What happens to the mobility of Group 1 elements going down the periodic table?

Whilst reading the book, So you want to go to Oxbridge (Third Edition), I noticed that it states that a past question given during an interview for the place on the undergraduate chemistry course (M.Chem) was:

What happens to the mobility of Group 1 elements going down the periodic table?

I do not understand the question. And so, I would both like to understand the question and know and understand its answer.

Preemptively, Thank you

• I am guessing they mean ionic mobilities (first, and only thing that comes to mind right now)? Simple explanation: The size of the cations increases down the group, and group elements form singly ionised cations (have a charge of +1), so the charge density decreases down the group. Extent of solvation is proportional to charge density, and consequently lithium ions would be the most solvated, hence least mobile. Whereas, $\ce{Cs^+}$ would the most mobile. If you want some simple math to support this, I can supply that later. – getafix Aug 28 '16 at 2:59
• @Karl Sorry, how is my question in comprehensible? – Saul McShane Aug 28 '16 at 14:42
• @Karl A simple grammatical mistake, which I have now corrected - although, I certainly would not say that it is incomprehensible, as the point I was trying to put across was and is obvious. Also, if the book gave an answer, I wouldn't be on here asking for the answer. – Saul McShane Aug 28 '16 at 17:01
• "Oxbridge, 3d edition" is some kind of collection of exam questions, without answers, context, anything ? They charge money for that? – Karl Aug 28 '16 at 17:07
• @Karl It is a commonly used portmanteau, I believe. – getafix Aug 28 '16 at 19:53

When you have charged particles in a liquid acted upon by a electric field, the experience a force, and said force accelerates them to a terminal, drift speed.

Note: The physical treatment of the problem, and the notion of terminal speed is quite similar to say a ball falling from the top of a sky scraper through air (a fluid medium)

Now, let's go back to our charged particles in fluids. Let's say we have applied a voltage $\Delta\phi$ between two large, planar electrodes, immersed in the liquid and separated by a distance $l$

The magnitude of the electric field, and the electric force experienced by an ion of charge $ze$ are given by the following relations:

$$E = \frac{\Delta\phi}{l}$$ $$F = zeE = (ze).\frac{\Delta\phi}{l}$$

A cation responds by accelerating towards the negative electrode, and an anion responds by accelerating towards the positive electrode. This acceleration, however, is short lived because the fluid medium offers a retarding force $F_{\text{friction}}$.

$$F_{\text{friction}}\propto s\text{, where s := speed}$$

Assuming we have spherical particles of radius $a$, let's say our retarding force is given by Stokes' Law

$$F_{\text{friction}} = (6\pi\eta a)s$$ $\eta$) is the viscosity of the medium.

When the electric force is exactly balaced by the viscous drag, the acceleration drops to zero and this final speed of the particle is called the drift speed ($s_d$). $$F_{\text{friction}} = F_{\text{electric}}$$ and solving for the drift speed we get, $$s_d = \frac{zeE}{6\pi\eta a}$$

At this point we note that the drift speed is proportional to the applied electric field and introduce the following definition:

$$s_d = \mu E$$ where, $$\mu = \frac{ze}{6\pi\eta a}$$

and we call $\mu$ the mobility.

From the aforementioned equations/definitions one can deduce that $\mu \propto \frac{1}{a}$. Thus, bigger/bulky ions ought to have lower mobilities, and lower drift speeds in solution.

However, $a$ is not the ionic radius, but the Stokes radius (or the hydrodyanmic radius), i.e the "effective radius" of an ion in solution taking into account its solvation shell (i.e the solvent molecules surrounding the ion). I believe the following illustration is instructive,

The size of this solvation shell depends on how "attractive" an ion appears to a solvent molecule, and that in turn depends on the electric field produced by the ions.

For spherical ions, we consider the electric field on the surface of a charged sphere: $$E _\text{surface of a charged sphere} \propto \frac{ze}{r^2}$$

here $r$ is the ionic radius. So smaller ions have a larger solvation shell than larger ions (assuming they carry the same charge).

This is the reason why mobilities of alkali metal ions increase from $\ce{Li^+}$ to $\ce{Cs^+}$ (and not the other way around); the lithium ions are effectively dragging a large amount of solvent molecules with them and thus, move slower.

• Hello. Thank you very much for your answer, it is extremely helpful and informative. I have one question though. In the equation: $F_{\text{friction}} = (6\pi\eta a)$ why is the flow velocity left out of the end of the equation as shown on the Wikipedia page on Stokes' Law you have referenced? I had never heard of Stokes' Law before, so the answer to my question is probably obvious. – Saul McShane Aug 28 '16 at 16:58
• A typo, sorry. I'll fix it right away – getafix Aug 28 '16 at 18:46
• Also, shouldn't all references to $(6\pi\eta a)$ be changed to $(6\pi\eta a)s$, such as in the drift speed equation and the mobility equation? – Saul McShane Aug 28 '16 at 19:44
• @SaulMcShane Oh no! we are solving for the drift speed, (i.e s) by equating $$F_{\text{friction}} = F_{\text{electric}}$$ I guess my notation isn't good; should've made it more clear. – getafix Aug 28 '16 at 19:47