Does electron shielding increase or stay constant moving LEFT to RIGHT across a period?

Question

Does electron shielding increase or stay constant moving left to right across a period?
I have read about both, and I just want to know which one is right.

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I believe that electron shielding remains constant because when you move across a period, you are essentially adding more valence electrons, not shielding electrons, in your valence shell. Therefore, your valence-electron-count increases from left to right in a period, but your shielding-electron-count stays the same.

So the overall shielding remains constant, just the amount of valence electrons increases

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However, my explanation has a flaw:
It does not support the reason why an atom's radius increases when it becomes an anion.

When an atom becomes an anion, it adds more valence electrons to its outermost shell. Now this is controversial to what I said previously, because a gain in valence electrons, but not shielding electrons, wouldn't increase the electron shielding of the atom. So the atomic radius wouldn't increase either!

The only way for an atom's radius to increase when it becomes an anion is if the electrons that are added to the valence shell have some sort of effect on the shielding, right?

This supports the first thought in my question that if we move left to right across a period, shielding increases.

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But the idea that electron shielding increases as we move left to right has a flaw as well:
It is controversial to another periodic trend: ionization energy

If electron shielding did increase across a period from left to right, then this would mean that ionization energy is lower as we move from left to right, which is not true.

The only way for ionization energy to increase across a period is if the only the number of protons and valence electrons have effect on the amount of energy required, not the shielding electrons. So the shielding-effect must stay the same all throughout a period.

This goes with the what I had initially said about electron-shielding remaining constant.

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So now we have a flaw for each possible answer to my question! Please correct me if I am wrong in any of my context.

Answer 1

To answer this question, it's important to define what you mean by shielding. Generally, shielding refers to a reduction in the effective nuclear charge experienced by an electron in a given orbital due to the other electrons on the same atom.

The quantitative degree of shielding for a given electron can be approximated by Slater's rules. According to those rules, electrons within the same group of orbitals (for example 4d) have a coefficient of 0.35 (except for 1s electrons, which have 0.30). So valence electrons do shield each other, just not as much as the lower level electrons shield the valence electrons.

For example, let's consider the elements with increasing numbers of 2p electrons (B, C, N, O, F, Ne). Going from left to right, each addition of a 2p electron reduces the effective nuclear charge experienced by another 2p electron by 0.35. So the amount of shielding is increasing as we move left to right.

The apparent contradiction with the ionization energy comes about because you have not considered the increase in the actual nuclear charge. Each time we add a 2p electron, we also add a proton to the nucleus. So even though 0.35 of that charge is shielded from a valence 2p electron, that electron still experiences an increase of 0.65 effective charge and is therefore bound more tightly than a 2p electron on the element to its left.

As a specific case, let's compare the shielding experienced by a 2p electron in carbon versus neon. The carbon electron configuration is $1s^22s^22p^2$. Using Slater's rules, we can calculate that the shielding coefficient for a 2p electron is 2(0.85) + 3(0.35) = 2.75. So the effective nuclear charge experienced by the 2p electron is 6-2.75=3.25.

For neon, the electron config is $1s^22s^22p^6$, so the shielding coefficient of a 2p electron is 2(0.85)+7(0.35)=4.15. So there's more shielding. But the effective charge is 10-4.15=5.85, so the effective charge is still greater and the energy of the coulombic interacation of the electron with the nuclear charge is greater in magnitude. Hence, the ionization energy is greater.

EDIT: Original version of this answer did not address the question about anion formation (presumably mostly interested in anion formation by addition to an already partially filled shell). When an anion is formed, the additional electron increases the shielding of the other valence electrons, but there is no change in the actual nuclear charge. Thus, each valence electron experiences a lower effective nuclear charge. Since the size of an orbital is a function of the energy of interaction with the nuclear charge, the atomic size increases.

Answer 2

Hopefully by explaining the 'flaws' may shed some light on your question.

It does not support the reason why an atom's radius increases when it becomes an anion.

The valence shell is not only comprised of 1 single orbital position of a fixed radius regardless of the number of electrons, the different orbitals with different geometries fill up as we move right across a period, and the more orbitals, there is more repulsion and range of radii for the orbitals, increasing the atomic radius. These orbitals are governed by the wavefunctions of these orbitals, which can be expressed in terms of the radius(as you usually use polar coordinates anyways) which increases as the number of electrons increase.

For anions it is the same, as the electrons occupy orbitals and pair repulsion and orbital repulsion increases.

But the idea that electron shielding increases as we move left to right has a flaw as well:It is controversial to another periodic trend: ionization energy

Lets take an example: Ionization energy of neon is higher than that of fluorine, which is one to the left across the same period, this is much because the valence shell with a full S and p orbitals has a higher stability due to something caused by symmetric repulsion (which is partly why palladium has a full d subshell despite the s orbitals of lower energy being available)and thus there will require more energy to ionize the atom as it moves right across a period