Final JEE - Main Exam September, 2020/02-09-2020/Morning Session
FINAL JEE–MAIN EXAMINATION – SEPTEMBER, 2020 (Held On Wednesday 02nd SEPTEMBER, 2020) TIME : 9 AM to 12 PM
TEST PAPER WITH ANSWER & SOLUTION
CHEMISTRY 1.
The IUPAC name for the following compound is:
Sol. Photoelectric effect (option 2), atomic spectrum
CHO
(option 3) and Black body radiations (option 4) may be explained by quantum theory.
CH3 H3 C
As on increasing temperature, all the values of internal energy becomes possible, it is not directly explained from quantum theory.
COOH (1) 2, 5-dimethyl-6-carboxy-hex-3-enal (2) 6-formyl-2-methyl-hex-3-enoic acid
3.
For the following Assertion and Reason, the
(3) 2, 5-dimethyl-5-carboxy-hex-3-enal
correct option is
(4) 2, 5-dimethyl-6-oxo-hex-3-enoic acid
Assertion (A) : When Cu (II) and sulphide ions are mixed, they react together
Official Ans. by NTA (4)
extremely quickly to give a solid.
CHO 3
Soll. H3C
5
2
EN
6
CH3
Reason (R) : The equilibrium constant of
IUPAC name
4
COOH
Cu2+ (aq) + S2– (aq) CuS(s)
is high because the solubility
1
2, 5-dimethyl-6-oxo-hex-3-enoic acid
The figure that is not a direct manifestation of the
Internal energy of Ar (1)
(1) Both (A) and (R) are true and (R) is the explanation for (A)
LL
quantum nature of atoms is :
300 400 500 600 Temperature (K)
Kinetic energy of (2) photoelectrons
Rb
K Na
A
2.
product is low.
Frequency of incident radiation
Increasing wavelength
(2) Both (A) and (R) are false (3) (A) is false and (R) is true (4) Both (A) and (R) are true but (R) is not the explanation for (A) Official Ans. by NTA (4)
Sol. Slow or fast process is kinetic parameter but extent less or more is thermodynamic parameter. 4.
If AB4 molecule is a polar molecule, a possible geometry of AB4 is : (1) Square pyramidal (2) Tetrahedral
(3)
(3) Square planar
Absorption spectrum
(4) Rectangular planar
T2 > T1
Intensity of black body (4) radiation
Official Ans. by NTA (1)
T1 Wavelength
Official Ans. by NTA (1) 1
Final JEE - Main Exam September, 2020/02-09-2020/Morning Session Sol. (1) If AB4 molecule is a square pyramidal then it has one lone pair and their structure should be
Sol.
A
D
B
Compound
B B
Gas + H2 catalyst
A :
B
C Basic gas
B and it should be polar because dipole moment of lone pair of 'A' never be cancelled by others.
Basic gas (C) must be ammonia (NH3). It means (B) gas should be N2 which is formed by heating of compound (A).
(2) If AB4 molecule is a tetrahedral then it has no lone pair and their structure should be
D
® N2 + Cr2O3 + 4H2O (1) (NH4)2Cr2O7 ¾¾
B B
B
D
® PbO + 2NO2 + (2) Pb(NO3)2 ¾¾
B
D ® 2Na + 3N2 (3) 2NaN3 ¾¾ D
and it should be non polar due to perfect symmetry.
® N2 + 2H2O (4) NH4NO2 ¾¾
(3) If AB4 molecule is a square planar then
So, (A) should not be Pb(NO3)2
B
6.
B
B
properties across a period is
B
(1) Electronegativity
LL
it should be non polar because vector sum of dipole moment is zero. (4) If AB4 molecule is a rectangular planar then
B
B
A
B
A
B
it should be non polar because vector sum of dipole moment is zero. On heating compound (A) gives a gas (B) which
(2) Electron gain enthalpy (3) Ionization enthalpy (4) Atomic radius
Official Ans. by NTA (4) Sol. In general across a period atomic radius decreases while ionisation enthalpy, electron gain enthalpy and electronegativity increases because effective nuclear charge (Zeff) increases. 7.
The statement that is not true about ozone is :
is constituent of air. This gas when treated with H2 in the presence of a catalyst gives another gas
(1) in the stratosphere, it forms a protective shield against UV radiation.
(C) which is basic in nature. (A) should not be:
(2) it is a toxic gas and its reaction with NO gives NO2.
(1) (NH4)2Cr2O7 (2) Pb(NO3)2 (3) NaN3 (4) NH4NO2 Official Ans. by NTA (2) 2
In general, the property (magnitudes only) that shows an opposite trend in comparison to other
A
5.
1 O 2 2
EN
A
(Haber's process)
(3) in the atmosphere, it is depleted by CFCs. (4) in the stratophere, CFCs release chlorine free & which reacts with O3 to give radicals (Cl) chlorine dioxide radicals. Official Ans. by NTA (4)
Final JEE - Main Exam September, 2020/02-09-2020/Morning Session Sol. In the stratosphere, CFCs release chlorine free & radical (Cl)
While with weak field ligands Mn(II) complexes can be high spin because they have more number of unpaired electron (unpaired electron = 5)
UV & Cl(g) CF2Cl2(g) ¾¾¾ ® Cl& (g) + CF 2
(IV) Aqueous solution of Mn(II) ions is pink in colour.
& which react with O3 to give chlorine oxide (ClO) & ) radical. radical not chlorine dioxide (ClO 2
10.
complex with aqua ligands, and the spin only
& + O (g) ® ClO(g) & Cl(g) + O2 (g) 3 8.
The metal mainly used in devising photoelectric
magnetic moment of the complex is 4.90 BM. The geometry and the crystal field stabilization energy
cells is:
of the complex is :
(1) Na
(2) Rb
(1) tetrahedral and –1.6 Dt + 1P
(3) Li
(4) Cs
(2) tetrahedral and –0.6 Dt (3) octahedral and –1.6 D0 (4) octahedral and –2.4 D0 + 2P
EN
Official Ans. by NTA (4) Sol. Cs used in photoelectric cell as it has least ionisation energy. 9.
Consider that a d 6 metal ion (M 2+) forms a
For octahedral Mn(II) and tetrahedral Ni(II) complexes, consider the following statements : (I) both the complexes can be high spin
Official Ans. by NTA (2) Sol. If spin only magnetic moment of the complex is 4.90 BM, it means number of unpaired electrons should be 4.
(II) Ni(II) complex can very rarely be low spin.
(A) In octahedral complex : [M(H 2O)6 ]2+
LL
(III) with strong field ligands, Mn(II) complexes can be low spin.
(IV) aqueous solution of Mn(II) ions is yellow in color.
d6
eg. 0.6 D0 Bary centre
The correct statements are :
A
(1) (I), (III) and (IV) only
(2) (II), (III) and (IV) only
't2g'
–0.4 D0
(3) (I), (II) and (III) only
C.F.S.E. = (–0.4 D0) × 4 + (+0.6 D0) × 2 + 0 × P = –0.4 D0
(4) (I) and (II) only
(B) In tetrahedral complex : [M(H 2O)4 ]2+
Official Ans. by NTA (3) Sol. (I) Under weak field ligand, octahedral Mn(II) and tetrahedral Ni(II) both the complexes are high spin complex. (II) Tetrahedral Ni(II) complex can very rarely be low spin because square planar (under strong ligand) complexes of Ni(II) are low spin complexes. (III) With strong field ligands Mn (II) complexes can be low spin because they have less number of unpaired electron (unpaired electron = 1)
d6
't2' 0.4 Dt Bary centre 'e'
–0.6 Dt
C.F.S.E. = (–0.6 Dt) × 3 + (+0.4 Dt) × 3 + 0 × P = –0.6 Dt 3
Final JEE - Main Exam September, 2020/02-09-2020/Morning Session 11.
(1)
of an organic compound showed presence of 0.08g of bromine. Which of these is the correct structure
CHO Br
(3)
(4) CHO Official Ans. by NTA (2)
NH2 (1)
O
(2) H3C–CH2–Br
CO2H OH
HBr (excess), ¾¾¾¾ ® D
Sol.
Br
Br
NH2 OH
(3)
(4) H3C–Br
Br O
(i) KOH (Alc.) ¾¾¾¾¾¾ ® (ii)H +
Br
3 ¾¾¾¾ ® Zn/H O+ 3
O CHO || + | + H–CH CHO CHO
EN
OH
Official Ans. by NTA (1) Sol. In Carius method
13.
mass of organic compound = 0.172 gm mass of Bromine = 0.08 gm
rate at which water molecules : (1) leaves the vapour increases (2) leaves the solution increases (3) leaves the solution decreases (4) leaves the vapour decreases Official Ans. by NTA (1) Official Ans. by ALLEN (3) Sol. With addition of solute in solvent, surface area for vapourisation decreases causes lowering in vapour pressure 14. Which one of the following graphs is not correct for ideal gas ?
LL
= 46.51%
An open beaker of water in equilibrium with water vapour is in a sealed container. When a few grams of glucose are added to the beaker of water, the
0.08 ´ 100 Hence % of Bromine = 0.172
NH2
80 é ù ´ 100 ú = 46.51% C6H6NBr ê%Br = 172 ë û
(1)
(2)
CO2H
Br
of the compound :
Br
OH
OH
In Carius method of estimation of halogen, 0.172g
A
Br (2) CH3CH2Br
C2H5Br %Br =
80 ´ 100 = 73.33% 109
NH2
(3)
Br
Br
d
C6H5NBr2
T (I)
(4) CH3Br 12.
T (II)
The major aromatic product C in the following reaction sequence will be :
O
HBr (excess), ¾¾¾¾ ® D
d (i) KOH(Alc.) (A) ¾¾¾¾¾¾ ®B (ii)H + O
3 ¾¾¾¾ ®C Zn/H O + 3
4
d
d
1/T P (III) (IV) d = Density, P = Pressure, T = Temperature
Final JEE - Main Exam September, 2020/02-09-2020/Morning Session (1) II
(2) III
(1) (iii) < (iv) < (ii) < (i)
(3) I
(4) IV
(2) (iii) < (iv) < (i) < (ii)
Official Ans. by NTA (1) Sol. PM = dRT Þ d µ 15.
(3) (iii) < (i) < (iv) < (ii)
1 T
(4) (i) < (iii) < (iv) < (ii)
Which of the following compounds will show
Sol. Increasing order of reactivity towards HCN addition
retention in configuration on nucleophic
O || Greater the electrophilicity on –C– group greater the reactivity in nucleophilic addition.
–
substitution by OH ion ? (1) CH3–CH–CH2Br
(2) CH 3–CH–Br
C2H 5
C6H5
O || CH
Br
OCH3
(4) CH3–C–H
(3) CH 3–CH–Br
+R +I
C6H 13
Official Ans. by NTA (1) SN 2
Br
SN2
Ph–CH–OH CH3
OH
17.
OH
SN 2
A
(3)
OH
NO2
LL
Ph
CHO
Et * (NCERT)
(2) CH3–CH–Br
(4)
CH3–CH–Br
OH SN 2
CH3–CH–OH
C6H 13
16.
CHO NO2 (ii)
CHO
O2 N
(iii)
(iii) < (i) < (iv) < (ii) While titrating dilute HCl solution with aqueous NaOH, which of the following will not be required?
(3) Burette and porcelain tile (4) Bunsen burner and measuring cylinder Official Ans. by NTA (4) Sol. Lab manual 18.
Consider the following reactions : (i) Glucose + ROH
CHO
OCH3
NO2
(–R) (–I)
(2) Pipette and distilled water
As language given, we have to go with option (1) as stereochemistry of chiral centre is not distortet. The increasing order of the following compounds
(i)
(–I) only
<
O || CH
(1) Clamp and phenolphthalein
C6 H 3
towards HCN addition is : H 3CO CHO
< OCH3
(–I)
* * OH CH3–CH–CH 2OH Sol. (1) CH3–CH–CH2–Br Et
<
EN
CH3
O || CH
dry HCl ¾¾¾¾ ®
Acetal
x eq.of ¾¾¾¾¾ ® acetyl derivative (CH 3CO)2 O
(iv) 5
Final JEE - Main Exam September, 2020/02-09-2020/Morning Session CH=O | (CH–OH)4 (ii) | CH2–OH
Ni/H 2 y eq.of (ii) Glucose ¾¾¾® A ¾¾¾¾¾ ® acetyl (CH CO) O 3
2
derivative z eq.of ® acetyl derivative (iii) Glucose ¾¾¾¾¾ (CH 3CO)2 O
(2) 4, 5 & 5
(3) 5, 4 & 5
(4) 4, 6 & 5
Ni H2
CH2–OAc | (CH2–OAc) 4 | CH2–OAc
'x', 'y' and 'z' in these reactions are respectively. (1) 5, 6, & 5
CH2–OH | 6 Eq. (CH–OH)4 Ac2O | CH2–OH
OH
Official Ans. by NTA (4)
O
Ac2O OH 5 Eq.
Sol. (iii)
ROH (i) Glucose + dry HCl ¾¾¾ ® Acetal
HO OH
x Eq ¾¾¾¾¾ ® acetyl derivative (CH 3CO)2 O
OAc
EN
Ni/H 2
y Eq ¾¾¾® A ¾¾¾¾¾ ® (CH 3CO)2 O
(ii) Glucose
derivative
(iii) Glucose
z Eq ¾¾¾¾¾ ® (CH 3CO)2 O
Glucose pentacetase
19.
OH
A
CH2OH
O
H OH
H
H
OH
dil.
OH
HCl
H
ROH
HO
H O+
3 ¾¾¾ ® Heat
H3C
CH3
(2)
H
H
OH
OR
CH3 (4)
CH3
CH3
Official Ans. by NTA (3) Å
H 3C
CH=CH2
Sol.
CH3 CH–CH3 H3O+ Heat
O
OAc
H
H
OAc
OR
ring expansion
major 6
OH
CH3
(1)
(3)
O
H OH
OAc
H3 C
CH3
Ac2O (4 Eq)
AcO
CH=CH2
O || R–O–C–CH 3
Acetyl derivative
H
OAc
The major product in the following reaction is :
H 3C
LL
O O || || R–OH + CH 3–C—O–C–CH 3
CH2OH
OAc
OAc
OAc
Acetyl derivative
due to presence of –OH group in Glucose the reaction is
so for (i)
O
Acetyl
D + –H
Å
CH3 CH3
Final JEE - Main Exam September, 2020/02-09-2020/Morning Session 20.
Which of the following is used for the preparation of colloids ?
Official Ans. by NTA (5.00) Sol. No. of chiral centres
(1) Ostwald Process (2) Van Arkel Method (3) Bredig's Arc Method (4) Mond Process Official Ans. by NTA (3) Sol. Bredig's Arc method is used to from metal colloids. 21. The Gibbs energy change (in J) for the given reaction at [Cu2+] = [Sn2+] = 1 M and 298K is: Cu(s) + Sn2+ (aq.) ® Cu2+ (aq.) + Sn(s) ; 0 0 (E Sn = -0.16V, E Cu = 0.34V, 2+ 2+ |Sn |Cu
OH
*
CH H3C *
O * CH3 N
24.
The oxidation states of iron atoms in compounds (A), (B) and (C), respectively, are x, y and z. The
Na 4 [Fe(CN)5 NOS)]
é Sn +2 ù Sol. DG = DGº + RT ln ê +2 ú ë Cu û
Na 4 [FeO 4 ]
(A)
(B)
[Fe 2 (CO) 9 ] (C)
EN
Official Ans. by NTA (6)
æ1ö = –2 × 96500 [(–0.16) – 0.34] + RT ln ç ÷ è1ø = 96500 J The mass of gas adsorbed, x, per unit mass of
x
Sol. (A) Na 4 [Fe(CN)5 (NOS)] (+1)4 + x + (–1)5 + (–1) 1 = 0
x = +2
y
(B) Na 4 [Fe O 4 ]
A
LL
adsorbate, m, was measured at various pressures, x p. A graph between log and log p gives a m straight line with slope equal to 2 and the intercept x equal to 0.4771. The value of at a pressure m of 4 atm is : (Given log 3 = 0.4771) Official Ans. by NTA (6.00) Official Ans. by ALLEN (48.00) x = k p x ...(1) Sol. m x log = log k + x log p Þ E55F m E55F E55F c + m
y
23.
*
N
sum of x,y and z is ____.
Take F = 96500 C mol–1) Official Ans. by NTA (96500.00)
22.
*
x
25.
(+1)4 + y + (–2)4 = 0 y = +4 z
(C) [Fe 2 (CO)9 ] 2z + 0 × 9 = 0
z=0 so (x + y + z) = +2 + 4 + 0 = 6 The internal energy change (in J) when 90g of water undergoes complete evaporation at 100ºC
Given c = log k = 0.4771 or k = 3 slope x = 2 x put in eq. (1) = 3 × (4)2 Þ 48 m The number of chiral carbons present in the
Official Ans. by NTA (189494.00)
molecule given below is _____ .
Official Ans. by ALLEN (189494.39)
H H3 C
(Given : DHvap for water at 373 K = 41 kJ/mol, R = 8.314 JK–1 mol–1)
Sol. H2O(l) H2O(g)
N
DH = DU + DngRT
O
90 gm of H2O Þ 5 moles of H2O
5 × 41000 J = DU + 1 × 8.314 × 373 × 5
C HO
is ____ .
CH3 N
DU = 189494.39 Joule 7
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