Question

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.

Answer 1

$Se^{-2}, Br^{-}, Rb^{+}, Sr^{+2}.$

Explanation:

• Atomic number of krypton [Kr] =36

• Electronic configuration of krypton [Kr] = [Ar]$3d^{10} 4p^{6}, 2s^{2}$
• 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]$(3d^{10}, 4s^{2}, 5p^{4})$

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

         Rb (Z) = [Kr]$(5s^{1})$

         Sr (Z) = [Kr]$(5s^{2})$

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

Hence the two anions are$Se^{-2}, Br{-}$

and, the two cations are$Rb^{+}, Sr{+2}$