# Van der Waals Radius clarifications

1. Would it be correct to say that in general, van der Waals radius decreases as we move from left to right in the periodic table?

2. The van der Waals radius of the noble gas in a period is larger than the radius of halogen that precedes it. Why is it so?

3. What is the difference between van der Waals radius for monoatomic (say neon) and non-monoatomic (say fluorine) gases?

Let me start by providing a reference to a nice collection of van der Waals' data (actually the site provides a lot of other interesting data about the elements, you might want to poke around). As to your questions,

1. Yes, that is a general trend, but as you can see there are many exceptions. The trend exists for the same reason that covalent radii generally shrink as we move from left to right across the periodic table, every time we add another proton and electron to an atom, the electrons do not effectively screen the outermost electrons from the positive nucleus so they become drawn in closer to the nucleus.
2. & 3. There are a lot of different ways to measure the van der Waals radius of an atom, but in recent times the preferred method is to examine the crystal structure of a molecule. Using $\ce{F2}$ as an example, the crystal would be composed of fluorine molecules. if we measure the crystallographic distance between two adjacent, but non-bonded fluorine atoms and divide by two, we would have a single value that could be used to estimate the van der Waals radius for fluorine (we would probably want to measure a number of similar values for fluorine from other molecules, maybe using $\ce{CH3F, NF3, ClF,}$ etc., for example, and average them together). Our crystal might look something like the drawing below.

Because diatomic and polyatomic molecules are polarizable, the bonded fluorine atoms will likely be slightly polarized when packed in the crystal in order to stabilize (lower the energy) the overall system. This polarization will cause non-bonded fluorines to draw a bit closer to each other then they would if there were no polarization. A crystal of a noble gas (let's say neon) will just be a collection of monatomic neon atoms, no bonds, no polarization. Therefor there is no "attractive" force to draw adjacent neon atoms closer together like there was in the case of fluorine. So even before we make our measurements we might expect the van der Waals radius for neon to be larger than that for fluorine, which turns out to be the case.

• Let me just point out, that there are intermolecular forces between noble gasses, usually the quite weak London dispersion interaction. Noble gasses are known to form clusters at low temperature. The energy to break apart these "bonds" (they are not really bonds in the goldbook definition) will increase in the group, as the polarisability will also increase. We had already a small discussion about this in the comments section to this question/answer. – Martin - マーチン Jul 16 '14 at 2:37

## protected by orthocresol♦Jul 6 '17 at 7:24

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).