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The p-Block Elements Test - 64

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The p-Block Elements Test - 64
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  • Question 1
    1 / -0
    $$Me_2SiCl_2$$ on hydrolysis will produce :
    Solution
    Dimethyldichlorosilane $$\displaystyle  Me_2SiCl_2 $$ on hydrolysis will produce $$\displaystyle [-O-(Me)_2Si-O-]_n $$, which is a silicone, a thermoplastic polymer.

    $$\displaystyle Me_2SiCl_2  \xrightarrow [\text{hydrolysis}]{H_2O} Me_2Si(OH)_2  $$ 

    $$\displaystyle n Me_2Si(OH)_2  \xrightarrow {\text{polymerisation}}[-O-(Me)_2Si-O-]_n  $$
  • Question 2
    1 / -0
    In borax the number of $$B-O-B$$ links and $$B-OH$$ bonds present are respectively:
    Solution
    There are $$5B-O-B$$ bonds and $$4B-O-H$$ bonds.

  • Question 3
    1 / -0
    The correct increasing order of the stability of $$Al^+,\,Ga^+,\,In^+,\,Tl^+$$ ions are:
    Solution
    The inert pair effect is the tendency of the electrons in the outermost atomic S orbital to remain unionized or unshared in compounds of post-transition metals. The term inert pair effect is often used in relation to the increasing stability of oxidation states that are two less than the group valency for the heavier elements of groups 13, 14, 15 and 16.
    As an example in group 13 the +1 oxidation state of $$Tl$$ is the most stable and $$Tl^{3+}$$ compounds are comparatively rare. The stability of the +1 oxidation state increases in the following sequence:
    $$Al^+<Ga^+<In^+<Tl^+$$
  • Question 4
    1 / -0
    Kieselguhr contains :
    Solution
    Keisulguhr is a naturally occurring soft silicons sedimentary rock that is easily crumbled into fine white-to-off-white powder. It contains $$80-90$$% silica, with $$2-4$$% alumina. So it has silicon.
  • Question 5
    1 / -0
    The hydrides of the first elements in groups 15 - 17, namely $$N{H}_{3}, {H}_{2}O$$ and $$HF$$ respectively, show abnormally high values for melting and boiling points. This is due to:
    Solution
    The hydrides of the first elements in groups 15 - 17, namely $$NH_3$$, $$H_2O$$ and HF respectively, show abnormally high values for melting and boiling points. This is due to the ability to form extensive inter-molecular H-bonding. In these hydrides, H atom is attached to highly electronegative N, O or F atoms resulting in stronger hydrogen bonds which lead to molecular associations. Large amount of energy is needed to break these hydrogen bonds.
  • Question 6
    1 / -0
    Group 13 elements are attacked by non-oxidising acids except:
    Solution
    Option $$(D)$$ is correct.
    Boron is not attacked by non-oxidising acids such as $$HCl$$ but reacts with oxidising acids such as conc. $$HNO_3$$ and conc.$$H_2SO_4$$ forming boric acid.
    Boron, being chemically a non-metal, is resistant to attack by non-oxidising acids but the other members of the group react with non-oxidising acids as typical metals and evolve hydrogen.
  • Question 7
    1 / -0
    $$(i)$$Its aqueous solution is alkaline to litmus.
    $$(ii)$$ It swells up to glassy material $$Y$$ on strong heating.
    $$(iii)$$When the concentrated solution of sulphuric acid is added to a hot solution of $$X$$.white crystal of an acid $$Z$$ separates out.

    The Compound $$Y$$ is :
    Solution
    When heated borax undergoes various transitions it first loses water molecules and swells. Then resulting in the formation of glassy material.

    $$Na_2B_4O_7.10H_2O\xrightarrow {\quad \triangle\quad   }Na_2B_4O_7\xrightarrow {\quad  \triangle\quad   } \underset{sodium\ metaborate}{2NaBO_2}+\underset{boric\ oxide}{B_2O_3}$$

    Sodium metaborate and boric oxide form the glassy material.
     Y is sodium metaborate and boric oxide.
  • Question 8
    1 / -0
    Read the following statements and identify compound $$Z$$
    $$(i)$$Its aqueous solution is alkaline to litmus.
    $$(ii)$$ It swells up to glassy material $$Y$$ on strong heating.
    $$(iii)$$When the concentrated solution of sulphuric acid is added to a hot solution of $$X$$.white crystal of an acid $$Z$$ separates out.
    Solution

  • Question 9
    1 / -0
    Element showing the phenomenon of allotropy is:
    Solution
    Among the given elements tin ($$Sn$$) shows the allotropy.
    β-tin (the metallic form, or white tin), which is stable at and above room temperature, is malleable. In contrast, α-tin (nonmetallic form, or gray tin), which is stable below 13.2 °C (55.8 °F), is brittle.
    Hence option $$C$$ is correct.
  • Question 10
    1 / -0
    The element that forms stable compounds in low oxidation state is:
    Solution
    The element that forms stable compounds in low oxidation state is $$Tl$$. The two electrons in the outer shell tend to remain paired and do not participate in compound formation (inert pair effect). 

    Due to this, $$Tl(I)$$ compounds are more stable than $$Tl(III)$$ compounds. 

    $$Tl(III)$$ ions act as oxidising agents and changes to $$Tl(I)$$ ions.

                  $$ \displaystyle Tl^{3+}+2e^- \rightarrow Tl^+$$

    Hence, option $$D$$ is correct.
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