# Trend in atomic radius for noble gases

In an exam, we were given the following graph and asked to explain why the slope of the change in r vs. Z changes dramatically at a point along the curve. I understand that the d block causes the change in slope for all of the series, but why is the slope from $\ce{He}$ to $\ce{Ne}$ less than that from $\ce{Ne}$ to $\ce{Ar}$? Why is this trend not present for any of the other series? I initially thought that this might have to do with the introduction of the p-block, but that would not explain the absence of the trend for the other series.

• I suspect that there exists an argument along the lines of the 2p orbital not having a radial node, therefore rapid stabilisation going across Period 2, leading to Ne being a tiny bit smaller than expected. But I'm not willing to flesh it out now (rather late and I want to sleep). – orthocresol Nov 2 '16 at 1:15
• Isn't this because He is the only element that has 1s orbital in this chart? Maybe if we included hydrogen, we would see a trend? All other elements start from the 2nd row and jump to the 3rd row except for Helium and Neon. – CoffeeIsLife Nov 4 '16 at 3:58
• @QuantumCAPUCCINO, $\ce{Li+}$, $\ce{Be2+}$, $\ce{B3+}$, and $\ce{C4+}$ all have the same electron configuration as He, so I don't think that is the case. – Niels Kornerup Nov 4 '16 at 5:20
• @Benzene Uh oh I missed that. Sorry~ – CoffeeIsLife Nov 4 '16 at 5:23
• They should’ve included hydride in the series beginning with fluoride … – Jan Nov 5 '16 at 20:05

I have a theory for the question.

First lets look at what the graph is telling us about the radius and the atomic number. For the Noble gases, as the atomic number is increasing the radius increases. It is because of the addition of more electron shells (simplification for now).

Helium has a full 1s orbital. Whereas Neon has a configuration of $1s^2,2s^2,2p^6$. Therefore the electronstatic force occurring on the electrons is slightly bigger than the electron on the helium atoms. Whereas Argon has a configuration of $[Ne]3s^23p^6$.

Now my theory. In He, we only had $1s^2$ and in Neon we added the extra shell and we went up til the 2p. Now the shells also act as barriers sometimes, therefore we have something called an effective nuclear charge. Now, since He has 1s2 and we added an extra shell, well the $1s2$ does not act much of a barrier and the nucleus is still able to attract the electrons in the second shell ($2s^2 2p^6$) electrons and the radii does not change drastically. But when you add a full shell on top of that, well then we have a problem. Because then the second shell electrons can act as a barrier in the electrostatic force, and the effective nuclear charge decreases and the repulsion between the electrons in the third shell will be greater, hence a greater atomic radii occurs between Ne and Ar, compared to He and Ne.

Now this trend will not be seen in the other ones. It is because the electron configuration is one thing but is not the only thing that is being shown on the graph. All of the other ones are ions. When something is an ion, for example the positive ones, then there are two things to look at, electron configuration and the charge. Meaning in a positive ion, you have more protons than electrons, meaning that the electrostatic force will much more greater in $Li^+$ than the one in $He$, even though they have the same electron configuration.

Whereas the trend in the noble gases is based on the fact that they all have $e^- = p^+$ and are neutral, whereas the others are not.

I hope I make sense here. But like I said, its just my theory for the question and I can be wrong :)