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I understand one thread of logic that adding electrons to create an anion creates more electron repulsion hence increasing the ionic radius.

What's wrong with the reasoning that there is now a larger negative charge and hence the positive nucleus and negative electron cloud have a stronger attraction so the ionic radius decreases. One thought was that the number of electrons increases and hence the nucleus has to pull more electrons toward it so the pull becomes weaker, but I'm not sure if a combination of charge and number of objects is a relevant metric.

Can someone explain the best way to think about this and how all of these concepts (electron repulsion, charge of electron cloud as a whole, and number of electrons) relate with each other and contribute to ionic radius? Thanks!

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    $\begingroup$ I think it's rather sad that this gets downvoted. Obviously OP's thought process is completely wrong, and I've pointed out as much in my answer; but if they were completely right, then they would not need to ask a question here. And having been here and having seen questions over many years, I must say that this confusion over electrostatic forces is actually quite common. $\endgroup$ Dec 20, 2021 at 2:57

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What's wrong with the reasoning that there is now a larger negative charge and hence the positive nucleus and negative electron cloud have a stronger attraction so the ionic radius decreases.

The problem is that you're treating it as if all the electrons can be collectively reasoned about as one single negative charge. This isn't true. The nucleus is considered as one single positive charge, because all of those positive charges are crammed within such a small space that we may as well consider them as being at the same point in space. On the other hand, this is not true of electrons. They don't glue themselves together like the nucleus does. Each electron must be considered individiually when describing the attractive/repulsive forces.

So, in other words, the negative charge has not gotten larger. There is no "the" negative charge. What has happened is that there are now more negative charges.

One thought was that the number of electrons increases and hence the nucleus has to pull more electrons toward it so the pull becomes weaker

This is not correct either; it is not even based on anything scientific. It sounds as if it's a human trying to lug a weight along, and a heavier weight is harder to move than a lighter one.

Electrostatic attraction simply doesn't work like that, though. The force between two charged objects is $F = kq_1q_2/r^2$, where $k$ is some constant, $q_1$ and $q_2$ are the charges, and $r$ is the distance between them. Clearly, if $q_1$ or $q_2$ become larger, the attractive force between them increases, not decreases as this line of argument would have one believe.

Having said that, note that this equation is not even applicable to the current situation. As mentioned previously, the addition of an electron does not increase the size of $q_1$ (or $q_2$). It is not "adding" to some existing basket of negative charge thereby making it more negative. The electron is a negative charge of its own. So there are at least two reasons why this line of logic is wrong.

Can someone explain the best way to think about this and how all of these concepts (electron repulsion, charge of electron cloud as a whole, and number of electrons) relate with each other

Of these three factors, only the first (electron repulsion) is actually useful. The "charge of electron cloud as a whole" has no useful meaning, as explained above, because the electrons are not to be considered as a collective entity.

Circling back to the original issue about why anions have a larger radius than the neutral atom: imagine that you have a neutral oxygen atom with 8 protons and 8 electrons, and this atom has found its equilibrium, most stable arrangement with an atomic radius of $r$. This means that all the electrons are within a sphere of radius $r$.*

What forces are at work here? Well, we have 8 different electrons, each of which are attracted to the nucleus and would like to be closer to the nucleus. That gives us 8 electron–nucleus attraction terms. Attraction is good, so we want to try to keep the electrons as close to the nucleus as possible. On the other hand, every electron is repelling every other electron. The number of pairs of electrons is (8 × 7 / 2) = 28, so we have 28 electron–electron repulsion terms. Repulsion is bad, so we want to try to keep the electrons as far apart as possible. Note that not all of these terms have the same size, so just because there are 28 versus 8 doesn't mean that the 28 is more important than the 8. I'm just listing this to drive home the point that electrons must be considered individually, and that it's not just some "nucleus versus all the electrons" tug-of-war.

Before adding in one more electron, let's consider why this radius $r$ is chosen. Why don't the existing 8 electrons, for example, cram themselves into a slightly smaller sphere, of radius $r - \delta$? Well, if they did, then the attractive forces would increase, and that's a good thing. On the other hand, because the electrons are now contained within an even smaller space, the repulsive forces would also increase, and that's a bad thing.

It turns out that $r$ is the special radius which perfectly balances these two factors. That is, if you take a slightly smaller radius $r - \delta$, the repulsions get too big; and if you take a slightly larger radius $r + \delta$, the attractions start to diminish.

Now consider adding one more electron to the mix. Where would this electron go? Of course, the new electron itself will want to be as close to the nucleus as possible. However, cramming it within the sphere of radius $r$ will also inevitably increase the electron repulsions inside; and recall that $r$ is already that radius which happens to nicely, snugly, balance the attractive and repulsive forces for 8 electrons. By adding this 9th electron within this sphere, we're totally destroying this balance as it introduces far more repulsions, while not changing the 8 original attractions. If the 8 other electrons had their own way, the 9th electron would simply be repelled; it wouldn't even be able to enter this sphere.

The expansion of the atomic/ionic radius is thus best viewed as a compromise between these two factors. The nucleus wants the 9th electron to be as close as possible, but the 8 existing electrons would rather not admit it into that same sphere. What happens in the end is something in-between: the sphere of electrons expands a little bit to fit in the 9th one, allowing it to stay somewhat close to the nucleus, but without increasing repulsions too much.

What about removing electrons?

Simple: when you remove an electron, there are fewer remaining repulsions, so the electrons happily move into a smaller sphere in order to maximise their attraction to the nucleus.

It would be wrong to think of this as being a "larger overall attraction per electron". The attraction between the nucleus and electron is not an averaged quantity. There is simply just one attractive force between the nucleus and every electron.

But books talk about "number of electrons"...

Yes, they do. The reason for this is because the number of electrons is an indirect proxy for the number of electron–electron repulsions. More electrons = more repulsions. If all else is equal, the most stable radius will be pushed outwards, in order to reduce the repulsions.

The "shielding" effect is a different way of expressing the same thing. Effectively, what it means is that the electron–nucleus attraction (i.e., the desire for the electrons to be as close to the nucleus as possible) is "countered" or "weakened" or "watered down" by the electron–electron repulsions (which make it very hard to cram electrons within that small space). This can be pictured as the nucleus not being "as positively charged" as it really is, which gives rise to the "effective nuclear charge" $Z_\mathrm{eff}$.

At the end of the day, fundamentally, everything boils down to electron–nucleus attractions and electron–electron repulsions.

Can I have an analogy

Yes, you may. If you want an analogy that is more true to the underlying physics, then think of this: the nucleus is a piece of tasty cake at the centre of a large room. The electrons are humans trying to get to the cake to have a nibble, but all while trying to maintain social distancing.


Footnote

* From a quantum mechanics point of view, this answer cuts a lot of corners, like in this phrase that electrons may be found within a sphere of radius $r$. But that's way, way beyond what you need to know at this point. I only mention this so that nobody complains about it in the comments.

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  • $\begingroup$ Thank you so much for the thorough response! This clarifies a lot of disparate concepts together. Appreciate it. $\endgroup$
    – user54689
    Dec 22, 2021 at 22:38

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