KVPY-SA 2019 Chemistry Paper with Solutions

KVPY-SA 2019 Chemistry paper with solutions is provided on this page. The question paper with solutions is a great study resource for aspirants who are preparing to write the exam. The KVPY paper will enable them to understand the questions asked in the paper and learn the different methods to solve them and also get a deeper understanding of the concepts. Candidates can download the KVPY SA 2019 question paper below. Kishore Vaigyanik Protsahan Yojana also known as KVPY is a scholarship program conducted annually by the Department of Technology & Science, Government of India. KVPY examination aims at encouraging the students to pursue a research career in basic sciences. Scholarships and Contingency are granted to the candidates qualifying the talent test (up to the pre PhD level). KVPY- SA is an examination conducted for students who have scored 75% aggregate marks in Maths and Science subjects in their Class 10 board exam and have enrolled in Class 11 Science stream.

KVPY SA 2019 - Chemistry

Question 1: The hybridizations of N, C and O shown in the following compound respectively, are –

KVPY-SA-2019 Solutions

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

Solution:

  1. Answer: a

    Hybridisation of

    O = 1 B.P. (σ bond) + 2 l.p. = 3 ⇒sp2

    C = 2 B.P (σ bond) = 2 ⇒ sp

    N = 2 B.P. (σ bond) + 1l.P. = 3 ⇒ sp2

    So, option (A) is correct


Question 2: The following compounds are

KVPY-SA-2019 Solved Paper

  1. a. geometrical isomers
  2. b. positional isomers
  3. c. optical isomers HCl
  4. d. functional group isomers

Solution:

  1. Answer: d

    a. CH3 – C ≡ C – CH2 – CH2 – CH2 – CH3 ⇒ C7H12 {– yne compound}

    b. CH2 = CH – CH2 – CH = CH – CH2 – CH3⇒ C7H12 {– diene compound}

    Both –yne and –diene compounds are having same molecular formula but different functional group. So, they are functional group isomers of each other.


Question 3: The major product of the following reaction is -

Solved Paper of KVPY-SA-2019

Solved Question Paper of KVPY-SA-2019

    Solution:

    1. Answer: a

      Solved Question Paper with Answer of KVPY-SA-2019


    Question 4: IUPAC name of the following compound is –

    KVPY-SA-2019 Solved Question Paper with Answer

    1. a. 1-hydroxycyclohex-4-en-3-one
    2. b. 1-hydroxycyclohex-3-en-5-one
    3. c. 3-hydroxycyclohex-5-en-1-one
    4. d. 5-hydroxycyclohex-2-en-1-one

    Solution:

    1. Answer: d

      KVPY-SA-2019 Question Paper with Answer

      OH → Hydroxy (prefix)

      = (double bond) →en (primary suffix)

      Ketone →one (secondary suffix)

      Root word →cyclo hex

      So, IUPAC name: 5–hydroxy cyclohex–2–en–1–one


    Question 5: In water-gas shift reaction, hydrogen gas is produced from the reaction of steam with -

    1. a. Methane
    2. b. Coke
    3. c. Carbon Monoxide
    4. d. Carbon Dioxide

    Solution:

    1. Answer: c

      The water-gas shift reaction is an equilibrium process. In which carbon monoxide and steam converting into hydrogen and carbon dioxide.

      CO + H2O → CO2 + H2

      (steam)


    Question 6: Treatment with lime can remove the hardness of water caused by -

    1. a. CaCl2
    2. b. CaSO4
    3. c. Ca(HCO3)2
    4. d. CaCO3

    Solution:

    1. Answer: c

      Lime (Ca(OH)2) is used to remove the temporary hardness of water causes by salts of Ca and Mg.

      KVPY-SA-2019 Paper with Answers


    Question 7: The most polarizable ion among the following is

    1. a. F
    2. b. I
    3. c. Na+
    4. d. Cl

    Solution:

    1. Answer: b

      According to Fajan’s rule

      Polarization ∝ size of anion ∝ 1/size of cation

      So, out of F, I, Na+ ,Cl. I- anion has the largest in size. So, it’s most polarizable.


    Question 8: For a multi-electron atom, the highest energy level among the following is -

    1. a. n =5, l = 0, m =0, s = +1/2
    2. b. n =4, l = 2, m =0, s = +1/2
    3. c. n =4, l = 1, m =0, s = +1/2
    4. d. n =5, l = 1, m =0, s = +1/2

    Solution:

    1. Answer: d

      Option

      (A) n = 5, l= 0, m = 0, s = + ½ ⇒ 5s orbital

      (B) n = 4, l = 2, m = 0, s = + ½ ⇒ 4d orbital

      (C) n = 4, l = 1, m = 0, s = + ½ ⇒ 4p orbital

      (D) n = 5, l = 1, m = 0, s = + ½ ⇒ 5p orbital

      So according to (n + l) rule, increasing order of energy level: higher the value of (n+l), higher will be the energy. If (n+l) value is same, then check the value of n. Higher the value of n, higher will be the energy.

      4p< 5s < 4d < 5p

      So, 5p is highest energy level.


    Question 9: The oxide which is neither acidic nor basic is -

    1. a. As2O3
    2. b. Sb4O10
    3. c. N2O
    4. d. Na2O

    Solution:

    1. Answer: c

      Some non-metal oxides do not show any acidic or basic nature. They are neutral oxide.

      Example: H2O, CO, NO, N2O etc.

      As2O3& Sb4O10 are amphoteric and Na2O is a basic oxide.


    Question 10: The element whose salts cannot be detected by the flame test is -

    1. a. Mg
    2. b. Na
    3. c. Cu
    4. d. Sr

    Solution:

    1. Answer: a

      Flame test:

      (A) Mg → None of the salts gives [but for burning Mg metal (intense white)]

      (B) Na → yellow.

      (C)
      KVPY-SA-2019 Sample Question Paper

      (D) Sr → Crimson red to scarlet


    Question 11: The plot of concentration of a reactant vs. time for a chemical reaction is shown below:

    KVPY-SA-2019 Practice Question Paper

    The order of this reaction with respect to the reactant is

    1. a. 0
    2. b. 1
    3. c. 2
    4. d. not possible to determine from this plot

    Solution:

    1. Answer: a

      Let, the reaction be zero order

       KVPY-SA-2019 Practice Question Paper

      So, according to law of mass action

      R ∝ [A]0 at time =t

      r=k [Since, [A]0=1] {k = rate constant}

      We know, rate= Change in concentration/change in time

      Therefore, -dA/dt =k ⇒ -dA = kdt

      On integrating both the side and applying limits

      A0AtdA=k0Atdt[A]0At=k[t]0t-\int_{A_{0}}^{A_{t}} dA=k \int_{0}^{A_{t}} dt \Rightarrow[A]_{0}^{A_{t}}=-k[t]_{0}^{t}

      At-A0 =-kt

      At = -kt+A0

      On comparing with straight line equation

      y=mx+c

      At = -kt+A0

      So, plot of At vs t

      Practice Question Paper of KVPY-SA-2019

      So order of reaction with respect to reactant is zero order.


    Question 12: During the free expansion of an ideal gas in an isolated chamber,

    1. a. internal energy remains constant
    2. b. internal energy decreases
    3. c. work done on the system is negative
    4. d. temperature increases

    Solution:

    1. Answer: a

      Free expansion of an ideal gas in an isolated chamber is irreversible adiabatic free expansion process.

      ⇒ In an adiabatic process → Heat is constant

      So, q = 0 (isolated chamber)

      ⇒ In irreversible process work done → w = – Pext. ΔV

      For free expansion →Pext. = 0

      So, w = 0

      We know, Internal energy (ΔU) = q + w = 0

      Hence, internal energy is constant.


    Question 13: The number of moles of water present in a spherical water droplet of radius 1.0 cm is [Given the density of water in the droplet = 1.0 g cm–3]

    1. a. π/18
    2. b. 2π/27
    3. c. 24 π
    4. d. 2π/9

    Solution:

    1. Answer: b

      Given:-

      Radius of water droplet = 1.0 cm

      Density of water droplet = 1.0 g/cm3

      We know, density = mass/ volume ⇒ mass = volume

      No. of moles = mass/molar mass {mass = volume}

      So, moles = volume/molar mass {volume of sphere = (4/3) πr3 }

      Mole = 43π(1.0)318\frac{\frac{4}{3}\pi (1.0)^{3}}{18} {molar mass of H2O = 18 g mol–1}

      No. of moles = 2π/27


    Question 14: Among the following, the correct statement about cathode ray discharge tube is

    1. a. the electrical discharge can only be observed at high pressure and at low voltages
    2. b. in the absence of external electrical or magnetic field, cathode rays travel in straight lines
    3. c. the characteristics of cathode rays depend upon the material of electrodes
    4. d.the characteristics of cathode rays depend upon the gas present in the cathode ray tube

    Solution:

    1. Answer: b

      Cathode rays or stream of electrons move in a straight line without any deflection in the absence of electric and magnetic field. But in the presence of the electric field, the cathode rays are deflected towards the positive pole.


    Question 15: For a spontaneous process

    1. a. enthalpy change of the system must be negative
    2. b. entropy change of the system must be positive
    3. c. entropy change of the surrounding must be positive
    4. d. entropy change of the system and surrounding (universe) must be positive

    Solution:

    1. Answer: d

      For a spontaneous process ΔG < 0

      So, we know, ΔG = ΔH – T ΔS

      So, for spontaneous Reaction, ΔH = –ve, ΔS = +ve

      Therefore, entropy change of the system and surrounding (universe) must be positive.


    Question 16: PbO2 is obtained from

    1. a. the reaction of PbO with HCl
    2. b. thermal decomposition of Pb(NO3)2 at 200 °C
    3. c. the reaction of Pb3O4 with HNO3
    4. d. the reaction of Pb with air at room temperature

    Solution:

    1. Answer: c

      Practice Question Paper with Answers of KVPY-SA-2019


    Question 17: For one mole of a van der Waals gas, the compressibility factor z(=PVRT)z(=\frac{PV}{RT}) at a fixed volume will certainly decrease if

    [Given: "a", "b" are the standard parameter for van der Waals gas]

    1. a."b" increases and "a" decreases at a constant temperature
    2. b. "b" decreases and "a" increases at a constant temperature
    3. c. temperature increases at constant "a" and "b" values
    4. d. "b" increases at constant "a" and temperature

    Solution:

    1. Answer: b

      a → intermolecular force of attraction

      b → volume occupied by molecules.

      We know, compressibility factor Z = Vreal/Videal = PV/RT

      To decreases Z value, Vreal (numerator) should decrease which decreases when ‘a’ increases & ‘b’ decreases at constant temperature


    Question 18: The correct statements among the following

    (i) E2s(H) > E2s (Li) > E2s (Na) > E2s (K)

    (ii)The maximum number of electrons in the shell with principal quantum number n is equal to 2n2

    (iii)Extra stability of half-filled subshell is due to smaller exchange energy.

    (iv)Only two electrons, irrespective of their spin, may exist in the same orbital.

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

    Solution:

    1. Answer: a

      (i) We know that, E = – 13.6 × (z2/n2)ev

      So, the higher the value of Z (atomic number) lower is the value of E2s

      (ii) Maximum number of electron in the shell = 2n2

      (iii) The extra stability of half-filled subshell is due to higher exchange energy.

      (iv) Only two electrons with opposite spin can exist in the same orbital. According to Pauli's exclusion principle. So, statement (i) and (ii) are correct.


    Question 19: An organic compound contains 46.78% of a halogen X. When 2.00 g of this compound is heated with fuming HNO3 in the presence of AgNO3, 2.21 g AgX was formed. The halogen X is

    [Given: atomic weight of Ag = 108, F = 19, Cl = 35.5, Br = 80, I = 127]

    1. a. F
    2. b. Cl
    3. c. Br
    4. d. I

    Solution:

    1. Answer: c

      Given: % if X in organic compound = 46.78%

      Weight of organic compound = 2.00g

      Weight of AgX formed = 2.21 g

      So, weight of X in organic compound =2 x (46.78/100) = 0.94 gm

      Let, the molecular weight of X = xg mol–1

      So, by the law of equivalence

      No. of eq. of X = no. of eq. of AgX

      Mole × valency factor = mole × valency factor

      0.94x×1=2.21108+x×1\frac{0.94}{x}\times 1=\frac{2.21}{108+x}\times 1

      (101.52 + 0.94x) =2.21x

      101.52 = 1.27x

      x = 79.9 ≈ 80

      So, halogen X is Bromine (Br)


    Question 20: An organic compound X with molecular formula C6H10, when treated with HBr, forms a gem dibromide. The compound X upon warming with HgSO4 and dil. H2SO4, produces a ketone which given a positive iodoform test. The compound X is

    Answers for KVPY-SA-2019

      Solution:

      1. Answer: d

        Given: molecular formula of organic compound (x) = C6H10 i.e. CnH2n–2

        So, terminal alkyne reacts with HgSO4 and dil. H2SO4

        Produce ketone with gives Iodoform test.

        Solved Question Paper for KVPY-SA-2019

        Therefore, Iodoform test is shown by those groups which contain
         KVPY-SA-2019 Solved Question Paper
        group in their structure. So, option (d) satisfies all the condition.


      Video Lessons -KVPY SA 2019 - Chemistry

      KVPY SA 2019 Chemistry Paper with Solutions

      KVPY SA 2019 Chemistry Paper with Solutions
      KVPY SA 2019 Chemistry Paper with Solutions
      KVPY SA 2019 Chemistry Paper with Solutions
      KVPY SA 2019 Chemistry Paper with Solutions
      KVPY SA 2019 Chemistry Paper with Solutions
      KVPY SA 2019 Chemistry Paper with Solutions
      KVPY SA 2019 Chemistry Paper with Solutions
      KVPY SA 2019 Chemistry Paper with Solutions
      KVPY SA 2019 Chemistry Paper with Solutions
      KVPY SA 2019 Chemistry Paper with Solutions
      KVPY SA 2019 Chemistry Paper with Solutions