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The values of the coefficient $b$ for helium and neon are

$$\begin{array}{cc} \hline \text{Element} & b\text{ / }\mathrm{dm^3\ mol^{-1}} \\ \hline \ce{He} & 2.38 \times 10^{-2} \\ \ce{Ne} & 1.67 \times 10^{-2} \\ \hline \end{array}$$ (source: Atkins & de Paula, Physical Chemistry, 9th ed., p 916)

This doesn't make sense, however, considering that neon should have a larger atomic radius than helium. Why is $b_\ce{He} > b_\ce{Ne}$?

orthocresol
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Math The Novice
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    It's a good question, but please, always include the units whenever you are discussing science! It may sound pedantic, but science is not science without units. – orthocresol Sep 27 '16 at 05:50
  • I'm guessing charge-to-volume ratio. – DHMO Sep 27 '16 at 10:37
  • So you mean if the charge density is higher than the atom would be "tougher" and any particle would therefore bounce back immediately. – Math The Novice Sep 27 '16 at 11:57
  • What does that even mean, there is no charge on the atoms. – orthocresol Sep 27 '16 at 14:20
  • I should be careful about the terminology. It should had been "electron density" here – Math The Novice Sep 27 '16 at 14:44
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    I suspect this depends very heavily neon's full $2p$ orbital and how this affects the volume actually excluded by the atom. I'd also be skeptical of values you find for van der waal's radii unless you trust the method by which they're found because there are quite a few different ways to get a guess experimentally. – jheindel Sep 29 '16 at 02:54
  • related: http://chemistry.stackexchange.com/a/37514/16683 – orthocresol Sep 30 '16 at 09:43

1 Answers1

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The question intregues me so I looked up the $a$ and $b$ values in Wikipedia on the web page Van der Waals constants. This doesn't answer the question, but adds more data to focus the question.

$$\begin{array}{cc} \hline \text{Element} & a\text{ / }\mathrm{L^2 \ bar\ mol^{-2}} & b\text{ / }\mathrm{L\ mol^{-1}}\\ \hline \ce{He} & 0.0346 & 0.0238 \\ \ce{Ne} & 0.2135 & 0.01709 \\ \ce{Ar} & 1.355 & 0.03201 \\ \ce{Kr} & 2.349 & 0.03978 \\ \ce{Xe} & 4.250 & 0.05105\\ \ce{Rn} & 6.601 & 0.06239 \\ \hline \end{array}$$

A plot of the Van der Waals $a$ and $b$ values. enter image description here

So is the b value for He high, or is the b value for Ne low ?!?

I also looked up for each element what was listed for the covalent radius and the Van der Waals radius on each elements individual Wikipedia webpage.

$$\begin{array}{cc} \hline \text{Element} & Covalent\text{ (pm) } & Van~der~Waals \text{ (pm)}\\ \hline \ce{He} & 28 & 140 \\ \ce{Ne} & 58 & 154 \\ \ce{Ar} & 106±10 & 188 \\ \ce{Kr} & 116±4 & 202 \\ \ce{Xe} & 140±9 & 216 \\ \ce{Rn} & 150 & 220 \\ \hline \end{array}$$

MaxW
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  • So Is the b value for He high, or is the b value for Ne low ? – MaxW Sep 27 '16 at 15:04
  • What is the plot for ? – Math The Novice Oct 05 '16 at 04:46
  • @MathTheNovice - As the text indicates "A plot of the Van der Waals a and b values [for the six noble gases]." The gist is to look at the curve and decide if the He Or Ne value is out of whack for the trend. – MaxW Oct 05 '16 at 06:50
  • A coefficient stands for the intereaction between molecules while b for molecule volume (In the interpretation from physics ) and I don't really get how it is related though the plot show a significant correlation between them – Math The Novice Oct 05 '16 at 08:59
  • I mean is the straight line appeared by coincidence – Math The Novice Oct 05 '16 at 09:01
  • @MathTheNovice - I don't know if helium or neon should be on the curve. The other point is that the "a" and "b" coefficients are not "pure" in the sense that both coefficients depend on both the interaction between particles and the volume of the particles. – MaxW Oct 05 '16 at 19:58
  • @MaxW Also the value of $a$ is very low for $\ce{He}$ is very low compare to even $\ce{H2}$. The only explanation I can think of is '$imperfection$ of van der Waals equation'. – Apurvium Aug 05 '21 at 06:01