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

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

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Answer 1 · 2017-10-25T07:38:01

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

Explanation:

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

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

$I^{-} :$ $I^(-):$ $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2} 5 p^{6} .$ $1s^(2)2s^(2)2p^6 3s^ 2 3p^ 6 3d^ 10 4s^ 2 4p^ 6 4d^ 10 5s^ 2 5p^ 6.$ 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, $I_{2}$ $I_2$ , as indeed are all of the halogens.....

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

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Explanation:

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

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Explanation:

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