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1. Several factors (steric, electronic, orbital interactions etc.) can affect the inversion barrier ofan amine. In the given pair which data is correctly placed ?
(a) (b)
(c) (d*) All of these
2. Circle represent most acidic hydrogens in these molecules. Which of the following is correctrepresentation ?
(a) (b)
(c) (d*) All of these
3. The pH at which maximum hydrate is present in an solution of oxaloacetic acid, is :
H O C||
C||
CH C||
O H
O O O
2-------
pK a =2 2. pK a =3 98.(a*) pH = 0 (b) pH = 12 (c) pH = 4 (d) pH = 6
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vs
Me|N
MeMe
DG = 7.9 kcal/mol
i – Pr
i – Pr
|N
Me
DG = 0.2 kcal/mol+ DG = 20.5 kcal/mol
NMe
DG = 7.0 kcal/molMe
Nvs
Me|
NMeMe
DG = 7.9 kcal/mol++
vs
Cl|
NClCl
DG = 22.9 kcal/mol++
OH
CF3
OH
O
MeO C2
CO H2HO
S
F
By : M.S. Chouhan
Advanced Problems
Organic Chemistryfor
IIT-JEE
in
4. Which of the following isomeric hydrocarbons is most acidic ?
(a) (b*) (c) (d)
5. Compound A and B, both were treated with NaOH, producing a single compound C.
HO
heat
-
¾®¾¾ C. Compound (C) is :
(a*) (b) (c) (d)
6. ¾¾¾®RCO H2 ; Product of above reaction is :
(a) (b)
(c*) both a and b (d) None of these
7. 150° ==¾®¾¾+¾ ®¾¾¾¾¾¾
C H C CH — CN2( ) ( )A B
Compound (B) reacts acrylonitrile to give (C), structure of compound (A) is :
(a) (b*) (c) (d)
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O
O
O
C — CH3O O
OTs
14C label
no label
OCOR OCOR
CH3O OCH3
CN
(C)CCH3CH3O
CH2
CH2
OCH3
O
CH3
OH
O
CH3
OH(A) (B)
+
8. Increasing order of rate of reaction with HNO H SO3 2 4 is :
(a) iii < ii< i (b) ii < iii < i(c) i < iii < ii (d*) i < ii < iii
9. Match the columns I, II and III (Matrix).
Col umn-I Col umn- II Column- IIII
Re ac tion Nature of product formed Num ber of chiral center pres - ent in product. (Consideronly one isomer in case of
racemic mixture orDiastereomer)
(a)Br
CCl2
4¾®¾¾
(p) Racemic mix ture (w) 0
(b) Br
CCl2
4¾®¾¾
(q) Meso (x) 1
(c) BrCCl
2
4¾®¾¾ (r) Diastereomer (y) 2
(d) (s) Vic i nal dihalide (z) 3
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(i)
O
O O
OOO
O
OOO OO
(ii) (iii)
CH3
H
CH3
CH H3 C ==C H CH3
Br2
CCl4
CH3
H
10.
The synthetic steroid ethynylestradiol (1) is a compound used in the birth control pill.
A. How many sp3 hybridised carbon atoms are present in compound (1)?
(a) 8 (b) 9 (c) 10 (d) 11 (e) 12
B. How many sp2 hybridised carbon atoms are present in compound (1) ?
(a) 4 (b) 5 (c) 6 (d) 7 (e) 8
C. How many sp hybridised carbon atoms are present in compound (1) ?(a) 2 (b) 4 (c) 6 (d) 8 (e) 10
D. Which of the following functional group is contained in compound (1) ?(a) A ketone (b) An alcohol (c) A carboxylic acid (d) An ester
E. How many asymmetric (stereogenic) centres are present in compound (1) ?(a) 2 (b) 3 (c) 4 (d) 5
1. In the prep aration of iron from haematite (Fe O )2 3 by the reaction with carbonFe O + C Fe+ CO2 3 2¾®
How much 80% pure iron could be produced from 120 kg of 90% pure Fe O ?2 3(a*) 94.5 kg (b) 60.48 kg(c) 116.66 kg (d) 120 kg
2. van der Waals constant b of helium is 24 mL mol-1. Find molecular diameter of helium.(a) 1.335 cm´-10 10 (b) 1.335 cm´-10 8
(c*) 2.67 10 cm8́- (d) 4.34 cm´-10 8
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Ethynylestradiol (1)HO
OH H
H H
H
H C3
3. There exist an equi librium between solid SrSO4 and Sr 2+ and SO42- ion in aque ous medium.
The possible equilibrium states are shown in figure as thick line. Now, if equilibrium isdist urbed b y a dd it ion of (a) Sr(NO )3 2 and (b) K SO2 4 and dotted line represent approachof system toward’s equilibrium. Match the columns given below :
(I) addition of Sr(NO )3 2 (II) addition of K SO2 4(a) (I) (iii), (II) (iv) (b*) (I)(iv),(II) (v) (c) (I) (vi), (II) (v) (d) (I) (iv), (II) (v)
4. Determine the potential of the following cell :
Pt|H bar H MnO2( , . )| ( , )|| ( , . ),– –g aq aq01 10 0134
+ M M Mn H Pt2 0 01 0 01+ +( , . ), ( , . )|aq aqM M
Given : EMnO4
–|Mn 2+1.51 V° =
(a) 1.54 V (b*) 1.48 V (c) 1.84 V (d) none of these
5. 0.1 M KI and 0.2 M AgNO3 are mixed in 3 1: volume ratio. The depression of freezing pointof the resulting solution will be [ ( )K f H O 1.862 = K kg mol-1] :
(a) 3.72 K (b) 1.86 K (c) 0.93 K (d*) 0.279 K
6. In an atomic bcc, what fraction of edge is not covered by atoms?(a) 0.32 (b) 0.16 (c*) 0.134 (d) 0.268
7. Select the correct statement(s) :(a*) A solution is prepared by addition of excess of AgNO3 solution in KI solution. The
charge likely to develop on colloidal particle is positive.(b*)The effects of pressure on physical adsorption is high if temperature is low(c) Ultracentrifugation process is used for preparation of lyophobic colloids.(d) Gold number is the index for extent of gold plating done
8. For a first order homogeneous gaseous reaction, A B C¾®+2
If the total pressure after time t was Pt and after long time ( )t ®¥ was P¥ then k in terms of P Pt , ¥ and t is :
(a) kt
PP Pt
=-
æ
èç
ö
ø÷
¥
¥
2.303log (b) k
tP
P Pt=
-
æ
èç
ö
ø÷
¥
¥
2.303log
2
(c*) k t
log P
(P P )t=
-
æ
èç
ö
ø÷
¥
¥
2.303 23
(d) none of these
9. Fixed mass of an ideal gas contained in a 24.63 L sealed rigid vessel at 1 atm is heated from -°73 C to 27°C. Calculate change in Gibb’s en ergy if entropy of gas is a function oftem per a ture as S =+-2 10 2 T (J/K): (Use 1 atm L 0.1= kJ)
(a) 1231.5 J (b) 1281.5 J (c*) 781.5 J (d) 0
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2+Sr
2–SO4
2–SO4
Sr2+ 2+Sr
2–SO4
2–SO4
2+Sr(iii) (iv) (v) (vi)
10. Column-I Col umn-II
(A) Or bital an gu lar mo men tum of anelec tron
(P) s sh
( )+12p
(B) An gu lar mo men tum of an elec tronin an orbit
(Q) n n( )+2
(C) Spin an gu lar mo men tum of anelec tron
(R) nh2p
(D) Magnetic moment of atom (S) l lh
( )+12p
* Answer : A – S, B – R, C – P, D – Q
11. Which is the correct order of ionization energies?(a) F > F> Cl > Cl– – (b*) F > Cl > Cl > F– –
(c) F > Cl > Cl > F– – (d) F > Cl > F > Cl– – –
12. Ionization energy of an element is :(a*) Equal in magnitude but opposite in sign to the electron gain enthalpy of
the cation of the element(b) Same as electron affinity of the element(c*) Energy required to remove one valence electron from an isolated
gaseous atom in its ground state(d) Equal in magnitude but opposite in sign to the electron gain enthalpy of the anion of the
element
13. O F2 2 is an unstable yellow orange solid and H O2 2 is a colourless liquid, both have O—Obond and O—O bond length in H O2 2 and O F2 2 respectively is :
(a) 1.22Å, 1.48Å (b*) 1.48 Å, 1.22 Å
(c) 1.22Å, 1.22Å (d) 1.48Å, 1.48Å
14. Which of the following equilibria would have highest and lowest value of K p at a commontemperature.(a*) BeCO 3 � BeO CO 2+ (b) CaCO3 � CaO CO2+
(c) SrCO3 � SrO CO2+ (d*) BaCO3 � BaO CO2+
15. In the structure of H CSF2 4, which of the following statement is/are correct?
(a*) Two C—H bonds are in the same plane of axial S—F bonds(b) Two C—H bonds are in the same plane of equatorial S—F bonds(c*) Total six atoms are in the same plane(d) Equatorial S—F plane is perpendicular to plane of p-bond
16. In the isoelectronic series of metal carbonyl, the CO bond strength is expected to increase inthe order :(a) [Mn(CO) ] [Cr(CO) ] [V(CO) ]6 6 6
+ -< < (b*)[V(CO)6] [Cr(CO)6]<[Mn(CO)6]-< +
(c) [V(CO) ] <[Mn(CO) ] [Cr(CO) ]6 6 6- +< (d) [Cr(CO) ]<[Mn(CO) ] [V(CO) ]6 6 6
+ -<
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17. [Fe(H O) ]2 62+ and [Fe(CN) ]6
4– differ in :
(a) geometry, magnetic moment (b) geometry, hybridization(c*) magnetic moment, colour (d) hybridization, number of d-electrons
18. Which of the following statements is true ?(a) In [PtCl (NH ) ]2 3 2
2+ the cis form is optically inactive while trans form is optically active
(b*)In [Fe(C O ) ]2 4 33– , geometrical isomerism does not exist while optical
isomerism exists(c) In Mabcd, square planar complexes show both optical as well as geometrical isomerism(d) In Mabcd tetrahedral complex, optical isomerism cannot be observed
19. The CFSE for [(CoCl)6] –4 complex is 18000cm –1. The D for [CoCl ]42– will be :
(a) 18000cm –1 (b) 16000 cm –1 (c*) 8000 cm–1 (d) 2000 cm –1
The sequential unknown reagents is/are:(a) H O2 2/neutral medium, H O /H , Cu2 2
+ 2+ salt solution(b*) H O / H , H O / OH , Fe2 2
+2 2
3+- salt solution(c) H O /OH , H O /H , Co2 2 2 2
2+- + salt solution(d) H O2 2/neutral medium, H O /OH Fe2 2
2+-, salt solution
21. Se lect cor rect state ment(s):(I) When excess FeCl3 solution is added to K [Fe(CN) ]4 6 solution, in addition to
Fe [Fe (CN) ]III II6
-, Fe [Fe (CN) ]II III6
- is also formed due to side redox reaction
(II) When FeCl2 is added to K [Fe(CN) ]3 6 solution, in addition to Fe [Fe (CN) ]II III6
-, Fe [Fe (CN) ]III II
6- is also formed due to side redox reaction
(III) Fe [Fe (CN) ]III II6
- is paramagnetic while Fe [Fe (CN) ]II III6
- is diamagnetic
(IV) Fe [Fe (CN) ]III II6
- is diamagnetic while Fe [Fe (CN) ]II III6
- is paramagnetic
(a*) I, II (b) III, IV (c) both (a) and (b) (d) None of these
22. Col umn-I(Pair of com plexes)
Col umn-II(Property which is similar in given pair)
(A) [Fe(CN) ]63– and [Co(NH ) ]3 6
2+ (P) Mag netic mo ment
(B) [Fe(H O) ]2 62+ and [Fe(CN) ]6
4– (Q) Ge om e try
(C) [Ni(CN) ]44– and [Ni(CO) ]4 (R) Hy bridi sa tion
(D) [Ni(H O) ]2 62+ and [NiCl ]4
2– (S) Num ber of d-electrons
* Answer : A – P, Q, R; B – Q, S; C – P, Q, R, S; D – P, S
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A B A CX Y Z20.
Pale yellowsolution
Yellowsolution
Deep bluesolution
Deep bluesolution
Regenerated
Na (comH O CO SO Na2S I2 Ag+ salt2 / /¾®¾®¾®¾¾¾®¾¾¾®A B C E2 2
DD plex)
a. The compound B and C are :(a) Na CO , Na SO2 3 2 4 (b) NaHCO , Na SO3 2 4(c*) Na CO , Na SO2 3 2 3 (d) None of these
b. The compound D is :(a) Na SO2 4 (b) Na S O2 4 6 (c) Na S O2 2 5 (d*) Na S O2 2 3
c. Oxidation number of each ‘S’ atom in compound D :(a) +2, +2 (b) +4, 0 (c*) +6, –2 (d) +5, –1
The molecule in which an atom is associated with more than 8 electrons is known ashypervalent molecule and less than 8 electrons is known as hypovalent molecule. Allhypervalent molecules must have dp-pp bonding but the molecules having back bondingneed not to have always dp-pp-bonding.
a. Which of the molecule is not hypovalent but completes its octet :(a) AlCl3 (b) AlBr2 (c*) AlF3 (d) BF3
b. Which of the following molecule is having complete octet :(a) BeCl2 (dimer) (b) BeH2 (dimer) (c) BeH2(s) (d*) BeCl (s)2
c. Which of the following molecule is not having dp-pp-bonding :(a) SO2 (b) P O4 10 (c) PF3 (d*) B N H3 3 6
An unknown mixture contains one or two of the following : CaCO , BaCl , AgNO ,3 2 3Na SO , ZnSO2 4 4 and NaOH. The mixture is completely soluble in water and solutiongives pink colour with phenolphthalein. When dilute hydrochloric acid is graduallyadded to the solution, a precipitate is formed which dissolves with further addition of theacid.
a. Which of the following combination of compounds is soluble in water?(a) BaCl and AgNO2 3 (b) AgNO and NaOH3(c) BaCl and Na SO2 2 4 (d*) ZnSO 4 and excess NaOH
b. The aqueous solution of mixture gives white precipitate with dil. HCl which dissolves inexcess of dil. HCl. It confirms:(a) BaCl NaOH2 + (b) Na SO NaOH2 4 +
(c*) ZnSO NaOH4 + (d) AgNO NaOH3 +
c. The white pre cip i tate is:(a) ZnSO4 (b) Na ZnO2 2(c*) Zn(OH)2 (d) ZnCl2
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P A S S A G E 2P A S S A G E
P A S S A G E 3P A S S A G E
P A S S A G E 1P A S S A G E
1. Initially blocks A and B are given impulse in opposite directions as shown in figure. Now wehave to calculate the :(i) Maximum stretch in spring(ii) Maximum velocity of block A and B in ground
frame.(iii)Minimum velocity of block A and B in ground
frame.
Sol. υυ υ
υCM =-
=( )( )2 2
30 0
0m m
m
At initial instant
v v v i®®®
=- =A A g gCM CM υ0$
v v v i®®®
=- =-B B g gCM CM 2 0υ$
From work energy equation in CM frameWe get wspring system CMKE=D
- =-+
12
012
22
320
2kXm mm mm
( )( )( )
( )υ
12
12
23
320
2kX mm =æèçö
ø÷( )υ
or Xmkm =
60υ
Get Ready for IIT-JEE 9
By : Er. Anurag Mishra
MechanicsIIT-JEE
for
Fig. 4.21
υ0m 2m
2υ0B A
(a)
2υ0m 2m
υ0
CM frame
Initial state
υ0
CM frame
At maximum stretch
υ0
υ= 0B/CM υ= 0A/CM
l + x0 m
Fig. 4.21
(b)
Vol. 1
When spring again regains its natural length in CM frame.
Most Important Concept
v v v® ®®
= +A Agroundframe
CMframe CM ground
frame
similarly v v v® ®®
= +B Bgroundframe
CMframe CM ground
frame
Note that velocity of any block in ground frame is superposition of two velocity vectors, velocity of block in CM frame and velocity of CM with respect to ground.
v®
A groundframe
is maximum when v®
A CMframe
has maximum magnitude and is in same direction as vector
v®
CM groundframe
.
Similarly v®
A groundframe
is minimum when v®
A CMframe
and v®
CM groundframe
vectors are in opposite direction.
| | | | | |v v v®® ®
= + =A AmaxCMframe CM ground
frame 2 0υ
Similarly v v v®® ®
= + =B Bmax CMframe CM ground
frame| | | | | 3 0υ
Minimum velocity of A is attained when block is at equilibrium
position (spring is relaxed) and v®
A CM and v®
CM g are
opposite to each other.
Thus | |min
v®
=A 0
Minimum velocity of B is attained at the instant B is moving toward left (opposite v®
CM ) and velocity
magnitude is υ0 (see figure)
Thus | |min
v®
=B 0
10 Get Ready for IIT-JEE
υ0
CM frame
|v = B/CM 0| υ0
|v = A/CM
υ|
AB
2
® ®
Fig. 4.22
Fig. 4.21
(c)
Blockdiagramsrepresentingsituation whenblock returnsto relaxedstate
What appears inCM frame
Blocks return to relaxed state
υ0
υ= 2υB/CM 0
m 2m
υ= υA/CM 0
What appears inground frame
υ= 3υA/g 0
m 2m
υ= 0B/g
ìíî
2. Two blocks A and B of masses 2m & 3m placed onsmooth horizontal surface are connected with a lightspring. The two blocks are given velocities as shownwhen spring is at natural length.
Col umn-I Col umn-II
(A) minimum magnitude of velocity of A ( )min
υA during motion (P) υ
(B) maximum magnitude of velocity of A ( )max
υA during motion (Q)υ
5
(C) maximum magnitude of velocity of B ( )max
υB during motion (R) 0
(D) velocity of centre of mass ( )υCM of the system comprised ofblocks A, B and spring
(S) 75υ
Sol. Step I:
υυ υυ
CM =-
=( )3 3
5 5m m
m
Step 2:
In COM frame
Initial velocity of A = --æèç ö
ø÷=-υ
υ υ
565
to left
Initial velocity of B =-==υυ
υυ5
45
45
to right
Blocks are executing SHM in CM frame with initial position as equilibrium position
Step III:
Velocity variation of B in ground frame, considering right as +ve
form 45 5υυ
υ+æèç ö
ø÷= to
45 5
35
υυυ+=-
so | |max
υ υB = & | |min
υB =0
Velocity variation A in ground frame
from 65 5
75
υυυ+æ
èç ö
ø÷= to
-+=-
65 5
υυυ
Thus minimum velocity of A is -υ
5 when spring is at maximum extension.
Get Ready for IIT-JEE 11
Fig. 4E.30
υ2m 3mK
υ
A B
3. The 10 kg block is resting on the horizontalsurface when the force 'F' is applied to it for7 second. The variation of 'F' with time isshown. Calculate the maximum velocityreached by the block and the total time 't'during which the block is in motion. Thecoefficient of static and kinetic friction areboth 0.50.
Sol. Block begins to move when
F =mN =́ =0 5 100 50. N
from t =0 to t =4sec F st=2 0 4££t =40N 4 7<£t
Block begins to move at t =2sec. after that
Fmddt
-=mNυ
25 50 10tddt
-=υ
or, d t dtυ0
υ
ò ò= -( . )2 5 52
4
υ= -2 5
25
2
2
4. t
t
=5m sAfter t = 4 sec.; block retards due to greater friction forceVelocity of block at t =7 sec(i.e. 3 sec. after force becomes 40 N)
υ=-́5 1 3 =2m s
At t =7 sec. force F is now only friction acts -=s aυ10
a =5m s2
0 2=-́( )s t t =4sec.
Total time interval for which block moves t =2sec. to t =7 9. sec
i.e., 5 4. sec.
12 Get Ready for IIT-JEE
(b)
N
FµN
mg
Fig. 2E.60
Fig. 2E.60
(c)
F = 40N
υ5m/s= initial
µ N = 5 0 Nk
Fig. 2E.60
(a)
100
40
F(N)
4 7t(s)
10 kg F
µ Nk
Fcos37°
Fsin37°
4. A force of 20 N is applied to a block at rest as shown in figure.After the block has moved a distance of 8m to the right thedirection of horizontal component of the force F is reversed indirection. Find the velocity with which block arrives at itsstarting point.
Sol. Stage 1: Motion till force reverses its direction
N =- °mg F sin 37
=-́20 2035
=8N F makcos37°-m=N
2045
0 25 8 2-́ =́́. a
a =7 m s2
Velocity of block after displacement of 8 m υ=́ =́2 7 8 12 m s
Concept
Kinetic friction is opposite to relative velocity it opposes relative motion. When horizontal componentforce is reversed, relative velocity is not changed therefore, direction of kinetic friction does notchange.
Stage 2: ¾®=υ112 m s
F cos37° is reversed, block continues along original direction, but due to retardetion createdby mkN and F cos37° block travels till it stops.
- °+=( cos )F N mak37 m
-́ + =́́( ( . )2045
0 25 7 8 2 a
or, a =-9m s2
Displacement of block in this phase
0 22=-υas;
sa
=υ
2
2
=´
=( )1122 9
569
m
Get Ready for IIT-JEE 13
Fig. 2E.61
(a)
m=2kg
37°
µ=0.25
Fig. 2E.61
(b)
µ Nk
N
Fcos37°
mg=20N
Fsin37°
Stage 3: Which block returns its a acceleration is: F N mgkcos37°-=m
a =7 m s2
Velocity of block when it returns to original position
υ2 2=as
=2 7 8569
´́ +æèç ö
ø÷
υ=16 7
3m s
5. A carriage of mass M and length l is joined to the end of aslope as shown in the figure. A block of mass m is releasedfrom the slope from height h. It slides till end of thecarriage (The friction between the body and the slope andalso friction between carriage and horizontal floor isnegligible) Coefficient of friction between block andcarriage is m. Find minimum h in the given terms.
(a) m1 +æèç ö
ø÷
Mm
l (b) 2 1m+æèç ö
ø÷
mM
l
(c) m2 +æèç ö
ø÷
mM
l (d) m1 +æèç ö
ø÷
mM
l
Sol.Concept:
Block slips relative to carriage, use relative motion equations of kinematics.
velocity of block, just before reaching carriage υ0 2=gh
Now acceleration of block
amgm
g1 =-=-m
m
acceleration of carriage amgM2 =
m
considering this moment as t =0, motion of block as seen from carriage
u ghrel ==υ0 2
a a a gmMrel =-=-+æ
èöø1 2 1m
Relative velocity of block when block moves through distance x with respect to carriage υυrel rel rel
2 2 2=+a x
when x l= =, υrel 0 Þ 2 2 1gh gmM
l= +æè
öø
m
Þ hmM
l=+æè
öø
m1
14 Get Ready for IIT-JEE
Fig. 2E.61
(c)
µ Nk
NFcos37°
mg
Fsin37°
m
h
Msmooth
Fig. 2E.102
N
A
µmg
mg
a1
N
A
µmg
mg
a2
µmg
Fig. 2E.102
(b)
Covering all the possible types of problems like numerical problems, objective problems, MCQ, matching type questions, integer answers type questions, previous year’s questions, comprehension based objective type problems & assertion reason type problems etc.
Covering all the possible types of problems like numerical problems, objective problems, MCQ, matching type questions, integer type questions, previous year’s questions, comprehension based objective type problems, etc.
Algebrafor
IIT-JEEBy : M. Uma Shankar & Dr. J.V. Rao
A Complete text book for IIT-JEE aspirants.
Objective Physics
Objective Physics
ConceptualInorganicChemistry
ConceptualPhysicalChemistry
By : Prabhat Kumar By : Prabhat Kumar & Adarsh Kumar
(IIT-JEE & AIEEE) (IIT-JEE & AIEEE)
Objective Physics
Entrance Exam.is
goal