6
$\begingroup$

In simple terms, the collision of two atoms $\ce{A}$ and $\ce{B}$ will result in ions $\ce{A^+}$ and $\ce{B^-}$ if $$I_a(\ce{A})+E_a(\ce{A})<I_a(\ce{B})+E_a(\ce{B})$$ where $I_a$ and $E_a$ are the ionisation energies and electron affinities, respectively. Let us view $\ce{A^+B^-}$ as a molecule of sorts, having ions of rigid, non-polarisable spheres. Say $r$ is the sum of the radii of such spheres. The potential energy $E_p$ of such a system would then be $$E_p=I_a(\ce{A})-E_a(\ce{B})-\frac{e^2}{r}+\frac{b}{r^n}.$$

It is intuitive that

  • the term $I_a(\ce{A})-E_a(\ce{B})$ characterises the energy required to form isolated ions $\ce{A^+}$ and $\ce{B^-}$ in the gas phase;
  • $-e^2/r$ takes into account culonic attraction between two ions.

It is not intuitive for me why ${b}/{r^n}$ is how it is.

The final term, introduced by Max Born, encompasses the repulsion generated by shells of electrons at either ion.

Of course, such effects cannot be ignored. But I am interested in why is it $b/r^n$. What is the justification? Feel free to provide the rigorous approach, even though it will probably$^{[1]}$ be above my abilities of comprehension.


Browsing on Wikipedia, the information might be given in the book Advanced Inorganic Chemistry by F. Albert Cotton, Geoffrey Wilkinson, Carlos A. Murillo, and Manfred Bochmann.

The pertinent section “1-6. Energetics of Ionic Crystals” starts at page 18$^{[2]}$, however I cannot find the derivation. There is only that the value of $n$ relies on a compressibility measurement, given as fractional change in volume per unit change in pressure, or

$$\frac{\Delta V}{V\Delta P}.$$

Also, this is seems somewhat contradictory to the earlier assumption of rigid spheres as written in another book$^{[3]}$. So, if anything, the linked book confused me further.

$^{[1]}$ Most certainly.

$^{[2]}$ Page 28 in the PDF.

$^{[3]}$ U. Palm, V. Past. Physical Chemistry. (1974) [To my knowledge, not available in English.]

$\endgroup$
  • 1
    $\begingroup$ Are you asking why the term is of the form $b/r^n$ or about the specifics of how to determine $b$ and $n$? $\endgroup$ – bon Jul 3 '16 at 19:39
  • $\begingroup$ @bon: The first option; why the term is of the form $b/r^n$. $\endgroup$ – Linear Christmas Jul 3 '16 at 20:38
3
$\begingroup$

The term $+b/r^n$ is a power law repulsion term between the ions, where $b$ is the bare ion radius (twice the van der Waals radius) and $r$ the separation.

The hard sphere model considers the ions as 'billiard balls' which means that $n=\infty$ since $+b/r^n$ is effectively zero when $r \gt b$ and infinite when $r \lt b$.

Two other types of potentials are used (a) the power law potential when $n$ is some integer value, usually between 9 and 16 and (b) the exponential potential $+a_0exp(-r/a_1)$ where $a_0$ and $a_1$ are adjustable parameters with $a_1 $ of the order of 0.02 nm.
The justification for using these potentials is weak (the power law particularly so) and they are used as they represent a better approximation than the hard sphere potential and are mathematically convenient. The exponential form is presumably justified since the combination of all orbitals decay exponentially with distance.

$\endgroup$
  • $\begingroup$ What exactly is meant by $n = \infty$? Could you clarify and/or provide a reference? $\endgroup$ – Linear Christmas Sep 5 '16 at 19:27
  • $\begingroup$ all it means is that the interaction is zero when r>b else infinity, i.e. hard sphere, so dividing by a huge power of $n$ makes $b/r^n$ zero. $\endgroup$ – porphyrin Sep 6 '16 at 15:35
  • $\begingroup$ I must do more research of available materials to readily accept the logic. So far, I have seen that even in the 'billiard ball' approximation $n \in [9, 15]$. Another states generally $n \approx 10$. From a physical chemistry textbook [3] $n$ depends on the ion. Thus saying $n=\infty$ seems odd... $\endgroup$ – Linear Christmas Sep 6 '16 at 16:48
  • $\begingroup$ If its a problem just ignore this part of the text and concentrate on what the hard sphere implies, namely that interaction is infinite at some cut-off distance and zero larger than this, just as with billiard balls bouncing elastically of one another. (By elastically I mean that no energy is retained within the balls themselves, all their energy is kinetic) $\endgroup$ – porphyrin Sep 6 '16 at 18:47
0
$\begingroup$

You may be aware of the general potential energy vs r curve? In which at r=0 energy is very repulsive, possibly infinitely so for some approximations, and then drops rapidly into a trough, the minimum of which is the 'equilibrium' distance r(eq) and as r increases rises back up? From my ed.3 Cotton&Wilkinson, pg 59.(Somewhat abbreviated) The reason the ions reside at r(eq) from each other and do not move closer is there are short-range repulsive forces due to overlapping electron clouds [so much for rigid spheres, huh?] Born proposed the simple assumption that the repulsive force between two ions could be represented as B'/rⁿ where B' and n are constants characteristic of the ion pair. We can therefor write E(repulsion) = B/rⁿ where B is related to B' by the crystal geometry. so, the reason has more to due with the shape of the P.E. curve (of solid state crystals, but perhaps you know a bit about how ions in the gas phase can be considered to occupy a lattice?). I have not researched this, but my w.a. guess is that it's a fudge factor, forcing r(eq) to the correct value. It may turn out, upon doing the quantum mechanics, that there's a deep reason that the form of this term works out so well. My guess is that Born just wanted something easy to differentiate or integrate - why make it more difficult than needs be? - and no one has seen a need to improve upon it, given it's empirical derivation. It turns out that n can be determined based on compressibility (C&W are talking about solid crystals, so...) as well as theoretically and B = {-A*(Z-)(Z+) e²*[r(eq)^(n-1)]}/n from the lattice energy. C&W go on to state that corrections to their equation are sometimes used (keep in mind Edition 3 is ancient! 1972 !) and specifically that this repulsion term is "not strictly correct from quantum-mechanical considerations. More refined expressions do not greatly change the results, however." where A is the Madelung constant, Zs are the ions charges.

$\endgroup$
  • 8
    $\begingroup$ Could you please edit this to include some paragraphing? I can't make heads or tails of this wall of text. $\endgroup$ – M.A.R. Jul 3 '16 at 20:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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