Let's imagine we have two polycrystalline samples: $\ce{Ca3(PO4)2}$ and $\ce{Na3PO4}$. A free electron is added to both. Some heat is released.

I wanted to check whether my reasoning about this scenario is correct. Could not find any data about this, though.

I already know that, for some other materials, surfaces' electron affinities depend on the crystal facet the electron went to. If this is the case here as well, I would accept this as a useful piece of information. However, maybe this is a negligible effect here.

Here's my attempt:

  • the free electron to be added to the surface cannot go to the negatively charged phosphate ion. It must be only attracted by $\ce{Na^+}$ or $\ce{Ca^2+}$ .
  • the ionic radii are similar.
  • $\ce{Ca^2+}$ has more charge per ion, therefore it may attract the added electron more.

Here's my issue:

  • There are twice as many $\ce{Na^+}$ ions per one phosphate compared to $\ce{Ca^2+}$ ions per phosphate. That makes me suspect, the added electron may rather be attracted to multiple $\ce{Na^+}$ simultaneously and occupy a larger volume or even conduction band instead. Larger orbital may lower its kinetic energy more, assuming this does not happen in $\ce{Ca3(PO4)2}$.

Which explanation is more plausible?

  • 1
    $\begingroup$ @Maurice There are two: 1) The electron goes to a local orbital in $\ce{Na3PO4}$. or 2) There are no such orbitals to go to in $\ce{Na3PO4}$, there is conduction band instead. $\endgroup$
    – Paul Kolk
    Commented Jul 9 at 8:59
  • $\begingroup$ The title has only one question : * Is the surface electron affinity of (tribasic) sodium or calcium phosphate higher ? * $\endgroup$
    – Maurice
    Commented Jul 9 at 12:24
  • $\begingroup$ @Maurice If anybody has a direct answer to the question in title, that would be great! Searching the internet for measured values of surface electron affinities usually gives nothing. $\endgroup$
    – Paul Kolk
    Commented Jul 9 at 17:07
  • 1
    $\begingroup$ Nobody has an answer to the question for two reasons : First, we don't know the meaning of the term "surface electron affinity" and second, we don't understand the end of the question which is : Higher. But higher than what ? $\endgroup$
    – Maurice
    Commented Jul 9 at 19:05
  • $\begingroup$ @Maurice OK. I see, this is an uncommon term. I mean the amount of heat released by a free electron upon attachment to the surface. By "higher", obviously is meant "which of the samples' -- $\ce{Na3PO4}$ or $\ce{Ca3(PO4)2}$ 's electron affinity is higher than the other's. $\endgroup$
    – Paul Kolk
    Commented Jul 9 at 19:44