KVPY-SX 2017 Chemistry Paper with Solutions

KVPY-SX 2017 Chemistry paper with solutions is a perfect resource which helps the candidates in understanding their weaker areas. Candidates can download the previous years’ question papers of KVPY-SX to figure out the easiest and fastest way to answer a particular question. Practising the KVPY-SX previous year papers will guide the students to attempt each question with the best possible approach and helps in building time management skills. Solving these papers with solutions will help the students understand the concepts and prepare them to face the main exam with more confidence.

KVPY SX 2017 - Chemistry

Question 1: The major product formed in the following reaction is

KVPY-SX 2017 Chemistry Paper with Solutions

    Solution:

    1. Answer: (a)

      KVPY-SX 2017 Chemistry Paper with Answers

      In the above molecule, lone pair of nitrogen is in conjugation with benzene as well as a carbonyl group. Therefore, the +M effect of the ‘NH’ group on benzene will be weak. The only mono substitution will take place.


    Question 2: Among the α-amino acids - threonine, tyrosine, methionine, arginine and tryptophan, those which contain an aromatic group in their side chain are

    1. a. threonine and arginine
    2. b. tyrosine and tryptophan
    3. c. methionine and tyrosine
    4. d. arginine and tryptophan

    Solution:

    1. Answer: (b)

      Solution for KVPY-SX 2017 Chemistry Paper


    Question 3: The number of stereoisomers possible for the following compound is CH3 – CH = CH – CH(OH) – CH3

    1. a. 1
    2. b. 2
    3. c. 3
    4. d. 4

    Solution:

    1. Answer: (d)

      KVPY-SX 2017 Chemistry Paper with Solutions Q3

      n = 2

      Number of possible stereoisomers = 2n = 22 = 4

      G.I. → Can be cis/trans.

      The chiral centre → can be R/S.

      Therefore possible combination

      Cis – R

      Cis – S

      Trans – R

      Trans – S


    Question 4: In electrophilic aromatic substitution reactions of chlorobenzene, the ortho/para directing ability of chlorine is due to its

    1. a. Positive inductive effects (+I)
    2. b. negative inductive effect (–I)
    3. c. Positive resonance effect (+R)
    4. d. negative resonance effect (–R)

    Solution:

    1. Answer: (c)

      KVPY-SX 2017 Chemistry Paper with Solutions Q4


    Question 5: Among the following,

    KVPY-SX 2017 Chemistry Paper with Solution

    the anti-aromatic compounds are

    1. a. I and IV
    2. b. III and V
    3. c. II and V
    4. d. I and III

    Solution:

    1. Answer: (b)

      KVPY-SX Chemistry 2017 Paper Solutions


    Question 6: Upon reaction with CH3MgBr followed by protonation, the compound that produces ethanol is

    1. a. CH3CHO
    2. b. HCOOH
    3. c. HCHO
    4. d. (CHO)2

    Solution:

    1. Answer: (c)

      KVPY-SX 2017 Chemistry Paper with Solutions Q6


    Question 7: Which of the following is NOT an oxidation-reduction reaction?

    1. a. H2 + Br2 → 2HBr
    2. b. NaCl + AgNO3→NaNO3 + AgCl
    3. c. 2Na2S2O3 + I2 → Na2S4O6 + 2NaI
    4. d. Cl2 + H2O → HCl + HOCl

    Solution:

    1. Answer: (b)

      KVPY-SX 2017 Chemistry Paper with Solutions Q7


    Question 8: The thermal stability of alkaline earth metal carbonates – MgCO3, CaCO3, SrCO3 and BaCO3, follows the order

    1. a. BaCO3 > SrCO3 > CaCO3 > MgCO3
    2. b. CaCO3 > SrCO3 > BaCO3 > MgCO3
    3. c. MgCO3 > CaCO3 > SrCO3 > BaCO3
    4. d. SrCO3 > CaCO3 > MgCO3 > BaCO3

    Solution:

    1. Answer: (a)

      Mg2+

      Ca2+

      Sr2+

      Ba2+

      Down the group, the size of cation increases and CO32– is also large in size.

      And large cation + large anion → more stability.

      Hence, more comparable the size of cation and anion, more will be the thermal stability of alkaline earth metal carbonates.

      Hence, BaCO3 > SrCO3 > CaCO3 > MgCO3.


    Question 9: When a mixture of diborane and ammonia is heated, the final product is

    1. a. BH3
    2. b. NH4BH4
    3. c. NH2NH2
    4. d. B3N3H6

    Solution:

    1. Answer: (d)

      KVPY-SX 2017 Chemistry Paper with Solutions Q9

      Borazole or Inorganic benzene


    Question 10: Among the following metals, the strongest reducing agent is

    1. a. Ni
    2. b. Cu
    3. c. Zn
    4. d. Fe

    Solution:

    1. Answer: (c)

      According to electrochemical series,

      SRP Value

      Ni

      – 0.25V

      Zn

      – 0.76V

      Fe

      – 0.44V

      Cu

      + 0.34V

      Lesser the value of E0RP, higher will be the tendency to get oxidized itself and reduce others and hence stronger will be the reducing agent.


    Question 11: The molecule which is NOT hydrolysed by water at 25°C is

    1. a. AlCl3
    2. b. SiCl4
    3. c. BF3
    4. d. SF6

    Solution:

    1. Answer: (a)

      a. AlCl3 + 3H2O → Al(OH)3 ↓ + 3HCl

      (Al has vacant p-orbital where H2O can attack.)

      b. SiCl4 + 4H2O → Si(OH)4 + 4HCl

      (Si has vacant d-orbital where H2O can attack.)

      c. BF3 + 3H2O → B(OH)3 + 3HF

      (B has vacant p-orbital where H2O can attack.)

      d. SF6 + H2O → (At 25o C) No reaction

      Sulphur has vacant d-orbital but due to steric hindrance created by 6F, SF6 cannot be hydrolysed at 25°C.


    Question 12: Among the following compounds, the one which does NOT produce nitrogen gas upon heating is

    1. a. (NH4)2Cr2O7
    2. b. NaN3
    3. c. NH4NO2
    4. d. (NH4)2(C2O4)

    Solution:

    1. Answer: (d)

      (NH4)2Cr2O7 → N2 + Cr2O3 + H2O

      2NaN3 → 2Na + 3N2

      NH4NO2 → N2 + H2O

      (NH4)2C2O4 → 2NH3 + H2C2O4


    Question 13: Chlorine has two naturally occurring isotopes, 35Cl and 37Cl. If the atomic mass of Cl is 35.45, the ratio of the natural abundance of 35Cl and 37Cl is closest to

    1. a. 3.5 : 1
    2. b. 3 :1
    3. c. 2.5 : 1
    4. d. 4 : 1

    Solution:

    1. Answer: (b)

      Atomic mass of Cl = 35.45 = Average atomic mass

      Average atomic mass = [n1M1 + n2M2] / [n1 + n2]

      35.45 = [n1 * 35 + n2 * 37] / [n1 + n2]

      35.45n1 + 35.45n2 = 35n1 + 37n2

      0.45n1 = 1.55n2

      n1 / n2 = 1.55 / 0.45 ≈ 3 / 1

      3 : 1


    Question 14: The reaction C2H6(g) → C2H4(g) + H2(g) is at equilibrium in a closed vessel at 1000 K. The enthalpy change (ΔH) for the reaction is 137.0 kJ mol–l. Which one of the following actions would shift the equilibrium to the right?

    1. a. Decreasing the volume of the closed reaction vessel
    2. b. Decreasing the temperature at which the reaction is performed
    3. c. Adding an inert gas to the closed reaction vessel
    4. d. Increasing the volume of the closed reaction vessel

    Solution:

    1. Answer: (d)

      C2H6(g) → C2H4(g) + H2(g); ΔH = +ve

      a. Volume ↓ → Reaction will shift in the direction where the number of moles ↓ according to Le-chatelier principle → backward direction.

      b. T ↓ → Reaction will shift in a direction where the temperature will increase (backwards) as the reaction is endothermic.

      c. Addition of inert gas at constant volume (no effect on equilibrium).

      d. Volume ↑ → Reaction will shift in a direction where the number of moles will increase i.e. forward direction.

      Correct option is (d).


    Question 15: The enthalpy (H) of an elementary exothermic reaction A -> B is schematically plotted against the reaction coordinate. The plots in the presence and absence of a catalyst are shown in dashed and solid lines, respectively. Identify the correct plot for the reaction.

    KVPY-SX 2017 Chemistry Paper with Solutions Q15

      Solution:

      1. Answer: (a)

        For exothermic reaction

        ΔH = EP – ER < 0

        EB – EA < 0

        EB < EA

        Catalyst does not change the initial and final position of the reaction. It will only decrease the activation energy, hence increase the rate of reaction.


      Question 16: Mg(OH)2 is precipitated when NaOH is added to a solution of Mg2+. If the final concentration of Mg2+ is 10–10M, the concentration of OH(M) in the solution is [Solubility product for Mg (OH)2 = 5.6 × 10–12]

      1. a. 0.056
      2. b. 0.12
      3. c. 0.24
      4. d. 0.025

      Solution:

      1. Answer: (c)

        Mg(OH)2(s) ⇌ Mg2+ + 2OH

        Equilibrium s 2s

        Ksp = [Mg2+][OH]2 = 5.6 × 10–12

        10–10 [OH]2 = 5.6 × 10–12

        [OH] = √5.6 × 10–2 = 0.24 M


      Question 17: A constant current (0.5 amp) is passed for 1 hour through

      (i) aqueous AgNO3,

      (ii) aqueous CuSO4 and

      (iii) molten AlF3, separately.

      The ratio of the mass of the metals deposited on the cathode is [MAg, MCu, MAl are molar masses of the respective metals]

      1. a. MAg: 2 MCu: 3 MA1
      2. b. MAg: MCu: MA1
      3. c. 6 MAg: 3 MCu: 2 MA1
      4. d. 3 MAg: 2MCu: MA1

      Solution:

      1. Answer: (c)

        KVPY-SX 2017 Chemistry Paper with Solutions Q17


      Question 18: A reaction has an activation energy of 209 kJ mol–1. The rate increases 10–fold when the temperature is increased from 27°C to X °C. The temperature X is closest to [Gas constant, R = 8.314 J mol–1K–1]

      1. a. 35
      2. b. 40
      3. c. 30
      4. d. 45

      Solution:

      1. Answer: (a)

        𝛆a = 209 kJmol–1

        Given:

        T1 = 27°C = 300 K

        T2 = x°C = (x + 273)K

        K2 / K1 = 10

        Apply

        log (K2 / K1) = [𝛆a / 2.303 R] [(1 / T1) - (1 / T2)]

        1 = log 10 = {[209 * 103] / [2.303 * 8.314]} {[(1 / 300) - (1 / (x + 273)]}

        x + 273 = 308.4 K

        x = 35.4°C ≈ 35°C


      Question 19: A mineral consists of a cubic close-packed structure formed by O2–ions, where half the octahedral voids are occupied by Al3+ and one-eighth of the tetrahedral voids, are occupied by Mn2+. The chemical formula of the mineral is

      1. a. Mn3Al2O6
      2. b. MnAl2O4
      3. c. MnAl4O7
      4. d. Mn2Al2O5

      Solution:

      1. Answer: (b)

        No. of atoms in ccp = N

        Then, No. of O.V. = N

        No. of T.V. = 2N

        No. of O–2 per unit cell = 8 × (1 / 8) + 6 × (1 / 2) = 4

        No. of Al+3 per unit cell = 4 × (1 / 2) = 2

        No. of Mn+2 per unit cell = 8 × (1 / 8) = 1

        MnAl2O4


      Question 20: For a 4p orbital, the numbers of radial and angular nodes, respectively, are

      1. a. 3, 2
      2. b. 1, 2
      3. c. 2, 4
      4. d. 2, 1

      Solution:

      1. Answer: (d)

        4p

        n = 4, l = 1

        No. of Radial nodes = n – l – 1 = 4 – 1 – 1 = 2

        No. of Angular nodes = l = 1


      Question 21: In the following reaction sequence

      KVPY-SX 2017 Chemistry Paper with Solutions Q21

      X and Y are

      KVPY-SX 2017 Chemistry Paper with answers Q21

        Solution:

        1. Answer: (b)

          KVPY-SX 2017 Chemistry Paper Q21

          [Hydroboration-Oxidation Reaction]

          Anti-markovnikov’s rule


        Question 22: In the following reactions

        Answers for KVPY-SX 2017 Chemistry Paper

        X and Y are

        KVPY-SX 2017 Chemistry Paper with answers Q22

          Solution:

          1. Answer: (a)

            KVPY-SX 2017 Chemistry Paper Q22


          Question 23: Which of the following alkenes can generate optically active compounds upon hydrogenation?

          KVPY-SX 2017 Chemistry Paper with Solutions Q23

          1. a. I, III and IV
          2. b. II and III
          3. c. I and III
          4. d. II and IV

          Solution:

          1. Answer: (c)

            KVPY-SX 2017 Chemistry Paper with answers Q23


          Question 24: When heated in air, brown copper powder turns black. This black powder would turn brown again when heated with

          1. a. CO b. O2c. H2d. NH3

            Answer: (c)

            KVPY-SX 2017 Chemistry Paper with Solutions Q24

          Solution:

          1. Answer: (c)

            KVPY-SX 2017 Chemistry Paper with Solutions Q24


          Question 25: The geometry and magnetic property of [NiCl4]2–, respectively, are

          1. a. Tetrahedral, paramagnetic
          2. b. tetrahedral, diamagnetic
          3. c. Square planar, paramagnetic
          4. d. square planar, diamagnetic

          Solution:

          1. Answer: (a)

            KVPY-SX 2017 Chemistry Paper with Solutions Q25

            Cl is a weak ligand

            No pairing of the electron will take place.

            • sp3 hybridisation, tetrahedral

            • 2 unpaired electrons → paramagnetic


          Question 26: Among

          (i) [Cr(en)3]3+,

          (ii) trans-[Cr(en)2Cl2]+,

          (iii) Cis-[Cr(en)2Cl2]+

          (iv) [Co(NH3)4Cl2]+

          the optically active complexes are

          1. a. i and ii
          2. b. i and iii
          3. c. ii and iii
          4. d. ii and iv

          Solution:

          1. Answer: (b)

            KVPY-SX 2017 Chemistry Paper with Solutions Q26


          Question 27: 227Ac has a half-life of 22 years with respect to radioactive decay. The decay follows two parallel paths: 227Ac → 227Th and 227Ac → 223Fr. If the percentage of the two daughter nuclides are 2.0 and 98.0, respectively, the decay constant (in year–1) for 227Ac → 227Th path is closest to

          1. a. 6.3 × 10–2
          2. b. 6.3 × 10–3
          3. c. 6.3 × 10–1
          4. d. 6.3 × 10–4

          Solution:

          1. Answer: (d)

            KVPY-SX 2017 Chemistry Paper with Solutions Q27

            k1 / k2 = 2 / 98

            t½ = 22 years

            k = 0.693 / t½ = 0.693 / 22 year = k1 + k2

            0.693 / 22 = k1 + (98 / 2) k1

            k1 = 6.3 × 10–4


          Question 28: A system consisting of 1 mol of an ideal gas undergoes a reversible process, A → B → C → A (schematically indicated in the figure below). If the temperature at the starting point A is 300 K and the work done in the process B → C is 1 L atm, the heat exchanged in the entire process in L atm is

          KVPY-SX 2017 Chemistry Paper with Solutions Q28

          1. a. 1.0
          2. b. 0.0
          3. c. 1.5
          4. d. 0.5

          Solution:

          1. Answer: (d)

            KVPY-SX 2017 Chemistry Paper with answers Q28

            Given: n = 1 mole, TA = 300 K, ωBC = 1 L-atm

            ωAB = –Pext.(V2 – V1) = –1(1.5 – 1) = –0.5 L-atm

            (Isobaric process)

            ωCA = 0 (dV = 0)

            (Isochoric process)

            ωtotal = ωAB + ωBC + ωCA

            = –0.5 + 1 + 0 = 0.5 L-atm

            ΔU calculation

            For complete cyclic process ABC → ΔU = 0 (As U → is a state function) According to first law,

            ΔU = q + w = 0

            q = –w = –0.5 L-atm

            Heat exchanged during the process = 0.5 L-atm.

            Correct option is (D).


          Question 29: A mixture of toluene and benzene boils at 100°C. Assuming ideal behaviour, the mole fraction of toluene in the mixture is closest to [Vapour pressures of pure toluene and pure benzene at 100°C are 0.742 and 1.800 bar respectively. 1 atm = 1.013 bar]

          1. a. 0.824
          2. b. 0.744
          3. c. 0.544
          4. d. 0.624

          Solution:

          1. Answer: (b)

            toluene = 0.742 bar

            benzene = 1.800 bar

            Mixture boils at 100°C → at 1 atm = 1.013 bar

            PT = 1.013 bar = PAXA + PBXB

            PT = Pot Xt + Pob Xb

            1.013 = 0.742 Xt + 1.8 (1 – Xt)

            Xt = 0.744

            Xb = 1 – 0.744 = 0.256

            Xt = mole fraction of toluene.


          Question 30: A two-dimensional solid pattern formed by two different atoms X and Y is shown below. The black and white squares represent atoms X and Y, respectively. The simplest formula for the compound based on the unit cell from the pattern is

          KVPY-SX 2017 Chemistry Paper with Solutions Q30

          1. a. XY8
          2. b. X4Y9
          3. c. XY2
          4. d. XY4

          Solution:

          1. Answer: (a)

            The unit cell of the above pattern will consist of 8 white squares and 1 black square i.e. it will form the centre unit cell.

            No. of white square Y = 8

            No. of black square X = 1

            Formula XY8.


          Video Lessons -KVPY SX 2017 - Chemistry

          KVPY SX 2017 Chemistry Paper with Solutions

          KVPY SX 2017 Chemistry Paper with Solutions
          KVPY SX 2017 Chemistry Paper with Solutions
          KVPY SX 2017 Chemistry Paper with Solutions
          KVPY SX 2017 Chemistry Paper with Solutions
          KVPY SX 2017 Chemistry Paper with Solutions
          KVPY SX 2017 Chemistry Paper with Solutions
          KVPY SX 2017 Chemistry Paper with Solutions
          KVPY SX 2017 Chemistry Paper with Solutions
          KVPY SX 2017 Chemistry Paper with Solutions
          KVPY SX 2017 Chemistry Paper with Solutions
          KVPY SX 2017 Chemistry Paper with Solutions
          KVPY SX 2017 Chemistry Paper with Solutions
          KVPY SX 2017 Chemistry Paper with Solutions
          KVPY SX 2017 Chemistry Paper with Solutions
          KVPY SX 2017 Chemistry Paper with Solutions
          KVPY SX 2017 Chemistry Paper with Solutions
          KVPY SX 2017 Chemistry Paper with Solutions