Chem Please Help

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.

asked by Sami
  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^-

    posted by DrBob222
  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?

    posted by Sami
  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.

    posted by DrBob222
  4. thanks

    posted by Sami

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