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What are the Hume-Rothery rules?

The Hume-Rothery rules are a set of observations that help predict whether two metals or two solid compounds will be highly soluble in each other. Established by the English metallurgist William Hume-Rothery, these rules are widely used in the study of the composition of alloys, which are nothing more than solid metallic solutions.

Thus, by taking a look at the Hume-Rothery rules, it is possible to predict how likely the solubility of two metals will be. Although they take into account various parameters such as the size of the atoms, the valences and the electronegativities, they are not always correct in all cases, with inexplicable exceptions: metals that are alloyed well even when in theory they should not.

The great solubility between silver and gold in the formation of their alloys obeys the Hume-Rothery rules

Gold and silver, two visually different metals, are actually very soluble in each other. Thanks to this solubility its atoms mix to form alloys. Said solubility is supported by the Hume-Rothery rules, which indicate that Au and Ag atoms will not have limited solubilities.

Rules

Rule 1: Size factor

For two solid metals, elements, or compounds to mix, their atoms must not differ too much in size. The predominant metal will be the solvent, which is where the solute, the metal of lower proportion, will dissolve.

The solvent atoms, also called hosts, will not be able to dissolve or host the solute atoms if the latter are very large or small. Why? Because it would involve deforming the solid structure of the solvent, which is undesirable if what you are looking for is an alloy.

Now, the first Hume-Rothery rule states that the difference between the atomic radii between the solute and solvent atoms should not be greater than 15%. That is, the solute atom must not be 15% larger or smaller than the solvent atoms.

The above can be easily calculated with the following equation:

% difference = (rsolute – solvent) / (solvent) x 100%

Where rsolute is the atomic radius of the solute, while rsolvent is the atomic radius of the solvent. This calculation should yield a value of% difference ≤ 15%.

Rule 2: Crystal Structure

The crystalline structures of the solute and the solvent must be the same or similar. Here the aforementioned reappears: the structure of the solvent cannot be greatly affected by the addition of the solute atoms.

For example, two metals with face-centered cubic (fcc) structures will mix without much trouble. While a metal with a compact hexagonal structure (hcp), it does not tend to mix very well with one with an fcc structure.

The solubilities are unlimited when the two metals have the same valencies. On the other hand, when these are different, the solvent tends to dissolve the solute with the highest valence.

The greater the valence, the smaller the solute atom will be, and the solid solution obtained will be of the interstitial type: the solute will be positioned within the hollow spaces or pores of the solvent’s crystal lattice.

For example, if a metal ordinarily has a valence of +2 (like copper), it will have limited solubility when mixed with a metal that has a valence of +3 (like aluminum).

Rule 4: Electronegativity

The solvent and the solute must not have very different electronegativities, otherwise their solubility will be limited. That is, a “very electronegative” metal will not be fully alloyed with a very electropositive metal; instead, the two combine to form an intermetallic compound, not an alloy.

Examples

The Hume-Rothery rules are correct in the following examples:

-Nickel-gold alloys, Au-Ni, in which nickel has good solubility in gold, since the crystal lattice of gold is only 1.15 times larger than that of nickel

-Solid solutions of hafnium and zirconium oxides, HfO 2 -ZrO 2 , where both ions mix perfectly due to having similar radii and valences, Hf 4+ and Zr 4+

-Absorption of hydrogen in palladium, since the radius of the hydrogen molecules does not differ by less than 15% from the atomic radii of palladium; otherwise, H 2 could never be retained interstitially in Pd crystals.

-Cadmium and magnesium alloys, Cd-Mg, for reasons similar to those stated for Au-Ni alloys. Also note that the valences of both metals are the same: Cd 2+ and Mg 2+ , which contributes to their solubility despite having relatively different atomic radii.

Solved exercises

Then and to finish, some simple exercises will be presented where the Hume-Rothery rules are put into practice.

Exercise 1

Having at hand the following data:

rAu: 0.1442 nm, fcc, +1

rAg: 0.1445 nm, fcc, +1

And according to the Hume-Rothery rules, would you expect unlimited solubility between the two metals?

Both gold and silver have fcc structures (rule 2), and the same valence number (+1, although gold can also have +3). So, we must base ourselves on the atomic radii before drawing superficial conclusions.

As the most expensive gold, we will assume that silver is the solvent, and gold, the solute. Taking their respective atomic radii expressed in nanometers (nm), we proceed to calculate the percentage of their differences:

% difference = (rsolute – solvent) / (solvent) x 100%

= (0.1442 – 0.1445) / (0.1445) x 100%

= 0.2076%

Note that we take a positive value, and that this is much less than 15%. Therefore, we can say that according to the Hume-Rothery rules, gold and silver will mix without any problem to form alloys.

Exercise 2

Having at hand the following data:

rCu: 0.128 nm, fcc, electronegativity 1.8, +2

rNi: 0.125 nm, fcc, electronegativity 1.8, +2

Would you wait for copper and nickel to form alloys without limitations?

Again, we repeat the previous calculation since it is the only parameter where they show differences. We assume that copper is the solvent and that nickel is the solute:

% difference = (rsolute – solvent) / (solvent) x 100%

= (0.125 – 0.128) / (0.128) x 100%

= 2.34%

This value is below 15%. Therefore, it is not surprising that both metals are alloyed without much difficulty.

Exercise 3

According to the following data:

rSi: 0.117 nm, cubic diamond, electronegativity 1.8, +4

rGe: 0.139 nm, cubic diamond, electronegativity 2.0, +4

Would you expect silicon and germanium to form solid solutions?

This time we noticed that germanium is slightly more electronegative than silicon, which can work against the solubility between the two. We calculate the difference between their atomic radii assuming that germanium is the solvent and that silicon is the solute:

% difference = (rsolute – solvent) / (solvent) x 100%

= (0.117 – 0.139) / (0.139) x 100%

= 15.82%

Note that the solubility between silicon and germanium crystals is limited: silicon atoms are 15.82% smaller than germanium atoms. In addition to this, we must add the difference between their electronegativities.

However, this does not mean that the two elements cannot be mixed, only that their Si-Ge alloys have limited percentages in the composition of one of the two elements; Outside of these values, the Si-Ge alloy does not exist.

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