I realize this should be a fairly basic question, but I'm still not quite satisfied with what I've been told from numerous sources. The general explanation seems to state that as we go down a group, there are more electron shells present to contribute to a "shielding" phenomenon, where inner electrons cancel out part of the inward force from the nucleus with a repulsive force. Thus with more energy levels present and more shielding, valence electrons begin to reside further and further away from the nucleus.

But at the same time, textbooks make a big deal about the concept of effective nuclear charge, which states that the "effective" force a given valence electron feels is a function of the net charge of the nucleus and non-valence electrons combined—the effective nuclear charge. I find this contradictory, since as we go down a group, the effective nuclear charge remains constant (equal numbers of protons and non-valence electrons are added). So what am I missing here? Why does shielding matter? Is the force on a single valence electron a function of the net nuclear charge or do we have to look at it from a more piecewise level, considering how other electrons are in between it and the nucleus and how it interacts with those individual electrons?


4 Answers 4


The description of $Z_\mathrm{eff}$ you gave is a bit too simplistic. As you correctly said, that would predict the same value of $Z_\mathrm{eff}$ for all elements in a group, which is not true.

In general we have $Z_\mathrm{eff} = Z - \sigma$ where $Z$ is the nuclear charge (that solely depends on the number of protons) and $\sigma$ is the shielding constant, which reflects the electron-electron repulsions but is not a simple function of the number of core electrons.

One of the earliest models to determine $\sigma$ was described by Slater in 1930.1 These are easily found online as "Slater's rules". These are quite simplistic, and so people tried to find better ways to calculate $\sigma$. Nowadays, one very popular source for values of $Z_\mathrm{eff}$ are the Clementi values.2 That's still quite long ago, so you can imagine that since then, we have come up with even more complicated ways to calculate it.

A peculiarity is that the values of $Z_\mathrm{eff}$ actually increase down the group. At the very least, that should dispel the myth that $Z_\mathrm{eff}$ only depends on the number of core electrons.

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(Source: Keeler & Wothers, Chemical Structure & Reactivity: An Integrated Approach, p 264)

"Surely that should mean that the atomic radii decrease down the group!" you say.

I don't fault you for thinking that at all. However, that's not what we see: the atomic radii clearly increase going down the group. The second factor is the principal quantum number, $n$.

In the hydrogen atom, an electron in the 1s orbital has a much smaller radius3 than an electron in the 2s orbital. That's something that comes right out of the quantum mechanical description of the hydrogen atom.4 The same can be said for other atoms: a 2s orbital in lithium is smaller than a 3s orbital in lithium.

Note that I didn't compare a 2s orbital in lithium with a 3s orbital in sodium. Why? That's because sodium has a larger effective nuclear charge, and this serves to contract all the orbitals in sodium relative to those in lithium. Essentially, the orbitals are pulled inwards by the larger $Z_\mathrm{eff}$.

So, as you can see, going from lithium to sodium, there are two competing factors:

  • Larger $n$ suggests that the outermost electron should be further away from the nucleus.
  • Larger $Z_\mathrm{eff}$ suggests that the outermost electron should be nearer to the nucleus.

Unsurprisingly, it turns out that the variation of atomic radius is dependent on a combination of the two. We know that the atomic radius increases going down the group, so that must mean that the increase in $n$ (going from 2 to 3) outweighs the increase in $Z_\mathrm{eff}$ (going from 1.28 to 2.51), although it's sadly difficult to find an exact mathematical formulation in any textbook. It's likely that the exact dependency is very complicated and nowadays, such atomic properties are mostly calculated computationally anyway, which removes the need for such a mathematical formula.

Notes and references

1 Phys. Rev. 1930, 36 (1), 57

2 J. Chem. Phys. 1963, 38, 2686

3 I'm talking about $\langle r \rangle$, not the radius of a Bohr orbit, although that does also increase going from 1s to 2s.

4 I'm avoiding the circular logic that "higher energy means further away". It is further away because that is just the form of the radial wavefunction obtained by solving the Schrodinger equation - period.

  • $\begingroup$ Okay, this was actually quite helpful - so I guess all I can really say without getting into quantum mechanics is that the increasing energy of the valence electrons (which necessitates that they be further from the nucleus) outweighs the slightly increasing effective nuclear charge. $\endgroup$
    – Nick
    Nov 8, 2016 at 17:32
  • $\begingroup$ Yeah, pretty much. You can see this trend in the first ionisation energy as well. $\endgroup$ Nov 8, 2016 at 17:51
  • $\begingroup$ How does the presence of electrons in the d-orbital affect this? This question popped up in my head when I saw the covalent radius of Group $15$ elements (you can check this: ptable.com). Now, here the covalent radius increases by almost $30pm$ when we go from N to P. But, from As onwards the change in covalent radius isn't that big. Why is it so? The only difference in the electronic configuration is that there are d and/or f electrons from As onwards. If anything, it should only increase it more because now the shielding effect is not only due to $4s$ but also due to $3d$ electrons. $\endgroup$
    – Kaushik
    Apr 6, 2021 at 16:57

I will try to be simple. There are 3 things which affect the atomic radius of an atom.

  1. Nuclear charge: nuclear charge tends to decrease the atomic radius of the atom. Higher the nuclear charge smaller will be the atomic radius.

  2. Shielding effect of inner shell electrons: the shielding effect of inner shell electrons decreases the effect of nuclear charge on outer shell electrons and hence an increase in shielding effect is associated with an increase in atomic radius.

  3. Number of shells: it is common that the higher the number of shells in an atom, the farther out the outer electron will be from the nucleus and the greater the atomic radius will be.

Now, if we combine all these three effects, we will arrive at the following conclusion:

On coming down the group,

  • Number of protons increases → nuclear charge increases → atomic size decreases.

  • Number of shells increases → atomic radius increases.

  • The shielding effect increases (since now electrons from previous shells also contribute to the shielding effect) → atomic radius increases.

As you can see that there are two effects which favour the increase in atomic radius and one which favour in its decrease. Often (in other words) we say that the combined effect of shielding effect and increase in number if shells outweigh the effect of increased nuclear charge and hence the atomic radius increases.


Atomic radius increases down the group and decreases from left to right in a period.

We have to keep in mind 3 main points while considering this trend:

  1. Elements of the same period have the same amount of shells and differ only in the number of electrons in the outer shell.
  2. The shielding effect of the outermost electrons on the electron in the same shell is almost negligible.
  3. The nuclear charge increases from left to right in a period and hence the effective nuclear charge increases on going left in a period.

Therefore the electrons get more attracted to the nucleus, hence the atomic radius contracts when we go from left to right in a period.


Down a group means that the valence electrons are on a higher energy level (they are further away from the nucleus). You would also expect that going down a group, nuclear charge increases (more protons), however, the electrostatic strength down a group is relatively the same. Why? well because there are more shielding due to more energy levels so overall nuclear attraction is similar.

Z eff = Z(no. of p) - S(shielding e) <-- finds effective/overall nuclear charge. Let's take oxygen and sulfur for example. Z eff for oxygen = 8-2=6 and Z eff for sulfur = 16 - 10 =6. This means that the overall electrostatic attraction between the nucleus and the valence electrons are similar, so the major factor must be that the electrons are on a higher energy level, so they are further away from the nucleus, that results in larger radius. So going down a group always mean larger radius.

  • $\begingroup$ But is saying that radius increases because the valence electrons are on a higher energy level and thus further away sort of answering the question with itself? $\endgroup$
    – Nick
    Nov 8, 2016 at 12:08
  • 1
    $\begingroup$ This is factually incorrect. $\endgroup$ Nov 8, 2016 at 13:24

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