JEE Main 2021 February 26 – Shift 2 Chemistry Question Paper with Solutions

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JEE Main 2021 26th February Shift 2 Chemistry Question Paper

SECTION A


Question 1. 2,4-DNP test can be used to identify

  1. a. aldehyde
  2. b. halogens
  3. c. ether
  4. d. amine

Solution:

  1. Answer: (a)

    JEE Main 26th Feb Shift 2 Chemistry Paper


Question 2. Identify A in the following chemical reaction.

JEE Main 26th Feb Shift 2 Chemistry Paper Question 2

    Solution:

    1. Answer: (c)

      JEE Main 26th Feb Shift 2 Chemistry Paper Question 2 solution


    Question 3. The nature of charge on resulting colloidal particles when FeCl3 is added to excess of hot water is:

    1. a. positive
    2. b. neutral
    3. c. sometimes positive and sometimes negative
    4. d. negative

    Solution:

    1. Answer: (a)

      If FeCl3 is added to excess hot water, a positively charged sol of hydrated ferric oxide is formed due to the adsorption of Fe3+ ions.


    Question 4. Match List-I with List-II

    JEE Main 26th Feb Shift 2 Chemistry Paper Question 4

    Choose the correct answer from the option given below:

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

    Solution:

    1. Answer: (c)

      JEE Main 26th Feb Shift 2 Chemistry Paper Question 4 solution


    Question 5. In molecule, the hybridization of carbon 1, 2, 3 and 4 respectively are:

    1. a. sp2, sp, sp2, sp3
    2. b. sp2, sp2, sp2, sp3
    3. c. sp2, sp3, sp2, sp3
    4. d. sp3, sp, sp3, sp3

    Solution:

    1. Answer: (a)


      JEE Main 26th Feb Shift 2 Chemistry Paper Question 5 solution


    Question 6. Match List-I with List-II.


    List-II 

    List-II 

    a. Sucrose

    i. b-D-Galactose and b-D-Glucose

    b. Lactose

    ii. a-D-Glucose and b-D-Fructose

    c. Maltose

    iii. a-D- Glucose and a-D-Glucose

    Choose the correct answer from the options given below:

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

    Solution:

    1. Answer: (d)

      Sucrose → a-D- Glucose and b-D- Fructose

      Lactose → b-D- Galactose and b-D- Glucose

      Maltose → a-D- Glucose and a-D- Glucose


    Question 7. Which pair of oxides is acidic in nature?

    1. a. N2O, BaO
    2. b. CaO, SiO2
    3. c. B2O3, CaO
    4. d. B2O3, SiO2

    Solution:

    1. Answer: (d)

      B2O3 and SiO2 both are oxides of non-metal and hence are acidic in nature.


    Question 8. Calgon is used for water treatment. Which of the following statements is NOT true about Calgon?

    1. a. Calgon contains the 2nd most abundant element by weight in the earth's crust
    2. b. It is also known as Graham's salt.
    3. c. It is a polymeric compound and is water-soluble.
    4. d. It does not remove Ca2+ ion by precipitation.

    Solution:

    1. Answer: (a)

      Na6(PO3)6 or Na6P6O18

      Order of abundance of element in earth crust is

      O > Si > Al > Fe > Ca > Na > Mg > K

      So, the second most abundant element in the earth’s crust is Si not Ca.


    Question 9. Ceric ammonium nitrate and CHCl3/alc. KOH are used for the identification of functional groups present in _________and________ respectively.

    1. a. alcohol, amine
    2. b. amine, alcohol
    3. c. alcohol, phenol
    4. d. amine, phenol

    Solution:

    1. Answer: (a)

      Alcohol gives a positive test with ceric ammonium nitrate and primary amines give carbonyl amine test with CHCl3, KOH


    Question 10. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.


    Assertion A: In TlI3, isomorphous to CsI3, the metal is present in +1 oxidation
    state.

    Reason R: Tl metals has fourteen f electrons in its electronic configuration.

    In the light of the above statements, choose the most appropriate Answer from the options given below:

    1. a. Both A and R are correct and R is the correct explanation of A
    2. b. A is not correct but R is correct
    3. c. Both A and R are correct R is NOT the correct explanation of A
    4. d. A is correct but R is not correct

    Solution:

    1. Answer: (c)

      TlI3 is Tl+ I3

      CsI3 is Cs+ I3


    Question 11. Thallium shows Tl+ state due to inert pair effect. The correct order of electron gain enthalpy is

    1. a. S > Se > Te > O
    2. b. O > S > Se > Te
    3. c. S > O > Se > Te
    4. d. Te > Se > S > O

    Solution:

    1. Answer: (a)

      Electron gain enthalpy of O is very low due to small size.


    Question 12. Identify A in the given chemical reaction.


    JEE Main 26th Feb Shift 2 Chemistry Paper Question 12

      Solution:

      1. Answer: (a)

        a.JEE Main 26th Feb Shift 2 Chemistry Paper Question 12 solution
        (Internal aldol condensation)


      Question 13. Match List-I with List-II


      List-I 

      List-II

      a. Siderite

      i. Cu

      b. Calamine 

      ii. Ca

      c. Malachite

      iii. Fe

      d. Cryolite 

      iv. Al

      -

      v. Zn

      Choose the correct answer from the options given below:

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

      Solution:

      1. Answer: (b)

        Siderite - FeCO3

        Calamine - ZnCO3

        Malachite - CuCO3.Cu(OH)2

        Cryolite - Na3AlF6


      Question 14. Identify A in the given reaction.


      JEE Main 26th Feb Shift 2 Chemistry Paper Question 14

      JEE Main 26th Feb Shift 2 Chemistry Paper Question 14

        Solution:

        1. Answer: (b)

          JEE Main 26th Feb Shift 2 Chemistry Paper Question 14 solution


        Question 15. Match List-I with List-II


        List-I 

        List-II

        a. Sodium Carbonate 

        i. Deacon

        b. Titanium 

        ii. Caster-Kellner

        c. Chlorine 

        iii. Van-Arkel

        d. Sodium hydroxide 

        iv. Solvay

        Choose the correct answer from the option given below

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

        Solution:

        1. Answer: (b)

          Sodium carbonate Na2CO3 and NaHCO3

          Titanium : Van arkel method

          Ti + I2 T1 → TiI4(g)

          TiI4(g) T2T2>T1 →TiRefined titanium +2I2(G)

          Chlorine: Decon's process

          HCl+O2 CuCl2 → H2O+Cl2

          Sodium hydroxide: Caster-Kellner cell


        Question 16. Match List-I with List-II.


        List-I 

        List-II

        (Molecule)

        (Bond order)

        a. Ne2

        i. 1

        b. N2

        ii. 2

        c. F2

        iii. 0

        d. O2 

        iv. 3

        Choose the correct answer from the options given below:

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

        Solution:

        1. Answer: (a)

          Ne2 BO = 0

          N2 BO = 3

          F2 BO = 1

          O2 BO = 2

          As per the molecular orbital theory.


        Question 17. Which of the following forms of hydrogen emits low energy b- particles?

        1. a. Proton H+
        2. b. Deuterium 12H_{1}^{2}\textrm{H}
        3. c. Protium 11H_{1}^{1}\textrm{H}
        4. d. Tritium 13H_{1}^{3}\textrm{H}

        Solution:

        1. Answer: (d)

          Tritium isotope of hydrogen is radioactive and emits low energy b particles. It is because

          of high n/p ratio of the tritium which makes the nucleus unstable.


        Question 18. Choose the correct order of the basic nature of the below given amines.


        A. Phenyl methanamine

        B. N, N-Dimethylaniline

        C. N-Methyl aniline

        D. Benzenamine

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

        Solution:

        1. Answer: (d)

          JEE Main 26th Feb Shift 2 Chemistry Paper Question 18 solution


        Question 19. Considering the given reaction, the major product among the following is:


        JEE Main 26th Feb Shift 2 Chemistry Paper Question 19

          Solution:

          1. Answer: (c)


            JEE Main 26th Feb Shift 2 Chemistry Paper Question 19 solution


          Question 20. Seliwanoff test and Xanthoproteic test are used for the identification of __________ and ___________ respectively

          1. a. ketoses, proteins
          2. b. proteins, ketoses
          3. c. aldoses, ketoses
          4. d. ketoses, aldoses

          Solution:

          1. Answer: (a)

            Seliwanoff test and Xanthoproteic test are used for identification of 'Ketoses' and proteins respectively.


          Question 1. The NaNO3 weighed out to make 50 mL of an aqueous solution containing 70.0 mg Na+ per mL is ________g. (Rounded off to the nearest integer)
          [Given: Atomic weight in g mol–1. Na: 23; N: 14; O: 16.

            Solution:

            1. Answer: 13

              Na+ = 70 mg/mL

              WNa in 50 mL solution = 70 × 50 mg = 3500 mg = 3.5 g

              Moles of Na+ in 50 ml solution =3.5/23

              Moles of NaNO3 = moles of Na+ = 3.5/23 mol

              Mass of NaNO3 =3.5 / 23 × 85 = 12.94 13 g Ans.


            Question 2. The number of stereoisomers possible for [Co(ox)2(Br)(NH3)]2- is ____________. [ox = oxalate]

              Solution:

              1. Answer: 3

                [Co(Ox)2Br(NH3)]2-

                JEE Main 26th Feb Shift 2 Chemistry Paper Question 2 solution

                Total stereoisomer = 2 (OI) + 1 POE (pair of enantiomers) = 3


              Question 3. The average S–F bond energy in kJ mol–1 of SF6 is __________. (Rounded off to the nearest integer)
              [Given : The values of standard enthalpy of formation of SF6(g), S(g) and F(g) are - 1100, 275 and 80 kJ mol–1 respectively.]

                Solution:

                1. Answer: 309

                  SF6(g) → S(g) + 6F(g)

                  JEE Main 26th Feb Shift 2 Chemistry Paper Question 3

                  6 × ES-F = 275 + 6×80 – (–1100) = 275 + 480 + 1100

                  6 × ES-F = 1855

                  ES-F = 1855/6 = 309.1667 = 309 kJ/mol


                Question 4. E.m.f of the following cell at 298 K in V is x ×10-2.

                Zn|Zn2+ (0.1 M)||Ag+(0.01 M)| Ag


                The value of x is _________. (Rounded off to the nearest integer)

                [Given:EZn2/Zn0=0.76;EAg+/Ag0=+0.80V;2.303RTF=0.059][Given: E_{Zn^{2}/Zn}^{0} = -0.76; E_{Ag^{+}/Ag}^{0} = +0.80V; \frac{2.303RT}{F} = 0.059]

                  Solution:

                  1. Answer: 147

                    Zn(s)|Zn+2(0.1M)||Ag+(0.01M)|Ag(s)

                    ( ) + 2 ⇄ 2Ag(s) + Zn+2

                    Eo = 0.80 + 0.76 = 1.56; Q = {Zn2+ / (Ag2)2}

                    E = Eo - 0.059/n log(Q)

                    E = 1.56 - 0.059 / 2 log [0.1 / (0.01)2]

                    E = 1.56 - 0.059 / 2 log [(10)3]

                    E = 1.4715 = 147.15 × 10-2 volt = x × 10-2

                    X = 147.15 ≃ 147


                  Question 5. A ball weighing 10g is moving with a velocity of 90ms-1. If the uncertainty in its velocity is 5%, then the uncertainty in its position is ________×10-33m. (Rounded off to the nearest integer)


                  [Given: h = 6.63×10-34 J s]

                    Solution:

                    1. Answer: 1

                      m = 10 g = 10-2 kg

                      v = 90 m/s

                      Δv = v × 5% = 90 × 5 / 100 = 4.5 m/s

                      JEE Main 26th Feb Shift 2 Chemistry Paper Question 5

                      Δx ≥ 1.17 × 10-33 = x × 10-33

                      x = 1.17≃1


                    Question 6. In a mildly alkaline medium, thiosulphate ion is oxidized by to "A". The oxidation state of sulphur in "A" is ________.

                      Solution:

                      1. Answer: +6

                        S2O2-3 + MnO-4  Alkaline Medium →A

                        A → SO2-4

                        Oxidation no. of 'S' = +6


                      Question 7. When 12.2 g of benzoic acid is dissolved in 100g of water, the freezing point of solution was found to be –0.93oC (Kf (H2O) = 1.86 K kg mol-1). Then the number (n) of benzoic acid molecules associated (assuming 100% association) is_____________.

                        Solution:

                        1. Answer: (2)

                          n PhCOOH (PhCOOH)n

                          N = 1/x = i{As, α=1}

                          ∆Tf = i × kf × m

                          0.93 = 1/n × 1.86 × 12.2 × 1000 / 122 × 100; n = 2


                        Question 8. If the activation energy of a reaction is 80.9 kJ mol–1, the fraction of molecules at 700K, having enough energy to react to form products is e–x. The value of x is ______. (Rounded off to the nearest integer)

                        [Use R = 8.31 JK-1 mol-1]

                          Solution:

                          1. Answer: (14)

                            Ea = 80.9 kJ / mol

                            Fraction of molecules able to cross energy barrier = e - Ea / RT = e-x

                            x = Ea / RT = 80.9 × 1000 / 8.31 × 700 = 13.91

                            x ≃ 14


                          Question 9. The pH of ammonium phosphate solution, if pka of phosphoric acid and pkb of ammonium hydroxide are 5.23 and 4.75 respectively, is_____________.

                            Solution:

                            1. Answer: 7

                              (NH4)3PO43NH4++PO43\left ( NH_{4} \right )_{3}PO_{4}\leftrightharpoons 3NH_{4}^{+} + PO_{4}^{3-}

                              [H+]=Ka×kwka×kb[H^{+}] = K_{a}\times \sqrt{\frac{kw}{k_{a}\times k_{b}}}

                              pH = pka + ½ {pkw - pka - pkb}

                              pH = 5.23 + ½ {14–5.23–4.75}

                              pH = 5.23 + ½ (4.02) = 7.24 = 7(Nearest integer)


                            Question 10. The number of octahedral voids per lattice site in a lattice is __________.


                            (Rounded off to the nearest integer)

                              Solution:

                              1. Answer: 1

                                Assuming FCC

                                No of lattice sites = 6 face centre + 8 corner = 14

                                No. of octahedral voids = 13

                                Ratio = 13/14 = 0.92857 = 1 (nearest integer)


                              Video Solutions- February 2021 Question Papers

                              JEE Main 2021 Chemistry Paper February 26 Shift 2

                              JEE Main 2021 Chemistry Paper With Solutions Feb 26 Shift 2
                              JEE Main 2021 Chemistry Paper With Solutions Feb 26 Shift 2
                              JEE Main 2021 Chemistry Paper With Solutions Feb 26 Shift 2
                              JEE Main 2021 Chemistry Paper With Solutions Feb 26 Shift 2
                              JEE Main 2021 Chemistry Paper With Solutions Feb 26 Shift 2
                              JEE Main 2021 Chemistry Paper With Solutions Feb 26 Shift 2
                              JEE Main 2021 Chemistry Paper With Solutions Feb 26 Shift 2
                              JEE Main 2021 Chemistry Paper With Solutions Feb 26 Shift 2
                              JEE Main 2021 Chemistry Paper With Solutions Feb 26 Shift 2
                              JEE Main 2021 Chemistry Paper With Solutions Feb 26 Shift 2
                              JEE Main 2021 Chemistry Paper With Solutions Feb 26 Shift 2
                              JEE Main 2021 Chemistry Paper With Solutions Feb 26 Shift 2
                              JEE Main 2021 Chemistry Paper With Solutions Feb 26 Shift 2
                              JEE Main 2021 Chemistry Paper With Solutions Feb 26 Shift 2
                              JEE Main 2021 Chemistry Paper With Solutions Feb 26 Shift 2