JEE Main & Advanced Chemistry The p-block Elements-II / p-ब्लॉक तत्व-II Question Bank Self Evaluation Test - The p-Block Elements (Group 15, 16, 17 & 18)

  • question_answer
    The formation of \[O_{2}^{+}{{[Pt{{F}_{6}}]}^{-}}\] is the basis for the formation of xenon fluorides. This is because

    A) \[{{O}_{2}}\] and Xe have comparable sizes.

    B) both \[{{O}_{2}}\] and Xe are gases.

    C) \[{{O}_{2}}\] and Xe have comparable ionization energies.

    D) Both [a] and [c]

    Correct Answer: D

    Solution :

    [d] (i) The first ionization energy of xenon \[\left( 1,170\text{ }kJ\text{ }mo{{l}^{-}} \right)\] is quite close to that of dioxygen \[\left( 1,180\,kJmo{{l}^{-1}} \right).\] (ii) The molecular diameters of xenon and dioxygen are almost identical. Based on the above similarities Barlett (who prepared \[O_{2}^{+}{{[Pt{{F}_{6}}]}^{-}}\] compound) suggested that since oxygen combines with \[Pt{{F}_{6}}\] so xenon should also form similar compound with \[Pt{{F}_{6}}\].


You need to login to perform this action.
You will be redirected in 3 sec spinner