Question

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

Answer 1

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.....