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The p-Block Elements Test - 56

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The p-Block Elements Test - 56
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  • Question 1
    1 / -0
    In which reaction ammonia acts as an acid?
    Solution
    NH3NH_3  is a strong base due to high electron density on nitrogen 
    So, it cannot act as an acid .
  • Question 2
    1 / -0
    Acidic nature of amino group is shown by the reaction :-
  • Question 3
    1 / -0
    Alkali metals dissolve in liquid ammonia and produces _______.
    Solution
    Alkali metal when reacts with liquid ammonia reaction is given by,

    M+(x+y)NH3M+(NH3)x+e(NH3)y M +(x+y) NH_3\rightarrow M^+(NH_3)x + e^{-}(NH_3)_y

     this electrons is free and gives very good conductivity as well blue colour.
  • Question 4
    1 / -0
    "N<P<As"" N < P < As"
    Above order is incorrect for:
    Solution
    N, P and AsN, \ P \ and \ As belong to the same group. On moving down the group the atomic radii of atoms increase. Due to the increase in atomic size, the covalent radii and ionic radii also increases. Also with an increase in the size of atom, the nuclear pull on the outermost electrons as well as the ability to attract electrons decreases, resulting in decrease in electronegativity. Hence, electropositivity increases. 
    M2O3M_2O_3 type of oxides have central atom in +3 oxidation state. On moving down the group the ability of atoms to loose electrons increases due to the weakening of nuclear pull on the outermost electrons, hence, basicity increases. Thus, the acidic nature of oxides decreases.
  • Question 5
    1 / -0
    The oxide of nitrogen responsible for acid rain is:
    Solution

  • Question 6
    1 / -0
    Which is the best nucleophile in polar solvent?
    Solution
    The small sized ion has greater charge density. Due to high charge density in polar solvent, they get easily hydrated. So II^{\circleddash} has greater size. So, it will be best nucleophile in polar solvent.
  • Question 7
    1 / -0
    Which of the following is NOT changed of Xe in given reaction?
    2XeOF4+SiO22XeO2F2+SiF42XeO{F}_{4} + Si{O}_{2} \rightarrow 2Xe{O}_{2}{F}_{2}+ Si{F}_{4}
    Solution
    2XeOF4+SiO22XeO2F2+SiF42XeOF_4+SiO_2\longrightarrow 2XeO_2F_2+SiF_4
    Covalency of XeOF4XeOF_4 is 6(refer to image 01)
    Covalency of XeO2F2XeO_2F_2 is 6(refer to image 02)
    So, covalency of XeXe is 6 which donot change.


  • Question 8
    1 / -0
    The pair with minimum difference in electronegativity  is :-
    Solution
        F=3.98     C-2.55       P-2.19         Na-0.93
        Cl=3.16     H-2.20       H- 2.20      Cs-0.79
    diff- 0.82       0.35           0.01             0.14
  • Question 9
    1 / -0
    Number of sigma bonds in P4O10P_{4}O_{10} is :
    Solution

  • Question 10
    1 / -0
    Which noble gas is most soluble in water?
    Solution
    Solubility of noble gases in water increase down the group due to increase in dipole interaction between H2OH_2O molecule and noble gas atom down the group .
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