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

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Answer 1 · 2017-10-25T07:38:01

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

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.....

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