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p-Block Elements Test - 4

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p-Block Elements Test - 4
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  • Question 1
    1 / -0

    PbCl4 exists while PbBr4 and PbI4 do not because

    Solution

    Due to the lower values of electronegativities of bromine and iodine they cannot oxidize Pb+2 to Pb+4.
    However, chlorine is more electronegative and a stronger oxidizing agent than bromine and iodine.

    Hence, PbCl4 exists while PbBr4 and PbI4 do not.

     

  • Question 2
    1 / -0

    Which of the following is the correct increasing order of oxidizing power?

    Solution

    The oxidizing power of XO2 (X = C, Si, Sn or Pb) increases as the group is descended. PbO2 can oxidize conc. HCl to Cl2 but SnO2 and SiO2 cannot. This is because the stability of the +2 oxidation state increases down the group due to the inert pair effect.

     

  • Question 3
    1 / -0

    The stability of the dihalides of Si, Ge, Sn and Pb increases steadily in the sequence

    Solution

    Among the elements of the p-block of the periodic table, stability of lower oxidation state increases down the group because of the inert pair effect.

    The two most common oxidation states exhibited by the elements of the group 14 are +2 and +4 and the stability of the +2 oxidation state increases down the group.

    Hence, the order of stability of the dihalides of Si, Ge, Sn and Pb is:
    SiX2 << GeX2 << SnX2 << PbX2.

     

  • Question 4
    1 / -0

    Which of the following anions is present in the chain structure of silicates?

    Solution

    Chain silicates are formed by the sharing of two oxygen atoms by each tetrahedron. Anion of chain silicates can be represented as (SiO3)n2-.

     

  • Question 5
    1 / -0

    Which of the following is not isostructural with SiCl4?

    Solution

    SO42-, NH4+ and PO43- are sphybridized having tetrahedral geometry, but SCl4 is sp3d hybridized having a distorted triangular bipyramidal geometry (see-saw shape).

     

  • Question 6
    1 / -0

    Which of these is not a monomer for a high molecular mass silicon polymer?

    Solution

    It can only form a dimer.

     

  • Question 7
    1 / -0

    Borax is prepared by treating colemanite with

    Solution

    Ca2B6O11 + 2Na2CO3 -----Heat-----» Na2B4O7 + 2NaBO+ 2CaCO3

     

  • Question 8
    1 / -0

    Which of the following compounds is/are obtained by heating boric acid?

    a. HBO2
    b. H2B4O7
    c. B2O3

    Solution

    Boric acid is soluble in boiling water. When heated above 170°C, it dehydrates, forming metaboric acid (HBO2).
    H3BO3 → HBO2 + H2O

    Metaboric acid is a white, cubic crystalline solid and is only slightly soluble in water. Metaboric acid melts at about 236°C and when heated above about 300°C, it further dehydrates, forming tetraboric acid or pyroboric acid (H2B4O7).
    4HBO2 → H2B4O7 + H2O

    Further heating leads to formation of boron trioxide.
    H2B4O7 → 2B2O3 + H2O

     

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