and III
b. I
c. II and IV
D
17. The most stable conformation of 2, 3-dimethylbutane is
CH3
CH3
CH3
H
CH3
H
CH3
H
CH3
a.
d. I and III
b.
H
H3C
CH3
H
CH3
H3C
c.
H
CH3
CH3
d.
CH3
H3C
H
H
B
18. In which the following molecules -electron density in ring is maximum?
NO2
a.
O
NH2
b.
CH3
CH3
OCH3
c.
d.
B
Θ
Sol. O is better donor than – NH2 and – OCH3
CH3
|
H SO
4
19. CH 3  C  CH 2  OH 2

|
P
; P is
(Major )
CH3
CH3
CH3
a. CH 3  C  CH  CH 3
CH3
b. CH 2  C  CH 2  CH 3
|
|
|
c. CH 3  C  CH 2
A
20. What is the order of stability of N2 and its ions?
d. none
b. N2  N2  N2  N22 
d. N22   N2  N2  N2
a. N2  N2  N2  N22
c. N2  N2  N2  N22 
A
Sol. Bond order of N2 = 3, N2 = 2.5, N2 = 2.5 and N22  is 2. Higher the bond order, more is the stability.
Dr. Sangeeta Khanna Ph.D
5
CHEMISTRY COACHING CIRCLE
G:\Mega Scholarship Test - MEDICAL.doc
Dr. Sangeeta Khanna Ph.D
21. In which of the following, the order is not in accordance with the property mentioned.
a. Li < Na < K < Rb – Atomic radius
b. F > N > O > C – Ionisation enthalpy
c. Si < P < S < Cl – Electronegativity
d. I < Br < F < Cl – Electronegativity
D
Sol. Electronegativity decreases down the group.
22. If the velocity of an electron in Bohr’s first orbit is 2.19 × 106 m s-1, what will be the de Broglie
wavelength associated with it?
a. 2.19 × 10-6 m
C
b. 4.38 × 10-6 m
c. 3.32 × 10-10 m
d. 3.32 × 1010 m
h
6.626  10 34 J s

m 9.11  10  31kg  2.19  10 6 m s -1
= 3.32 × 10-10 m
23. 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:
Sol.  
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%
24. 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
No. of moles of BaSO4 =

 0.0429
M 233
No. of moles of Na2SO4 needed = 0.0429
MV
5 V
0.0429 
or 0.0429 
1000
1000
V = 8.58 mL
25. 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?
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
Sol. H2  Cl 2  2HCl
2g
71 g
73 g
2g H2 reacts with 71 g Cl2
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  102.8 g HCl
100 g Cl2 will produce
71
Dr. Sangeeta Khanna Ph.D
6
CHEMISTRY COACHING CIRCLE
G:\Mega Scholarship Test - MEDICAL.doc
Dr. Sangeeta Khanna Ph.D
26. 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
X
27.
Residue + Colourless gas
heating
water
excess of
Z
CO2
Y
Identify X, Y and Z
X
a. Ca(HCO3)2
c. CaCO3
B
Sol.
CaCO3
(X)
Y
CaCO3
CaO
heat
(Z)
X
b. CaCO3
d. CaCO3
Y
Ca(OH)2
CaO
Z
Ca(HCO3)2
Ca(HCO3)2
CaO + CO2
H2O
heating
Ca(HCO3)2
Z
Ca(OH)2
Ca(OH)2
excess of
CO2
Ca(OH)2
(Y)
28. 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
29. A solution containing Mn2+, Fe2+, Zn2+ and Hg2+ with a molar concentration of 10–3 M each is treated
with 10-16 M sulphide ion solution. Which ions will precipitate first if Ksp of MnS, FeS, ZnS and HgS are
10-19, 10-23, 10-20 and 10-54 respectively?
a. FeS
b. MnS
c. HgS
C
Sol. M2+ + S2–  MS
Lower the value Ksp, lower will be solubility. Hence HgS will precipitate first.
Dr. Sangeeta Khanna Ph.D
7
d. ZnS
CHEMISTRY COACHING CIRCLE
G:\Mega Scholarship Test - MEDICAL.doc
Dr. Sangeeta Khanna Ph.D
30. What will be the solubility of AgCl in 0.05 M NaCl aqueous solution if solubility product of AgCl is
1.5 × 10-10?
a. 3 × 10–9 mol L–1
A
Sol. Ksp = [Ag+] [Cl– ]
[Cl–] = NaCl = 0.05 M
b. 0.05 mol L–1
c. 1.5 × 10–5 mol L–1
d. 3 × 109 mol L–1
1.5  10 10
 3  10  9 M
0.05
[Ag+] = solubility = 3 × 10–9 M
31. 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?
Ag 
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
32. Graphs between pressure and volume are plotted at different temperatures. Which of the following
isotherms represents Boyle’s law as PV = constant?
After dilution =
T1
T2
T3
P
P
T3
T2
T1
PV
T1
T2
T3
1/V
P
V
(i)
log P
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
33. Given below is the table showing shapes of some molecules having lone pairs of electrons. Fill up the
blanks left in it
Molecule type
AB2E2
bp
2
lp
P
Shape Example
Bent
H2O
AB3E2
3
2
Q
ClF3
AB5E
5
R
S
BrF5
AB4E2
4
2
T
U
a.
b.
c.
d.
C
P
2
4
2
3
Q
Square pyramidal
T-shaped
T-shaped
Square planar
Dr. Sangeeta Khanna Ph.D
R
2
5
1
2
S
T-shaped
Square planar
Square pyramidal
T-shaped
8
T
Square planar
Square pyramidal
Square planar
Square pyramidal
U
H2O2
SO3
XeF4
BrCl3
CHEMISTRY COACHING CIRCLE
G:\Mega Scholarship Test - MEDICAL.doc
Dr. Sangeeta Khanna Ph.D
34. Match the molecules given in column I with their shapes given in column II and mark the appropriate
choice.
Column I (Molecule)
Column II (Shape)
(A) SF6
(i)
(B)
SiCl4
(ii)
(C) AsF5
(iii)
(D) BCl3
(iv)
a. (A)  (iv), (B)  (ii), (C)  (iii), (D)  (i)
b. (A)  (iv), (B)  (i); (C)  (ii), (D)  (iii)
c. (A)  (iii), (B)  (i), (C)  (ii), (D)  (iv)
d. (A)  (ii), (B)  (iii), (C)  (i), (D)  (iv)
B
Sol. (A) SF6
 Octahedral
(B) SiCl4
 Tetrahedral
(C) AsF5
 Trigonal bipyramidal
(D) BCl3
 Trigonal planar
35. In the given graph, a periodic property (R) is plotted against atomic numbers (Z) of the elements.
Which property is shown in the graph and how is it correlated with reactivity of the elements?
a(520)
(R)
b(496)
c(419)
d(403)
e(374)
(Z)
a. Ionisation enthalpy in a group, reactivity decreases from a  e.
b. Ionisation enthalpy in a group, reactivity increases from a  e
c. Atomic radius in a group, reactivity decreases from a  e.
d. Metallic character in a group, reactivity increases from a  e.
B
Sol. I.E. in a group decreases and reactivity increases.
36. Match the column I with column II and mark the appropriate choice.
Column – I
Column – II
(Atom)
(No. of unpaired electrons)
(A) 15P
(i) 6 unpaired electrons
(B) 24Cr
(ii) 2 unpaired electrons
(C) 26Fe
(iii) 3 unpaired electrons
(D) 14Si
(iv) 4 unpaired electrons
a. (A)  (ii), (B)  (i), (C)  (iii), (D)  (iv)
b. (A)  (i), (B)  (iii), (C)  (ii), (D)  (iv)
c. (A)  (iii), (B)  (i), (C)  (iv), (D)  (ii)
d. (A)  (iv), (B)  (ii), (C)  (i), (D)  (iii)
C
Dr. Sangeeta Khanna Ph.D
9
CHEMISTRY COACHING CIRCLE
G:\Mega Scholarship Test - MEDICAL.doc
Dr. Sangeeta Khanna Ph.D
Sol. 15P = [Ne] 3s2 3p3
24Cr
= [Ar] 4s1 3d5
26Fe
= [Ar] 4s2 3d6
14Si =
[Ne] 3s2 3p2
37. Match the mass of elements given in column I with the no. of moles given in column II and mark the
appropriate choice.
Column – I
(A) 28 g of He
(B) 46 g of Na
(C) 60 g of Ca
(D) 27 g of Al
Column – II
(i) 2 moles
(ii) 7 moles
(iii) 1 mole
(iv) 1.5 moles
a. (A)  (iv), (B)  (iii), (C)  (ii), (D)  (i)
b. (A)  (i), (B)  (iii), (C)  (ii), (D)  (iv)
c. (A)  (iii), (B)  (ii), (C)  (i), (D)  (iv)
d. (A)  (ii), (B)  (i), (C)  (iv), (D)  (iii)
D
28
46
Sol. (A) : 28 g of He =
= 7 mol
(B) : 46 g of Na =
 2 mol
4
23
60
27
(C) : 60 g of Ca =
(D) : 27 g of Al =
= 1 mol
 1.5 mol
40
27
38. What is the relationship between the two structures shown?
CH3
Cl
CH3
Cl
a. Constitutional isomers
b. Stereoisomers
c. Different drawing of the same conformation of the same compound
d. Different conformation of the same compound
A
Sol. Position isomers
CH3
C = CH2
39.
+ HBr
A
(Major)
Product (A) of the above reaction is:
Br
Br
a.
Br
b.
Br
c.
d.
A
CH3
Sol.
C – CH3
ring expansion
Dr. Sangeeta Khanna Ph.D
CH3
CH3
shift
10
Br
CH3
–
CH3
Br
–
CHEMISTRY COACHING CIRCLE
G:\Mega Scholarship Test - MEDICAL.doc
Dr. Sangeeta Khanna Ph.D
40. Among the following in which pair, the second ion is more stable than first?
a.
b.
and
c.
CH2
and
d.
and
and
D
Section – B (Assertion & Reason)
a.
b.
c.
d.
1.
Both assertion and reason are true and reason is the correct explanation of assertion.
Both assertion and reason are true but reason is not the correct explanation of assertion.
Assertion is true but reason is false.
Both assertion and reason are false.
Assertion: On moving down the group, ionisation enthalpy decreases.
Reason: With increase in size of the atom the force of attraction between the nucleus and valence
electrons decreases
a. (A)
b. (B)
c. (C)
d. (D)
A
Sol. I.E. decreases down the group due to increase in atomic size and the outermost electron being
increasingly farther from the nucleus, there is an increased shielding of the nuclear charge by the
electrons of the inner levels.
2. Assertion: The energy of the electron in a hydrogen atom has a negative sign for all possible orbits.
Reason: When the electron is attracted by the nucleus and is present in orbit n, the energy is emitted
and its energy is lowered.
3.
a. (A)
b. (B)
c. (C)
d. (D)
A
Assertion: Hydrogen peroxide acts as an oxidizing agent as well as a reducing agent
Reason: It contain both hydrogen and oxygen
4.
a. (A)
b. (B)
c. (C)
d. (D)
B
Assertion: The carbonate of lithium decomposes easily on heating to form lithium oxide and CO2
Reason: Lithium being very small in size polarises large carbonate ion leading to the formation of more
stable Li2O and CO2
5.
a. (A)
b. (B)
c. (C)
d. (D)
A
Assertion: A solution of NH4Cl in water is acidic in nature
Reason: Ammonium ions formed undergo hydrolysis to form NH4OH which is a weak base and
remains unionized in solution
a. (A)
b. (B)
A
Sol. NH4Cl
NH 4  Cl 
NH4Cl + H2O
NH 4 OH 
HCl
weak base
6.
c. (C)
d. (D)
strong acid
HCl
H+ + Cl– (Acidic)
Assertion: Soft water lathers with soap but not hard water.
Reason: Hard water contains calcium and magnesium salts which react with soap to form insoluble
salts which form scum and not lather
a. (A)
b. (B)
Dr. Sangeeta Khanna Ph.D
c. (C)
11
d. (D)
CHEMISTRY COACHING CIRCLE
G:\Mega Scholarship Test - MEDICAL.doc
Dr. Sangeeta Khanna Ph.D
A
Sol. 2RCOONa + Ca2+  (RCOO)2Ca + 2Na+
2RCOONa Mg2+  (RCOO)2Mg + 2Na+
7. Assertion: 1 mole of water is equal to 6.023 × 1023 molecules.
Reason: The mass of one mole of a substance in grams is called the molar mass
a. (A)
b. (B)
c. (C)
d. (D)
B
Sol. Both statements are true but independent of each other.
8. Assertion: The gases show ideal behaviour when the volume occupied is large so that the volume of
the molecules can be neglected in comparison to it.
Reason: The behaviour of the gas becomes more ideal when pressure is very low
9.
a. (A)
b. (B)
c. (C)
d. (D)
B
Assertion: Heat of neutralization of HNO3 and NaOH is same as that of HCl and KOH.
Reason: Both HNO3 and HCl are strong acids and NaOH and KOH are strong bases.
a. (A)
b. (B)
c. (C)
d. (D)
A
Sol. Heat of a strong acid and strong base is always same since neutralization involves reaction between
H+ and OH– ions.
10. Assertion: At high altitudes, liquids boil at lower temperatures in comparison to that of sea level.
Reason: At high altitudes, atmospheric pressure is high
a. (A)
b. (B)
c. (C)
C
Sol. Due to low atmospheric pressure, the liquid boils at lower temperature.
d. (D)
Section – C (Assertion & Reason)
a.
b.
c.
d.
1.
Both assertion and reason are true and reason is the correct explanation of assertion.
Both assertion and reason are true but reason is not the correct explanation of assertion.
Assertion is true but reason is false.
Both assertion and reason are false.
Assertion: Viscosity of liquids decreases as the temperature rises.
Reason: At high temperature, molecules have high kinetic energy and can overcome the inter
molecular forces to flow faster.
a. (A)
b. (B)
c. (C)
d. (D)
A,B
Sol. Viscosity of liquids decreases as the temperature rises.
2. Assertion: An exothermic process which is non–spontaneous at high temperature may become
spontaneous at low temperature.
Reason: Spontaneity is also favoured by increase in entropy.
3.
a. (A)
b. (B)
c. (C)
d. (D)
A
Assertion: Kp can be less than, greater than or equal to Kc.
Reason: Relation between Kp and Kc depends on the change in number of moles of gaseous reactants
and products (n).
a. (A)
A
Sol. Kp = Kc (RT)n
b. (B)
Dr. Sangeeta Khanna Ph.D
c. (C)
12
d. (D)
CHEMISTRY COACHING CIRCLE
G:\Mega Scholarship Test - MEDICAL.doc
Dr. Sangeeta Khanna Ph.D
4.
Assertion: HNO3 acts only as oxidizing agent while HNO2 can act both as a reducing agent and an
oxidizing agent.
Reason: In HNO3, oxidation state of nitrogen is +5 which is maximum. In HNO 2, oxidation state of
nitrogen is +3 which can change from –3 to +5.
a. (A)
b. (B)
c. (C)
d. (D)
A
Sol. Since in HNO2 oxidation state of nitrogen can increases or decrease it can act as an oxidizing as well
as a reducing agent.
5. Assertion: Carbon monoxide is a poisonous gas.
Reason: Carbon monoxide combines with haemoglobin to form carboxyhaemoglobin with stops
absorption of oxygen.
6.
a. (A)
b. (B)
c. (C)
d. (D)
A
Assertion: Carbon atom is tetravalent though it has two unpaired electrons.
Reason: Carbon has unique ability to form p - p multiple bonds like C = C, C  C.
a. (A)
b. (B)
c. (C)
d. (D)
B
Sol. Carbon is tetravalent due to empty p-orbitals which hybridise in excited state to give sp3 hybridisation
with one electron each in 2s and 3p-orbitals showing tetravalency.
C (ground state)
1s
2s
1s
2s
2p
C (ground state)
2p
Hybrid orbitals
7.
Assertion: CO2 is a gas at room temperature while SiO2 is crystalline solid.
Reason: SiO2 is a network of silicon and oxygen atoms joined by multiple bonds.
a. (A)
b. (B)
c. (C)
d. (D)
C
Sol. SiO2 is a network of Si and O atoms joined by single bonds in a tetrahedral manner.
8. Assertion: The reaction, C2H5Br + 2Na + C2H5Br  C2H5C2H5 + 2NaBr is known as Wurtz reaction.
Reason: The reaction is carried out in presence of dry ether.
9.
a. (A)
b. (B)
c. (C)
d. (D)
B
Assertion: cis-Form of alkene is found to be more polar than the trans-form.
Reason: Since the groups are in opposite directions, the dipole moments of bonds cancel each other
making trans form almost non-polar
a. (A)
A
b. (B)
CH3
CH3
H
d. (D)
CH3
H
C=C
Sol.
c. (C)
C=C
H
CH3
cis-But-2-ene
( = 0.33 D)
H
Trans-But-2-ene
( = 0)
10. Assertion : Tropylium ion has delocalised -electron cloud.
Reason: It has 12 electron in conjugation therefore it is antiaromatic.
a. (A)
C
b. (B)
Dr. Sangeeta Khanna Ph.D
c. (C)
13
d. (D)
CHEMISTRY COACHING CIRCLE
G:\Mega Scholarship Test - MEDICAL.doc