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.
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^-
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?
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.
thanks
To understand the solution, let's break it down step by step.
First, we have the element Iodine with the atomic number Z=53. The electron configuration of Iodine is 1s22s22p63s23p64s23d104p65s24d105p5.
The question asks for the formation of the common ions of Iodine. To form ions, atoms gain or lose electrons to achieve a stable electron configuration.
In this case, we start with Iodine in its neutral state, which has 53 electrons. We want to write the electron configurations for the common ions of Iodine.
Iodine forms ions by gaining one electron to achieve a stable electron configuration. Let's consider this process step by step:
1. Start with the electron configuration of Iodine: 1s22s22p63s23p64s23d104p65s24d105p5.
2. Locate the highest energy level of the atom that contains valence electrons. In this case, it's the 5p orbital.
3. Remove one electron from the 5p orbital, so the electron configuration becomes: 1s22s22p63s23p64s23d104p65s24d105p6.
4. Finally, simplify the electron configuration by writing the symbol of the noble gas that comes before Iodine in the periodic table, which is Krypton (Kr). The condensed electron configuration for the Iodine ion is [Kr]5s24d105p6.
So, the condensed electron configuration of Iodine when it gains one electron to form the common ion I- is [Kr]5s24d105p6, and it is isoelectronic with Xenon (Xe).
To summarize:
Iodine ion (I-) has the electron configuration [Kr]5s24d105p6, which is equivalent to the noble gas Xenon (Xe).
I hope this explanation helps you understand how to determine the electron configuration of main group ions.