Is #"iodide anion"#, #I^(-)#, isoelectronic with #"xenon"#?

1 Answer
Oct 25, 2017

Answer:

You are correct it would be isoelectronic with the Noble Gas, #Xe#...

Explanation:

And thus elemental iodine, the atom, is a reasonably strong oxidant, and we use the #"old aufbau principle"#, and distribute 54 electrons.....

#I:# #1s^(2)2s^(2)2p^6 3s^ 2 3p^ 6 3d^ 10 4s^ 2 4p^ 6 4d^ 10 5s^ 2 5p^ 5.#

#I^(-):# #1s^(2)2s^(2)2p^6 3s^ 2 3p^ 6 3d^ 10 4s^ 2 4p^ 6 4d^ 10 5s^ 2 5p^ 6.# Are there 54 electrons? If not, there should be.

Of course, we must simply remember that when we use iodine, we work with the diatom, #I_2#, as indeed are all of the halogens.....