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List four common ions with a valence electron configuration of [Kr]. List the ions in the order of increasing atomic number. Hint: two anions and two cations.

Respuesta :

[tex]Se^{-2}, Br^{-}, Rb^{+}, Sr^{+2}.[/tex]

Explanation:

  • Atomic number of krypton [Kr] =36
  • Electronic configuration of krypton [Kr] = [Ar][tex]3d^{10} 4p^{6}, 2s^{2}[/tex]
  • Common ions with valence electronic configuration of [Kr] are bromine ion, strontium ion, selenite ion and rubidium ion.
  • The Atomic number of the following ions in increasing order are;

       Se < Br < Rb < Sr

       Se (Z) = 34

       Br (Z) = 35

       Rb (Z) = 37

       Sr (Z) = 38

  • Electronic configuration of the following elements are:

         Se (Z) = [Ar][tex](3d^{10}, 4s^{2}, 5p^{4})[/tex]

         Br (Z) = [Ar][tex](3d^{10}, 4s^{2}, 5p^{5})[/tex]

         Rb (Z) = [Kr][tex](5s^{1})[/tex]

         Sr (Z) = [Kr][tex](5s^{2})[/tex]

Therefore, the respective ions formed according to theabove electronic configuration are [tex]Se^{-2}, Br^{-}, Rb^{+}, Sr^{+2}[/tex]

Hence the two anions are[tex]Se^{-2}, Br{-}[/tex]

and, the two cations are[tex]Rb^{+}, Sr{+2}[/tex]

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