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RReessoonnaannccee EEdduuvveennttuurreess LLttdd..Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
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TThhiiss ssoolluuttiioonn wwaass ddoowwnnllooaadd ffrroomm RReessoonnaannccee JJEEEE ((MMAAIINN)) 22002200 SSoolluuttiioonn ppoorrttaall
7340010333
PPAAPPEERR--33 ((BB..EE..//BB.. TTEECCHH..))
JJEEEE ((MMaaiinn)) 22002200
COMPUTER BASED TEST (CBT)
Questions & Solutions
DDaattee:: 0099 JJaannuuaarryy,, 22002200 ((SSHHIIFFTT--11)) || TTIIMMEE :: (9.30 am to 12.30 pm)
DDuurraattiioonn:: 33 HHoouurrss || MMaaxx.. MMaarrkkss:: 330000
SSUUBBJJEECCTT :: CCHHEEMMIISSTTRRYY
RReessoonnaannccee EEdduuvveennttuurreess LLttdd..Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : U80302RJ2007PLC024029
TThhiiss ssoolluuttiioonn wwaass ddoowwnnllooaadd ffrroomm RReessoonnaannccee JJEEEE ((MMAAIINN)) 22002200 SSoolluuttiioonn ppoorrttaall
7340010333
FOR 2020–21
| JEE MAIN-2020 | DATE : 09-01-2020 (SHIFT-1) | PAPER-3 | OFFICAL | CHEMISTRY
Resonance Eduventures Ltd.Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : 80302RJ2007PLC024029
This solution was download from Resonance JEE (MAIN) 2020 Solution portal
PAGE # 1
7340010333
PART : CHEMISTRY
SECTION – 1 : (Maximum Marks : 80) Straight Objective Type (lh/ks oLrqfu"B izdkj)
This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its answer, out of which Only One is correct.
bl [k.M esa 20 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. Which of these will produce the highest yield in Friedel Craft reaction ?
ÝhMy Øk¶V vfHkfØ;k esa buesa ls dkSu vf/kdre mRikn nsxk \
(1)
CONH2
(2)
OH
(3)
NH2
(4)
Cl
Ans. (4)
Sol. Aniline and phenol form complex with lewis acid so most reactive among the given compounds for Friedel
Craft reaction is chlrobenzene.
Sol. ,uhfyu rFkk fQukWy yqbZl vEy ds lkFk ladqy cukrs gSA blfy;s fn;s x;s ;kSfxdks es ls Dyksjkscsathu ÝhMy Øk¶V
vfHkfØ;k ds izfr vf/kdre fØ;k'khy gksxkA
2. The correct order of heat of combustion for following alkadienes is :
fuEu ,sYdkMkbUl ds fy, ngu dh Å"ek dk lgh Øe gS &
(A)
(B)
(C)
(1) (b) < (c) < (a) (2) (c) < (b) < (a) (3) (a) < (b) < (c) (4) (a) < (c) < (b)
Ans. (3)
Sol. In isomers of hydrocarbon heat of combustion depends upon their stabilities.
As the stability increases heat of combustion decreases.
Stability order
>
>
Sol. gkbMªksdkcZu ds leko;oh;ksa es ngu dh Å"ek muds LFkkf;Ro ij fuHkZj djrh gSA
LFkkf;Ro c<+us ds lkFk] ngu dh Å"ek es deh gksrh gSA
LFkkf;Ro dk Øe
>
>
3. 'X' melts at low temperature and is a bad conductor of electricity in both liquid and solid state. X is :
(1) Carbon tetrachloride (2) Zinc sulphide (3) Silicon carbide (4) Mercury
'X' fuEu rki ij fi?kyrk gS rFkk nzo rFkk Bksl nksuksa voLFkkvksa esa fo|qr dk dqpkyd gSA X gS %
(1) dkcZu VsVªkDyksjkbM (2) ftad lYQkbM (3) flfydkWu dkckZbM (4) edZjh
Ans. (1)
Sol. CCl4 Non-conductor in solid and liquid phase.
CCl4 Bksl rFkk æo izkoLFkk esa dqpkyd gSA
| JEE MAIN-2020 | DATE : 09-01-2020 (SHIFT-1) | PAPER-3 | OFFICAL | CHEMISTRY
Resonance Eduventures Ltd.Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : 80302RJ2007PLC024029
This solution was download from Resonance JEE (MAIN) 2020 Solution portal
PAGE # 2
7340010333
4. A chemist has 4 samples of artificial sweetener A,B, C and D. To identify these samples, he performed certain experiments and noted the following observations :
(i) A and D both form blue-violet colour with ninhydrin. (ii) Lassaigne extract of C gives positive AgNO3 test and negative Fe4[Fe(CN)6]3 test. (iii) Lassaigne extract of B and D gives positive sodium nitroprusside test. Based on these observations which option is correct ? (1) A : Alitame ; B : Saccharin ; C : Aspartame; D : Sucralose (2) A : Aspartame; B : Alitame ; C : Saccharin ; D : Sucralose (3*) A : Aspartame; B : Saccharin ; C : Sucralose ; D : Alitame (4) A : Saccharin ; B : Alitame ; C : Sucralose ; D : Aspartame
,d dsfeLV ds ikl Ñf=ke e/kqjdksaa A,B, C rFkk D dk 4 izfrn'kZ gSaA bu izfrn'kksZ dks igpkuus ds fy, mlus dqN iz;ksx fd;s x;s rFkk fuEu izs{k.kksa dks uksV fd;k %
(i) A rFkk D nksuksa fuugkbfMªu ds lkFk uhyk cSaxuh jax nsrs gSA
(ii) C dk ySlsa lkjdÙk (Lassaigne extract) AgNO3 ds lkFk ldkjkRed rFkk Fe4[Fe(CN)6]3 ds lkFk udkjkRed ijh{k.k nsrk gSA
(iii) B rFkk D dk ySlsa lkjdÙk lksfM;e ukbVªksiqzlkbM ds lkFk ldkjkRed ijh{k.k nsrk gSA bu izs{k.kksa ds vk/kkj ij dkSu lk fodYi lgh gS\
(1) A : ,syhVse ; B : lSdjhu ; C : ,sLijVse ; D : lqØkykst
(2) A : ,sLijVse; B : ,syhVse ; C : lSdjhu ; D : lqØkykst
(3*) A : ,sLijVse; B : lSdjhu ; C : lqØkykst ; D : ,syhVse
(4) A : lSdjhu ; B : ,syhVse ; C : lqØkykst ; D : ,sLijVse Ans. (3)
Sol. A – Aspartame
B – Saccharine
C – Sucralose
O
D – Alitame
| JEE MAIN-2020 | DATE : 09-01-2020 (SHIFT-1) | PAPER-3 | OFFICAL | CHEMISTRY
Resonance Eduventures Ltd.Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : 80302RJ2007PLC024029
This solution was download from Resonance JEE (MAIN) 2020 Solution portal
PAGE # 3
7340010333
(i) A & D give positive test with ninhydrin because both have free carboxylic and amine groups. (ii) C form precipitate with AgNO3 in the lassaigne extract of the sugar because it has chlorine atoms.
(iii) B & D give positive test with sodium nitroprusside because both have sulphur atoms.
Sol. A – ,sLikVsZe
B – lSdjhu
C – lqØkykst
O
D – ,sfyVse
(i) A vkSj D fuugkbMªhu ds lkFk ldkjkRed ijh{k.k nsrs gSa] D;ksafd nksuksa es isIVkbM cU/k ik;s tkrs gSA
(ii) C 'kdZjk ds yslkus fu"d"kZ eas, AgNO3 ds lkFk vo{ksi nsrk gS] D;ksafd bles Dyksjhu ijek.kq gksrs gSA
(iii) B vkSj D lksfM;e ukbVªksizqlkbM ds lkFk ldkjkRed ijh{k.k nsrk gSA D;ksafd nksuksa es lYQj ijek.kq gksrs gSA
5. The electronic configurations of bivalent europium and trivalent cerium are :
(atomic number : Xe = 54, Ce = 58, Eu = 63)
(1) [Xe] 4f4 and [Xe] 4f9 (2) [Xe] 4f7 and [Xe] 4f1
(3) [Xe] 4f7 6s2 and [Xe] 4f2 6s2 (4) [Xe] 4f2 and [Xe] 4f7
f}la;kstd ;wjksfi;e rFkk f=kla;kstd lhfj;e ds bysDVªkWfud foU;kl gSa %
(ijek.kq la[;k : Xe = 54, Ce = 58, Eu = 63)
(1) [Xe] 4f4 rFkk [Xe] 4f9 (2) [Xe] 4f7 rFkk [Xe] 4f1
(3) [Xe] 4f7 6s2 rFkk [Xe] 4f2 6s2 (4) [Xe] 4f2 rFkk [Xe] 4f7
Ans. (2)
Sol. Eu2+ : [Xe]4f7
Ce3+ : [Xe]4f1
| JEE MAIN-2020 | DATE : 09-01-2020 (SHIFT-1) | PAPER-3 | OFFICAL | CHEMISTRY
Resonance Eduventures Ltd.Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : 80302RJ2007PLC024029
This solution was download from Resonance JEE (MAIN) 2020 Solution portal
PAGE # 4
7340010333
6. Complex X of composition of Cr(H2O)6Cln has a spin only magnetic moment of 3.83 BM. It reacts with
AgNO3 and shows geometrical isomerism. The IUPAC nomenclature of X is :
(1) Dichloridotetraaqa chromium(IV) chloride dehydrate
(2) Hexaaqua chromium(III)chloride
(3) Tetraaquadichlorido chromium(III) chloride dehydrate
(4) Tetraaquadichlorido chromium(IV) chloride dihydrate
Cr(H2O)6Cln la?kVu ds ladqy X dk fLiu ek=k dk pqEcdh; vk?kw.kZ 3.83 BM gSA ;g AgNO3 ds lkFk vfHkfØ;k djrk
gS vkSj T;kferh; leko;ork iznf'kZr djrk gSA X dk IUPAC uke gS %
(1) MkbDyksjkbMkVsVªk,sDok Øksfe;e(IV) DyksjkbM MkbgkbMªsV
(2) gsDlk,DokØksfe;e(III) DyksjkbM
(3) VsVªk,sDokMkbDyksjkbMks Øksfe;e(III) DyksjkbM MkbgkbMªsV
(4) VsVªk,sDokMkbDyksjkbMks Øksfe;e(IV) DyksjkbM MkbgkbMªsV Ans. (3)
Sol. Cr(H2O)6Cln (complex)spin = 3.83 B.M.
From data of magnetic moment oxidation number of Cr should be +3
Hence complex is Cr(H2O)6Cl3.
Complex shows geometrical isomerism therefore formula of complex is [Cr(H2O)4Cl2]Cl2H2O.
It's IUPAC Name: Tetraaquadichloridochromium(III) chloride dihydrate
Cr(H2O)6Cln (ladqy)pØ.k = 3.83 B.M.
pqEcdh; vk?kw.kZ ds eku ls Cr dk vkWDlhdj.k vad +3 gksuk pkfg,A
bl izdkj ladqy Cr(H2O)6Cl3 gSA
ladqy T;kferh; leko;ork n'kkZrk gSA blfy, ladqy dk lw=k [Cr(H2O)4Cl2]Cl2H2O gSA
bldk IUPAC uke VsVªk,DOkkMkbZDyksjkbMksaØksfe;e (III) DyksjkbM MkbZgkbMªsV gSA
7. The major product Z obtained in the following reaction scheme is
fuEu vfHkfØ;k Ldhe esa izkIr gksus okyk eq[; mRikn Z gS %
NaNO2 + HCl
273-278 K
NH2
Br
X Cu2Br2
Y HNO3
H2SO4 Z
(1)
NO2
Br Br
(2)
Br
Br
O2N
(3)
Br Br
NO2
NO2
(4)
NO2
Br
Br
Ans. (4)
Sol.
NH2
Br
(1) NaNO2 + HCl (2) Cu2Br2
N2+Cl–
Br
Br
Br
(3)HNO3
Conc. H2SO4
Br
Br
NO2
| JEE MAIN-2020 | DATE : 09-01-2020 (SHIFT-1) | PAPER-3 | OFFICAL | CHEMISTRY
Resonance Eduventures Ltd.Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : 80302RJ2007PLC024029
This solution was download from Resonance JEE (MAIN) 2020 Solution portal
PAGE # 5
7340010333
8. The major product (Y) in the following reactions is :
fuEu vfHkfØ;kvksa esa eq[; mRikn (Y) gS %
H3C–CH—CCH3
CH3
HgSO4, H2SO4
H3O X
(i) C2H5MgBr, H2O
(ii) Conc. H2SO4Y
(1)
CH3–CH–C=CH2
CH2–CH3
CH3
(2)
CH3–C=C–CH3
CH2–CH3
CH3
(3)
CH3–C–CH–CH3
CH2–CH3
CH2
(4)
CH3–CH–C=CH–CH3
CH3
CH3
Ans. (2)
Sol.
H3C–CH—C–CH3
O CH3
(1) EtMgBr/Ether
(2) H3O
CH3–CH—C–CH3
CH2–CH3
CH3 OH H+/
CH3–C=C–CH3
CH2–CH3
CH3
Sol.
H3C–CH—C–CH3
O CH3
(1) EtMgBr/bZFkj
(2) H3O
CH3–CH—C–CH3
CH2–CH3
CH3 OH H+/
CH3–C=C–CH3
CH2–CH3
CH3
9. B has a smaller first ionization enthalpy than Be. Consider the following statements :
(I) It is easier to remove 2p electron than 2s electron
(II) 2p electron of B is more shielded from the nucleus by the inner core of electrons than the 2s electrons
of Be
(III) 2s electron has more penetration power than 2p electron
(IV) atomic radius of B is more than Be
(atomic number B = 5, Be = 4)
The correct statement are :
(1) (I), (III) and (IV) (2) (II), (III) and (IV)
(3) (I), (II) and (IV) (4) (I), (II) and (III)
B dh izFke vk;uu ,UFkSYih Be ls de gSA fuEu dFkuksa ij fopkj dhft,%
(I) 2s bysDVªkWu dh rqyuk esa 2p bysDVªkWu gVkuk vklku gSA
(II) Be ds 2s bysDVªkWuksa dh rqyuk esa B ds 2p bysDVªkWu vkarfjd dksj bysDVªkWuksa }kjk ukfHkd ls T;knk ifjjf{kr gSaA
(III) 2p bysDVªkWuksa dh rqyuk esa 2s bysDVªkWu dh izos'kh lkeF;Z (penetration power) T;knk gSA
(IV) B dh ijek.kq f=kT;k] Be ls T;knk gSA
(ijek.kq la[;k B = 5, Be = 4)
lgh dFku gS %
(1) (I), (III) rFkk (IV) (2) (II), (III) rFkk (IV)
(3) (I), (II) rFkk (IV) (4) (I), (II) rFkk (III) Ans. (4)
Sol. Theory Based.
lS)kfUrd
| JEE MAIN-2020 | DATE : 09-01-2020 (SHIFT-1) | PAPER-3 | OFFICAL | CHEMISTRY
Resonance Eduventures Ltd.Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
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This solution was download from Resonance JEE (MAIN) 2020 Solution portal
PAGE # 6
7340010333
10. Identify (A) in the following reaction sequence :
(A) (i) CH3MgBr
(ii) H+ , H2O
(3) Conc. H2SO4 /
O3/Zn, H2O
Gives Positive Iodoform test
(B)
C
H3C C=O
O
CH3
H
CH3
fuEu vfHkfØ;k vuqØe esa (A) dh igpku dhft, &
(A) (i) CH3MgBr
(ii) H+ , H2O
(3) lkUnz H2SO4/
O3/Zn, H2O
ldkjkRed vkbMksQkeZ
ijh{k.k nsrk gSA
(B)
C
H3C C=O
O
CH3
H
CH3
(1)
O
CH3 CH3
CH3
(2)
O
CH3
CH3
(3)
O CH3
(4)
O
CH3
CH3
Ans. (4)
Sol. (1) CH3MgBr
(2) H3O+
O3
HO O
H,
Zn, H2O
CH=O
C
H3C CH3
O
CH3
11. According to the following diagram, A reduces BO2 when the temperature is :
200 400 600 800 1000 1200 1400 1600
–1200
–1000
–800
–600
A + O2 AO2
T(ºC)
G
º/kJ
mol–
1
B + O2 BO2
(1) < 1200ºC (2) > 1200ºC but < 1400ºC (3) < 1400ºC (4) > 1400ºC
| JEE MAIN-2020 | DATE : 09-01-2020 (SHIFT-1) | PAPER-3 | OFFICAL | CHEMISTRY
Resonance Eduventures Ltd.Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : 80302RJ2007PLC024029
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PAGE # 7
7340010333
fuEu fp=k ds vuqlkj A, BO2 dk vip;u djrk gS tc rki gS %
200 400 600 800 1000 1200 1400 1600
–1200
–1000
–800
–600
A + O2 AO2
T(ºC)
G
º/kJ
mol–
1
B + O2 BO2
(1) < 1200ºC (2) > 1200ºC ijUrq < 1400ºC
(3) < 1400ºC (4) > 1400ºC Ans. (4)
Sol. A + BO2 B + AO2
G = –ve Only above 1400°C
dsoy 1400°C ds Åij
12. The acidic, basic and amphoteric oxides, respectively, are :
vEyh;] {kkjh; rFkk mHk;/kehZ vkWDlkbMsa Øe'k% gSa & (1) Na2O, SO3, Al2O3 (2) N2O3, Li2O, Al2O3 (3) Cl2O, CaO, P4O10 (4) MgO, Cl2O, Al2O3
Ans. (2) Sol. Non-metal oxides are acidic in nature alkali metal oxides are basic in nature Al2O3 is amphoteric.
vèkkfRod vkWDlkbM vEyh; izdf̀r ds gksrs gSA {kkjh; èkkrq vkWDlkbM {kkjh; izd̀fr ds gksrs gSA Al2O3 mHk;èkehZ gSA
13. If the magnetic moment of a dioxygen species is 1.73 B.M., it may be :
(1) 2O or
2O (2) O2 or
2O (3) O2 or
2O (4) O2,
2O or
2O
,d MkbZvkWDlhtu Lih'kht dk pqEcdh; vk?kw.kZ 1.73 B.M. gS] ;g gks ldrh gS %
(1) 2O vFkok
2O (2) O2 vFkok 2O (3) O2 vFkok
2O (4) O2, 2O vFkok
2O
Ans. (1)
Sol. O2 = 1s2 *1s22s2*2s22pz22px
2 = 2py2 *2px
1 = *2py1
O2– = 1s2 *1s22s2*2s22pz
22px2 = 2py
2 *2px2 = *2py
1
O2+ = 1s2 *1s22s2*2s22pz
22px2 = 2py
2 *2px1 = *2py
0
14. The Ksp for the following dissociation is 1.6 × 10–5
PbCl2(s) aq
2aq Cl2Pb
Which of the following choices is correct for a mixture of 300 mL 0.134M Pb(NO3)2 and 100 mL 0.4 M NaCl ?
(1) Q = Ksp (2) Q > Ksp (3) Not enough data provided (4) Q < Ksp
| JEE MAIN-2020 | DATE : 09-01-2020 (SHIFT-1) | PAPER-3 | OFFICAL | CHEMISTRY
Resonance Eduventures Ltd.Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
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PAGE # 8
7340010333
fuEu fo;kstu ds fy;s Ksp dk eku 1.6 × 10–5 gS]
PbCl2(s) aq
2aq Cl2Pb
0.134M Pb(NO3)2 ds 300 mL rFkk 0.4 M NaCl ds 100 mL dks feykdj cuk;s x;s feJ.k ds fy, fuEu eas ls dkSu lk fodYi lgh gS \
(1) Q = Ksp (2) Q > Ksp
(3) i;kZIr vk¡dM+k miyC/k ugha (4) Q < Ksp
Ans. (2) Sol. Q = [Pb2+][Cl–]2
2300 0.134 100 0.4
400 400
23 0.1340.1
4
= 0.105 × 10–2 =1.005 ×10–3
Q > Ksp
15. If enthalpy of atomisation for Br2(I) is x kJ/mol and bond enthalpy for Br2 is y kJ/mol the relation between them : (1) is x = y (2) is x < y (3) does not exist (4) is x > y
;fn Br2(I) ds fy, d.ku ,UFkSYih (enthalpy of atomisation) x kJ/mol gks rFkk Br2 ds fy, vkcU/k ,UFkSYih y kJ/mol gks] rks muds chp lEcU?k & (1) x = y gksxk (2) x < y gksxk (3) curk ugha gSA (4) x > y gksxk Ans. (4)
Sol.
Br2()
Br2(g)
HVap.
Hatomisation = x kJ/mole 2Br(g)
Bond energy = y kJ/mole
Br2()
Br2(g)
HVap.
ijek.oh;dj.k = x kJ/mole 2Br(g)
caèk ÅtkZ = y kJ/mole
Hatomisation = Hvap + Bond energy Hence x > y
Hijek.oh;dj.k = Hvap + caèk ÅtkZ bl izdkj x > y
16. [Pd(F)(Cl)(Br)(I)]2– has n number of geometrical isomers. Then, the spin-only magnetic moment and crystal field stabilisation energy [CFSE] of [Fe(CN)6]n–6, respectively, are :
[Note : Ignore the pairing energy]
(1) 5.92 BM and 0 (2) 0 BM and –2.4 0
(3) 1.73 BM and –2.0 0 (4) 2.84 BM and –1.6 0
[Pd(F)(Cl)(Br)(I)]2– ds T;kferh; leko;oksa dh la[;k n gSA rc [Fe(CN)6]n–6 dk fLiu ek=k pqEcdh; vk?kw.kZ rFkk fØLVy {ks=k LFkk;hdj.k ÅtkZ [CFSE] Øe'k% gSa%
[uksV : ;qXeu ÅtkZ dks NksM+ nhft,A]
(1) 5.92 BM rFkk 0 (2) 0 BM rFkk –2.4 0
(3) 1.73 BM rFkk –2.0 0 (4) 2.84 BM rFkk –1.6 0
Ans. (3)
| JEE MAIN-2020 | DATE : 09-01-2020 (SHIFT-1) | PAPER-3 | OFFICAL | CHEMISTRY
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7340010333
Sol. Number of Geometrical Isomers in square planar [PdFClBrI]2– are = 3 Hence, n = 3 [Fe(CN)6]3–
Fe3+ = 3d5 , According to CFT configuration is 221 00
2g gt e
2n n
=1.73 B.M.
0 2 0–0.4 0.6g egCFSE nt n
= –0.40 × 5 = –2.00
Sol. oxZ leryh; [PdFClBrI]2– esa T;kfefr; leko;oh;ksa dh la[;k = 3 gSA blfy;s, n = 3
[Fe(CN)6]3–
Fe3+ = 3d5 , CFT ds vuqlkj foU;kl 221 00
2g gt e
2n n
=1.73 B.M.
0 2 0–0.4 0.6g egCFSE nt n = –0.40 × 5 = –2.00
17. The de Broglie wavelength of an electron in the 4th Bohr orbit is
pkSFkh cksj d{kk esa ,d bysDVªkWu dh Mh&czksXyh rjaxnS/;Z gksxh %
(1) 4a0 (2) 2a0 (3) 6a0 (4) 8a0
Ans. (4)
Sol. 2 r = n
2 × 2n
Za0 = n
2 × 24
1a0 = 4
= 8 a0
18. The compound that cannot act both as oxidising and reducing agent is :
og ;kSfxd tks mipk;d rFkk vipk;d nksuksa dh rjg dk;Z ugha dj ldrk gS &
(1) HNO2 (2) H3PO4 (3) H2SO3 (4) H2O2
Ans. (2) Sol. As in H3PO4 Phosphorous is present it's maximum oxidation number state hence it cannot act as reducing
agent.
pwafd H3PO4 esa QkWLQksjl bldh mPpre vkWDlhdj.k voLFkk esa mifLFkr gSA blfy, ;g vipk;d ds leku dk;Z ugha dj ldrk gSA
19. The increasing order of basicity for the following intermediates is (from weak to strong)
fuEu e/;ofrZ;ksa ds fy, {kkjh;rk dk c<+rk Øe gS (nqcZy ls izcy) :
CH3
H3C–C
CH3
(i)
H2C=CH–CH2
(ii)
HCC
(iii)
CH3
(iv)
CN
(v)
(1) (iii) < (iv) < (ii) < (i) < (v) (2) (iii) < (i) < (ii) < (iv) < (v) (3) (v) < (iii) < (ii) < (iv) < (i) (4) (v) < (i) < (iv) < (ii) < (iii) Ans. (3) Sol. Basicity is inversely proportional to electronegativity.
Sol. {kkjh;rk] fo|qr_.krk ds O;qRØekuqikrh gksrh gSA
| JEE MAIN-2020 | DATE : 09-01-2020 (SHIFT-1) | PAPER-3 | OFFICAL | CHEMISTRY
Resonance Eduventures Ltd.Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
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PAGE # 10
7340010333
20. For following reactions
A K700 product
A catalyst
K500 product
it was found that the Ea is decreased by 30 kJ/mol in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assumer pre exponential factor is same) :
(1) 105 kJ/mol (2) 195 kJ/mol (3) 135 kJ/mol (4) 75 kJ/mol
fuEu vfHkfØ;k ds fy,
A K700 mRikn
A mRizsjd
K500 mRikn
;g ik;k x;k gS fd mRizsjd dh mifLFkfr esa Ea , 30 kJ/mol ls ?kV xbZA ;fn nj vifjofrZr jgs rks mRiszfjr vfHkfØ;k ds fy, lafØ;.k ÅtkZ gksxh (eku yhft;s iwoZ pj?kkrkadh xq.kd ogh jgrk gS) :
(1) 105 kJ/mol (2) 195 kJ/mol (3) 135 kJ/mol (4) 75 kJ/mol Ans. (4) Sol. Kcat = K
1
1
Ea
RTAe
=
2
2
Ea
RTAe
1 2
1 2
Ea Ea
T T Ea1 = Ea2 – 30
2Ea 30
500
= 2Ea
700
5Ea2 = 7Ea2 – 210
Ea2 = 210
2 = 105 kJ/mole
Activation energy of the catalysed reaction = 105 – 30 = 75 kJ/mole
mRizsfjr vfHkfØ;k dh lafØ;.k ÅtkZ = 105 – 30 = 75 kJ/mole
SECTION – 2 : (Maximum Marks : 20)
This section contains FIVE (05) questions. The answer to each question is NUMERICAL VALUE with two
digit integer and decimal upto one digit.
If the numerical value has more than two decimal places truncate/round-off the value upto TWO decimal
places.
Full Marks : +4 If ONLY the correct option is chosen.
Zero Marks : 0 In all other cases
[kaM 2 ¼vf/kdre vad% 20)
bl [kaM esa ik¡p (05) iz'u gSA izR;sd iz'u dk mÙkj la[;kRed eku (NUMERICAL VALUE) gSa] tks f}&vadh; iw.kkZad
rFkk n'keyo ,dy&vadu eas gSA
;fn la[;kRed eku esa nks ls vf/kd n’'keyo LFkku gS ] rks la[;kRed eku dks n'keyo ds nks LFkkuksa rd VªadsV@jkmaM
vkWQ (truncate/round-off) djsaA
vadu ;kstuk :
iw.kZ vad % +4 ;fn flQZ lgh fodYi gh pquk x;k gSA
'kwU; vad % 0 vU; lHkh ifjfLFkfr;ksa esaA
| JEE MAIN-2020 | DATE : 09-01-2020 (SHIFT-1) | PAPER-3 | OFFICAL | CHEMISTRY
Resonance Eduventures Ltd.Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : 80302RJ2007PLC024029
This solution was download from Resonance JEE (MAIN) 2020 Solution portal
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7340010333
21. 108 g of silver (molar mass 108 g mol–1) is deposited at cathode from AgNO3(aq) solution by a certain quantity of electricity. The volume (in L) of oxygen gas produced at 273 K and 1 bar pressure from water by the same quantity of electricity is _________________
,d fuf'pr fo|qr ek=kk }kjk AgNO3(tyh;) ls 108 g flYoj (eksyj nzO;eku 108 g mol–1) dSFkksM+ ij fu{ksfir fd;k x;kA fo|qr dh mlh ek=kk }kjk 273 K rFkk 1 ckj nkc ij cuk;h xbZ vkWDlhtu dk vk;ru (L esa) gksxk _________________
Ans. 5.66 to 5.67
Sol. 1081
108Ag deposit
n mole
–Ag e Ag
1F charge is required to deposit 1 mole of Ag
Ag ds ,d eksy fu{ksi.k ds fy, 1 F vkos'k dh vko';drk gksrh gSA
–2 2
12 2
2
H O O H e
2F charge deposit 1
2 mole
2F vkos'k fu{ksfir djrk gS1
2mole
1F charge will deposit 1
4mole
1F vkos'k fu{ksfir djrk gS 1
4mole
2O
nRTV
P
=1 0.08314 273
4 1
2OV = 5.674 L
22. How much amount of NaCl should be added to 600 g for water ( = 1.00 g/mL) to decrease the freezing point of water to –0.2ºC ? __________. (The freezing point depression constant of water = 2 K kg mol–1)
600g ikuh ( = 1.00 g/mL) esa NaCl dh fdruh ek=kk feyk;h tk;s fd mldk fgekad ?kVdj –0.2ºC gks tk;s ?
__________. (ikuh ds fy, fgekad voueu fLFkjkad = 2 K kg mol–1)
Ans. 1.74 to 1.76
Sol. Tf = 0.2°C
Tf = ikfm
0.2 = 2×2×600
1000
5.58
w
41000
6005.582.0w
1.2 58.5
40
= 1.76g
| JEE MAIN-2020 | DATE : 09-01-2020 (SHIFT-1) | PAPER-3 | OFFICAL | CHEMISTRY
Resonance Eduventures Ltd.Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : 80302RJ2007PLC024029
This solution was download from Resonance JEE (MAIN) 2020 Solution portal
PAGE # 12
7340010333
23. The hardness of a water sample containing 10–3 M MgSO4 expressed as CaCO3 equivalents (in ppm) is _____________.
(molar mass of MgSO4 is 120.37 g/mol)
10–3 M MgSO4 okys ty ds izfrn'kZ dh dBksjrk ftldks CaCO3 lerqY; (ppm esa) vfHkO;Dr fd;s tkus ij] gksxh _____________.
(MgSO4 dk eksyj lagfr (molar mass) 120.37 g/mol)
Ans. 100.00 to 100.00
Sol. 10–3 molar MgSO4 10–3 moles of MgSO4 present in 1 L solutions.
10–3 eksyj MgSO4 1 L foy;u esa mifLFkr MgSO4 ds 10–3 eksy
3 4CaCO MgSOn n
3(in term of CaCO )ppm = 310 100
1000
× 106
3(in term of CaCO )ppm = 100 ppm
24. The mass percentage of nitrogen in histamine is _____________
fgLVSfeu esa ukbVªkstu dh nzO;eku izfr'krrk gS _____________
Ans. 37.80 to 38.20
Sol. Structure of Histamine is
N
NH
NH2
Molecular formula of Histamine is C5H9N3
Molecular mass of Histamine is 111
Percentage nitrogen by mass in Histamine = 100111
42 = 37.84%
Sol. fgLVkfeu dh lajpuk
N
NH
NH2
fgLVkfeu dh vkf.od lajpuk C5H9N3
fgLVkfeu dk vkf.od nzO;eku 111
fgLVkfeu es nzO;eku ls ukbVªkstu dk izfr'kr = 100111
42 = 37.84%
25. The molarity of HNO3 in a sample which has density 1.4 g/mL and mass percentage of 63% is __________. (Molecular Weight of HNO3 = 63 )
ml izfrn'kZ esa] ftldk ?kuRo 1.4 g/mL rFkk nzO;eku izfr'krrk 63% dh gks] HNO3 dh eksyjrk gksxh __________.
(HNO3 dk v.kqHkkj = 63 )
Ans. 14.00 to 14.00
Sol. 63% w/w HNO3 solution foy;u
M = 63 1.4
63 100
× 1000 mole/L = 14 mole/L
RReessoonnaannccee EEdduuvveennttuurreess LLttdd..Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
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7340010333