Isoelectronic ions

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William Lan 2l
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Joined: Fri Sep 29, 2017 7:07 am

Isoelectronic ions

Postby William Lan 2l » Wed Oct 25, 2017 8:23 pm

For the isoelectronic ions Na+, F-, and Mg2+, why is it that Mg2+ has the largest nuclear charge of the 3 (it has the strongest attrition for the electrons and therefore the smallest radius of the 3)?? I thought that as we go down a group, there is shielding which causes electrons to be further from the nucleus, increasing the radius,

Likewise, why does F- ion have the lowest nuclear charge of the 3 (it has the largest radius of the 3)? When we go down a period, doesn't the effective nuclear charge increase, causing the atomic radius to be smaller?

Im really confused.

Austin Ho 1E
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Joined: Fri Sep 29, 2017 7:04 am
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Re: Isoelectronic ions

Postby Austin Ho 1E » Wed Oct 25, 2017 8:47 pm

Effective nuclear charge increases as you move right across a period and it decreases as you move down a group. You were correct in this assumption. However, for these isoelectric atoms, we know they all have the same valence electron configuration; in this case, for F-, Na+, and Mg2+ they would all share [Ne], or Neon's valence structure.

Since they all have the same number of electrons, we need to look at the number of protons to determine the magnitude of nuclear charge. F- has 9 protons, Na has 11, and Mg has 12. Keep in mind that they still all have the same number of electrons and the same valence configuration. Thus, since Mg has 12 protons to just 10 electrons, it has the highest nuclear charge as the # of protons (12) outweighs the # of electrons (10). Likewise, since F- has only 9 protons to 10 electrons, it has the lowest nuclear charge as the # of protons is outweighed by the # of electrons.

Your assumptions are correct, but that only applies to neutral atoms. If you were comparing F, Na, and Mg, your assumptions would be correct. However, since they have the same number of electrons (isoelectronic), you cannot use those assumptions.

Hope this helps!

Julie Steklof 1A
Posts: 50
Joined: Thu Jul 13, 2017 3:00 am

Re: Isoelectronic ions

Postby Julie Steklof 1A » Thu Oct 26, 2017 7:03 pm

The above response is correct. Because each ion has the same amount of electrons, to determine size we need to look at the number of protons within the ion. Because Magnesium 2+ has the highest nuclear charge, and each ion in the series have the same number of electrons, it will have the smallest ionic radius because there is less repulsion between electrons and more attraction to the nucleus. On the opposite end, there is Fluorine - which experiences more electron repulsion and with a smaller nuclear charge, a larger radius.


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