Question

I know how to write elctron configurations but I am having problems with electron config. of main group ions.

Problem: Using condensed electron configurations, write reactions for the formation of the common ions of the following element: Iodine (Z=53)

Solution: I ([Kr] 5s24d105p5)+ e- arrow I- ([Kr]5s24d105p6) (same as Xe)

I have went over it many times and I don't understand what is being asked and how the answer was gotten. Please help.

Answer 1

What do you not understand about the answer? The question asks for the formation of I^- from I and to do it with electron configuration. The reaction is
I + e ===> I^-
and the question asks for you to use the electron configuration for I and I^-. That's what the solution does. Since the questions specifically says "ions" (or is that a typo) they may also refer to the formation of I3^-

Answer 2

I know the short electron config for I is 5s24d105p5....but why do I have to write e- in the first part and also isnt Xe the closest noble gas, why is Kr used instead? In the second part since did an electron become added because Iodine is -1?

Answer 3

You add an electron to the first part because adding an electron to I is what fills the outside 5s2 5p5 to 5s2 5p8 and that makes it an I^-. Not using electron configurations would give you an equation I wrote in my first reponse such as I + e ==> I^- OR
4d10 5s2 5p5 + e ==> 4d10 5s2 5p6
AND you DON't use [Xe] in the last part because you want people to KNOW that YOU KNOW the the I neutral atom as simply added an electron to become a filled shell of 5s2 5p6. Writing [Xe] is correct with respect to the number of electrons, (both of them have the same number) but that doesn't show tha the extra electron has been added to the p shell to form 5p6 from 5p5.

Answer 4

thanks