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The p-Block Elements Test - 58

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The p-Block Elements Test - 58
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  • Question 1
    1 / -0
    Which is not true about borax ?
    Solution
    In borax $$(Na_{2}B_{4}O_{7} . 10H_{2}O),$$ among $$10$$ water molecules $$2$$ molecules are part of structure, i.e., exists $$Na_{2}[B_{4}O_{5} (OH)_{4}] . 8H_{2}O$$
    $$Na_{2}[B_{4}O_{5}(OH)_{4}] . 8H_{2}O + 2HCl \rightarrow 2NaCl + 4H_{3}BO_{3} + 5H_{2}O$$
    Methyl orange $$(pH = 3.7)$$ is used to detect end point. Aq. solution of borax acts as buffer, as borax is salt of strong base $$NaOH$$ and weak acid $$H_{3}BO_{3}.$$

    Hence, Option "B" is the correct answer.

  • Question 2
    1 / -0

    Directions For Questions

    $$A$$ is a white crystalline solid. Its aqueous solution is alkaline in nature. It is used in water softening. On heating, it swells up to form a puffy mass, $$B$$. Strong heating of $$B$$ gives $$C$$. Heating of $$C$$ with nickel oxide gives a brown bead, $$D$$.

    ...view full instructions

    The aqueous solution of 'A' is alkaline due to
    Solution
    A is borax or sodium tetraborate $$(Na_2B_4O_7.10H_2O)$$.

    One mole of A contains 10 moles of water of crystallization.

    It is a white crystalline solid and is used in water softening. Its aqueous solution is alkaline due to the formation of sodium hydroxide upon hydrolysis of $$Na_2B_4O_7$$. The reaction is as follows:

    $$Na_2B_4O_7+7H_2O \rightleftharpoons  2NaOH + 4 H_3BO_3$$

  • Question 3
    1 / -0
    Which one of the following statements is incorrect about the structural properties of $$ BF_{3}$$?
    Solution
    $$B{ F }_{ 3 }$$ has a trigonal planar structure, all three $$B-F$$ bonds lie in the same plane. Thus, the p orbitals of Boron and Fluorine become parallel.
    p-orbital of Boron is empty while fluorine has a lone pair. Fluorine donates its lone pair and this is called back bonding, or $$p\pi -p\pi $$ bond. Resonating structures can be obtained.
    Since the $$B{ F }_{ 3 }$$ molecule is highly symmetrical, it has no dipole moment. All bonds are of equal length, thus option (B) is correct.
    In Boron trihalides, the length of $$B-F$$ bond would be shorter than expected for a single bond, due to $$p\pi -p\pi$$ bonding.
    $$B{ Cl }_{ 3 }$$ is also a trigonal planar molecule. Back bonding does not change the structure of the molecule. It simply decreases bond length $$B-X$$ and decreases Lewis acidity of $$B{ X }_{ 3 }$$.
    Thus, option $$(D)$$ is incorrect.
  • Question 4
    1 / -0
    The tendency to show inert pair effect follows the order :
    Solution
    Down the group inert pair effect increases so the correct order is $$B < Al < Ga < In < Tl$$.
  • Question 5
    1 / -0
    Which of the following is incorrect regarding the structural features of borax?
    Solution
    In the structure of borax, none of the boron atoms have lone pairs of electrons.

  • Question 6
    1 / -0
    Order of stability of the ions of Ge, Sn and Pb is:
    Solution
    On moving down group IV A, the stability of the +2 oxidation state increases due to the inert pair effect which becomes more pronounced in heavier elements.

    Hence, the correct order of the stability of oxidation states is $$Ge^{2+} < Sn^{2+} < Pb^{2+}$$.

    So, the correct option is ' A'.
  • Question 7
    1 / -0
    Consider the following reactions:

    i)$$Cr_{2}O_{3}+2Al\rightarrow Al_{2}O_{3}+2Cr+heat$$
    ii)$$Al_{2}O_{3}+2Cr\rightarrow Cr_{2}O_{3}+2Al+heat$$
    iii)$$2Al+6NaOH\rightarrow 2Na_{3}AlO_{3}+3H_{2}$$

    Out of these, the possible reactions are:
  • Question 8
    1 / -0
    The number of $$-OH$$ units directly bonded to boron atoms in borax is :
    Solution
    As can be seen from the image 4 −OH units are directly bonded to boron atoms in borax.

  • Question 9
    1 / -0
    Match the ores from List-I with their corresponding formulae from List-II.

    List -I                                                        List -II
    A. Borax                                                   1) $$Al_{2}O_{3}.2H_{2}O$$
    B. Razorite                                               2) $$Ca_{2}B_{6}O_{11}.5H_{2}O$$ 
    C. Bauxite                                                3) $$Na_{2}B_{4}O_{7}.4H_{2}O$$
    D.Colemanite                                           4) $$Na_{2}B_{4}O_{7}.10H_{2}O$$
    Solution
    Borax: $$Na_{2} B_{4} O_{7} .10H_{2} O $$

    Razorite: $$Na_{2} B_{4} O_{7} .4H_{2} O $$

    Bauxite: $$Al_{2} O_{3} .2H_{2} O $$

    Colemanite: $$Ca_{2} B_{6} O_{11} .5H_{2} O $$

    So, the correct option is 'C'.
  • Question 10
    1 / -0
    An alkyl halide reacts with a group 14 element $$Y$$ at 570K with $$Cu$$ as a catalyst and produces a dialkyl chloro compound $$Z$$. The compound $$Z$$ on hydrolysis gives another compound which is a strong water repellent and quite inert chemically. The dioxide of $$Y$$ is acidic in nature. The alkyl halide can also be obtained from methane after monosubstitution. Identify $$Y$$ and $$Z$$.
    Solution
    Monosubstitution of methane gives methyl chloride. It reacts at $$570K$$ with $$Si$$ having $$Cu$$ as catalyst to produce $$(CH _{3})_{2} SiCl_{2}$$, which is the compound $$Z$$. 

    $$ (CH _{3})_{2} SiCl_{2}$$ on hydrolysis gives $$ (CH _{3})_{2} Si(OH)_{2}$$, which is a strong water repellent and quite inert chemically. Also $$SiO_2$$ is acidic in nature.

    Option 'A' is correct.
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