Ionic trends

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Rebecca Altshuler 1D
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Joined: Fri Sep 28, 2018 12:15 am

Ionic trends

Postby Rebecca Altshuler 1D » Mon Oct 22, 2018 1:54 pm

I understand the trends for the periodic table, but why do the trends for ions not behave in the same way? For example, for phosphorous, sulfur, and chlorine, the order of increasing atomic size would go Cl, S, P, but for the ionized forms of these elements, the order is reversed. Why is this?

Laura Gong 3H
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Joined: Fri Sep 28, 2018 12:26 am
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Re: Ionic trends

Postby Laura Gong 3H » Mon Oct 22, 2018 2:36 pm

The reason is because when those elements are ionized, the electrostatic attraction experienced between the electrons and the nucleus change. Chlorine has the greatest amount of protons compared to S and P which means that when it loses an electron, it will pull the remaining electrons in even tighter, decreasing the ionic radius. The same logic goes for S and P. In their ionized form, the nucleus (# of protons) will play a bigger role in the electrostatic attraction equation.

Max Hayama 4K
Posts: 63
Joined: Fri Sep 28, 2018 12:16 am

Re: Ionic trends

Postby Max Hayama 4K » Mon Oct 22, 2018 8:47 pm

Because they all reside in the same row, all of the valence electrons will experience the same shielding. The only difference is the number of protons, which will pull the electrons inwards, thus decreasing atomic radius. As a result, chlorine has more protons than sulfur and phosphorus, so it has the smallest radius. I don't believe the order is reversed if these elements are ionized, because they will effectively gain electrons, which will increase the atomic radius. Therefore phosphorus will still be the largest in radius because it gains 3 electrons, causing more electron-electron repulsion and less electrostatic attraction between the nucleus and the electrons. Sulfur gains 2 and chlorine gains 1, keeping the order the same.


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