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NEET-XII-Chemistry

Previous Year Paper year:2017

with Solutions - page 7

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  • Qstn #76
    Identify the incorrect statement related to
    ``PCl_5`` from the following:
    (1) Three equatorial P-Cl bonds make an
    angle of 120° with each other
    (2) Two axial P-Cl bonds make an angle of
    180° with each other
    (3) Axial P-Cl bonds are longer than
    equatorial P-Cl bonds
    (4) ``PCl_5``
    molecule is non-reactive
    digAnsr:   4
    Ans : ( 4 )
    Sol. P
    Cl
    Cl
    Cl
    Cl
    Cl
    120°
    202 pm
    90°
    240 pm
    Equatorial bond
    Axial bond
    (1) True
    (2) True P 180°
    Cl
    Cl
    (3) True
    Axial bond : 240 pm
    Equatorial bond : 202 pm
    (4) False
    Due to longer and hence weaker axial
    bonds, PCl
    5
    is a reactive molecule.
  • Qstn #77
    Which will make basic buffer?
    (1) 50 mL of 0.1 M NaOH + 25 mL of 0.1 M ``CH_3COOH``
    (2) 100 mL of 0.1 M ``CH_3COOH`` + 100 mL of 0.1 M NaOH
    (3) 100 mL of 0.1 M HCl + 200 mL of 0.1 M ``NH_4OH``
    (4) 100 mL of 0.1 M HCl + 100 mL of 0.1 M NaOH
    digAnsr:   3
    Ans : ( 3 )
    Sol.
    (1) +  +
    3 3 2
    CH COOH NaOH CH COONa H O
    Before 25 mL 50 mL
    0
    × 0.1M× 0.1 M
    = 2.5 mmol = 5 mmol
    After 0 2.5 mmol 2.5 mmol
    This is basic solution due to NaOH.
    This is not basic buffer.
    (2) +  +
    = =
    3 3 2
    CH COOH NaOH CH COONa H O
    Before 100 mL 100 mL
    0
    × 0.1 M × 0.1 M
    10 mmol 10mmol
    After 0 0 10 mmol
    Hydrolysis of salt takes place.
    This is not basic buffer.19
    (3) +  +
    = =
    4 4 2
    HCl NH OH NH Cl H O
    Before 100 mL 200 mL
    0
    × 0.1 M × 0.1 M
    10 mmol 20 mmol
    After 0 10 mmol 10 mmol
    This is basic buffer
    (4)
    +  +
    = =
    2
    HCl NaOH NaCl H O
    100 mL 100 mLBefore
    0
    × 0.1 M × 0.1 M
    10 mmol 10 mmol
    After 0 0 10 mmol
    &implies; Neutral solution
  • Qstn #78
    The major product of the following
    reaction is:
    m78.png
    (1) m78a.png
    (2) m78b.png
    (3) m78c.png
    (4) m78d.png
    digAnsr:   4
    Ans : ( 4 )

    Sol.
    COOH
    COOH
    + NH
    3
    COO NH
    - +
    4
    COO NH
    - +
    4
    CONH
    2
    CONH
    2
    - 2H O
    2
    ▵
    - NH
    3
    Strong heating
    Phthalimide
    NH
    C
    C
    O
    O
  • Qstn #79
    Match the Xenon compounds in Column-I
    with its structure in Column-II and assign the
    correct code:
    Column-I Column-II
    (a )`` XeF_4`` (i) Pyramidal
    (b) ``XeF_6`` (ii) Square planar
    (c) ``XeOF_4`` (iii) Distorted octahedral
    (d) ``XeO_3`` (iv) Square pyramidal
    Code:
    ( a ) (b) ( c ) (d)
    (1) (i) (ii) (iii) (iv)
    (2) (ii) (iii) (iv) (i)
    (3) (ii) (iii) (i) (iv)
    (4) (iii) (iv) (i) (ii)
    digAnsr:   2
    Ans : ( 2 )
    Sol. (a) XeF
    4
    : Xe
    F
    FF
    F
    &implies; Square planar
    (b) XeF
    6
    : Xe
    F
    F
    F
    FF
    F
    &implies; Distorted
    octahedral
    (c) XeOF
    4
    : Xe
    F
    FF
    F
    O
    &implies; Square
    pyramidal
    (d) XeO
    3
    :
    Xe O
    O
    O
    &implies;Pyramidal
  • Qstn #80
    The manganate and permanganate ions are
    tetrahedral, due to :
    (1) The ``\pi``-bonding involves overlap of
    p-orbitals of oxygen with d-orbitals of
    manganese
    (2) There is no ``\pi``-bonding
    (3) The ``\pi``-bonding involves overlap of
    p-orbitals of oxygen with p-orbitals of
    manganese
    (4) The ``\pi``-bonding involves overlap of
    d-orbitals of oxygen with d-orbitals of
    manganese
    digAnsr:   1
    Ans : ( 1 )
    20
    Sol. • Manganate (MnO
    4
    2-) : Mn
    O
    O
    -
    O
    O
    -
    &implies; -bonds are of d-p type
    • Permanganate (MnO
    4
    -) : Mn
    O
    O
    -
    O
    O
    &implies; -bonds are of d-p type
  • Qstn #81
    Which of the following species is not stable?
    (1) ``[SiF_6]^{2-}``
    (2) ``[GeCl_6]^{2-}``
    (3) ``[Sn(OH)_6]^{2-}``
    (4) ``[SiCl_6]^{2-}``
    digAnsr:   4
    Ans : ( 4 )
    Sol. • Due to presence of d-orbital in Si, Ge and
    Sn they form species like SiF
    6
    2-, [GeCl
    6
    ]2-,
    [Sn(OH)
    6
    ]2-
    • SiCl
    6
    2- does not exist because six large
    chloride ions cannot be accommodated
    around Si4+ due to limitation of its size.
  • Qstn #82
    For a cell involving one electron ``E°_{cell}``= 0.59 V
    at 298 K, the equilibrium constant for the cell
    reaction is :[Given that ``\frac{2.303 RT}{F}``=0.059V at T = 298 K.
    (1) ``1.0 × 10^2``
    (2) ``1.0 × 10^5``
    (3) ``1.0 × 10^{10}``
    (4) ``1.0 × 10^{30}``
    digAnsr:   3
    Ans : ( 3 )
    Sol.
    E
    cell
    = E°
    cell
    -
    0.059
    log Q
    n
    ...(i)
    (At equilibrium, Q = K
    eq
    and E
    cell
    = 0)
    0 = E°
    cell
    - eq
    0.059
    logK
    1
    (from equation (i))
    cell
    eq
    E° 0.59
    logK = = = 10
    0.059 0.059
    K
    eq
    = 1010 = 1 × 1010
  • Qstn #83
    Which of the following is an amphoteric
    hydroxide?
    (1) ``Sr(OH)_2``
    (2)`` Ca(OH)_2``
    (3) ``Mg(OH)_2``
    (4) ``Be(OH)_2``
    digAnsr:   4
    Ans : ( 4 )
    Sol. Be(OH)
    2
    amphoteric in nature, since it can
    react both with acid and base
    Be(OH)
    2
    + 2HCl  BeCl
    2
    + 2H
    2
    O
    Be(OH)
    2
    + 2NaOH  Na
    2
    [Be(OH)
    4
    ]
  • Qstn #84
    A gas at 350 K and 15 bar has molar volume
    20 percent smaller than that for an ideal gas
    under the same conditions. The correct
    option about the gas and its compressibility
    factor (Z) is :
    (1) Z > 1 and attractive forces are dominant
    (2) Z > 1 and repulsive forces are dominant
    (3) Z < 1 and attractive forces are dominant
    (4) Z < 1 and repulsive forces are dominant
    digAnsr:   3
    Ans : ( 3 )
    Sol. • Compressibility factor(Z) = real
    ideal
    V
    V
    ∵ V
    real
    < V
    ideal
    ; Hence Z < 1
    • If Z < 1, attractive forces are dominant
    among gaseous molecules and
    liquefaction of gas will be easy.
  • Qstn #85
    A compound is formed by cation C and
    anion A. The anions form hexagonal close
    packed (hcp) lattice and the cations occupy
    75% of octahedral voids. The formula of the
    compound is :
    (1)``C_2A_3``
    (2)``C_3A_2``
    (3) ``C_3A_4``
    (4) ``C_4A_3``
    digAnsr:   3
    Ans : ( 3 )21
    Sol. • Anions(A) are in hcp, so number of anions
    (A) = 6
    Cations(C) are in 75% O.V., so number of
    cations (C)
    =
    3
    6×
    4
    =
    18
    4
    =
    9
    2
    • So formula of compound will be
    &implies;
    9 6 9 12
    2
    C A C A
    C
    9
    A
    12
    &implies; C
    3
    A
    4
  • Qstn #86
    In which case change in entropy is negative?
    (1) Evaporation of water
    (2) Expansion of a gas at constant
    temperature
    (3) Sublimation of solid to gas
    (4) 2H(g) ``\longrightarrow `` ``H_2``(g)
    digAnsr:   4
    Ans : ( 4 )
    Sol. •     ▵ ⇀↽2 2H O H O , S 0v
    • Expansion of gas at constant
    temperature, ▵S > 0
    • Sublimation of solid to gas, ▵S > 0
    • 2H(g)  H
    2
    (g), ▵S < 0 (∵▵n
    g
    < 0)
  • Qstn #87
    Which of the following series of transitions in
    the spectrum of hydrogen atom fall in visible
    region?
    (1) Lyman series
    (2) Balmer series
    (3) Paschen series
    (4) Brackett series
    digAnsr:   2
    Ans : ( 2 )
    Sol. In H-spectrum, Balmer series transitions fall
    in visible region.
  • Qstn #88
    The method used to remove temporary
    hardness of water is :
    (1) Calgon's method
    (2) Clark's method
    (3) Ion-exchange method
    (4) Synthetic resins method
    digAnsr:   2
    Ans : ( 2 )
    Sol. Clark's method is used to remove temporary
    hardness of water, in which bicarbonates of
    calcium and magnesium are reacted with
    slaked lime Ca(OH)
    2
    Ca(HCO
    3
    )
    2
    + Ca(OH)
    2
     2CaCO
    3
    + 2H
    2
    O
    Mg(HCO
    3
    )
    2
    + 2Ca(OH)
    2
     2CaCO
    3
     + Mg(OH)
    2

    + 2H
    2
    O
  • Qstn #89
    Which one is malachite from the following?
    (1)`` CuFeS_2``
    (2) ``Cu(OH)_2``
    (3)`` Fe_3O_4``
    (4) ``CuCO_3 .Cu(OH)_2``
    digAnsr:   4
    Ans : ( 4 )
    Sol. Malachite : CuCO
    3
    .Cu(OH)
    2
    (Green colour)
  • Qstn #90
    The correct order of the basic strength of
    methyl substituted amines in aqueous solution
    is :
    (1) ``(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N``
    (2) ``(CH_3)_3N > CH_3NH_2 > (CH_3)_2NH``
    (3) ``(CH_3)_3N > (CH_3)_2NH > CH_3NH_2``
    (4) ``CH_3NH_2 > (CH_3)_2NH > (CH_3)_3N``
    digAnsr:   1
    Ans : ( 1 )
    Sol. In aqueous solution, electron donating
    inductive effect, solvation effect (H-bonding)
    and steric hindrance all together affect basic
    strength of substituted amines
    Basic character :
    (CH
    3
    )
    2
    NH > CH
    3
    NH
    2
    > (CH
    3
    )
    3
    N
    2° 1° 3°
    22
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