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What I know is, that in a substitutional alloy, the atoms of the solute metal take positions normally occupied by the solvent metal atoms, as they are more or less the same size. In an interstitial alloy, the solute metal atoms occupy the "holes" between the solvent metal atoms. My question is now, which type of alloy is harder and how is this connected to the amount of solute added? I think that interstitial alloys are generally harder and less malleable and ductile than substitutional alloys, as the smaller solute atoms in the holes prevent the ability of the metal cations to slide past each other. Am I right or wrong?

Actually, this was a question that our teacher gave us as homework, but I thought that it was a little poorly worded, so I tried to reword it. But I'm not sure if I understood the task correctly. Maybe someone will understand it, so I took a picture of it: (there are even some grammar mistakes in it and he says alloy A and alloy B, and I thought he means interstitial and substitutional alloys).

enter image description here

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closed as too broad by Tyberius, Mithoron, Jon Custer, airhuff, Todd Minehardt Sep 5 '17 at 1:06

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Sounds about right. $\endgroup$ – Ivan Neretin Apr 10 '16 at 20:03
  • $\begingroup$ Given the huge range of alloys out there, I'm not sure that such a simple minded approach has a great deal of value. In particular, there are many 'interstitial' alloys where the crystal structure of the alloy is not the same as that of the pure materials - how do you determine their hardness then? For example, if the pure material is fcc, but the alloy is bcc (there are many such systems out there), than that crystal change has dramatic effects on the possible dislocation-glide plane system. Personally, I think it is a bad question. $\endgroup$ – Jon Custer Apr 11 '16 at 16:06

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