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The p-Block Elements Test - 65

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The p-Block Elements Test - 65
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  • Question 1
    1 / -0
    Which of the following chemicals contains chlorine?
    Solution
    Spirit of salt an old name for hydrochloric acid ($$HCl$$).
    Hence option D is correct,
  • Question 2
    1 / -0
    $$XeF_6$$ on partial hydrolysis with water produces a compound $$X$$. The same compound $$X$$ is formed when $$XeF_6$$ reacts with silica. The compound $$X$$ is:
    Solution
    Partial hydrolysis of  $$\displaystyle XeF_6$$ gives $$\displaystyle XeOF_4$$ (compound $$X$$)
    $$\displaystyle XeF_6 + H_2O \rightarrow XeOF_4+2HF$$

    $$\displaystyle XeF_6$$ reacts with silica $$\displaystyle SiO_2$$ to form  $$\displaystyle XeOF_4$$ (compound $$X$$)
    $$\displaystyle 2XeF_6 + SiO_2 \rightarrow 2XeOF_4+SiF_4$$
  • Question 3
    1 / -0
    The noble gas mixture is cooled in a coconut bulb at $$173 K$$. The gases that are not adsorbed are:
    Solution
    When the mixture of noble gas is cooled in a coconut bulb at $$173\ K$$ then argon, krypton and xenon are adsorbed on charcoal while Helium and Neon are not adsorbed.
    Hence option $$A$$ is correct.
  • Question 4
    1 / -0
    In the Ostwald's process for the manufacture of $$HNO_3$$, the catalyst used is:
    Solution
    Ostwal's process is a chemical process for the manifacture of nitric acid in large scale.In the ostwalds process, during the preparation of $$HNO_3$$, which can use for fertilisers, vanadium pentaoxide ($$V_2O_5$$) is used as catalyst. 
  • Question 5
    1 / -0
    An atom has $$2K, 8L$$ and $$5M$$ electrons.
    Choose the correct statement(s) regarding it.
    A. Trivalent anion of this atom will have $$12$$ protons in its nucleus.
    B. Trivalent cation of this atom will have six p-electrons in it.
    C. This atom form an amphoteric oxide of formula $$X_{2}O_{3}$$.
    D. One of its allotrope is tetra atomic $$(X_{4})$$
    Solution

    In given atom the structure is, K(2) L(8) M(5)

    Configuration will be:

    $$X=1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$$

    $$X – 3e \rightarrow X^{3+}$$

    So, it will be trivalent metal cation.

    $$X^{3+} = 1s^{2}2s^{2}2p^{6}3s^{2}$$

    So, p electrons will be 6.

    As per the configuration, this element is the phosphorous (atomic number 15). It exists as allotrope in tetra atomic i.e.  $$X_{4}$$ format.

    The correct option is D. 

  • Question 6
    1 / -0
    Ammonia is not produced in the reaction of:
    Solution
    Reaction between $$NH_4Cl,NaNO_2$$ do not produce $$NH_3$$ instead it produces $$N_2$$

     $$NH_4Cl+NaNO_2\rightarrow N_2+2H_2O+NaCl$$
  • Question 7
    1 / -0
    Which of the following pairs are correctly matched? 
    1. Haber process - Manufacture of $$NH_3$$ 
    2. Leblanc process - Manufacture of $$H_2SO_4$$
    3. Birkeland-Eyde process - Manufacture of $$HNO_3$$
    4. Solvay process - Manufacture of $$Na_2CO_3$$ 
  • Question 8
    1 / -0
    Nitrogen forms stable $$N_{2}$$ molecule but phosphorus is converted to $$P_{4}$$ from $$P_{2}$$ because:
    Solution
    $$P$$ has empty $$d$$ orbital, that can be filled by an additional electron. And since its size is bigger it is least likely to handle triple bond, as size increases $$\pi$$ overlapping decreases. 
  • Question 9
    1 / -0
    Fluorine is the best oxidising agent because it has:
    Solution

    Correct Option: $$B$$

    Hint: Fluorine has the least bond enthalpy 

    Explanation:
    Fluorine is an element that has high electronegativity.     

     Fluorine has low bond enthalpy and highest reduction potential so more ability to loose the electrons.

     Fluorine can also be easily reduced. 

     Fluorine is thus one of the best oxidizing agents. 

     

    Henceforth Correct  option is the highest reduction potential (B) 

  • Question 10
    1 / -0
    Ammonia is not produced in the reaction of ________.
    Solution


    $$N{ H }_{ 4 }Cl+KOH\rightarrow KCl+{ H }_{ 2 }O+N{ H }_{ 3 }$$ 

    $$AlN+3{ H }_{ 2 }O\rightarrow N{ H }_{ 3 }+Al{ (OH) }_{ 3 }$$ 

    $$N{ H }_{ 4 }Cl+NaN{ O }_{ 2 }\rightarrow { N }_{ 2 }+2{ H }_{ 2 }O+NaCl$$

    $$2N{ H }_{ 4 }Cl+Ca{ (OH) }_{ 2 }\rightarrow Ca{ Cl }_{ 2 }+2N{ H }_{ 3 }+{ H }_{ 2 }O$$

    Option C is correct.
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