1. 1
CHEMISTRY[IRTPF-FT-V-P2]
Read the paragraph and answer the questions (18 – 20) given below:
E1 elimination reactions are normally accompanied by SN1 substitution as both have a common
carbocationic intermediate. Similarly, E2 elimination is accompanied by SN2 substitution, though in this
case the parallel, concerted processes involve entirely separate pathways. Thus, elimination vs
substitution considers
(a) factors influencing E1/SN1 product ratios
(b) factors influencing E2/SN2 product ratios
(c) factors influencing change of pathway, i.e. E1/SN1 to E2/SN2 or vice-versa,
as such a shift often changes the proportion of elimination to substitution product. In both (a) and (b), the
carbon structure of the substrate is of considerable importance as the proportion of elimination rising on
going from primary to secondary to tertiary.
18. What are the possible products for given reaction?
CH3
OH
OH
C
H3
2 4
H SO


(i)
CH3
CH3
(ii)
CH3
O
CH3
(iii)
CH3
O
CH3
(iv)
CH2
CH2
(A) (i) and (ii) (B) (iii) and (iv)
(C) (ii) and (iv) (D) (i) and (iv)
19. Which of the following compounds are optically active?
(i)
CH3
Br
H
H
Br I
(ii)
CH3
H
Br
CH3
Br H
(iii) C C
H
C
H3
C
CH3
H
(iv) C C
H
C
H3
C C
CH3
H
(A) (i) and (ii) (B) (i) and (iii)
(C) (ii) and (iii) (D) (i) and (iv)
20. Factors which favour substitution product are
(i) a tertiary alkyl boride (ii) a primary alkyl boride
(iii) concentrated alcoholic potash (iv) dilute aqueous potash
(A) (i) and (ii) (B) (ii) and (iii)
COMPREHENSION-I

2. IRTPF-FT-V/(P2)-PCM-IITJEE/07
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(C) (iii) and (iv) (D) (ii) and (iv)
18. C
19. B
20. D
Read the paragraph and answer the questions (21 – 23) given below:
Silicon is second most abundant element occurring in earth crust it always found in combined state. It is
obtain by reduction of silica it exists in two allotropic forms amorphous and crystalline. Amorphous forms
is obtain by heating dry powdered silica with magnesium. Crystalline forms is obtained by heating sand or
quartz in an furnaces. Silicon is used in manufacture of certain alloys and polymers such as ferrosilicon,
silicon bronze etc.
The compounds of silicon are carborundum, silicons, silicates, glass etc.
Silicons are organosilicon polymer and silicates are metal derivatives of silicic acid. While carborundum
are obtained by heating mixture of sand carbon and salt.
21. Which type of linkage present in silicones?
(A)
NH
O
(B)
Si
O
(C) Si–O–Si (D)
R Si
O
22. The structural unit of silicates is:
(A) SiO2 (B) 4
4
SiO
(C) 2
3
SiO
(D) 2
2
SiO
23. (SiH3)3N is weaker base than (CH3)3N because:
(A) In (SiH3)3N lone pair of electron on nitrogen is involve in d p
   bonding
(B) In 3 3
(CH ) N

steric effect of three alkyl group makes it least basic
(C) Absence of d p
   back bonding in 3 3
(CH ) N

(D) All are correct
21. C
22. B
23. A
Read the paragraph and answer the questions (24 – 26) given below:
COMPREHENSION-II
COMPREHENSION-III

3. 3
If a discharge is passed through hydrogen gas (H2) at low pressure, some hydrogen atoms are formed,
which emit light in the visible region. The light can be studied with a spectrometer and is found to
comprise a series of lines of different wavelength. The lines get increasingly closer to each other as the
wavelength decreases until the continuum is reached. Sommerfield explained the splitting of lines by
assuming that some of the orbits were elliptical, and that they precessed in space around the nucleus.
For the orbit closest to the nucleus, the principle quantum number n = 1, and there is a circular orbit. For
the next orbit, n = 2 and both circular and elliptical orbits are possible. To define an elliptical orbit a
second quantum number k is needed. The shape of ellipse is defined by the ratio of the major and minor
axis.
Thus,
major axis n
minor axis k
 , where k is called azimuthal quantum number and may have values from 1, 2, …n.
Thus, for n = 2, n/k may have values 2/2 (circular) and 2/1 (elliptical orbit). The presence of these extra
orbits which have slightly different energies from each other accounts for the extra lines in the spectrum
revealed under high resolution. The original quantum number k now has been replaced by a new
quantum number where = k – 1.
24. When the value of principle quantum number, n = 3, how many orbitals are possible in the
valence shell of the atom?
(A) 8 (B) 9
(C) 3 (D) 4
25. The splitting of spectral lines in the presence of magnetic field is called
(A) Stark effect (B) Zeeman effect
(C) Thomson’s effect (D) None of the above
26. The lines become increasingly closer as the wavelength () decreases, it explains that
(A) the distance between any two orbits increases as the principle quantum number increases.
(B) the distance between any two orbits decreases as the principle quantum number increases.
(C) distance between consecutive orbits is always same.
(D) None of the above
24. B
25. B
26. B
Read the paragraph and answer the questions (27 – 29) given below:
Stable equilibrium is of various types. Mechanical equilibrium is achieved when all particles are at rest
and total potential energy of the system is minimum. At any stage where particles are at rest but the
system is not at stable equilibrium as it can reduce its potential energy by reverting to another position, is
called metastable equilibrium. Thermal equilibrium results from the absence of temperature gradients in
the system. Chemical equilibrium is obtained when no further reaction occurs between reacting
substances, i.e. forward and reverse rates of reaction are equal. When steam reacts with solid iron at
high temperature, Fe3O4(s) and hydrogen gas are produced. But the reaction never goes to completion.
This is because as the products are formed, the reaction proceeds in reverse direction and when rate of
reverse reaction is equal to rate of forward reaction, the concentration of reactants and products become
constant and equilibrium is reached.
27. Which of the following is correct?
(A) Chemical equilibrium can be stable, unstable or metastable.
(B) The example given in the passage refers to thermodynamic equilibrium.
COMPREHENSION-IV

4. IRTPF-FT-V/(P2)-PCM-IITJEE/07
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(C) Addition of catalyst to a reaction in chemical equilibrium increases the amount of heat
evolved in the reaction.
(D) Addition of catalyst to a reaction in chemical equilibrium gives a new reaction path to the
system.
28. The correct expression for equilibrium constant of the reaction, taken as an example of chemical
equilibrium in the above passage is
(A)
2
2
2
2
(pH )
(pH O)
(B)
4
2
4
2
(pH )
(pH O)
(C)
4
2 3 4
4
2
(pH ) [Fe O ]
(pH O) [Fe]
(D) 3 4
[Fe O ]
[Fe]
29. When two reactants (A) and (B) are mixed to give products (C) and (D), the concentration
quotient (Q) at initial stage of the reaction
(A) is zero (B) decreases with time
(C) is independent of time (D) increases with time
27. D
28. B
29. D
30. The heat of combustion of (1) C6H6 is -3302 kJ/mol and that of (2) C8H8 in – 4581 kJ/mol which is
more stable?
30. 1
30. The heat of combustion per carbon atom for
(1) i.e. Benzene C6H6
3302
550.3 kJ/mol
6

  
and for (2) i.e. Cyclooctatetraene C8H8
4581
572.6 kJ/mol
8

  
Since energy released is more in C8H8 and not in C6H6 hence stability of C6H6 > C8H8
1 2
 
31. A hydrocarbon with vapour density = 27 containing C = 88.88% decolourizes KMnO4 and Br2
water without evolving HBr. It gave no ppt. with ethereal ammonical AgNO3 or Cu2Cl2. When
reacted with dil. H2SO4 in the presence of mercuric sulphate, methyl ethyl ketone is formed. What
is the molecular weight of hydrocarbon?
31. Element % %/At. wt. Relative
number of
atoms
Simplest ratio
C 88.88 7.4 1 2
H 11.12 11.12 1.50 3
Empirical formula = C2H3
Mol. mass = 2 × 27 = 54
Mol. mass 54
n 2
Eq.wt. 27
  
 Molecular formula = 2(C2H3) = C4H6
Degree of unsaturation = 2
Hence can be CH2 = CH – CH = CH2
NUMERICAL BASED QUESTIONS

5. 5
or
3 3
CH C C CH
  
or
3 2
CH CH C CH

But there is no acetylenic H atom because of – ve ammonical AgNO3 or ammonical Cu2Cl2 test.
Also only 3 3
CH C C CH
   can give 3 2 5
||
CH C C H
O
  with HgSO4 + H2SO4
Compound is 3 3
CH C CCH

Molecular weight = 54
31. 54
32. An organic compound (A) gives positive Liebermann’s nitroso reaction and on treatment with
CHCl3/KOH followed by hydrolysis gives (B) and (C). Compound (B) gives pink colour with
Schiff’s reagent but not (C) which is steam volatile. (C) on treatment with LiAlH4 gives (D)
(C7H8O2) which on oxidation gives E. ‘E’ reacts with acetic anhydride and gives compound F.
Calculate molecular weight of ‘F’.
32. Mol. wt. = 9 ×12 + 8 × 1 + 4 × 16 = 108 + 8 + 64 = 180 mol-1
32. 180
33. Calculate the minimum weight (in gm) of NaOH required to be added in R.H.S. to consume all the
H+ present in R.H.S. of cell of e.m.f. + 0.701V at 25°C before its use. Give your answer upto three
decimal points.
Zn | Zn+2 (0.1 M) || HCl 1 litre | H2 | Pt (1 atm)
2
0
Zn/ Zn
E 0.76 volt
  
33. The given cell is
2
(0.1M) 1 lit 1 atm
Zn | Zn || HCl | H | Pt

So, the cell reaction can be written as
At anode : Zn(s) 
(0.1 M)
Zn 2e
 

At cathode: 2H+ + 2e–  2
(1 atm)
H
––––––––––––––––––––––––––
Zn(s) + 2H+  2
(0.1M) 1 atm
Zn H (g)



2
0 0 0
cell Zn/ Zn H /H
E E E
 
  = 0.76 + ) = 0.76 volt
From Nernst’s equation
Ecell = 0 2
cell 2
[Zn ] pH
0.0592
E log
2 [Zn(s)][H ]



= 0
cell
0.0592
E
2
 log 2
[Zn ]
[H ]

 2
[Zn ]
[H ]


= 0.76 –
0.0592
2
log 2
0.1
[H ]

= 0.76 +
0.0592
2
log
[H ]
0.1

 [H+] = 0.0316 mol lit–1
or, [H+] = 0.0316 equivalent lit–1 [ here mole = equivalent]
From law of equivalent
Equivalent of NaOH = Equivalent of HCl
W
40
= 0.0316, where W = Wt. of NaOH
 W = 1.264 gm
33. 1.264

6. IRTPF-FT-V/(P2)-PCM-IITJEE/07
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34. A V dm3 flask contains gas A and another flask of 2V dm3 contains B at the same temperature. If
density of gas A is 3.0 gm dm-3 and that of B is 1.5 g dm-3 and mol wt. of A = ½ mole wt of B, then
calculate the ratio of pressure exerted by gas.
34. 4
34. For gas A, PA =
3
RT
mA

For gas B, PB =
B
1.5
RT
m

 A B
B A
P 2 m 2 2 mA
4
P m mA
  
  