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Topic: Radii of ions??  (Read 10356 times)

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Offline meeztered

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Radii of ions??
« on: September 07, 2006, 05:12:38 PM »
I know you probably cant help me by tommorow. But I can try anyway. I am completely lost and cannot figure out this question. If anyone could please help me that would be great. Here it is:

The ions S2-, Cl-, K+, Cas2+, and Sc3+ have the same total number of electrons as the noble gas argon. How would you expect the radii of these ions to vary? Would you expect to see the same variation in the series O2-, F-, Na+, Mg2+, and Al3+, in which each ion has the same total number of electrons as the noble gas neon? Explain your answer.

Online Borek

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Re: Radii of ions??
« Reply #1 on: September 07, 2006, 05:56:17 PM »
What's the difference between these ions? Why do they have different charges if they have the same number of electrons?

Opposite charges are atracted.
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Offline meeztered

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Re: Radii of ions??
« Reply #2 on: September 07, 2006, 06:07:34 PM »
I dont think that it has anything to do with charges but, I really dont know. That is the exact question, its asking how the radii varies not the charge, right?

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Re: Radii of ions??
« Reply #3 on: September 07, 2006, 06:13:35 PM »
I dont think that it has anything to do with charges

It has. Think about the questions I have posted.
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Offline meeztered

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Re: Radii of ions??
« Reply #4 on: September 07, 2006, 06:40:41 PM »
Ok, Im pretty lost. I've been looking on the internet and the only thing I can find about radii is that it decreases from left to right in the rows of the periodic table. I can't figure anything out about the charges and what they have to do with it. So... I dont know where to go from here.  ???

Offline meeztered

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Re: Radii of ions??
« Reply #5 on: September 07, 2006, 06:57:31 PM »
Ok I just found this, I don't know if its right or not maybe you could tell me. The radii of these ions vary because positive ions are smaller than the atoms they come from, and negative ions are bigger than the atoms they come from. So, the negative ones (S2-, Cl-) would be bigger, right? And K+, Ca2+, and Sc3+ would be smaller.

Offline Yggdrasil

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Re: Radii of ions??
« Reply #6 on: September 07, 2006, 10:55:43 PM »
The answer has to do with the effective nuclear charge of the outer-shell electrons in each species.  Since they all have the same number of electrons, the shielding is the same in all species.  What does change though is the charge of the nucleus.  Does this information help explain the trends you see?

Offline english

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Re: Radii of ions??
« Reply #7 on: September 08, 2006, 09:25:49 AM »
As you add or remove charge from the outer shells of those elements, you are affecting the radii of the atoms by adding or removing repulsions.

Since they all have the same Zeff, adding an electron(s) will increase electron-electron repulsions.  Removing them will decrease electron-electron repulsions and should increase Zefffor the remaining electrons.

What do you think is happening when we add charge (increase repulsions)?  What happens when we remove charge (reduce these repulsions)?

Two simple rules should help you understand, but I hope you will ask yourself these questions first.

"Cations are smaller than their parent atoms.  When a cation forms, electrons are removed from the outer shell.  The resulting decrease in electron repulsions allows the nuclear charge to pull the remaining electrons closer.
Anions are larger than their parent atoms.  When an anion forms, electrons are added to the outer shell.  The increase in repulsions causes the [outer] electrons to occupy more space" (Silberberg, 2006: 320).


And Borek please point if I have missed something.
« Last Edit: September 08, 2006, 09:40:57 AM by k.V. »

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Re: Radii of ions??
« Reply #8 on: September 08, 2006, 09:36:10 AM »
In general you are correct, but we are talking about constant number of electrons here. Simple planetary (Bohr's) model is enough to understand what is happening and why.
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Offline meeztered

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Re: Radii of ions??
« Reply #9 on: September 08, 2006, 02:55:46 PM »
I had to turn that in today. hope i got better than a 50. thx for the help though.

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