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Different isotopes of the same element have the same number of electrons, so the nature of bonding between different isotopes of the same element should be the same, yet their melting points and boiling points vary. This would suggest that the interparticle forces and distances are different for different isotopes of the same element. Why is this exactly?

My best guess is that is has something to do with the kinetic energy (KE) formula

$$E_k = mv^2/2$$

If two samples of different isotopes of the same element were kept under the same pressure and temperature (so the KE of the particles in the different samples would the same), the particles of the heavier isotope would have a lower velocity than the particles of the lighter isotope, as a greater mass means a lower velocity (where KE is equal). Does it have something to do with the fact particles of a heavier isotope move more slowly when KE is equal?

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    $\begingroup$ Sounds about right, though your reasoning would look better with some quantum hand-waving around it. $\endgroup$ – Ivan Neretin Aug 15 '17 at 11:15
  • $\begingroup$ Related: chemistry.stackexchange.com/questions/81235/… $\endgroup$ – paracetamol Aug 15 '17 at 11:42
  • $\begingroup$ There's probably a primary kinetic isotope effect as well. $\endgroup$ – Zhe Aug 15 '17 at 13:51
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As you already mention in your question, melting and boiling of a substance is related to the inter-particle forces. You also correctly assumed that the same number of electrons results in identical (electronic) bonding properties, which basically is a different formulation of the Born-Oppenheimer Approximation (as in the BOA the nuclear mass is assumed to be infinite, different isotopes should have the same electronic structure).

So what is the difference then?

Well, it's in the vibration between the different particles. Consider a number of (neutral) particles in a certain phase (it can be solid, liquid or gas) at a certain distance with respect to each other. Because of dispersion forces between the particles, they interact and to a first approximation, every particle experiences a harmonic force $F(x)=-k(x-x_e)$. The magnitude of this force depends on the distance between the particles and on the electronic structure of the particles and in principle not on the mass. However, as you might know, a harmonic force in quantum mechanics results in quantized energy levels given by

$$ E_v=h\nu_e(v+\frac{1}{2}), $$ where $E_v$ is the vibrational energy of the $v$th level and $h$ is Planck's constant. What is important to realize is that the energy of the lowest state is not equal to zero but has a value of $h\nu_e/2$, which is referred to as the zero point energy. Of course the potential is not really harmonic but rather anharmonic so that at a certain distance between the particles the force becomes negligible and the particles dissociate (or melt or evaporate). The lower the zero point energy, the larger the binding energy. The constant $\nu_e$ is given by $\nu_e=\frac{1}{2\pi}\sqrt\frac{k}{\mu}$, where $k$ is the harmonic force constant and $\mu$ is the reduced mass. The constant $k$ is the same for the different isotopes as it depends on the electronic potential between the particles but the reduced mass is of course different for the different isotopes. As you can see from the definition of $\nu_e$, the heavier species have a smaller zero point energy and thus a larger binding energy and will require higher temperatures to melt and boil.

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  • $\begingroup$ This isn't always true, at least not for molecules (maybe it is for monoatomic species?) 13C-methanol has a higher vapor pressure than 12C-methanol, for example. $\endgroup$ – Curt F. Feb 15 at 17:06
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    $\begingroup$ My answer assumes that isotopic substitution does not change the vibrational frequencies in the molecules too much. Methanol is an interesting molecule due to its internal rotation which is very sensitive to a mass substitution. I found this paper which explains the strange effect for the 13C methanol (the same happens when substituting the H from the hydroxyl group). $\endgroup$ – Paul Feb 16 at 19:24
  • $\begingroup$ Thanks for that link, it is very interesting and relevant to a somewhat related question over here $\endgroup$ – Curt F. Feb 16 at 21:49

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