# Why are Pd and Pt completely soluble in Cu, but Ag only partly in Cu when forming a substitutional solid solution?

Why are Pd and Pt completely soluble in Cu, but Ag only partly in Cu when forming a substitutional solid solution? If we look at the Hume-Rothery rules, they all are within the 15% atomic radius factor with respect to Cu and all have the same crystal structure:

+----+---------------+-----------------+-------------------+---------+
|    | Atomic Radius | Crystal Struct. | Electronegativity | Valence |
+----+---------------+-----------------+-------------------+---------+
| Cu | 0.1278        | FCC             | 1.9               | +2      |
| Pd | 0.1376        | FCC             | 2.2               | +2      |
| Pt | 0.1387        | FCC             | 2.2               | +2      |
| Ag | 0.1445        | FCC             | 1.9               | +1      |
+----+---------------+-----------------+-------------------+---------+


The last two rules are:

• The solvent and solute should have similar electronegativity. Intermetallics tend to form when the difference is large.
• Other factors being equal, a metal will have more of a tendency to dissolve another metal of higher valency than of a lower valency.

Then Pd and Pt have an electronegativity difference of 0.3 while Ag has 0. Pd and Pt have same valency while Ag has one less.

I would have expected that Ag would form a complete solid solution while Pd and Pt would only partly form a solid solution with Cu because difference of 0.3 in EN is quite large, because the valence difference between Ag and Cu is only 1 electron and because of the words: 'Other factors being equal,' it sounds like the other rules are more important.

To sum up, my thought process was:

Pd, Pt and Ag meet rule 1 and 2. Pd and Pt don't meet rule 3, so only form partly solid solutions. Then Ag meets rule 3 and rule 4 is less important and yet it is still close, so Ag will completely dissolve in Cu.

Why is this thought process wrong? If a difference in EN of 0.3 isn't big, what is then the border? Like 15% is the border for rule 1?

• Hume-Rothery's rules aren't hard-and-fast rules, they are guidelines. Not every binary metallic mixture (more than 2200 possible) has been deeply investigated, and the ability to do so via computational modeling is only just now becoming possible. Any deeper understanding goes beyond the scope of materials engineering into chemistry or, more likely, physics. To understand fully why Hume-Rothery doesn't always work as expected would require detailed analysis or modeling of electron orbital interactions, and for transition metals it can get fairly complicated. Or so I understand. Commented Dec 14, 2015 at 21:50
• And let's not throw Ni in there as well. The answer is: it is complex, and simple rules can't address it with any precision. Commented Dec 14, 2015 at 23:48
• I think it's apparent from this that the difference in valence state is a more significant attribute than electronegativity. The electronegativity does narrowly fall within the 15% which is only an approximation ($\mathrm{0.3/2.2 = 0.136}$; $\mathrm{0.3/1.9 = 0.158}$).
– A.K.
Commented Dec 14, 2015 at 23:51
• Although a question on liquid phases, the point remains: chemistry.stackexchange.com/questions/35014/… - if you want I could summarize the fcc noble / near noble metal situation, but it is similarly complex. For example, Cu-Ni shows a broad range of full miscibility, with a solid fcc miscibility gap starting to form below 400C. In Au-Ni that miscibility gap starts around 600C. For Ag-Ni, there is no temperature where there is no miscibility gap in the solid, and the miscibility gap extends even in to the liquid! Commented Dec 15, 2015 at 15:01