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)
K( ) + ( ) e- arrow ( ) ( ) + ( ) e-
I (Z = 53)
I ( ) + ( ) e- arrow ( ) ( ) + ( ) e-
(c) Pb (Z = 82)
Pb ( ) + ( ) e- arrow ( ) ( ) + ( ) e-
Pb ( ) + ( ) e- arrow ( ) ( ) + ( ) e-
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To determine the formation of common ions for the elements K, I, and Pb using condensed electron configurations, we need to first write the condensed electron configuration for each element.
K (Z = 19) - Potassium:
The electron configuration for potassium is [Ar] 4s1.
To form a cation, K+ (potassium ion), we need to remove one electron.
K ( ) + 1 e- arrow K+ ( )
To form an anion, K- (potassium ion), we need to add one electron.
K ( ) + 1 e- arrow K- ( )
I (Z = 53) - Iodine:
The electron configuration for iodine is [Kr] 5s24d105p5.
To form a cation, I+ (iodine ion), we need to remove one electron.
I ( ) + 1 e- arrow I+ ( )
To form an anion, I- (iodine ion), we need to add one electron.
I ( ) + 1 e- arrow I- ( )
Pb (Z = 82) - Lead:
The electron configuration for lead is [Xe] 6s24f145d106p2.
To form a cation, Pb2+ (lead ion), we need to remove two electrons.
Pb ( ) + 2 e- arrow Pb2+ ( )
To form an anion, Pb4- (lead ion), we need to add four electrons.
Pb ( ) + 4 e- arrow Pb4- ( )