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

Oct 25, 2017

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

Explanation:

And thus elemental iodine, the atom, is a reasonably strong oxidant, and we use the $\text{old aufbau principle}$, and distribute 54 electrons.....

$I :$ $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{10} 4 {s}^{2} 4 {p}^{6} 4 {d}^{10} 5 {s}^{2} 5 {p}^{5.}$

${I}^{-} :$ $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{10} 4 {s}^{2} 4 {p}^{6} 4 {d}^{10} 5 {s}^{2} 5 {p}^{6.}$ 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.....