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The p-Block Elements Test - 26

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The p-Block Elements Test - 26
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  • Question 1
    1 / -0
    Mono atomic element among the following is:
    Solution
    Oxygen is diatomic gas.
     
    Phosphorous is tetra atomic.

    Sulphur is polyatomic.

    Whereas being an inert gas, Krypton is monoatomic.
  • Question 2
    1 / -0
    The elements with atomic numbers $$2, 10, 18, 36, 54$$ and $$86$$ are collectively known as:
    Solution
    Zero group of Mendeleev's periodic table or the group $$18$$ of long form of periodic table is comprised of six elements and these eleements are called as inert gases as trhese elements do not react (inert) with other elements due to their octet outer electronic configaration which is stable configaration. The electronic configuration is listed in the following table.
    S.No
    Element
    Symbol
    Atomic number
    Electronic configuration
    1
    Helium He
    21s2
    2
    Neon
    Ne
    10
    [He] 2s2sp6
    3
    Argon
    Ar
    18
    [Ne] 3s23p6
    4
    Krypton
    Kr
    36
    [Ar] 3d104s24p6
    5
    Xenon
    Xe
    54
    [Kr] 4d105s25p6
    Radon Rn
    86
    [Xe] 4f145d106s26p6
  • Question 3
    1 / -0
    Which of the following electronic configuration corresponds to an inert gas?
    Solution
    Noble gases have full-filled configuration i.e the subshell are completely filled.

    They have the general electronic configuration as $$ns^2 np^6$$. 

    Thus option B represents a noble gas.
  • Question 4
    1 / -0
    Elements with atomic numbers 9, 17, 35, 53 are collectively known as:
    Solution
    Elements having the atomic number:

    9 is fluorine: $$(He)2s^22p^5$$
    17 is chlorine: $$(Ne)3s^23p^5$$
    35 is bromine: $$(Ar)4s^24p^5$$
    53 is iodine: $$(Kr)5s^25p^5$$

    We can observe here that all these elements having the general electronic configuration of $$ns^2np^5$$, hence these elements belong to $$17^{th}$$ group elements.
  • Question 5
    1 / -0
    The inert gas present in the second long period is:
    Solution
    Periods $$4$$ and $$5$$ are called long periods as they contain $$18$$ elements each.

    $$4$$th period is called the first long period and the inert gas in this period is $$Kr$$.

    $$5$$th period is the second long period where $$Xe$$ is the inert gas element.
  • Question 6
    1 / -0
    $$Al+X\overset{\Delta}{\rightarrow} Y \overset{H_{2}O}{\rightarrow} Z+NH_{3}$$
    Here $$X, Y$$ and $$Z$$ are respectively:
    Solution
    $$2Al+N_{2}\overset{\Delta}{\rightarrow}2AlN\overset{H_{2}O}{\rightarrow}Al(OH)_{3}+NH_{3}$$
  • Question 7
    1 / -0
    If a new noble gas is discovered then, what will be its atomic number?
    Solution
    $$Rn = 86+ 32 = 118$$

    The atomic number of next inert gas to be discovered will be $$118$$.
  • Question 8
    1 / -0
    The outer most electronic configuration of the most electronegative element is:
    Solution
    Halogens are most electronegative.

    Therefore outer electronic configuration is $$ns^{2}\ np^{5}.$$

    Option C is correct.
  • Question 9
    1 / -0
    Which of the following does not contain fluorine?
    Solution
    Chile saltpeter - $$NaNO_{3}$$ does not contain Fluorine.
    All contain Fluorine $$\left\{\begin{matrix}
    Fluorspar - CaF_{2}\\Cryolite - Na_{3}AlF_{6}
    \\Flourospatite - Ca_{5}(PO_{4})_{3}F \end{matrix}\right.$$
  • Question 10
    1 / -0
    To prepare $$XeF_6$$ at 573 K, 50-60 atm pressure, the ratio of xenon and fluorine used is :
    Solution
    The ratio of xenon and fluorine is 1:20 at 573K, 50-60 atm pressure to prepare $$XeF_{6}$$.
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