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p-Block Elements Test - 6

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p-Block Elements Test - 6
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Weekly Quiz Competition
  • Question 1
    4 / -1

     

    Which is incorrect statement?

     

    Solution

     

     

    In tellurium, differentiating electron enters in to 5p subshell. Its valence shell configuration is 5s25p4.

     

     

  • Question 2
    4 / -1

    In earth’s crust, sulphur exists as

    Solution

    Sulphur exists in elemental form which is extracted by Frasch process.
    Sulphide ores are zinc blende (ZnS), galena (PbS), copper pyrites (CuFeS2), cinnabar (HgS),
    Sulphate ores are gypsum (CaSO4. 2H2O), epsom (MgSO4 .7 H2O) .

  • Question 3
    4 / -1

    The element used in the production of photocells and solar cells and also used in zerox machines is

    Solution

    Se is non-conductor in dark but acts as conductor when exposed to light.

  • Question 4
    4 / -1

    The number of bond pairs and lone pairs in rhombic sulphur molecule are

    Solution

  • Question 5
    4 / -1

    Which has lowest bond energy (single bond)?

    Solution

    Presence of lone pairs on bonded atoms generally lowers the bond energy.
    Bond energy order is
    O—O (146) < (S—S) (266) < S—H(366) < O—H(464) kJ/mol
     

  • Question 6
    4 / -1

    Harmful UV radiations emitted from the sun are prevented from reaching the earth by the presence of ozone in the

    Solution

    Stratosphere layer strongly absorb harmful UV-radiations (λ. 255 nm).

  • Question 7
    4 / -1

    Point out the incorrect statement among the following.

    Solution

    It is 369 K or 96°C. The stable form at room temperature is rhombic sulphur which transforms to monoclinic sulphur when heated about 369 K.

  • Question 8
    4 / -1

    The correct order of thermal stability of the hydrides of group 16 elements is

    Solution



    Thermal stability ∝ bond dissociation energy 
    (where, E = group 16 elements)
    On moving down the group, bond dissociation energy decreases due to increase in bond length. Thus, the order of bond dissociation energy or thermal stability is 
    H2O > H2S > H2Se > H2Te > H2PO

  • Question 9
    4 / -1

    When O2 changes to 

    Solution


  • Question 10
    4 / -1

    Increasing order of oxidation number of oxygen is found in

    Solution


  • Question 11
    4 / -1

     

    Q.  Which of the following property do not increase down the group from oxygen to polonium?

     

    Solution

     

    Electron affinity is decreases down the group due to increase in size.

     

  • Question 12
    4 / -1

     

    Which trend enthalpy is correct ?

    Solution

    Bonds → Bond enrgy
    N-O → 201 kJ/mol
    P-O → 340 kJ/mol
    N-F → 272 kJ/mol
    P-F → 490 kJ/mol
    N≡N → 946 kJ/mol
    P≡P → 490 kJ/mol
    N-N → 160 kJ/mol
    P-P → 209 kJ/mol

    Hence option C is correct.

     

  • Question 13
    4 / -1

     

    Boiling Point of liquid nitrogen is

    Solution

     

    Liquid nitrogen is a cryogenic fluid that can cause rapid freezing on contact with living tissue. 

    The temperature is held constant at 77 K by slow boiling of the liquid, resulting in the evolution of nitrogen gas.

     

  • Question 14
    4 / -1

     

    Which is the correct order w.r.t the given property?

    Solution

     

    Only electronegativity of the given 4 decreases down the group whilst all the others increase or stay the same down the group.

     

  • Question 15
    4 / -1

     

    Which among the following forms basic oxide?

    Solution

     

    As we move down the group metallic character increases so Bi is a metal thus its oxide is basic.

     

  • Question 16
    4 / -1

     

    Passage

    Sulphur and the rest of the elements of 16th group are less electronegative than oxygen. They can acquire ns2np6 by sharing two electrons with the atoms of other elements and thus, exhibit +2 oxidation state in their compounds, in addition to this, atoms have vacant d-orbitals in their valence shell due to which electrons I can be promoted from the s- and p-orbitals of the same shell. As a result, they can show +4 and +6 oxidation states. The oxidation states of elements never exceed +6.

    Q.

    Oxidation states of sulphur in H2S2O7 and H2S2O8 respectively are

     

    Solution

     

    H2S2O7 has no peroxy bond while H2S2O8 has one peroxy bond.

     

  • Question 17
    4 / -1

    Passage

    Sulphur and the rest of the elements of 16th group are less electronegative than oxygen. They can acquire ns2np6 by sharing two electrons with the atoms of other elements and thus, exhibit +2 oxidation state in their compounds, in addition to this, atoms have vacant d-orbitals in their valence shell due to which electrons I can be promoted from the s- and p-orbitals of the same shell. As a result, they can show +4 and +6 oxidation states. The oxidation states of elements never exceed +6.

    Q. 

    Like sulphur, oxygen does not show +4 and +6 oxidation states .The reason is

    Solution

    Oxygen does not show +4 and +6 oxidation states because oxygen has no cf-orbitals in its valence shell.

  • Question 18
    4 / -1

     

    Match the Column I with Column II and mark the correct option from the codes given below.

     

    Solution

     

     

  • Question 19
    4 / -1

    Match the Column I with Column II and mark the correct option from the codes given below.

    Solution

  • Question 20
    4 / -1

    Match the Column i with Column II and mark the correct option from the codes given below.

    Solution

  • Question 21
    4 / -1

     

    Direction (Q. Nos. 21-24) This section contains 4 questions. When worked out will result in an integer from 0 to 9 (both inclusive).

    Q. 

    Number of chemicals that produce dioxygen on heating

    KCIO3, (NH4)2 Cr2O7, NH4NO3, NaNO3, KMnO4, K2Cr2O7, H2O2, Pb3O4, NH4NO2

     

    Solution

     

    (NH4)2 Cr2O7, NH4NO2 gives dinitrogen while NH4NO3 gives N2O.

     

  • Question 22
    4 / -1

     

    In the fully excited state, number of unpaired electrons in tellurium are

     

    Solution

     


     

  • Question 23
    4 / -1

     

    Number of chemical species having bent shape in 

     

    Solution

     

    O3, NO2, SO2 and H2S show bent shape.

     

  • Question 24
    4 / -1

     

    Among the following compounds, the number of compounds which have oxidation states of S is +4 ?

    PbS, SO2, SF6, Na2S2O3, H2SO3

     

    Solution

     

    The correct answer is 2.
    Let S is in x oxidation state
    In PbS, 2 + x = 0 =>  x = -2
    In SO2 , x + 2 × (-2) = 0 => x= +4
    In SF6 , x + 1 × (-6) = 0 => x = +6
    In Na2S2O3, 2 ×(+1) + 2 × x + 3 × (-2) = 0 => x= +2

     

  • Question 25
    4 / -1

    Direction (Q. No. 25) This section is based on Statement I and Statement II. Select the correct answer from the codes given below.

    Q. 

    Statement I : Oxygen is more electronegative than sulphur, yet H2S is acidic, while H2O is neutral.

    Statement II : H— S bond is weaker than O—H bond.

     

    Solution

     

    H—S bond is weaker than H—O bond hence, H2S is more acidic than H2O.

     

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