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]
