paper 1 questions & solutions

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22-05-2016JEE (Advanced) 2016

®

DATE: 03-10-2021

JEE (ADVANCED) 2021

PAPER 1

Questions & Solutions

Time : 9:00 am – 12:00 pm | Duration : 3 Hrs. | Total Marks : 180

SSUUBBJJEECCTT:: CCHHEEMMIISSTTRRYY

mailto:[email protected]

http://www.resonance.ac.in/reso/results/jee-main-2014.aspx





PAPER-1 : INSTRUCTIONS TO CANDIDATES

Question Paper-1 has three (03) parts: Physics, Chemistry and Mathematics.

Each part has a total Nineteen (19) questions divided into four (04) sections (Section-1, Section-2, Section-3 & Section-4)

Total number of questions in Question Paper-1 are Fifty Seven (57) and Maximum Marks are One Hundred and Eighty (180).

Type of Questions and Marking Schemes

SECTION-1

• This section contains FOUR (04) questions. • Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct

answer. • For each question, choose the option corresponding to the correct answer. • Answer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If ONLY the correct option is chosen; Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered); Negative Marks : −1 In all other cases.

SECTION 2 • This section contains THREE (03) question stems. • There are TWO (02) questions corresponding to each question stem. • The answer to each question is a NUMERICAL VALUE. • For each question, enter the correct numerical value corresponding to the answer in the designated place

using the mouse and the on-screen virtual numeric keypad. • If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal

places. • Answer to each question will be evaluated according to the following marking scheme: Full Marks : +2 If ONLY the correct numerical value is entered at the designated place; Zero Marks : 0 In all other cases.

SECTION 3

• This section contains SIX (06) questions. • Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four

option(s) is (are) correct answer(s). • For each question, choose the option(s) corresponding to (all) the correct answer(s). • Answer to each question will be evaluated according to the following marking scheme: Full Marks : +4 If only (all) the correct option(s) is(are) chosen; Partial Marks : +3 If all the four options are correct but ONLY three options are chosen; Partial Marks : +2 If three or more options are correct but ONLY two options are chosen, both of

which are correct; Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a correct

option; Zero Marks : 0 If unanswered; Negative Marks : −2 In all other cases. • For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct

answers, then choosing ONLY (A), (B) and (D) will get +4 marks; choosing ONLY (A) and (B) will get +2 marks; choosing ONLY (A) and (D) will get +2marks; choosing ONLY (B) and (D) will get +2 marks; choosing ONLY (A) will get +1 mark; choosing ONLY (B) will get +1 mark; choosing ONLY (D) will get +1 mark; choosing no option(s) (i.e. the question is unanswered) will get 0 marks and choosing any other option(s) will get −2 marks.



JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-1 | CHEMISTRY

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PAGE # 1

7340010333

PART : CHEMISTRY

SECTION-1

• This section contains FOUR (04) questions. • Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct

answer. • For each question, choose the option corresponding to the correct answer. • Answer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If ONLY the correct option is chosen; Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered); Negative Marks : −1 In all other cases.

1. The major product formed in the following reaction is

NaNH2

NA liq.NH3

(A)

(B)

(C)

(D)

Ans. (B) Sol. It is a birch reduction. In Birch reduction only internal Alkyne can reduced to trans alkene. While terminal alkyne can’t be reduced to alkene. 2. Among the following, the conformation that corresponds to the most stable conformation of meso-

butane-2,3-diol is

(A)

H Me

H Me

OH

OH

(B)

H OH

MeH

Me

OH

(C)

H

OH

Me

Me

H

OH

(D)

H OH

Me H

Me

OH

Ans. (B) Sol. C & D are not meso compound but A & B are meso compound and out of A and B. B is more stable due

to intramolecular hydrogen bonding.




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7340010333

3. For the given close packed structure of a salt made of cation X and anion Y shown below (ions of

only one face are shown for clarity), the packing fraction is

approximately (packing fraction = 100

efficiencypacking)

(A) 0.74 (B) 0.63 (C) 0.52 (D) 0.48 Ans. (B)

Sol. 2[rx + ry] = 2 a {a = 2ry ; ry = 0.59}

2[rx + ry] = 2 (2ry)

2rx = 2 2 ry – 2ry

rx = ry [ 2 – 1]

rx = 0.414 ry = 0.207 a

Packing fraction = 3

3y

3x

a

r3

4r

3

43

= 0.63

4. The calculated spin only magnetic moments of [Cr(NH3)6]3+ and [CuF6]3– in BM, respectively, are

(Atomic numbers of Cr and Cu are 24 and 29, respectively) (A) 3.87 and 2.84 (B) 4.90 and 1.73 (C) 3.87 and 1.73 (D) 4.90 and 2.84

Ans. (A)

Sol. [Cr(NH3)6]3+ Cr3+ 3d3 unpaired e– = 3

= 3.87 BM [CuF6]3– Cu3+ 3d8 unpaired e– = 2

= 2.87 BM

SECTION 2 • This section contains THREE (03) question stems. • There are TWO (02) questions corresponding to each question stem. • The answer to each question is a NUMERICAL VALUE. • For each question, enter the correct numerical value corresponding to the answer in the designated place

using the mouse and the on-screen virtual numeric keypad. • If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal

places. • Answer to each question will be evaluated according to the following marking scheme: Full Marks : +2 If ONLY the correct numerical value is entered at the designated place; Zero Marks : 0 In all other cases.




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PAGE # 3

7340010333

Question Stem for Question Nos. 5 to 6 Question Stem

For the following reaction scheme, percentage yields are given along the arrow:

H2O Mg2C3 P

(4.0 g)

T 80%

NaNH2 MeI

Q

red hot iron tube

873 K

40%

Hg2+/H+ 333 K

Ba(OH)2 NaOCl U

(y g)

R (x g) 75%

100%

S heat 80%

(decolourises) Baeyer's reagent)

x g and y g are mass of R and U, respectively.

(Use: Molar mass (in g mol–1) of H, C and O as 1, 12 and 16, respectively) 5. The value of x is ______. Ans. (1.62) 6. The value of y is ______. Ans. (3.2) Sol. (Q.5 & Q.6)

H3O+

Mg2C3 CH3–CCH NaNH2

MeI

– –

Hg2+/H+

–

– – –

– – –

CH3–C–CH3

CH3–CC–CH3

0.1 mol 0.1 × 0.75 = 0.075

O

Ba(OH)2

CH3–C–CH=C–CH3

O

NaOCl

CH3

T 0.05 × 0.08 = 0.04 moles

CHCl3 + HO–C–CH=C

O

CH3

CH3

0.04×0.8 ×100 = 3.2 gm

CH3

CH3 CH3

CH3

CH3 CH3

Red hot tube

0.075 × 0.4 ×1/3 = 1.62

So X = 1.62 ; Y = 3.2




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PAGE # 4

7340010333

Question Stem for Question Nos. 7 and 8

Question Stem

For the reaction, 𝐗(𝑠) 𝐘(𝑠) + 𝐙(𝑔), the plot of ln op

pzversus

T

104

is given below (in solid line), where 𝑝𝐙 is the pressure (in bar) of the gas Z at temperature T and op = 1 bar.

(Given, R

H

T

1d

)K(lnd o

, where the equilibrium constant, op

pzK and the gas constant, R = 8.314 J K–1

mol–1)

7. The value of standard enthalpy, oH (in kJ mol–1) for the given reaction is ___. Ans. (166.28)

8. The value of oS (in J K–1 mol–1) for the given reaction, at 1000 K is ___.

Ans. (141.34)

Sol. (7 & 8)

Gº =Hº – TSº

–RTlnK = Hº – TSº

lnk =

R

ºS

T

1

R

ºH

y = mx + c

For given reaction K = (P2)

Slope = 21012

)3(710

R

ºH 4

= Hº = (2 × 104)R = 16.628 × 104 J = 166.28 kJ

(ii) y = mx + c

– 3 = –2

T

104

+ c

– 3 = –2 [10] + c

c = 17

But c = R

ºS = 17

Sº = 17 × 8.314 = 141.338 J/K




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PAGE # 5

7340010333

Question Stem for Question Nos. 9 and 10 Question Stem

The boiling point of water in a 0.1 molal silver nitrate solution (solution A) is x C. To this solution A, an

equal volume of 0.1 molal aqueous barium chloride solution is added to make a new solution B. The

difference in the boiling points of water in the two solutions A and B is y × 10–2 C.

(Assume: Densities of the solutions A and B are the same as that of water and the soluble salts

dissociate completely.

Use: Molal elevation constant (Ebullioscopic Constant), 𝐾𝑏 = 0.5 K kg mol–1; Boiling point of pure water

as 100 C.)

9. The value of x is ___.

Ans. (100.1)

10. The value of |y| is ___.

Ans. (2.5)

Sol. (9 & 10)

Initially for solution A [0.1 m AgNO3]

Tb = i Kb m

Tb = 2 × 0.5 × 0.1

T’b – 100 = 0.1

T’b = 100.1°C

x = 100.1°C

(II) 2AgNO3 + BaCl2 Ba(NO3)2 + 2AgCl

0.1V 0.1V

ion milimole

Ag+ 0.1V

NO3– 0.1V

Ba2+ 0.1V

Cl– 0.2V

Tb = i Kb × m

= 0.5

V2

V1.0V1.0V1.0

= 0.5 × 2

3.0

T’b – 100 = 0.075

T’b = 100.075

Differece in Tb = 100.1 – 100.075 = 0.025 = 2.5 × 10–2

y = 2.5




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PAGE # 6

7340010333

SECTION 3

• This section contains SIX (06) questions. • Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four

option(s) is (are) correct answer(s). • For each question, choose the option(s) corresponding to (all) the correct answer(s). • Answer to each question will be evaluated according to the following marking scheme: Full Marks : +4 If only (all) the correct option(s) is(are) chosen; Partial Marks : +3 If all the four options are correct but ONLY three options are chosen; Partial Marks : +2 If three or more options are correct but ONLY two options are chosen, both of

which are correct; Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a correct

option; Zero Marks : 0 If unanswered; Negative Marks : −2 In all other cases. • For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct

answers, then choosing ONLY (A), (B) and (D) will get +4 marks; choosing ONLY (A) and (B) will get +2 marks; choosing ONLY (A) and (D) will get +2marks; choosing ONLY (B) and (D) will get +2 marks; choosing ONLY (A) will get +1 mark; choosing ONLY (B) will get +1 mark; choosing ONLY (D) will get +1 mark; choosing no option(s) (i.e. the question is unanswered) will get 0 marks and choosing any other option(s) will get −2 marks.

11. Given :

OH H

H

CH2OH

CHO

OH H

OH H

HO HNO3

P

[D = +52.7º]

D-Glucose

The compound(s), which on reaction with HNO3 will give the product having degree of rotation,

[]D = –52.7º is (are)

(A)

H

H

CH2OH

CHO

OH

OH

HO

HO

H

H

(B)

H

H

CH2OH

CHO

OH

H

HO

HO

H

HO




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PAGE # 7

7340010333

(C)

H

OH

CHO

CH2OH

H

H

H

HO

HO

HO

(D)

OH

OH

CH2OH

CHO

H

OH

H

H

HO

H

Ans. (CD) 12. The reaction of Q with PhSNa yields an organic compound (major product) that gives positive Carius

test on treatment with Na2O2 followed by addition of BaCl2. The correct option(s) for Q is (are)

(A) O2N F

NO2

(B)

O2N

I

O2N

(C)

MeS

Br

O2N

(D) O2N

Cl

MeS

Ans. (AD)

13. The correct statement(s) related to colloids is(are)

(A) The process of precipitating colloidal sol by an electrolyte is called peptization.

(B) Colloidal solution freezes at higher temperature than the true solution at the same concentration.

(C) Surfactants form micelle above critical micelle concentration (CMC). CMC depends on temperature.

(D) Micelles are macromolecular colloids

Ans. (C) Sol. Surfactants form micelle above critical micelle concentration (CMC). CMC depends on temperature.

14. An ideal gas undergoes a reversible isothermal expansion from state I to state II followed by a reversible adiabatic expansion from state II to state III. The correct plot(s) representing the changes from state I to state III is(are) (p: pressure, V: volume, T: temperature, H: enthalpy, S: entropy)

(A)

(B)

(C) (D)




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PAGE # 8

7340010333

Ans. (ABD)

Sol. (i) For isothermal process T constant and for reversible adiabatic process entropy is constant.

(ii) For isothermal process H = 0

(iii) Slope of adiabatic process is times of isothermal process

So, graph AD is correct.

15. The correct statement(s) related to the metal extraction processes is(are)

(A) A mixture of PbS and PbO undergoes self-reduction to produce Pb and SO2.

(B) In the extraction process of copper from copper pyrites, silica is added to produce copper silicate.

(C) Partial oxidation of sulphide ore of copper by roasting, followed by self-reduction produces blister

copper.

(D) In cyanide process, zinc powder is utilized to precipitate gold from Na[Au(CN)2].

Ans. (ACD)

Sol. (A) PbS + PbO Pb + SO2

(Self reduction of Pb)

(B) In extraction process of Cu SiO2 is added to make FeSiO3

(C) Zn + Na[Au(Cu)2] Au(s) + Na2[Zn(CN)4]

16. A mixture of two salts is used to prepare a solution S, which gives the following results:

Dilute NaOH(aq)

Room temperature

white precipitate(s)

only Room temperature

White precipitate(s)

only

S (aq solution of the salts)

Dilute HCl(aq)

The correct option(s) for the salt mixture is (are)

(A) Pb(NO3)2 and Zn(NO3)2 (B) Pb(NO3)2 and Bi(NO3)3

(C) AgNO3 and Bi(NO3)3 (D) Pb(NO3)2 and Hg(NO3)2

Ans. (ABC)

Sol.

PbCl2

Pb+2

HCl Ag+

White

AgCl White

Zn(OH)2 Zn+2

NaOH Hg+2

White

HgO Red brown

Bi(OH)3 White Bi+3




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PAGE # 9

7340010333

SECTION 4

• This section contains THREE (03) questions.

• The answer to each question is a NON-NEGATIVE INTEGER.

• For each question, enter the correct integer corresponding to the answer using the mouse and the on-

screen virtual numeric keypad in the place designated to enter the answer.

• Answer to each question will be evaluated according to the following marking scheme:

Full Marks : +2 If ONLY the correct numerical value is entered at the designated place;

Zero Marks : 0 In all other cases.

17. The maximum number of possible isomers (including stereoisomers) which may be formed on mono-bromination of 1-methylcyclohex-1-ene using Br2 and UV light is _____ .

Ans. 13

Sol. This is allylic substitution at – carbon

Cl2

UV (553K)

Cl

+

Cl

+

•

Cl

1 2

2

+Cl

2

+

Cl

2

+

2

+Cl

2

18. In the reaction given below, the total number of atoms having sp2 hybridization in the major product P is

______. (1) O3 (excess)

then Zn/H2O

(2) NH2OH (excess) P

Ans. 12 Sol.

NH2 – OH

O O

O O

N – OH HO – N

N – OH N – OH




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PAGE # 10

7340010333

19. The total number of possible isomers for [Pt(NH3)4Cl2]Br2 is ______ . Ans. (6)

Sol. Total number of isomer of complex [Pt(NH3)4Cl2]Br2

(1) Cis - [Pt(NH3)4Cl2]Br2, (2) Trans - [Pt(NH3)4Cl2]Br2,

NH3

Pt

NH3 Cl

Cl

Cis form

NH3

NH3

NH3

Pt

NH3 NH3

NH3

Cl

Cl

Trans form

(3) Cis - [Pt(NH3)4BrCl]ClBr (4) Trans - [Pt(NH3)4BrCl]ClBr

NH3

Pt

NH3 Cl

Br

Cis form

NH3

NH3

NH3

Pt

NH3 NH3

NH3

Br

Cl

Trans form

(5) Cis - [Pt(NH3)4Br2]Cl2 (6) Trans - [Pt(NH3)4Br2]Cl2

NH3

Pt

NH3 Br

Br

Cis form

NH3

NH3

NH3

Pt

NH3 NH3

NH3

Br

Br

Trans form





SECTION 4

• This section contains THREE (03) questions.

• The answer to each question is a NON-NEGATIVE INTEGER.

• For each question, enter the correct integer corresponding to the answer using the mouse and the on-

screen virtual numeric keypad in the place designated to enter the answer.

• Answer to each question will be evaluated according to the following marking scheme:

Full Marks : +2 If ONLY the correct numerical value is entered at the designated place;

Zero Marks : 0 In all other cases.



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