Using condensed electron configurations, give reactions showing the formation of the common ions of the following elements. (Type your answers in the following order. In the first box enter your answer using the format [Ar] 4s2 3d10 4p2 for [Ar]4s23d104p2. For second box enter the number of electrons and if there are no electrons enter 0. For the third box type your answer in the format of [NH4]+ for NH4+ or [Mg]2+ for Mg2+. Boxes four and five have the same format as the first and second box. Give the equation for the smaller charged ion first.)
K (Z = 19)
I (Z = 53)
(c) Pb (Z = 82)
You can find electron configurations here.
http://www.webelements.com/
To determine the condensed electron configuration of each element, we need to know the atomic number and the number of electrons for each ion.
For K (Z = 19), the atomic number is 19. The common ion formed is K+ (potassium ion), which means it loses one electron. Therefore, the condensed electron configuration for K+ is [Ar] 4s1.
For I (Z = 53), the atomic number is 53. The common ion formed is I- (iodide ion), which means it gains one electron. Therefore, the condensed electron configuration for I- is [Kr] 5s24d105p6.
For Pb (Z = 82), the atomic number is 82. The common ion formed is Pb2+ (lead ion), which means it loses two electrons. Therefore, the condensed electron configuration for Pb2+ is [Xe] 6s24f145d10.
Here are the answers in the requested format:
1. [Ar] 4s1, -1, [Ar] (K+)
2. [Kr] 5s24d105p6, +1, [Kr] (I-)
3. [Xe] 6s24f145d10, -2, [Xe] (Pb2+)