JEE Advanced Question Paper 2021 Chemistry Paper 2

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JEE Advanced 2021 Paper 2 Chemistry Question Paper

Question 1. The reaction sequence(s) that would lead to o-xylene as the major product is(are).


JEE Advanced Question Paper 2021 Chemistry Paper 2 Question 1

    Solution:

    1. Answer: (a, b)

      JEE Advanced Question Paper 2021 Chemistry Paper 2 Question 1 solution

      JEE Advanced Question Paper 2021 Chemistry Paper 2 Question 1 solution


    Question 2. Correct option(s) for the following sequence of reactions is(are)


    JEE Advanced Question Paper 2021 Chemistry Paper 2 Question 2

    1. a. Q = KNO2, W = LiAlH4
    2. b. R = benzenamine, V = KCN
    3. c. Q = AgNO2, R = phenylmethanamine
    4. d. W = LiAlH4, V = AgCN

    Solution:

    1. Answer: (c, d)

      JEE Advanced Question Paper 2021 Chemistry Paper 2 Question 2 solution

      Therefore, correct options are

      Q = AgNO2, R = phenylmethanamine

      W = LiAlH4, V = AgCN


    Question 3. For the following reaction;


    2X+YkP2X + Y\overset{k}{\rightarrow}P

    The rate of reaction is d[P]dt=k[X]\frac{d[P]}{dt} = k[X]. Two moles of X are mixed with one mole of Y to make 1.0 L of solution. At 50 s, 0.5 mole of Y is left in the reaction mixture. The correct statement(s) about the reaction is(are). (Use: ln 2 = 0.693)

    a. The rate constant, k, of the reaction is 13.86 × 10-4 s-1.

    1. a. The rate constant, k, of the reaction is 13.86 × 10-4 s-1.
    2. b. Half-life of X is 50 s.
    3. c. At 50 s, - d[X] / dt = 13.86 × 10-3 mol L-1 s-1.
    4. d. At 100 s, d[Y] / dt = 3.46 × 10-3 mol L-1 s-1.

    Solution:

    1. Answer: (b, c, d)

      rate = d[P]dt=k[X]\frac{d[P]}{dt} = k[X]

      2X + Y → P

      2 mole 1 mole

      1 mole 0.5 mole 0.5 mole

      - d[X] / dt = k1[X] = 2k[X] ⇒ 2k = k1

      - d[Y] / dt = k2[X] = 2k[X] ⇒ k2 = k

      2K = 1/50 In2

      K = 1 / 100 In2 = 0.6 93 / 100 = 6.93 × 10-3 × s-1 = 50 sec

      At 50 sec,

      d[X] / dt = 2k[X] = 2 × 0.693 / 100 × 1

      = 13.86 × 10-3 mol L-1 s-1

      At 100 sec

      -d[Y] / dt = k2[X] = k[X] × 0.693 / 100 × ½

      (Concentration of X after 2 half-lives = ½ M)

      = 3.46 × 10-3 mol L-1 s-1


    Question 4. Some standard electrode potentials at 298 K are given below:


    Pb2+/Pb –0.13 V

    Ni2+/Ni –0.24 V

    C2+/Cd –0.40 V

    Fe2+/Fe –0.44 V

    To a solution containing 0.001 M of X2+ and 0.1 M of Y2+, the metal rods X and Y are inserted (at 298 K) and connected by a conducting wire. This resulted in the dissolution of X.

    The correct combination(s) of X and Y, respectively, is(are)

    (Given: Gas constant, R = 8.314 J K-1 mol-1, Faraday constant, F = 96500 C mol-1)

    1. a. Cd and Ni
    2. b. Cd and Fe
    3. c. Ni and Pb
    4. d. Ni and Fe

    Solution:

    1. Answer: (a, b, c)

      X + Y2+ X2+ + Y

      E=E00.062log10(103101)E = E^{0} - \frac{0.06}{2} log_{10}\left ( \frac{10^{-3}}{10^{-1}} \right )

      E = Eo + 0.06

      (a) E° = –(– .4) + (– .24) = .16 > 0

      (b) E° = –(– .4) + (– .44) = –.04 < 0 and Ecell = –0.04 + 0.06 = +0.02 > 0

      (c) E° = –(– .24) + (– .13) = .11 > 0

      (d) E° = –(– .24) + (– .44) = –.2 < 0

      ∴ Ecell = –0.2 + 0.06 = –0.14 < 0

      ∴ If Ecell > 0 then the cell construction is possible.


    Question 5. The pair(s) of complexes wherein both exhibit tetrahedral geometry is(are) (Note: py = pyridine, Given: Atomic numbers of Fe, Co, Ni and Cu are 26, 27, 28 and 29, respectively)

    1. a. [FeCl4] and [Fe(CO)4]2–
    2. b. [Co(CO)4] and [CoCl4]2–
    3. c. [Ni(CO)4] and [Ni(CN)4]2–
    4. d. [Cu(py)4]+ and [Cu(CN)4]3–

    Solution:

    1. Answer: (a, b, d)

      [FeCl4] → Fe3+, 3d5 (weak field ligand) = sp3

      [Fe(CO)4]–2 → Fe2–, 3d10 →sp3

      [Co(CO)4] → Co, 3d10 → sp3

      [CoCl4]2– → Co2+, 3d7 (weak field ligand) → sp3

      [Ni(CO)4] → Ni, 3d10 → sp3

      [Ni(CN)4]2– → Ni2+, 3d8 (strong field ligand) → dsp2

      [Cu(py)4]+ → Cu+, 3d10 →sp3

      [Cu(CN)4]3– → Cu+, 3d10 → sp3

      In 3d10 electronic configuration, only sp3 hybridisation and tetrahedral geometry are possible.


    Question 6. The correct statement(s) related to oxoacids of phosphorous is(are).

    1. a. Upon heating, H3PO3 undergoes a disproportionation reaction to produce H3PO4 and PH3.
    2. b. While H3PO3 can act as a reducing agent, H3PO4 cannot.
    3. c. H3PO3 is a monobasic acid.
    4. d. The H atom of the P-H bond in H3PO3 is not ionizable in water.

    Solution:

    1. Answer: (a, b, d)

      4H3PO3ΔPH3+3H3PO44H_{3}PO_{3} \overset{\Delta }{\rightarrow}PH_{3} + 3H_{3}PO_{4}

      In H3PO4, phosphorous is present in the highest oxidation state, i.e., +5. So H3PO4 cannot act as a reducing agent. Structure of H3PO3,

      JEE Advanced Question Paper 2021 Chemistry Paper 2 Question 6 solution

      It is a dibasic acid.

      H atom present in the P–H bond is not ionizable.

      These P-H bonds are not ionisable to give H+ and do not play any role in basicity. Only those H atoms which are attached with oxygen in P-OH form are ionisable and cause the basicity. Thus, H3PO3 and H3PO4 are dibasic and tribasic, respectively as the structure of H3PO3 has two P – OH bonds and H3PO4 three.


    Question Statement for Questions 7 and 8.


    At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 × 102 S cm2 mol–1. At 298 K, for an aqueous solution of the acid, the degree of dissociation is a and the molar conductivity is y × 102 S cm2 mol–1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y × 102 S cm2 mol–1.

    Question 7. The value of α is _______.

      Solution:

      1. Answer: (0.215)


      Question 8. The value of y is _______.

        Solution:

        1. Answer: (0.86)

          Solution for Questions 7 and 8.

          Molar conductivity of HX at infinite dilution

          Λm\Lambda _{m}^{\infty } = 4 × 102 S cm2 mol-1

          Molar conductivity of HX at conc. c1 = y × 102 S cm2 mol-1

          JEE Advanced Question Paper 2021 Chemistry Paper 2 Question 7 and 8 solution

          11α1=920(13α1)\frac{1}{1-\alpha _{1}} = \frac{9}{20(1-3\alpha _{1})}

          20 - 60α1 = 9 - 9α1

          ⇒ α1 = 11/51 = 0.215

          Y = 4α1 = 0.086


        Question Statement for Questions 9 and 10.


        The reaction of x g of Sn with HCl quantitatively produced a salt. The entire amount of the salt reacted with y g of nitrobenzene in the presence of the required amount of HCl to produce 1.29 g of an organic salt (quantitatively).

        (Use Molar masses (in g mol–1) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively).

        Question 9. The value of x is ________.

          Solution:

          1. Answer: (3.57)

            Sn + HCl → SnCl2

            ⇒ Moles of ammonium salt = 1.29 / 129 = 0.01

            ⇒ Moles of nitrobenzene = 0.01

            No. of eq. of nitrobenzene = No. of eq. of SnCl2

            6 × (0.01) = 2 × nSnCl2n_{SnCl_{2}}

            nSnCl2n_{SnCl_{2}} = 0.03

            ⇒ nSn = 0.03

            wSn = 0.03 × 119

            x = 3.57


          Question 10. The value of y is ________.

            Solution:

            1. Answer: (1.23)

              Solution of Question Nos. 9 and 10

              Sn + HCl → SnCl2

              ⇒ Moles of ammonium salt = 1.29 / 129 = 0.01

              ⇒ Moles of nitrobenzene = 0.01

              ⇒ y = 0.01 × Molar mass of nitrobenzene

              = 0.01 × 123

              = y = 1.23


            Question Statement for Questions 11 and 12.


            A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03 M KMnO4 solution to reach the endpoint. Number of moles of Fe2+ present in 250 mL solution is x × 10–2 (consider complete dissolution of FeCl2). The amount of iron present in the sample is y% by weight.

            (Assume: KMnO4 reacts only with Fe2+ in the solution Use: Molar mass of iron as 56 g mol–1)

            Question 11. The value of x is ________.

            1. a.swer: (1.875)

            Solution:


              Question 12. The value of y is ________.

                Solution:

                1. Answer: (18.75)

                  Solution of Question Nos. 11 and 12

                  8H+ + 5Fe2+ + MnO → 5Fe3+ + Mn2+ + 4H2O

                  For 25 ml,

                  meq of Fe2+ = meq of MnO

                  = 12.5 × 0.03 × 5

                  For 250 ml,

                  mmoles of Fe2+ = 12.5 × 0.03 × 5 × 250 / 25

                  moles of Fe2+ = 18.75 / 1000 mol

                  = 18.75 × 10–3 mol

                  = 1.875 × 10–2 mol

                  x = 1.875

                  Weight of Fe2+ = 1.875 × 10–2 × 56 = 1.05 g

                  % purity of Fe2+ y = 18.75%

                  = 1.05 / 5.6 ×100


                Statement: The amount of energy required to break a bond is the same as the amount of energy released when the same bond is formed. In a gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by the s-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below:


                JEE Advanced Question Paper 2021 Chemistry Paper 2 Question 12 solution

                Question 13. The correct match of the C-H bonds (shown in bold) in Column J with their BDE in Column K is;

                Column JColumn K
                MoleculeBDE (kcal mol–1)
                (P) H-CH(CH3)2(i) 132
                (Q) H-CH2Ph(ii) 110
                (R) H-CH=CH2(iii) 95
                (S) H-CºCH(iv) 88

                1. a. P – iii, Q – iv, R – ii, S – i
                2. b. P – i, Q – ii, R – iii, S – iv
                3. c. P – iii, Q – ii, R – i, S – iv
                4. d. P – ii, Q – i, R – iv, S – iii

                Solution:

                1. Answer: (a)

                  JEE Advanced Question Paper 2021 Chemistry Paper 2 Question 13

                  Order of stability of free radical

                  Q > P > R > S

                  Stability of free radical α 1 / Bond energy

                  ∴ Order of bond energy :

                  S > R > P > Q


                Question 14. For the following reaction,


                CH4(g)+Cl2(g)lightCH3Cl(g)+HCl(g)CH_{4} (g) + Cl_{2} (g) \overset{light}{\rightarrow}CH_{3}Cl(g) + HCl(g)

                the correct statement is;

                1. a. Initiation step is exothermic with DH° = –58 kcal mol–1
                2. b. Propagation step involving ·CH3 formation is exothermic with DH° = –2 kcal mol–1
                3. c. Propagation step involving CH3Cl formation is endothermic with DH° = +27 kcal mol–1
                4. d. The reaction is exothermic with DH° = –25 kcal mol–1

                Solution:

                1. Answer: (d)

                  JEE Advanced Question Paper 2021 Chemistry Paper 2 Question 14 solution

                  Step (1) → Endothermic (bond breaking)

                  Step (2) → ∆H = 105 – 103 = 2 kcal/mol (Endothermic)

                  Step (3) → ∆H = 58 – 85 = –27 kcal/mol (Exothermic)

                  For complete reaction

                  CH4(g)+Cl2(g)lightCH3Cl(g)+HCl(g)CH_{4} (g) + Cl_{2} (g) \overset{light}{\rightarrow}CH_{3}Cl(g) + HCl(g)

                  ∆H = 58 + 105 – 85 – 103

                  = –25 kcal/mol


                Question Statement for Questions 15 and 16.


                The reaction of K3[Fe(CN)6] with freshly prepared FeSO4 solution produces a dark blue precipitate called Turnbull's blue. The reaction of K4[Fe(CN)6] with the FeSO4 solution in the complete absence of air produces a white precipitate X, which turns blue in the air. Mixing the FeSO4 solution with NaNO3, followed by slow addition of concentrated H2SO4 through the side of the test tube produces a brown ring.

                Question 15. Precipitate X is

                1. a. Fe4[Fe(CN)6]3
                2. b. Fe4[Fe(CN)6]
                3. c. K2Fe[Fe(CN)6]
                4. d. KFe[Fe(CN)6]

                Solution:

                1. Answer: (c)


                Question 16. Among the following, the brown ring is due to the formation of

                1. a. [Fe(NO)2(SO4)2]2–
                2. b. [Fe(NO)2(H2O)4]3+
                3. c. [Fe(NO)4(SO4)2]
                4. d. [Fe(NO)(H2O)5]2

                Solution:

                1. Answer: (d)

                  Solution of Question Nos. 15 and 16

                  JEE Advanced Question Paper 2021 Chemistry Paper 2 Question 16 solution


                Question 17. One mole of an ideal gas at 900 K, undergoes two reversible processes, I followed by II, as shown below. If the work done by the gas in the two processes are the same, the value of Inv3v2In\frac{v_{3}}{v_{2}} is ____.


                JEE Advanced Question Paper 2021 Chemistry Paper 2 Question 17

                (U: internal energy, S: entropy, p: pressure, V: volume, R: gas constant)

                (Given: molar heat capacity at constant volume, C of the gas is 5 / 2 R)

                  Solution:

                  1. Answer: (10)

                    Process I is adiabatic reversible

                    Process II is a reversible isothermal process

                    Process I - (Adiabatic Reversible)

                    ∆U / R = 450 - 2250

                    ∆U = -1800 R

                    WI = ∆U = -1800R

                    Process II - (Reversible Isothermal Process)

                    T1 = 900 K

                    Calculation of T2 after the reversible adiabatic process

                    –1800R = nCv(T2 – T1)

                    –1800R 1 × 5/2 R(T2 – 900)

                    T2 = 180 K

                    WII = –nRT2 In = W

                    –1 × R × 180ln v3 / v2 –1800R

                    ln v3 / v210


                  Question 18. Consider a helium (He) atom that absorbs a photon of wavelength 330 nm. The change in the velocity (in cm s-1) of the He atom after the photon absorption is_____.


                  (Assume: Momentum is conserved when the photon is absorbed.

                  Use: Planck constant = 6.6 × 10-34 J s, Avogadro number = 6 × 1023 mol-1, Molar mass of He = 4 g mol-1)

                    Solution:

                    1. Answer: (30)

                      Momentum of photon = hλ=6.6×1027330×107\frac{h}{\lambda} = \frac{6.6\times 10^{-27}}{330\times 10^{-7}} gm cm s-1

                      Momentum of 1 mole of He-atoms = m∆v

                      ∴ m∆v = NA × h / λ

                      4×Δv=6×1023×6.6×1027330×1074\times \Delta v = \frac{6\times10^{23} \times 6.6\times 10^{-27}}{330\times 10^{-7}}

                      Δv=6×6.6×10233×4\Delta v = \frac{6\times 6.6\times 10^{2}}{33\times 4} = 30 cm s-1

                      ∴ Change in velocity of He-atoms = 30 cm s-1


                    Question 19. Ozonolysis of ClO2 produces oxide of chlorine. The average oxidation state of chlorine in this oxide is ____.

                      Solution:

                      1. Answer: (6)

                        ClO2 contains an odd electron and is paramagnetic. It reacts with ozone to give O2 and Cl2O6.

                        2ClO2 + 2O3 → Cl2O6 + 2O2

                        In Cl2O6, the average oxidation state of Cl is +6.


                      JEE Advanced 2021 Chemistry Paper 2 Solutions

                      JEE Advanced 2021 Chemistry Paper 2 Question and Solutions
                      JEE Advanced 2021 Chemistry Paper 2 Question and Solutions
                      JEE Advanced 2021 Chemistry Paper 2 Question and Solutions
                      JEE Advanced 2021 Chemistry Paper 2 Question and Solutions
                      JEE Advanced 2021 Chemistry Paper 2 Question and Solutions
                      JEE Advanced 2021 Chemistry Paper 2 Question and Solutions
                      JEE Advanced 2021 Chemistry Paper 2 Question and Solutions
                      JEE Advanced 2021 Chemistry Paper 2 Question and Solutions
                      JEE Advanced 2021 Chemistry Paper 2 Question and Solutions
                      JEE Advanced 2021 Chemistry Paper 2 Question and Solutions
                      JEE Advanced 2021 Chemistry Paper 2 Question and Solutions
                      JEE Advanced 2021 Chemistry Paper 2 Question and Solutions