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Classification of Elements and Periodicity in Properties Test - 63

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Classification of Elements and Periodicity in Properties Test - 63
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  • Question 1
    1 / -0
    An element X has electronic configuration 2, 8, 2. It forms ionic compounds with ions like nitrate, sulphate and phosphate. What will be the formulae of these compounds?
  • Question 2
    1 / -0
    The amount of energy when a million atoms of iodine are completely converted into $$I^{-}$$ ions in the vapor state according to the equation,
    $$I_{(g)} + e^{-} \rightarrow I^{-}_{(g)}$$ is 4.9 x 10$$^{-13}$$ J. What would be the electron gain enthalpy of iodine in terms of $$kJ\ mol^{-1}$$ and $$eV$$ per atom?
    Solution
    The electron gain enthalpy ($$E_g$$) of an element is defined as the energy released when an atom in gaseous state gains an electron.
    $$I(g)+e^- \rightarrow I^-(g)$$
     In case of iodine
    $$E_g$$ for 1 million atoms ($$10^6 atoms$$)
    $$E_g=4.9 \times 10^{−13} J$$ for $$10^6 atoms$$
    $$E_g=\dfrac{4.9 \times 10^{−13}}{10^6} J/atom$$
    For one mol of $$I$$ atoms, amount of energy released in the formation of $$I^−$$ ions by gaining 1 mol of electrons will be:
    $$E_g=\dfrac{4.9 \times 10^{−13}}{10^6} \times N_A$$
    where $$N_A=6.023 \times 10^{23} atoms/mol$$
    $$E_g=\dfrac{4.9 \times 10^{−13}}{10^6} \times 6.023 \times 10^{23}J/mol$$
    $$=29.51 \times 10^{4} J/atom= 295.1kJ/mol$$
    Since halogens have negative $$E_g$$
    Thus, electron gain enthalpy for one mol iodine $$= -295 kJ mol^{−1}$$
    In eV units we have the relation:
    $$1 eV= 1.6021 \times 10^{-19}J=1.6021 \times 10^{-22}kJ$$
    $$-295 kJ/mol= \dfrac{-295}{1.6021 \times 10^{-22}}eV/mol=184.13 \times 10^{22}eV/mol$$
    For one atom, $$E_g$$ in eV is:= $$\dfrac{-184.13 \times 10^{22}}{N_A}eV$$
    =$$\dfrac{-184.13 \times 10^{22}}{6.023 \times 10^{23}}eV$$
    $$=-3.06eV/atom$$
  • Question 3
    1 / -0
    Which of the following represent correct order of decreasing $${\text{E}}_{{\text{ox}}}^{\text{O}}$$ or reducing nature?
    Solution
    Oxidation potential is the potential at which oxidation (removal of $$e^-$$) occurs at anode in an electrochemical cell.
    $$Li>Na>K>Rb$$
    $$Li$$ has high due to its' small size and as we move down size increases and hence oxidation potential decreases.
  • Question 4
    1 / -0
    The second electron gain enthalpy of O and S (in $$KJ{ mol }^{ -1 }$$) respectively are:
  • Question 5
    1 / -0
    Which of the following elements will gain one electron more readily in comparison to other elements of their group?
  • Question 6
    1 / -0
    Which one of the following statement is incorrect?
    Solution

  • Question 7
    1 / -0
    Electron gain enthalpy of $$17th$$ group elements are $$-349, -328, -325,-296, -270\ kJ/mol$$ (not in order). The element which has electron gain enthalpy $$-296 kJ/mol$$ is:
    Solution
    On moving down the group atomic size and nuclear charge increases. The effect of increase in atomic size is more pronounced than that of nuclear charge.
    $$X+e^{-}\longrightarrow X^-$$
    Electron gain enthalpy becomes less negative on going down the group.
    $$F=-328$$ $$KJ/mol$$
    $$Cl=-349$$ $$KJ/mol$$
    $$Br=-325$$ $$KJ/mol$$
    $$I=-296$$ $$KJ/mol$$
  • Question 8
    1 / -0
    The electronic configuration of an element is $$1s^22s^22p^63s^23p^3$$ . The atomic number of the element, which is just below the above element in the periodic table is__
  • Question 9
    1 / -0
    Which one of the following statements is incorrect?
    Solution
    Electron gain enthalpy is the energy released when an electron is added to an isolated gaseous atom. 

    Its value is negative and the ease with which an atom accepts electrons its electron gain enthalpy becomes more negative that is, its value increases. 

    With an increase in nuclear charge, the electron gain enthalpy also increases.

    Due to the extra stability of nitrogen (it has half-filled valence shell $$[He]1s^22p^3$$, which provides extra stability to it), it is very difficult to add an electron to it. 

    Therefore, it has almost zero electron gain enthalpy.

    In general, on moving down the group, the electron gain enthalpy decreases due to an increase in the atomic size, but on moving from $$F$$ to $$Cl$$ the electron gain enthalpy increases. 

    Due to the very small size of $$F$$ atom and absence of $$3^{rd}$$ shell, the incoming electron experiences repulsion due to the valence electrons and its electron gain enthalpy is less than $$Cl$$. After $$Cl$$ the electron gain enthalpy follows the general order and decreases on moving down to $$Br$$ and $$I$$.

    Therefore, the option is C.
  • Question 10
    1 / -0
    Which of the following possess highest electron affinity:
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