Based on the octet rule, iodine most likely forms what kind of an ion?

1 Answer
anor277
Jul 24, 2018

Well, for a start iodine is a non-metal....and therefore likely to form an anion.....(Why? because non-metals are oxidizing, whereas metals are reducing)

Explanation:

But let us look at it terms of electronic structures. We can go thru the rigmarole of writing the structure of the iodine ATOM, #Z=53#.

And so....

#I:underbrace(1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10)4s^(2)4p^(6))_"configuration of krypton"4d^(10)5s^(2)5p^(5)#

And thus the iodine ATOM only needs to be SINGLY REDUCED to attain an electronic configuration that is isoelectronic with the nearest Noble Gas, i.e. #Kr, Z=54#. Of course we know that iodine is a diatomic molecule, as are all the halogens, and we would write the reduction reaction as...

#1/2I_2 + e^(-) rarr I^(-)#

Of course the given electronic configuration reflects the old octet rule....#2:8:18:18:7# for the iodine atom, and #2:8:18:18:8# iodide ion.

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