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Topic: Why are positive ions smaller than their respective neutral atoms?  (Read 2503 times)

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Offline habbababba

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This question sounds easy to answer. Well so I thought before a colleague brought to my attention one point that I did not critically think of. We split in explaining why this is so when he specified the ion in question: Mg2+. So why is Mg2+ smaller than an Mg atom?

My answer consists of two entries:

1) Considering that the radius of the atom is the distance between the nucleus and the outermost shell (I'm using the Van der Waals radius because it serves a better purpose in answering this question), this distance becomes smaller in Mg2+ since the third shell is not occupied anymore. Since the outermost occupied shell in Mg2+ is the second one, and since the second shell is closer to the nucleus that the third, this gives Mg2+ a smaller radius than Mg.

2) The repulsion between the electrons in the third shell and the electrons in the second shell lowers the average force of attraction each electron in the second shell feels from the nucleus in an Mg atom. Since 2 electrons are removed in Mg2+, then we can say that the average repulsion the electrons in the second shell feel now is less and so the average force of attraction the electrons in the second shell are feeling now in Mg2+ increases. This results in 'pulling' the electron cloud closer to the nucleus in Mg2+.

My colleague argues that the first explanation is correct while the second is nonsense. He asks me the following question:
If my second point was correct, then would the distance between the nucleus and the second shell in an Mg atom (let's not forget that there is the third shell occupied in an Mg atom) be larger than the distance between the nucleus and the second shell in Mg2+?

Offline Corribus

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Re: Why are positive ions smaller than their respective neutral atoms?
« Reply #1 on: May 03, 2015, 10:40:59 AM »
First, just so you're clear, there really is no such thing as a hard radius for an atom or ion. It's a purely experimental parameter.

Of your two potential explanations, #1 is certainly the most important for most ions. But I don't regard #2 as nonsense at all, although in most cases it's not a large contributor to the effective ionic radius. If you have two electrons in the same shell, they will shield each other. For instance, in the case of Mg, each 2s electron partially blocks the other from the full power of the nuclear core charge (because each electron spends some of its time closer to the nucleus than the other). If you were able to remove a single electron, the effective core charge felt by the other would be increased, which would lead to some degree of contraction. But Mg+ doesn't really persist for any length of time, and since the ionic radius is basically determined by measuring the inter-nuclear distance and dividing by two, you won't really find any values for the ionic radius of Mg+ to demonstrate this effect.

Technically speaking, the 2s electrons of Magnesium also shield the 1s electrons because of penetration effects, but the degree of it is very tiny. Nevertheless, this does mean that removing the 2s electrons will also cause contraction of the 1s electrons. Therefore some of the observed smaller atomic radius of Mg2+ compared to Mg is resulting from your explanation #2. But, it's such a small effect compared to the loss of an entire shell that it's almost not worth mentioning. I can't put any exact numbers on it off the top of my head but of the ~45 pm difference in radius between Mg and Mg2+, I'd guess your explanation #2 accounts for no more than ~2 pm of it. To find out for sure you'd have to do some kind of calculation of the atomic wavefunctions to see what kind of contraction you get for the inner shell electrons when you remove those in the outer shell, or look for indirect evidence (photoelectron spectra, possibly).
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline habbababba

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Re: Why are positive ions smaller than their respective neutral atoms?
« Reply #2 on: May 03, 2015, 01:30:12 PM »
Okay Thank you.

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