NEET-XII-Chemistry

Previous Year Paper year:2018

with Solutions - page 2

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  • Qstn #60
    Which one is a wrong statement ?
    (1) Total orbital angular momentum of electron in
    's' orbital is equal to zero
    (2) An orbital is designated by three quantum
    numbers while an electron in an atom is
    designated by four quantum numbers.
    (3) The electronic configuration of N atom is
    n60.png
    (4) The value of m for ``dz^2`` is zero
    digAnsr:   3
    Ans : (3)
    Sol. The correct configuration of 'N' is
  • Qstn #61
    Consider the following species:
    ``CN^+, CN^-, NO and CN``
    Which one of these will have the highest bond order?
    (1)`` NO ``
    (2)`` CN^-``
    (3) ``CN^+ ``
    (4) CN
    digAnsr:   2
    Ans : (2)
    Sol. Ion/Species Total electron Bond order
    NO 15 2.5
    CN- 14 3
    CN+ 12 2
    CN 13 2.5
  • Qstn #62
    Which of the following statements is not true for
    halogens ?
    (1) All form monobasic oxyacids.
    (2) All are oxidizing agents.
    (3) All but fluorine show positive oxidation states.
    (4) Chlorine has the highest electron-gain enthalpy.
    digAnsr:   B
    Ans : (Bonus)
  • Qstn #63
    Which one of the following elements is unable to
    form ``MF_6^{3-}``?
    (1) Ga (2) AI (3) B (4) In
    digAnsr:   3
    Ans : (3)
    Sol. MF
    6
    -3
    Boron belongs to 2nd period and it does not have
    vacant d-orbital.
  • Qstn #64
    In the structure of ``ClF_3``, the number of lone pairs
    of electrons on central atom 'Cl' is
    (1) one
    (2) two
    (3) four
    (4) three
    digAnsr:   2
    Ans : (2)
    Sol. Cl - F
    F
    F
    2 lone pair at equitorial position.
  • Qstn #65
    Considering Ellingham diagram, which of the
    following metals can be used to reduce alumina ?
    (1) Fe
    (2) Zn
    (3) Mg
    (4) Cu
    digAnsr:   3
    Ans : (3)
    Sol. Mg has more -▵G value then alumina. So it will be
    in the lower part of Ellingham diagram. Metals
    which has more -▵G value can reduce those metals
    oxide which has less -▵G value.
  • Qstn #66
    The correct order of atomic radii in group 13
    elements is
    (1) B < Al < In < Ga < Tl
    (2) B < Al < Ga < In < Tl
    (3) B < Ga < Al < Tl < In
    (4) B < Ga < Al < In < Tl
    digAnsr:   4
    Ans : (4)
    Sol. In group 13 due to transition contraction [Al > Ga]
  • Qstn #67
    The correct order of N-compounds in its decreasing
    order of oxidation states is
    ``\ce (1) HNO3, NO, N2, NH4Cl``
    ``\ce (2) HNO3, NO, NH4Cl, N2``
    ``\ce (3) HNO3, NH4Cl, NO, N2``
    ``\ce (4) NH4Cl, N2, NO, HNO3``
    digAnsr:   1
    Ans : (1)
    Sol.
    5
    3HNO
    +
    ,
    2
    NO
    +
    ,
    0
    2N ,
    3
    4NH Cl
  • Qstn #68
    On which of the following properties does
    coagulating power of an ion depend ?
    (1) The magnitude of the charge on the alone
    (2) Size of the ion alone
    (3) Both magnitude and sign of the charge the ion
    (4) The sign of charge on the ion alone
    digAnsr:   3
    Ans : (3)
    Sol. According to Hardy Schulze rule : The coagulating
    power of an ion depend on both magnitude and sign
    of the charge of the ion.
  • Qstn #69
    Following solutions were prepared by mixing
    different volumes of NaOH and HCl of different
    concentrations :
    a. 60mL ``\frac{M}{10}``HCl +
    40mL ``\frac{M}{10}`` NaOH
    b. 55mL ``\frac{M}{10}``HCl +
    45mL ``\frac{M}{10}`` NaOH
    c. 75mL ``\frac{M}{5}``HCl +
    25mL ``\frac{M}{5}`` NaOH
    d. 100mL ``\frac{M}{10}``HCl +
    100mL ``\frac{M}{10}`` NaOH
    pH of which one of them will be equal to 1 ?
    (1) b
    (2) a
    (3) d
    (4) c
    digAnsr:   4
    Ans : (4)
    Sol. As N1V1 > N2V2
    So acid is left at the end of reaction
    Nfinal solution = [H
    +] =
    1 1 2 2
    1 2
    N V N V
    V V

    +
    =
    1 1
    75 25
    5 5
    75 25
      
    +
    =
    1
    0.1
    10
    =
    pH = -log[H+] = 1
    
    
    14
    NEET(UG)-2018
  • Qstn #70
    The solubility of ``BaSO_4`` in water
    2.42 × 103 ``gL^{-1}``at 298 K. The value
    of solubility product (``K_{sp}``) will be
    (Given molar mass of BaSO4 = 233 g ``mol ^{-1}``
    (1) ``1.08 × 10^{-10} mol^2 L^{-2}``
    (2)`` 1.08 × 10^{-12} mol^2 L^{-2}``
    (3)`` 1.08 × 10^{-14} mol^2 L^{-2}``
    (4)`` 1.08 × 10^{-8} mol^2 L^{-2}``
    digAnsr:   1
    Ans : (1)
    Sol. solubility of BaSO4 = 2.42 × 10-3 gL-1
    ∴ s =

      = 
    3
    5 12.42 10 1.038 10 mol L
    233
    Ksp = s
    2 = (1.038 × 10-5)2
    = 1.08 × 10-10 mol2 L-2
  • Qstn #71
    Given van der Waals constant for ``\ce NH3, H2`` and ``CO_2``
    are respectively 4.17, 0.244, 1.36 and 3.59, which
    one of the following gases is most easily liquefied?
    (1) ``\ce NH3``
    (2) ``H_2 ``
    (3)`` O_2 ``
    (4)`` CO_2``
    digAnsr:   1
    Ans : (1)
    Sol. Critical temperature  vanderwaal constant(a)
    maximum "a" &implies; gas with maximum TC &implies; easiest
    liquification = NH3
  • Qstn #72
    The compound A on treatment with Na gives B, and
    with ``PCl_5 ``gives C. B and C react together to give
    diethyl ether. A, B and C are in the order
    (1) ``\ce C2H5OH, C2H6, C2H5Cl``
    (2) ``\ce C2H5OH, C2H5Cl, C2H5ONa``
    (3) ``\ce C2H5Cl, C2H6, C2H5OH``
    (4) ``\ce C2H5OH, C2H5ONa, C2H5Cl``
    digAnsr:   4
    Ans : (4)
    Sol. C H OH 2 5 C H ONa2 5
    Na
    A B
    C H OH 2 5 C H Cl2 5
    PCl5
    A C

    C H ONa + 2 5 C H -Cl2 5
    A C
    SN
    2
    Williamson's
    synthesis of ether
    C H -O-C H2 5 2 5
    diethylether
  • Qstn #73
    Hydrocarbon (A) reacts with bromine by substitution
    to form an alkyl bromide which by Wurtz reaction
    is converted to gaseous hydrocarbon containing less
    than four carbon atoms. (A) is
    (1) CH ``\equiv `` CH
    (2)`` CH_2=CH_2``
    (3)`` CH_3-CH_3 ``
    (4) ``CH_4``
    digAnsr:   4
    Ans : (4)
    Sol.
    CH4 CH -Br3
    Br2 CH -CH 3 3
    Na
    ether
    (less than four 'C')
    h
  • Qstn #74
    The compound ``\ce C7H8 ``undergoes the following
    reactions :
    ``\ce{$C7H8$ ->[\ce{ \frac {3Cl_2}{\lambda}}] $A$} ``
    ``\ce{$$ ->[\ce{\frac {Br_2}{Fe}}] $B$} \\``
    ``\ce{$$ ->[\ce{\frac {Zn}{HCl}}] $C$} \\``
    The product 'C' is
    (1) m-bromotoluene
    (2) o-bromotoluene
    (3) 3-bromo-2,4,6-trichlorotoluene
    (4) p-bromotoluene
    digAnsr:   1
    Ans : (1)
    Sol.
    CH3
    
    3Cl2

    CCl3
    
    Br2
    Fe
    CCl3
    Zn
    HCl
    Br
    CH3
    Br
    m-bromotoluene