JEE Main 2021 Question Paper Chemistry Feb 24 Shift-1

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Feb 24 Shift 1 - Chemistry Question Paper and Solutions

Question 1. What is the reason for the formation of a meta product in the following reaction?

JEE Main 24th Feb Shift 1 Chemistry Paper Question 1

  1. a. Aniline is ortho/para directing
  2. b. Aniline is meta directing
  3. c. In acidic medium, aniline is converted into anilinium ion, which is ortho/para directing
  4. d. In acidic medium, aniline is converted into anilinium ion which is meta directing

Solution:

  1. Answer: (d)

    JEE Main 24th Feb Shift 1 Chemistry Paper Question 1 solution

    In an acidic medium, aniline is converted into anilinium ion, which is meta directing.


Question 2. The missing reagent P is:

JEE Main 24th Feb Shift 1 Chemistry Paper Question 2

JEE Main 24th Feb Shift 1 Chemistry Paper Question 2

    Solution:

    1. Answer: (a)

      JEE Main 24th Feb Shift 1 Chemistry Paper Question 2 solution

      (Pink Colour)


    Question 3. Which force is responsible for the stacking of the α-helix structure of protein? Hydrogen (1H, 2H, 3H) is ________

    1. a. H-bond
    2. b. Ionic bond
    3. c. Covalent bond
    4. d. Van der Waals forces

    Solution:

    1. Answer: (a)

      Hydrogen bond is responsible for the stacking of α-helix structure of protein. α-helix is helical in shape because of hydrogen bonding between >C=O and -NH group.


    Question 4. The gas evolved due to anaerobic degradation of vegetation causes:

    a. Global warming and cancer

    1. a. Global warming and cancer
    2. b. Acid rain
    3. c. Ozone hole
    4. d. Metal corrosion

    Solution:

    1. Answer: (a)

      Anaerobic degradation means the breakdown in the absence of oxygen. During anaerobic degradation of vegetation, CH4 is produced, which is responsible for global warming as it absorbs sun’s heat. It is also known to cause cancer.


    Question 5. Match the following:

    (i) Caprolactam

    (ii) Acrylonitrile

    (iii) 2-chlorobuta-1,3-diene

    (iv) 2-Methylbuta-1,3-diene

    p>(a) Neoprene

    (b) Buna N

    (c) Nylon – 6

    (d) Natural rubber

    1. a. (i) →(b), (ii) → (c), (iii) → (a), (iv) → (d)
    2. b. (i) → (a), (ii) → (c), (iii) → (b), (iv) → (d)
    3. c. (i) → (c), (ii) → (b), (iii) → (a), (iv) → (d)
    4. d. (i) → (c), (ii) → (a), (iii) → (b), (iv) → (d)

    Solution:

    1. Answer: (c)

      (i) Caprolactum is a monomer of nylon-6.

      (ii) Acrylonitrile is one of the monomers of BUNA-N. BUNA-N is a polymer of Buta-1,3-diene and acrylonitrile.

      (iii) 2-Chlorobuta-1,3-diene also known as chloroprene, is a monomer of neoprene.

      (iv) 2-Methyl-buta-1,3-diene upon polymerisation gives natural rubber.

      Hence, the correct matches are (i) → (c), (ii) → (b), (iii) → (a), (iv) → (d).


    Question 6. What is the major product of the following reaction

    JEE Main 24th Feb Shift 1 Chemistry Paper Question 6

    JEE Main 24th Feb Shift 1 Chemistry Paper Question 6

      Solution:

      1. Answer: (a)

        JEE Main 24th Feb Shift 1 Chemistry Paper Question 6 solution

        1. The lone pair of oxygen attacks H+ from HCl

        2. Removal of H2O takes place, forming a 2o carbocation intermediate.

        3. 1,2 Hydride shift takes place, giving a more stable 3o carbocation.

        4. Cl- from HCl acts as a nucleophile and attacks the 3o carbocation to form the product.


      Question 7. What is the major product of the following reaction?

      JEE Main 24th Feb Shift 1 Chemistry Paper Question 7

      JEE Main 24th Feb Shift 1 Chemistry Paper Question 7

        Solution:

        1. Answer: (c)

          JEE Main 24th Feb Shift 1 Chemistry Paper Question 7 solution

          Since HI is an acid, the pi bond attacks H+ and a carbocation intermediate is formed. Since this is a secondary carbocation, 1,2 methyl shift takes place to give a more stable tertiary carbocation. Since I- is a weak base, it will act as a nucleophile and forms the product (c).


        Question 8. Identify the major product:

        JEE Main 24th Feb Shift 1 Chemistry Paper Question 8

        JEE Main 24th Feb Shift 1 Chemistry Paper Question 8

          Solution:

          1. Answer: (b)

            JEE Main 24th Feb Shift 1 Chemistry Paper Question 8 solution


          Question 9. The products A and B are:

            Solution:

            1. Answer: (a)

              Alkenes reacting with cold, dilute KMnO4 will give 1,2-diols (vicinal diols). This is called syn- dihydroxylation. CrO3 cannot oxidise tertiary alcohols. So, the secondary -OH group gets oxidised to ketone.

              JEE Main 24th Feb Shift 1 Chemistry Paper Question 9 solution


            Question 10. Which reagent (A) is used for the following given conversion?

            JEE Main 24th Feb Shift 1 Chemistry Paper Question 10

            1. a. Cu / ∆ / high pressure
            2. b. Molybdenum Oxide
            3. c. Manganese Acetate
            4. d. Potassium Permanganate

            Solution:

            1. Answer: (b)

              Molybdenum Oxide acts as a catalyst and catalyzes the conversion of propane to propanal.

              JEE Main 24th Feb Shift 1 Chemistry Paper Question 10 solution

              This is an oxidative dehydrogenation.


            Question 11. Find A and B.

            JEE Main 24th Feb Shift 1 Chemistry Paper Question 11

            JEE Main 24th Feb Shift 1 Chemistry Paper Question 11

              Solution:

              1. Answer: (c)

                JEE Main 24th Feb Shift 1 Chemistry Paper Question 11 solution


              Question 12. Which of the following pairs are isostructural

              a) TiCl4, SiCl4

              b) SO2-3, CrO2-3

              c) NH3, NO-3

              d) ClF3, BCl3

              1. a. A, B
              2. b. A, C
              3. c. B, C
              4. d. A, D

              Solution:

              1. Answer: (a)

                Isostructural compounds are those that have the same structure as well as same hybridisation.

                In case (B), the structure of SO2-4 is tetrahedral as it has 4σ bonds and zero lone pairs on the central atom. Similarly, CrO2-4 also has 4σ bonds and zero lone pairs on the central atom. According to VSEPR, the structures of SO2-4 and CrO2-4 are:

                JEE Main 24th Feb Shift 1 Chemistry Paper Question 12 solution

                Hence, we can say, SO2-4 and CrO2-4 are isostructural.

                In case (A), TiCl4 has 4σ bonds and zero lone pairs on the central atom. Hence, it is tetrahedral. Similarly, SiCl4 is also tetrahedral.

                JEE Main 24th Feb Shift 1 Chemistry Paper Question 12 solution

                Hence, TiCl4 and SiCl4 are also isostructural.

                In case (C), NH3 has 3σ bonds and 1 lone pair of electrons around nitrogen. Thus, according to VSEPR, NH3 has a pyramidal structure. In NO-3, the nitrogen atom has 3 bonds and no lone pair of electrons. Thus, it is trigonal planar. According to VSEPR, the structures of NH3 and NO-3 are:

                JEE Main 24th Feb Shift 1 Chemistry Paper Question 12 solution

                Hence, NH3 and NO-3 are not isostructural.

                In case (D), ClF3 has 3 σ and 2 lone pairs on the central atom. Its structure is distorted T shaped (but geometry is trigonal bipyramidal). BCl3 has 3 σ bonds and no lone pair on the B atom. Thus, it is trigonal planar.

                JEE Main 24th Feb Shift 1 Chemistry Paper Question 12 solution

                Hence, ClF3 and BCl3 are not isostructural.


              Question 13. Which of the following ores are concentrated by cyanide of group 1st element?

              1. a. Sphalerite
              2. b. Malachite
              3. c. Calamine
              4. d. Siderite

              Solution:

              1. Answer: (a)

                Cyanide acts as a depressant and is therefore used to concentrate the galena-sphalerite ore through the froth floatation process. Hence, option (a) is the correct answer.


              Question 14. S-1: Colourless cupric metaborate is converted into cuprous metaborate in a luminous flame.

              1. a. S-1 is true and S-2 is false
              2. b. S-1 is false and S-2 is true
              3. c. Both are false
              4. d. Both are true

              Solution:

              1. Answer: (c)

                Cupric metaborate (Cu(BO2)2 is blue-green in color and is not colorless, but it does convert into cuprous metaborate in a luminous flame(reducing flame)

                2Cu(BO2)2 + C → 2CuBO2 + B2O3 + CO↑

                Thus, S-1 is incorrect.

                Cuprous metaborate is formed by reacting copper sulphate with boric anhydride in a luminous flame.

                CuSO4 + B2O3 → Cu(BO2)2 + SO3

                Hence, S-2 is also incorrect.


              Question 15. In the given reactions,

              1) I2 + H2O2 + 2OH- → 2I- + 2H2O + O2

              2) H2O2 + HOCl → Cl- + H3O+ + O2

              1. a. H2O2 acts as an oxidising agent in both the reactions
              2. b. H2O2 acts as a reducing agent in both the reactions
              3. c. H2O2 acts as an oxidising agent in reaction (1) and as a reducing agent in reaction (2)
              4. d. H2O2 acts as a reducing agent in reaction (1) and as an oxidizing agent in reaction (2)

              Solution:

              1. Answer: (b)

                In the first reaction, I is reduced from 0 to -1.

                In the second reaction, Cl is reduced from +1 to -1

                Hence, in both the reactions, H2O2 acts as a reducing agent.


              Question 16. EoM2+/ M has a positive value for which of the following elements of 3d transition series?

              1. a. Cu
              2. b. Zn
              3. c. Cr
              4. d. Co

              Solution:

              1. Answer: (a)

                According to the electrochemical series, elements present above H have positive Eo value. For M2+(aq)/ M(s),out of the given options, Cu2+(aq)/Cu(s) will be the answer.

                Cu2+(aq) + 2e- → Cu(s) ; Eo = +0.34 V

                Zn2+(aq) + 2e- → Zn(s) ; Eo = -0.763 V

                Co2+(aq) + 2e- → Co(s) ; Eo = -0.28 V

                Cr2+(aq) + 2e- → Cr(s) ; Eo = -0.9 V


              Question 17. Identify X, Y, Z in the given reaction sequence.

              Al+NaOHXY(g)ZAl + NaOH \rightarrow X\overset{Y(g)}{\rightarrow}Z

              1. a. X = Na[Al(OH)4] ; Y = CO2 ; Z = Al2O3.xH2O
              2. b. X = Na[Al(OH)4] ; Y = SO2 ; Z = Al2O3.xH2O
              3. c. X = Al(OH)3 ; Y = CO2 ; Z = Al2O3
              4. d. X = Al(OH)3 ; Y = SO2 ; Z = Al2O3

              Solution:

              1. Answer: (a)

                Al+NaOHNa[Al(OH)4]CO2NaHCO3+Al(OH)3+H2OAl + NaOH \rightarrow Na[Al(OH)_{4}]\overset{CO_{2}}{\rightarrow}NaHCO_{3} + Al(OH)_{3} + H_{2}O

                Thus, X = Na[Al(OH)4] ; Y = CO2 ; Z = Al2O3.xH2O


              Question 18. The slope of the straight line given in the following diagram for adsorption is:

              JEE Main 24th Feb Shift 1 Chemistry Paper Question 18

              1. a. 1/n (0 to 1)
              2. b. 1/n (0.1 to 0.5)
              3. c. log n
              4. d. log (1/n)

              Solution:

              1. Answer: (a)

                According to Freundlich adsorption isotherm, log (x/m) = log K + 1/n log P.

                So, the slope of the graph of log (x/m) vs log P is 1/n. Where, 1/n is from 0 to 1.


              Question 19. The composition of gun metal is:

              1. a. Cu, Zn, Sn
              2. b. Al, Mg, Mn, Cu
              3. c. Cu, Ni, Fe
              4. d. Cu, Sn, Fe

              Solution:

              1. Answer: (a)

                The components of gun metal are:

                Copper ~ 88%

                Tin ~ 8%
                Zinc ~ 4%
                Majorly, copper


              Question 20. Arrange Mg, Al, Si, P and S in the correct order of their ionisation potentials.

                Solution:

                1. Answer: P > S > Si > Mg > Al

                  In the third period of the periodic table,

                  Element

                  Valence Shell Configuration

                  Na

                  3s1

                  Mg

                  3s2

                  Al

                  3s23p1

                  Si

                  3s23p2

                  P

                  3s23p3

                  S

                  3s23p4

                  Ideally, from left to right, ionization enthalpy increases. But we see exceptions in (Mg, Al) & (P, S) where the element to the left, Mg and P, have higher ionization enthalpies compared to their respective immediate neighbors due to stable electronic configuration.

                  So, the order will be: P > S > Si > Mg > Al


                Section B

                Question 1. Cl2(g)⇌ 2Cl(g)

                For the given reaction at equilibrium, moles of Cl2(g) is equal to the moles of Cl(g) and the equilibrium pressure is 1atm. If Kp of this reaction is x ×10–1, find x.

                  Solution:

                  1. Answer: 5

                    According to the question: Cl2 ⇌ 2Cl [Ptotal = 1 atm].

                    Moles of Cl2 = Moles of Cl (at equilibrium).

                    Given: Kp = x ×10-1

                    nCl2 = nCl.

                    Therefore, PCl2 = PCl

                    Hence, PCl2 = PCl = 0.5 atm

                    kp= (PCl)2/ PCl2 = 0.5 = 5 x 10-1

                    So, x = 5


                  Question 2. S8 + bOH- → cS2- + sS2O2-3 + H2O. Find the value of c.

                    Solution:

                    1. Answer: 4

                      Let us look at the half reactions i.e. oxidation and reduction separately.

                      Oxidation:

                      S8 → S2O2-3

                      S8 + 24OH- → 4S2O2-3 + 12H2O +16e- ….(1)

                      Reduction:

                      S8 → S2-

                      S8 + 16e- → 8S2- ….(2)

                      Adding both the reactions (1) and (2),

                      2S8 + 24OH- → 4S2O2-3 + 8S2- + 12H2O

                      Dividing the whole equation by 2,

                      S8 + 12OH- → 2S2O2-3 + 4S2- + 6H2O

                      So, c = 4


                    Question 3. Calculate the time taken in seconds for 40% completion of a first order reaction, if its rate constant is 3.3× 10-4 sec-1.

                      Solution:

                      1. Answer: 1518

                        t=1kln[Ao][A]=1×2.3033.33×104log[5][3]=1×2.3033.33×104×0.22t = \frac{1}{k}ln\frac{[A_{o}]}{[A]} = \frac{1\times 2.303}{3.33\times 10^{-4}}log\frac{[5]}{[3]} = \frac{1\times 2.303}{3.33\times 10^{-4}}\times 0.22

                        = 30 x 103 x 0.506 = 1518 sec


                      Question 4. For a chemical reaction, Keq is 100 at 300K, the value of ΔGo is –xR Joule at 1 atm pressure. Find the value of x. (Use ln 10 = 2.3)

                        Solution:

                        1. Answer: 1382

                          Given: Keq= 100 at 300K and ΔGo = –xR Joule 

                          ΔGo = -RTln(Keq) = -2.303RTlog(Keq) = -1381.8R

                          Therefore, x = 1381.8 or 1382


                        Question 5. Cu2+ + NH3 ⇌ [Cu(NH3)]2+ K1 = 104

                          Solution:

                          1. [Cu(NH3)]2+ + NH3 ⇌ [Cu(NH3)]2+ K2 = 1.58 × 103

                            [Cu(NH3)2]2+ + NH3 ⇌ [Cu(NH3)3]2+ K3 = 5 × 102

                            [Cu(NH3)3]2+ + NH3 ⇌ [Cu(NH3)4]2+ K4 = 102

                            If the dissociation constant of [Cu(NH3)4]2+ is X × 1012. Determine X.

                            Answer: 1.26

                            Overall reaction constant (β):

                            β = K1× K2 × K3 × K4

                            = 104 (1.58 ×103) × 5 × 102 × 102 = 7.9 × 1011

                            Cu2+ + 4NH3 ⇌ [Cu(NH3)4]2+ ; β = 7.9 × 1011

                            So, the dissociation constant (KDisso.) will be:

                            1β=17.9×1011\frac{1}{\beta } = \frac{1}{7.9 \times 10^{11}}

                            KDisso = 1 / β = 1.26 × 10-12

                            Hence, the value of X = 1.26


                          Question 6. 9.45g of CH2ClCOOH is dissolved in 500 mL of H2O solution and the depression in freezing point of the solution is 0.5°C. Find the percentage dissociation.

                          (Kf)H2O = 1.86K kg mole-1.

                            Solution:

                            1. Answer: 34.4%

                              CH2Cl COOH CH2ClCOO- + H+

                              Initial 1 0 0

                              Dissociated α α α

                              Left (1-α) α α

                              i = final moles / initial moles = 1 - α + α + α / 1 = 1 + α

                              ΔTf = i x Kf × m

                              0.5 = (1 + α) × 1.86 × m

                              Molality (m) = nB / WA (kg) = WBMB x WA (kg) = 9.45 × 1000 / 94.5 × 500 = 0.2

                              Here, A is for solute and B is for solvent

                              WA=WH2O = 500g (Density of H2O = 1g/mL)

                              0.5 = (1+ α) ×1.86 × 0.2

                              α = 0.344

                              Percentage of dissociation = 34.4%


                            Question 7. What is the coordination number in Body Centered Cubic (BCC) arrangement of identical particles?

                              Solution:

                              1. Answer: 8

                                The easiest way is to look at the atom at the body center. It lies at the center of the body diagonal and touches the eight corner atoms. So, coordination number = 8.


                              Question 8. Among the following compounds, how many are amphoteric in nature?

                              Be(OH)2, BeO, Ba(OH)2, Sr(OH)2

                                Solution:

                                1. Answer: 2

                                  The oxide and hydroxide of Be are amphoteric. Hence BeO and Be(OH)2 is amphoteric. Ba(OH)2 and Sr(OH)2 are basic.


                                Question 9. 4.5 g of a solute having molar mass of 90 g/mol is dissolved in water to make a 250 mL solution. Calculate the molarity of the solution.

                                  Solution:

                                  1. Answer: 0.2

                                    WB (given weight of solute) = 4.5 g

                                    MB (Molar mass of solute) = 90 g/mol

                                    VS (Volume of solution) = 250 mL

                                    = 250 / 1000 L = 1/4 L

                                    Molarity (M) = nB / VS(L) = WB / MB.VS(L = 4.5 / (90 ×1/4) = 0.2 molar


                                  Question 10. The mass of Li3+ is 8.33 times the mass of a proton. If Li3+ and proton are accelerated through the same potential difference, then the ratio of de Broglie's wavelength of Li3+ to proton is x ×10–1. Find x

                                    Solution:

                                    1. m (Li3+) = 8.33 × m p+ (given)

                                      Debroglie's wavelength (λ) = h / p

                                      KE = ½ mv2

                                      Multiplying by m on both sides

                                      m. KE = ½ mv2

                                      2m. KE = (mv)2 = p2

                                      P = √2m. KE

                                      Also KE= q × V

                                      So, P = 2mq.V

                                      JEE Main 24th Feb Shift 1 Chemistry Paper Question 10 solution

                                      Comparing with x × 10 -1 = 2 × 10 -1

                                      x = 2


                                    Video Solutions- February 2021 Question Papers

                                    JEE Main 2021 Chemistry Paper February 24 Shift 1

                                    JEE Main 2021 Chemistry Paper With Solutions Feb 24 Shift 1
                                    JEE Main 2021 Chemistry Paper With Solutions Feb 24 Shift 1
                                    JEE Main 2021 Chemistry Paper With Solutions Feb 24 Shift 1
                                    JEE Main 2021 Chemistry Paper With Solutions Feb 24 Shift 1
                                    JEE Main 2021 Chemistry Paper With Solutions Feb 24 Shift 1
                                    JEE Main 2021 Chemistry Paper With Solutions Feb 24 Shift 1
                                    JEE Main 2021 Chemistry Paper With Solutions Feb 24 Shift 1
                                    JEE Main 2021 Chemistry Paper With Solutions Feb 24 Shift 1
                                    JEE Main 2021 Chemistry Paper With Solutions Feb 24 Shift 1
                                    JEE Main 2021 Chemistry Paper With Solutions Feb 24 Shift 1
                                    JEE Main 2021 Chemistry Paper With Solutions Feb 24 Shift 1
                                    JEE Main 2021 Chemistry Paper With Solutions Feb 24 Shift 1
                                    JEE Main 2021 Chemistry Paper With Solutions Feb 24 Shift 1
                                    JEE Main 2021 Chemistry Paper With Solutions Feb 24 Shift 1
                                    JEE Main 2021 Chemistry Paper With Solutions Feb 24 Shift 1
                                    JEE Main 2021 Chemistry Paper With Solutions Feb 24 Shift 1
                                    JEE Main 2021 Chemistry Paper With Solutions Feb 24 Shift 1
                                    JEE Main 2021 Chemistry Paper With Solutions Feb 24 Shift 1
                                    JEE Main 2021 Chemistry Paper With Solutions Feb 24 Shift 1