Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers and students. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can anyone explain why the melting and boiling points of the noble gases increase as you go down the periodic table?

Edit: What causes the melting and boiling points to rise when the atomic number increases? What role do the valence electrons play in this?

share|improve this question

The melting and boiling points of noble gases are very low in comparison to those of other substances of comparable atomic and molecular masses. This indicates that only weak van der Waals forces or weak London dispersion forces are present between the atoms of the noble gases in the liquid or the solid state.

The van der Waals force increases with the increase in the size of the atom, and therefore, in general, the boiling and melting points increase from $\ce{He}$ to $\ce{Rn}$.

Helium boils at $-269\ \mathrm{^\circ C}$. Argon has larger mass than helium and have larger dispersion forces. Because of larger size the outer electrons are less tightly held in the larger atoms so that instantaneous dipoles are more easily induced resulting in greater interaction between argon atoms. Therefore, its boiling point ($-186\ \mathrm{^\circ C}$) is more than that of $\ce{He}$.

Similarly, because of increased dispersion forces, the boiling and melting points of monoatomic noble gases increase from helium to radon.

enter image description here

For more data of melting and boiling points of noble gas compounds, read this page

share|improve this answer

Valence electrons have not much to do with this, as their outer shell is closed. As the other answer mentioned, dispersion forces are the ones responsible for any interaction between these atoms.

The size dependence therefore is directly coming from the size dependence of dispersion forces:

In a very simplistic way, a random charge fluctuation can polarize the otherwise perfectly apolar atoms. This induced dipole moment is then responsible for the dispersion interactions. The polarizibility of an atom increases (easier to polarize) if the atomic number increases, therefore the interactions in nobel gases will reflect this behavior.

share|improve this answer

Other answers have mentioned that dispersion forces are the key to answering the question but not how they increase from helium to radon (or let’s take xenon because that’s not radioactive so I feel safer breathing it).

The larger the mass of a nucleus the more protons are in there, and the more protons in a nucleus the more electrons are around the outside. Traditionally, one thought of electrons orbiting the nucleus rather like satellites orbiting a planet on more or less fixed orbits at certain heights. But that picture is wrong. It is much better to consider electrons as waves that completely surround the nucleus. If one were to translate these waves into particles (because of the wave-particle dualism at quantum levels), all one would get would be probabilities of finding specific electron $e$ at specific location $x$.

For a neutral atom that is surrounded by nothing, these probabilities depend only on the wave function, and are inherently centrosymmetric or anticentrosymmetric, leading to a net charge distribution of zero. In a sample of xenon gas, however, other atoms approach the atom we are observing. Consider an atom approaching our xenon atom at from ‘above’ (i.e. perpendicular to our viewpoint). While the atom as a whole is neutral, the could of electrons surrounding the nucleus is negatively charged. This new negative charge changes the potential energy the electrons in our observed atom are perceiving: While we originally had a centrosymmetric potential distribution (decreasing positive charge intensity from the nucleus) we now have a second negative source at 12 o’clock. Therefore, the probability distributions will shift ever so slightly and it will be slightly more likely to detect our electron $e$ at position $y$ below the nucleus rather than at position $z$ above the nucleus. Note that this simplification relies on the freezing of time at a certain moment when the other atom is approaching and a certain controlled environment that I have invented for example purposes.

With the electrons now more likely to be at $y$ rather than $z$, we can say that we have created a spontaneous or an induced dipole. A mild positive charge is now pointing towards the other atom and ever so slightly attracting it. This will, if you advance the time flow by one spec, create another induced dipole in the originally approaching atom plus further in every other atom that is close around. We cannot freeze this picture though. Every infinitesimal change in position, movement direction or rotation will change the entire picture, meaning that our induced dipole is extremely short-lived.

It is only the combination of all these induced dipoles and their slight attraction that attracks the different atoms together. Since they rely on electron distribution changing, the more electrons we have the stronger these induced dipoles can be and the more force can be excersized between atoms. Since the number of electrons loosely correlates with mass (and strictly correlates with nuclear charge), larger atoms are said to display stronger van der Waals forces than smaller ones.

share|improve this answer

protected by Community May 10 at 16:12

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).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.